TANGENTS & NORMALS MCQs for JEE MATHEMATICS — Medium, Page 3

61 Medium

61 of 97

Application +4 | -1

Angle between the tangents to the curve

y = {x^2} - 5x + 6

at the point (2, 0) and (3, 0) is

View answer and explanation

Correct option: A

\frac{\pi }{2}

Option A (

\frac{\pi }{2}

) is correct because the product of the slopes of the tangents at the given points is -1, indicating perpendicularity.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{2}$ . The angle between the tangents is found by calculating the slopes and using the condition for perpendicularity.

Step-by-Step Solution:

Step 1: Given the curve $y = x^2 - 5x + 6$ .

Step 2: Find the derivative of $y$ with respect to $x$ : $\frac{dy}{dx} = 2x - 5$ .

Step 3: Evaluate the derivative at the point (2, 0): $m_1 = \left( \frac{dy}{dx} \right)_{(2, 0)} = 2(2) - 5 = -1$ .

Step 4: Evaluate the derivative at the point (3, 0): $m_2 = \left( \frac{dy}{dx} \right)_{(3, 0)} = 2(3) - 5 = 1$ .

Step 5: Check if the tangents are perpendicular by multiplying their slopes: $m_1 \times m_2 = (-1) \times (1) = -1$ .

Step 6: Since the product of the slopes is -1, the tangents are perpendicular, and the angle between them is $\frac{\pi}{2}$ .

Step 7: Final Answer: $\frac{\pi }{2}$

Why other options are wrong:
- Option B: $\frac{\pi }{6}$ This is incorrect because the slopes do not correspond to an angle of $\frac{\pi }{6}$ .
- Option C: $\frac{\pi }{4}$ This is incorrect because the slopes do not correspond to an angle of $\frac{\pi }{4}$ .
- Option D: $\frac{\pi }{3}$ This is incorrect because the slopes do not correspond to an angle of $\frac{\pi }{3}$ .

Key Concepts:
- Derivatives and slopes of tangents
- Condition for perpendicular lines: $m_1 \times m_2 = -1$

Memory Aids:
- Perpendicular lines have slopes that are negative reciprocals of each other.

Quick Tips:
- Find the derivative to get the slope of the tangent.
- Check the product of slopes to determine if lines are perpendicular.

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Application +4 | -1

The equation of a normal to the curve,

\sin \mathrm{y}=\mathrm{x} \sin \left(\frac{\pi}{3}+y\right)

at

\mathrm{x}=0

, is :

View answer and explanation

Correct option: B

2 x+\sqrt{3} y=0

Option B (

2 x+\sqrt{3} y=0

) is correct because it represents the equation of the normal to the curve at x=0, derived from the derivative and normal slope.

Solution approach

Correct Answer + Core Reason: $2 x+\sqrt{3} y=0$ . The equation of the normal is found by calculating the derivative at x=0 and using the negative reciprocal slope.

Step-by-Step Solution:

Step 1: Given the curve equation $\sin y = x \sin \left(\frac{{\pi}}{{3}} + y\right)$ .

Step 2: Differentiate the equation with respect to $x$ : $\cos y \frac{{dy}}{{dx}} = \sin \left(\frac{{\pi}}{{3}} + y\right) + x \cos \left(\frac{{\pi}}{{3}} + y\right) \frac{{dy}}{{dx}}$ .

Step 3: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = \frac{{\sin \left(\frac{{\pi}}{{3}} + y\right)}}{{\cos y - x \cos \left(\frac{{\pi}}{{3}} + y\right)}}$ .

Step 4: Evaluate $\frac{{dy}}{{dx}}$ at $x = 0$ . When $x = 0$ , $\sin y = 0$ , so $y = 0$ . Thus, $\frac{{dy}}{{dx}}|_{{(0,0)}} = \frac{{\sin \left(\frac{{\pi}}{{3}}\right)}}{{\cos 0 - 0 \cos \left(\frac{{\pi}}{{3}}\right)}} = \frac{{\frac{{\sqrt{3}}}{{2}}}}{{1}} = \frac{{\sqrt{3}}}{{2}}$ .

Step 5: The slope of the normal is the negative reciprocal of the tangent's slope: $-\frac{{2}}{{\sqrt{3}}}$ .

Step 6: The equation of the normal at (0, 0) is $y - 0 = -\frac{{2}}{{\sqrt{3}}}(x - 0)$ , which simplifies to $2x + \sqrt{3}y = 0$ .

Why other options are wrong:
- Option A: $2 x-\sqrt{3} y=0$ - This is incorrect because it uses the slope of the tangent instead of the normal.
- Option C: $2 y-\sqrt{3} x=0$ - This is incorrect because it uses the reciprocal of the normal's slope and swaps x and y.
- Option D: $2 y+\sqrt{3} x=0$ - This is incorrect because it uses the reciprocal of the normal's slope and swaps x and y with incorrect signs.

Key Concepts:
- Differentiation
- Equation of normal
- Finding slope of normal

Memory Aids:
- Remember the relationship between tangent and normal slopes: $m_1 * m_2 = -1$

Quick Tips:
- Always find the derivative first to get the slope of the tangent.
- Remember to take the negative reciprocal to find the slope of the normal.

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Analyzing +4 | -1

The normal to the curve represented parametrically by

x = a\left( {\cos \theta + \theta \sin \theta } \right)\,and\,y = a\left( {\sin \theta - \theta \cos \theta } \right)

at any point

\theta

, is such that it

View answer and explanation

Correct option: B

Is at a constant distance from the origin

Option B (Is at a constant distance from the origin) is correct because the equation of the normal simplifies to a form where the distance from the origin is constant.

Solution approach

Correct Answer + Core Reason: Is at a constant distance from the origin. The normal's equation simplifies to a form where the distance from the origin is constant.

Step-by-Step Solution:

Step 1: Given $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$ .

Step 2: Find $\frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta$ .

Step 3: Find $\frac{dy}{d\theta} = a(\cos \theta - \cos \theta + \theta \sin \theta) = a\theta \sin \theta$ .

Step 4: Find $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta$ .

Step 5: The slope of the normal is $-\frac{1}{\frac{dy}{dx}} = -\cot \theta$ .

Step 6: The equation of the normal at any point is $y - a(\sin \theta - \theta \cos \theta) = -\cot \theta (x - a(\cos \theta + \theta \sin \theta))$ .

Step 7: Simplify the equation: $y - a\sin \theta + a\theta \cos \theta = -\cot \theta x + a\cot \theta \cos \theta + a\cot \theta \theta \sin \theta \Rightarrow y\sin \theta - a\sin^2 \theta + a\theta \cos \theta \sin \theta = -x\cos \theta + a\cos^2 \theta + a\theta \sin \theta \cos \theta \Rightarrow x\cos \theta + y\sin \theta = a(\sin^2 \theta + \cos^2 \theta) \Rightarrow x\cos \theta + y\sin \theta = a$ .

Step 8: The equation $x\cos \theta + y\sin \theta = a$ represents a line at a constant distance $|a|$ from the origin.

Why other options are wrong:
- Option A: Incorrect because the angle the normal makes with the x-axis is not constant; it depends on $\theta$ .
- Option C: Incorrect because the normal does not pass through the origin unless $a = 0$ , which is not generally true.
- Option D: Incorrect because not all three conditions are satisfied.

Key Concepts:
- Parametric differentiation
- Equation of a normal to a curve
- Distance of a line from the origin

Memory Aids:
- Remember the formula for the slope of the normal: $-\frac{1}{\frac{dy}{dx}}$ .
- Recall the equation of a line in normal form: $x\cos \theta + y\sin \theta = p$ , where $p$ is the distance from the origin.

Quick Tips:
- Simplify the equation of the normal as much as possible.
- Recognize the standard forms of lines and their properties.

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Application +4 | -1

The curve

x+y-\ell \mathrm{n}(x+y)=2 x+5

has a vertical tangent at the point

(\alpha, \beta)

. Then

\alpha+\beta

is equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct because for a vertical tangent,

\frac{{dy}}{{dx}}

must be undefined, which occurs when the denominator of

\frac{{dy}}{{dx}}

is zero, leading to

\alpha + \beta = 1

.

Solution approach

Correct Answer + Core Reason: 1. A vertical tangent occurs when $\frac{{dx}}{{dy}} = 0$ , which implies $\frac{{dy}}{{dx}}$ is undefined. This happens when $\alpha + \beta = 1$ .

Step-by-Step Solution:

Step 1: Differentiate the given equation with respect to x.\nx + y - \ln(x+y) = 2x + 5
1 + $\frac{{dy}}{{dx}}$ - $\frac{{1}}{{x+y}}$ (1 + $\frac{{dy}}{{dx}}$ ) = 2

Step 2: Solve for $\frac{{dy}}{{dx}}$ .
$\frac{{dy}}{{dx}}$ (1 - $\frac{{1}}{{x+y}}$ ) = 1 + $\frac{{1}}{{x+y}}$
$\frac{{dy}}{{dx}}$ = \frac{{1 + $\frac{{1}}{{x+y}}$ }}{{1 - $\frac{{1}}{{x+y}}$ }} = \frac{{ $\frac{{x+y+1}}{{x+y}}$ }}{{ $\frac{{x+y-1}}{{x+y}}$ }} = $\frac{{x+y+1}}{{x+y-1}}$

Step 3: For a vertical tangent, $\frac{{dx}}{{dy}} = 0$ , which means $\frac{{dy}}{{dx}}$ is undefined.\nThis occurs when the denominator is zero: x + y - 1 = 0\nTherefore, x + y = 1

Step 4: Evaluate at the point $(\alpha, \beta)$ .
α + β = 1

Final Answer: 1

Why other options are wrong:
- Option B (2): This is incorrect because it doesn't satisfy the condition for a vertical tangent.
- Option C (3): This is incorrect because it doesn't satisfy the condition for a vertical tangent.
- Option D (4): This is incorrect because it doesn't satisfy the condition for a vertical tangent.

Key Concepts:
- Implicit Differentiation
- Vertical Tangents

Memory Aids:
- Vertical Tangent: $\frac{{dx}}{{dy}} = 0$ or $\frac{{dy}}{{dx}}$ is undefined.

Quick Tips:
- Remember that a vertical tangent occurs where the derivative $\frac{{dy}}{{dx}}$ is undefined.

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Application +4 | -1

If

y = 4x - 5

is tangent to the curve

{y^2} = p{x^3} + q

at (2, 3) then (p,q) is

View answer and explanation

Correct option: D

\left( {2,\; - 7} \right)

Option D

\left( {2,\; - 7} \right)

is correct because these values of p and q satisfy both the tangency condition and the point lying on the curve.

Solution approach

Correct Answer + Core Reason: $\left( {2,\; - 7} \right)$ . The values of p and q are found by using the tangency condition and the fact that the point lies on the curve.

Step-by-Step Solution:

Step 1: Given the curve equation ${y^2} = p{x^3} + q$ and the tangent $y = 4x - 5$ at the point (2, 3).

Step 2: Differentiate the curve equation with respect to x: $2y\frac{{dy}}{{dx}} = 3p{x^2}$ .

Step 3: Find the slope of the tangent to the curve at (2, 3): $\frac{{dy}}{{dx}} = \frac{{3p{x^2}}}{{2y}} \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{(2,3)}} = \frac{{3p{{(2)}^2}}}{{2(3)}} = \frac{{12p}}{6} = 2p$ .

Step 4: The slope of the given tangent line $y = 4x - 5$ is 4. Therefore, $2p = 4 \Rightarrow p = 2$ .

Step 5: Since the point (2, 3) lies on the curve, substitute x=2 and y=3 into the curve equation: ${3^2} = p{(2)^3} + q \Rightarrow 9 = 8p + q$ .

Step 6: Substitute p=2 into the equation $9 = 8p + q$ : $9 = 8(2) + q \Rightarrow 9 = 16 + q \Rightarrow q = -7$ .

Step 7: Therefore, (p, q) = (2, -7).

Why other options are wrong:
- Option A (2, 7): This does not satisfy the curve equation at the given point.
- **Option B ( $\left( { - 2,\;7} \right)$ ): This does not satisfy the tangency condition.
- Option C ( $\left( { - 2,\; - 7} \right)$ ): This does not satisfy the tangency condition.

Key Concepts:
- Differentiation
- Tangent to a Curve

Memory Aids:
- Remember that the slope of the tangent line is equal to the derivative of the curve at the point of tangency.

Quick Tips:**
- Use both the tangency condition and the point lying on the curve to solve for the unknowns.

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Application +4 | -1

The abscissa of the points, where the tangent to the curve

y = {x^3} - 3{x^2} - 9x + 5

is parallel to

x

-axis, are

View answer and explanation

Correct option: D

x = - 1

and 3

Option D (

x = - 1

and 3) is correct because these are the x-coordinates where the derivative of the curve is zero, indicating a horizontal tangent line.

Solution approach

Correct Answer + Core Reason: $x = -1$ and 3. The abscissas are found by setting the derivative equal to zero.

Step-by-Step Solution:

Step 1: Given the curve $y = x^3 - 3x^2 - 9x + 5$ .

Step 2: Differentiate with respect to x: $\frac{{dy}}{{dx}} = 3x^2 - 6x - 9$ .

Step 3: Since the tangent is parallel to the x-axis, $\frac{{dy}}{{dx}} = 0$ : $3x^2 - 6x - 9 = 0$ .

Step 4: Divide by 3: $x^2 - 2x - 3 = 0$ .

Step 5: Factor the quadratic equation: $(x - 3)(x + 1) = 0$ .

Step 6: Solve for x: $x = 3$ and $x = -1$ .

Why other options are wrong:
- Option A: This answer is incorrect. It may be a result of incorrect calculation of the derivative or not solving for x correctly.
- Option B: This answer is incorrect. It likely arises from misinterpreting the derivatives or making an arithmetic error.
- Option C: This answer is incorrect. It may be a result of not differentiating the equations correctly or not solving for x.

Key Concepts:
- Derivatives as slopes of tangents.
- Finding critical points.

Memory Aids:
- A horizontal tangent has a slope of zero.

Quick Tips:
- Always set the derivative equal to zero to find the x-coordinates of horizontal tangents.

67 Medium

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Application +4 | -1

If

y = 4x - 5

is a tangent to the curve

{y^2} = p{x^3} + q

at (2, 3), then

View answer and explanation

Correct option: A

p = 2,\;q = - 7

Option A (

p = 2,\;q = - 7

) is correct because these values satisfy both the tangency condition and the point lying on the curve.

Solution approach

Correct Answer + Core Reason: $p = 2,\;q = - 7$ . The values of p and q are found by using the tangency condition and the fact that the point lies on the curve.

Step-by-Step Solution:

Step 1: Given the curve equation ${y^2} = p{x^3} + q$ and the tangent $y = 4x - 5$ at the point (2, 3).

Step 2: Differentiate the curve equation with respect to x: $2y\frac{{dy}}{{dx}} = 3p{x^2}$ .

Step 3: Find the slope of the tangent to the curve at (2, 3): $\frac{{dy}}{{dx}} = \frac{{3p{x^2}}}{{2y}} \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{(2,3)}} = \frac{{3p{{(2)}^2}}}{{2(3)}} = \frac{{12p}}{6} = 2p$ .

Step 4: The slope of the given tangent line $y = 4x - 5$ is 4. Therefore, $2p = 4 \Rightarrow p = 2$ .

Step 5: Since the point (2, 3) lies on the curve, substitute x=2 and y=3 into the curve equation: ${3^2} = p{(2)^3} + q \Rightarrow 9 = 8p + q$ .

Step 6: Substitute p=2 into the equation $9 = 8p + q$ : $9 = 8(2) + q \Rightarrow 9 = 16 + q \Rightarrow q = -7$ .

Step 7: Therefore, p = 2 and q = -7.

Why other options are wrong:
- **Option B ( $p = - 2,\;q = 7$ ): These values do not satisfy the curve equation at the given point.
- Option C ( $p = - 2,\;q = - 7$ ): These values do not satisfy the tangency condition.
- Option D ( $p = 2,\;q = 7$ ): This does not satisfy the curve equation at the given point.

Key Concepts:
- Differentiation
- Tangent to a Curve

Memory Aids:
- Remember that the slope of the tangent line is equal to the derivative of the curve at the point of tangency.

Quick Tips:**
- Use both the tangency condition and the point lying on the curve to solve for the unknowns.

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Application +4 | -1

A tangent is drawn at the point

M

with abscissa unity on the curve

y=\sqrt{\left(5-x^{\frac{2}{3}}\right)^3}

. If length of segment of tangent intercepted between the co-ordinate axes is

\sqrt{{N}}

, then

{N}

is equal to

View answer and explanation

Correct option: D

125

Option D (125) is correct. By finding the derivative, the tangent equation, and then the intercepts, the length of the segment is calculated to be \

\sqrt{{125}}

\$.

Solution approach

Correct Answer + Core Reason: 125. The length of the tangent segment intercepted between the coordinate axes is \ $\sqrt{{125}}$ \ $, thus N = 125.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Find the derivative of the curve$ y=\sqrt{\left(5-x^{ $\frac{{2}}{{3}}$ }\right)³} = (5-x^{ $\frac{{2}}{{3}}$ })^{ $\frac{{3}}{{2}}$ } $.\n$ $\frac{{dy}}{{dx}}$ = $\frac{{3}}{{2}}$ (5-x^{ $\frac{{2}}{{3}}$ })^{ $\frac{{1}}{{2}}$ } · (- $\frac{{2}}{{3}}$ x^{- $\frac{{1}}{{3}}$ }) = -\sqrt{{5-x^{ $\frac{{2}}{{3}}$ }}} · x^{- $\frac{{1}}{{3}}$ }

Step 2: Evaluate the derivative at x=1:
\ $\frac{{dy}}{{dx}}|_{{x=1}} = -\sqrt{{5-1}} \cdot 1^{-\frac{{1}}{{3}}} = -2$ \ $\n\n**Step 3:** Find the y-coordinate at x=1: \n$ y = $\sqrt{{(5-1)^3}}$ = $\sqrt{{4^3}}$ = 8. So, the point M is (1, 8).

Step 4: Find the equation of the tangent at (1, 8):
\ $y - 8 = -2(x - 1) \Rightarrow y = -2x + 10 \Rightarrow 2x + y = 10$ \ $\n\n**Step 5:** Find the x-intercept (y=0): 2x = 10, x = 5. The x-intercept is (5, 0).\n\n**Step 6:** Find the y-intercept (x=0): y = 10. The y-intercept is (0, 10).\n\n**Step 7:** Calculate the length of the segment between the intercepts: \n$ L = $\sqrt{{5^2 + 10^2}}$ = $\sqrt{{25 + 100}}$ = $\sqrt{{125}}$

Step 8: Determine N: Since the length is \ $\sqrt{{N}}$ \ $, N = 125.\n\n**Why other options are wrong:**\n- **Option A:** 25 is incorrect. It likely results from only considering one of the intercepts or miscalculating the distance.\n- **Option B:** 50 is incorrect. This value doesn't correspond to the correct length of the intercept calculated using the tangent line.\n- **Option C:** 75 is incorrect. This is not the correct squared length of the tangent segment.\n- **Option D:** 125 is correct.\n\n**Key Concepts:**\n- Differentiation\n- Equation of a tangent line\n- Finding intercepts\n- Distance formula\n\n**Memory Aids:**\n- Remember the formula for the tangent line:$ y - y_1 = m(x - x_1) $\n- Distance formula:$ d = $\sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

Quick Tips:
- Always double-check your differentiation and algebraic manipulations.
- Draw a diagram to visualize the problem.

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Application +4 | -1

A curve passes through the point

(2,0)

and the slope of the tangent at any point

(x, y)

is

x^2-2 x

for all value of

x

. The point of maximum ordinate on the curve is

View answer and explanation

Correct option: A

(0,4 / 3)

Option A ((0,4/3)) is correct because the point of maximum ordinate is (0, 4/3), which is found by integrating the slope function and finding the critical points.

Solution approach

Correct Answer + Core Reason: (0, 4/3). The point of maximum ordinate is found by integrating the given slope and finding critical points.

Step-by-Step Solution:

Step 1: Given the slope $\frac{dy}{dx} = x^2 - 2x$ , integrate with respect to $x$ to find the equation of the curve: $y = \int (x^2 - 2x) dx = \frac{x^3}{3} - x^2 + C$ .

Step 2: The curve passes through the point $(2, 0)$ . Substitute this point into the equation to find the constant $C$ : $0 = \frac{2^3}{3} - 2^2 + C \Rightarrow 0 = \frac{8}{3} - 4 + C \Rightarrow C = 4 - \frac{8}{3} = \frac{4}{3}$ .

Step 3: The equation of the curve is $y = \frac{x^3}{3} - x^2 + \frac{4}{3}$ .

Step 4: To find the maximum ordinate, find the critical points by setting $\frac{dy}{dx} = 0$ : $x^2 - 2x = 0 \Rightarrow x(x - 2) = 0 \Rightarrow x = 0$ or $x = 2$ .

Step 5: Find the second derivative to determine whether the critical points are maxima or minima: $\frac{d^2y}{dx^2} = 2x - 2$ .

Step 6: Evaluate the second derivative at $x = 0$ : $\frac{d^2y}{dx^2}|_{x=0} = 2(0) - 2 = -2 < 0$ . This indicates a maximum at $x = 0$ .

Step 7: Evaluate the second derivative at $x = 2$ : $\frac{d^2y}{dx^2}|_{x=2} = 2(2) - 2 = 2 > 0$ . This indicates a minimum at $x = 2$ .

Step 8: The maximum ordinate occurs at $x = 0$ . Substitute $x = 0$ into the equation of the curve to find the corresponding $y$ -value: $y = \frac{0^3}{3} - 0^2 + \frac{4}{3} = \frac{4}{3}$ .

Step 9: The point of maximum ordinate is $(0, \frac{4}{3})$ .

Why other options are wrong:
- Option B ((0,2/3)): This is incorrect. It might be obtained by incorrectly calculating the constant of integration or the y-value at the critical point.
- Option C ((1,2/3)): This is incorrect. It might be a random guess or a misunderstanding of the problem.
- Option D ((2,4/3)): This is incorrect. It corresponds to a minimum, not a maximum.

Key Concepts:
- Integration
- Derivatives
- Critical points
- Second derivative test

Memory Aids:
- Remember that a maximum occurs when the first derivative is zero and the second derivative is negative.

Quick Tips:
- Always check the second derivative to confirm that you have a minimum or maximum.

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Application +4 | -1

If the sum of the squares of the intercepts on the axes cut off by the tangent to the curve

{x^{1/3}} + {y^{1/3}} = {a^{1/3}}

(with

a > 0

) at

P\left( {a/8,\;a/8} \right)

is 2, then

a =

View answer and explanation

Correct option: C

4

Option C (

\frac{x}{b}

) is correct because it represents the acceleration of the particle, derived by differentiating the given equation with respect to time.

Solution approach

Correct Answer + Core Reason: $\frac{x}{b}$ . The acceleration is found by differentiating the given equation with respect to time.

Step-by-Step Solution:

Step 1: Given the equation $a + bv^2 = x^2$ .

Step 2: Differentiate both sides with respect to time $t$ : $\frac{d}{dt}(a + bv^2) = \frac{d}{dt}(x^2)$ .

Step 3: Apply the chain rule: $0 + 2bv\frac{dv}{dt} = 2x\frac{dx}{dt}$ .

Step 4: Simplify: $2bv\frac{dv}{dt} = 2x\frac{dx}{dt}$ .

Step 5: Recall that $v = \frac{dx}{dt}$ (velocity) and $a = \frac{dv}{dt}$ (acceleration). Substitute these into the equation: $2bva = 2xv$ .

Step 6: Solve for $a$ : $a = \frac{2xv}{2bv} = \frac{x}{b}$ .

Step 7: Final Answer: $\frac{x}{b}$ .

Why other options are wrong:
- **Option A ( $bx$ ):** This is incorrect because the acceleration is inversely proportional to $b$ , not directly proportional.
- **Option B ( $\frac{x}{a}$ ):** This is incorrect because the acceleration is related to $b$ , not $a$ .
- **Option D ( $\frac{x}{{ab}}$ ):** This is incorrect because the acceleration is not inversely proportional to the product of $a$ and $b$ .

Key Concepts:
- Differentiation with respect to time
- Chain rule
- Velocity and acceleration

Memory Aids:
- Velocity is the rate of change of position with respect to time: $v = \frac{dx}{dt}$
- Acceleration is the rate of change of velocity with respect to time: $a = \frac{dv}{dt}$

Quick Tips:
- Remember to use the chain rule when differentiating with respect to time.
- Double-check your algebra when solving for the acceleration.

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Application +4 | -1

If the line

y=\sqrt{2} x-4 k \sqrt{2}

is normal chord of parabola

y^2=2 x

whose length is

a \sqrt{a}(a \in N)

, then the value of

\left(a+\frac{1}{k}\right)

is

View answer and explanation

Correct option: A

5

Option A (5) is correct because by finding the normal chord and its length, we can determine the values of a and k, and then calculate (620/(p+q)) where p/q is the probability that a < b.

Solution approach

Correct Answer + Core Reason: 5. The value of $\frac{620}{(p+q)}$ is calculated based on the probability that a < b.

Step-by-Step Solution:

Step 1: Define the probabilities.\nGiven that the probability of rolling n is proportional to $n^2$ , we have $P(n) = Kn^2$ for $n = 1, 2, 3, 4, 5, 6$ .\nSince the sum of probabilities must be 1, we have:
$K(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$
$K(1 + 4 + 9 + 16 + 25 + 36) = 1$
$91K = 1 \Rightarrow K = \frac{1}{91}$ \nSo, $P(1) = \frac{1}{91}, P(2) = \frac{4}{91}, P(3) = \frac{9}{91}, P(4) = \frac{16}{91}, P(5) = \frac{25}{91}, P(6) = \frac{36}{91}$

Step 2: Calculate the probability that a < b.\nLet A be the event that a < b, B be the event that a = b, and C be the event that a > b.\nBy symmetry, $P(A) = P(C)$ . Also, $P(A) + P(B) + P(C) = 1$ .
$P(B) = \sum_{i=1}^6 [P(i)]^2 = \left(\frac{1}{91}\right)^2 + \left(\frac{4}{91}\right)^2 + \left(\frac{9}{91}\right)^2 + \left(\frac{16}{91}\right)^2 + \left(\frac{25}{91}\right)^2 + \left(\frac{36}{91}\right)^2$
$P(B) = \frac{1 + 16 + 81 + 256 + 625 + 1296}{91^2} = \frac{2275}{8281} = \frac{25}{91}$
$2P(A) + P(B) = 1 \Rightarrow 2P(A) = 1 - \frac{25}{91} = \frac{66}{91}$
$P(A) = \frac{33}{91}$

Step 3: Express the probability in lowest form.\nThe probability that a < b is $\frac{33}{91}$ , so $p = 33$ and $q = 91$ .

Step 4: Calculate $\frac{620}{p+q}$ .
$p + q = 33 + 91 = 124$
$\frac{620}{p+q} = \frac{620}{124} = 5$

Why other options are wrong:
- Option B (-5): The value cannot be negative.
- Option C (6): Incorrect calculation.
- Option D (-6): The value cannot be negative.

Key Concepts:
- Probability
- Weighted Dice
- Symmetry

Memory Aids:
- Remember to normalize the probabilities so that their sum is 1.

Quick Tips:
- Use symmetry to simplify the calculation of probabilities.

72 Medium

72 of 97

Analyzing +4 | -1

If

\beta

is one of the angles between the normals to the ellipse,

x^2+3 y^2=9

at the points

(3 \cos \theta, \sqrt{3} \sin \theta)

and

(-3 \sin \theta, \sqrt{3} \cos \theta) ; \in\left(0, \frac{\pi}{2}\right) ;

then

\frac{2 \cot \beta}{\sin 2 \theta}

is equal to

View answer and explanation

Correct option: B

\frac{2}{\sqrt{3}}

Option B (

\frac{2}{\sqrt{3}}

) is correct. By finding the slopes of the normals at the given points and using the tangent formula, the expression simplifies to

\frac{2}{\sqrt{3}}

.

Solution approach

Correct Answer + Core Reason: $\frac{2}{\sqrt{3}}$ . The slopes of the normals are calculated, and the tangent of the angle between them is used to simplify the given expression.

Step-by-Step Solution:

Step 1: Find the derivative of the ellipse equation.
$x^2 + 3y^2 = 9$
$2x + 6y \frac{{dy}}{{dx}} = 0$
$\frac{{dy}}{{dx}} = -\frac{{x}}{{3y}}$

Step 2: Find the slope of the normal.\nSlope of the normal $= -\frac{{dx}}{{dy}} = \frac{{3y}}{{x}}$

Step 3: Calculate the slopes of the normals at the given points.
$m_1 = \left(-\frac{{dx}}{{dy}}\right)_{(3\cos\theta, \sqrt{3}\sin\theta)} = \frac{{3\sqrt{3}\sin\theta}}{{3\cos\theta}} = \sqrt{3}\tan\theta$
$m_2 = \left(-\frac{{dx}}{{dy}}\right)_{(-3\sin\theta, \sqrt{3}\cos\theta)} = \frac{{3\sqrt{3}\cos\theta}}{{-3\sin\theta}} = -\sqrt{3}\cot\theta$

Step 4: Use the formula for the tangent of the angle between two lines.
$\tan\beta = \left| \frac{{m_1 - m_2}}{{1 + m_1 m_2}} \right| = \left| \frac{{\sqrt{3}\tan\theta + \sqrt{3}\cot\theta}}{{1 - 3\tan\theta\cot\theta}} \right| = \left| \frac{{\sqrt{3}(\tan\theta + \cot\theta)}}{{1 - 3}} \right| = \frac{{\sqrt{3}}}{{2}} |\tan\theta + \cot\theta|$

Step 5: Simplify the expression.
$\frac{{1}}{{\cot\beta}} = \frac{{\sqrt{3}}}{{2}} \left| \frac{{\sin\theta}}{{\cos\theta}} + \frac{{\cos\theta}}{{\sin\theta}} \right| = \frac{{\sqrt{3}}}{{2}} \left| \frac{{1}}{{\sin\theta\cos\theta}} \right| = \frac{{\sqrt{3}}}{{\sin 2\theta}}$
$\frac{{2\cot\beta}}{{\sin 2\theta}} = \frac{{2}}{{\sqrt{3}}}$

Why other options are wrong:
- Option A: $\sqrt{2}$ : This is incorrect due to an error in the simplification of the trigonometric expression.
- Option C: $\frac{{1}}{{\sqrt{3}}}$ : This is incorrect due to an error in the simplification of the trigonometric expression.
- Option D: $\frac{{\sqrt{3}}}{{4}}$ : This is incorrect due to an error in the simplification of the trigonometric expression.

Key Concepts:
- Equation of an ellipse
- Slope of a normal
- Angle between two lines
- Trigonometric identities

Memory Aids:
- $\tan(\theta) = \frac{{\sin(\theta)}}{{\cos(\theta)}}$
- $\cot(\theta) = \frac{{\cos(\theta)}}{{\sin(\theta)}}$

Quick Tips:
- Remember the formula for the angle between two lines.
- Simplify trigonometric expressions carefully.

73 Medium

73 of 97

Application +4 | -1

If the tangent at the point

(1,1)

of the curve

y^2-x(2-x)^2

meets the curve again at the point

\mathrm{P}

, then

\mathrm{P}

is ...... .

View answer and explanation

Correct option: C

\left(\frac{9}{4}, \frac{3}{8}\right)

Option C ((9/4, 3/8)) is correct because it is the point where the tangent at (1,1) intersects the curve again, found by solving the system of equations.

Solution approach

Correct Answer + Core Reason: $\left(\frac{{9}}{{4}}, \frac{{3}}{{8}}\right)$ . The point of intersection is found by solving the equations of the curve and the tangent.

Step-by-Step Solution:

Step 1: Find the derivative of the curve $y^2 = x(2-x)^2$ with respect to $x$ .\nUsing implicit differentiation, $2y \frac{{dy}}{{dx}} = (2-x)^2 + x \cdot 2(2-x)(-1) = (2-x)^2 - 2x(2-x) = (2-x)(2-x - 2x) = (2-x)(2-3x)$ .\nSo, $\frac{{dy}}{{dx}} = \frac{{(2-x)(2-3x)}}{{2y}}$ .

Step 2: Evaluate the derivative at the point (1,1): $\frac{{dy}}{{dx}}|_{(1,1)} = \frac{{(2-1)(2-3(1))}}{{2(1)}} = \frac{{(1)(-1)}}{{2}} = -\frac{{1}}{{2}}$ .

Step 3: Find the equation of the tangent line at (1,1) with slope $-\frac{{1}}{{2}}$ : $y - 1 = -\frac{{1}}{{2}}(x - 1)$ , which simplifies to $y = -\frac{{1}}{{2}}x + \frac{{3}}{{2}}$ .

Step 4: Find the intersection points of the tangent line and the curve by substituting the equation of the tangent line into the equation of the curve: $(\frac{{-x}}{{2}} + \frac{{3}}{{2}})^2 = x(2-x)^2$ .

Step 5: Simplify the equation: $\frac{{x^2}}{{4}} - \frac{{3x}}{{2}} + \frac{{9}}{{4}} = x(4 - 4x + x^2) = 4x - 4x^2 + x^3$ .\nMultiply by 4: $x^2 - 6x + 9 = 16x - 16x^2 + 4x^3$ .\nRearrange: $4x^3 - 17x^2 + 22x - 9 = 0$ .

Step 6: Since (1,1) is a point of intersection, $(x-1)$ is a factor. Perform polynomial division: $(4x^3 - 17x^2 + 22x - 9) / (x-1) = 4x^2 - 13x + 9$ .\nFactor the quadratic: $4x^2 - 13x + 9 = (4x - 9)(x - 1)$ .\nSo, $4x^3 - 17x^2 + 22x - 9 = (x-1)(x-1)(4x-9) = 0$ .\nThe roots are $x = 1$ (twice) and $x = \frac{{9}}{{4}}$ .

Step 7: Find the y-coordinate for $x = \frac{{9}}{{4}}$ : $y = -\frac{{1}}{{2}}(\frac{{9}}{{4}}) + \frac{{3}}{{2}} = -\frac{{9}}{{8}} + \frac{{12}}{{8}} = \frac{{3}}{{8}}$ .

Final Answer: $\left(\frac{{9}}{{4}}, \frac{{3}}{{8}}\right)$

Why other options are wrong:
- Option A: Incorrect. This point does not lie on the curve and the tangent line.
- Option B: Incorrect. This point does not lie on the curve and the tangent line.
- Option D: Incorrect. This point does not lie on the curve and the tangent line.

Key Concepts:
- Implicit differentiation
- Equation of a tangent line
- Finding intersection points of curves

Memory Aids:
- Remember the equation of a tangent line: $y - y_1 = m(x - x_1)$ .

Quick Tips:
- When finding intersection points, substitute one equation into the other and solve for the variables.

74 Medium

74 of 97

Application +4 | -1

If the curve

y = 2{x^3} + a{x^2} + bx + c

passes through the origin and the tangents drawn to it at

x = - 1

and

x = 2

are parallel to the

x

-axis, then the values of

a,b

and care respectively

View answer and explanation

Correct option: B

- 3, - 12\mathrm{ }and\mathrm{ }0

Option B ( - 3, - 12 and 0) is correct because these values satisfy the given conditions: the curve passes through the origin, and the tangents at x = -1 and x = 2 are parallel to the x-axis.

Solution approach

Correct Answer + Core Reason: $- 3, - 12\mathrm{ }and\mathrm{ }0$ . These values satisfy the given conditions.

Step-by-Step Solution:

Step 1: Given the curve equation: $y = 2{x^3} + a{x^2} + bx + c$ .

Step 2: Since the curve passes through the origin (0, 0), substitute x = 0 and y = 0 into the equation:
$0 = 2{\left( 0 \right)^3} + a{\left( 0 \right)^2} + b\left( 0 \right) + c \Rightarrow c = 0$ .

Step 3: Find the derivative of the curve equation:
$\frac{{dy}}{{dx}} = 6{x^2} + 2ax + b$ .

Step 4: Since the tangents at x = -1 and x = 2 are parallel to the x-axis, $\frac{{dy}}{{dx}} = 0$ at these points.\nAt $x = - 1$ : $6{\left( { - 1} \right)^2} + 2a\left( { - 1} \right) + b = 0 \Rightarrow 6 - 2a + b = 0$ .\nAt $x = 2$ : $6{\left( 2 \right)^2} + 2a\left( 2 \right) + b = 0 \Rightarrow 24 + 4a + b = 0$ .

Step 5: Solve the system of equations:
$6 - 2a + b = 0$ and $24 + 4a + b = 0$ .\nSubtracting the first equation from the second gives: $18 + 6a = 0 \Rightarrow a = - 3$ .\nSubstituting a = -3 into the first equation: $6 - 2\left( { - 3} \right) + b = 0 \Rightarrow 6 + 6 + b = 0 \Rightarrow b = - 12$ .

Step 6: The values are $a = - 3$ , $b = - 12$ , and $c = 0$ .

Why other options are wrong:
- Option A: $12, - 3\mathrm{ }and\mathrm{ }0$ - These values do not satisfy the derivative conditions at x = -1 and x = 2.
- Option C: $- 3,12\mathrm{ }and\mathrm{ }0$ - The value of b does not satisfy the derivative conditions.
- Option D: $3, - 12\mathrm{ }and\mathrm{ }0$ - The value of a does not satisfy the derivative conditions.

Key Concepts:
- Derivatives and tangents
- Solving systems of equations

Memory Aids:
- A tangent parallel to the x-axis has a slope of 0.

Quick Tips:
- Find the derivative, set it to zero at the given points, and solve for the unknowns.

75 Medium

75 of 97

Application +4 | -1

The point on the curve

y = 2{x^2} - 6x - 4

at which the tangent is parallel to the

x

-axis, is

View answer and explanation

Correct option: D

\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)

Option D ((

\frac{3}{2}

,\\; -

\frac{{17}}{2}

)) is correct because it represents the point on the curve where the derivative (slope of the tangent) is zero, indicating a horizontal tangent.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{3}{2},\\; - \frac{{17}}{2}} \right)$ . The point is found by setting the derivative equal to zero.

Step-by-Step Solution:

Step 1: Find the derivative: $y = 2{x^2} - 6x - 4 \Rightarrow \frac{{dy}}{{dx}} = 4x - 6$ .

Step 2: Set the derivative equal to zero: $4x - 6 = 0 \Rightarrow x = \frac{3}{2}$ .

Step 3: Substitute x = 3/2 into the original equation: $y = 2(\frac{3}{2})^2 - 6(\frac{3}{2}) - 4 = 2(\frac{9}{4}) - 9 - 4 = \frac{9}{2} - 13 = \frac{{9 - 26}}{2} = - \frac{{17}}{2}$ .

Step 4: The point is (3/2, -17/2).

Why other options are wrong:
- **Option A (( $\frac{3}{2}$ , $\frac{{13}}{2}$ )): This is incorrect because the y-coordinate is calculated wrong.
- Option B (( $\frac{{ - 5}}{2}$ ,\\; - $\frac{{17}}{2}$ )): This is incorrect because the x-coordinate is calculated wrong.
- Option C (( $\frac{3}{2}$ , $\frac{{17}}{2}$ )): This is incorrect because the y-coordinate is calculated wrong.

Key Concepts:
- Derivatives
- Tangent to a curve

Memory Aids:
- Tangent parallel to x-axis means derivative is zero.

Quick Tips:**
- Always find the derivative first.
- Substitute the x-value back into the original equation to find the y-value.

76 Medium

76 of 97

Analyzing +4 | -1

The normal to the curve

x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)

at any point

\theta

is such that

View answer and explanation

Correct option: A

It is at a constant distance from the origin

Option A (It is at a constant distance from the origin) is correct because the equation of the normal can be simplified to a form where the distance from the origin is constant.

Solution approach

Correct Answer + Core Reason: It is at a constant distance from the origin. The equation of the normal can be simplified to a form where the distance from the origin is constant.

Step-by-Step Solution:

Step 1: Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ .
$x = a(\cos \theta + \theta \sin \theta) \Rightarrow \frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta$
$y = a(\sin \theta - \theta \cos \theta) \Rightarrow \frac{dy}{d\theta} = a(\cos \theta - \cos \theta + \theta \sin \theta) = a\theta \sin \theta$

Step 2: Find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta$

Step 3: Find the equation of the normal.\nThe slope of the normal is $-\frac{1}{\tan \theta} = -\cot \theta = -\frac{\cos \theta}{\sin \theta}$ .\nThe equation of the normal is $y - a(\sin \theta - \theta \cos \theta) = -\frac{\cos \theta}{\sin \theta}(x - a(\cos \theta + \theta \sin \theta))$ .

Step 4: Simplify the equation of the normal.
$y \sin \theta - a\sin^2 \theta + a\theta \cos \theta \sin \theta = -x \cos \theta + a\cos^2 \theta + a\theta \sin \theta \cos \theta$
$x \cos \theta + y \sin \theta = a(\sin^2 \theta + \cos^2 \theta) = a$

Step 5: Determine the distance from the origin.\nThe equation $x \cos \theta + y \sin \theta = a$ represents a line whose perpendicular distance from the origin is $|a|$ . Since $a$ is a constant, the distance from the origin is constant.

Why other options are wrong:
- Option B: It passes through $\left( {\frac{{a\pi }}{{2}},\; - a} \right)$ : Substituting the point into the equation of the normal does not satisfy the equation.
- Option C: It makes angle $\frac{\pi }{2} - \theta$ with the $x$ -axis: The slope of the normal is $-\cot \theta$ , so the angle it makes with the x-axis is not $\frac{\pi }{2} - \theta$ .
- Option D: It passes through the origin: The equation of the normal is $x \cos \theta + y \sin \theta = a$ . If it passes through the origin (0,0), then $a = 0$ , which is not generally true.

Key Concepts:
- Parametric differentiation
- Equation of a normal to a curve
- Distance of a line from the origin

Memory Aids:
- Remember the parametric differentiation formulas.
- Remember the equation of a line in normal form.

Quick Tips:
- Simplify the equation of the normal to determine its properties.

77 Medium

77 of 97

Application +4 | -1

If the curves

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{12}} = 1\;

and

{y^3} = 8x\;

intersect at right angle then the value of

{a^2}

is equal to

View answer and explanation

Correct option: D

4

Option D (4) is correct because the value of

{a^2}

is 4 when the given curves intersect at right angles, which is found by equating the product of their slopes to -1.

Solution approach

Correct Answer + Core Reason: 4. The value of ${a^2}$ is found by using the condition for orthogonal intersection of two curves.

Step-by-Step Solution:

Step 1: Given curves are $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{12}} = 1$ and ${y^3} = 8x$ .

Step 2: Differentiate the first equation with respect to $x$ : $\frac{{2x}}{{{a^2}}} + \frac{{2y}}{{12}}\frac{{dy}}{{dx}} = 0$ .

Step 3: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = - \frac{{12x}}{{{{a^2}y}}} = {m_1}$ (slope of the first curve).

Step 4: Differentiate the second equation with respect to $x$ : $3{y^2}\frac{{dy}}{{dx}} = 8$ .

Step 5: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = \frac{8}{{3{y^2}}} = {m_2}$ (slope of the second curve).

Step 6: Apply the condition for orthogonal intersection: ${m_1}{m_2} = -1$ .

Step 7: Substitute the expressions for ${m_1}$ and ${m_2}$ : $\left( { - \frac{{12x}}{{{{a^2}y}}} }\right)\left( {\frac{8}{{3{y^2}}}} \right) = -1$ .

Step 8: Simplify the equation: $\frac{{96x}}{{3{a^2}{y^3}}} = 1$ .

Step 9: Substitute ${y^3} = 8x$ into the equation: $\frac{{96x}}{{3{a^2}(8x)}} = 1$ .

Step 10: Simplify and solve for ${a^2}$ : $\frac{{96x}}{{24{a^2}x}} = 1 \Rightarrow 24{a^2} = 96 \Rightarrow {a^2} = \frac{{96}}{{24}} = 4$ .

Why other options are wrong:
- Option A: 16: This is incorrect because it does not satisfy the condition for orthogonal intersection after substituting back into the equations.
- Option B: 12: This is incorrect because it does not satisfy the condition for orthogonal intersection after substituting back into the equations.
- Option C: 8: This is incorrect because it does not satisfy the condition for orthogonal intersection after substituting back into the equations.

Key Concepts:
- Differentiation
- Orthogonal intersection of curves
- Slopes of curves

Memory Aids:
- Condition for orthogonal intersection: ${m_1}{m_2} = -1$

Quick Tips:
- Remember to differentiate implicitly when dealing with equations involving both $x$ and $y$ .
- Always check your answer by substituting it back into the original equations.

78 Medium

78 of 97

Application +4 | -1

Which one of the following curves is the orthogonal trajectory of straight lines passing through a fixed point

(a, b)

?

View answer and explanation

Correct option: D

(x-a)^{2}+(y-b)^{2}=k

Option D (

(x-a)^{2}+(y-b)^{2}=k

) is correct because straight lines passing through a fixed point (a, b) are normals to a circle centered at (a, b), making the circle the orthogonal trajectory.

Solution approach

Correct Answer + Core Reason: $(x-a)^{2}+(y-b)^{2}=k$ . Lines passing through a fixed point (a,b) are normals to a circle centered at (a,b).

Step-by-Step Solution:

Step 1: Consider the family of straight lines passing through the fixed point (a, b). The equation of such a line can be written as $y - b = m(x - a)$ , where m is the slope.

Step 2: To find the orthogonal trajectory, we need to find a family of curves that intersect these lines at right angles. This means the slope of the tangent to the orthogonal trajectory at any point (x, y) is the negative reciprocal of the slope of the line passing through (a, b).

Step 3: Differentiate the equation of the line with respect to x: $\frac{dy}{dx} = m$ .

Step 4: The slope of the orthogonal trajectory is $-\frac{1}{m}$ .

Step 5: Replace m with $\frac{y-b}{x-a}$ in the expression for the slope of the orthogonal trajectory: $-\frac{1}{m} = -\frac{x-a}{y-b}$ .

Step 6: Consider the equation of a circle centered at (a, b): $(x - a)^2 + (y - b)^2 = k$ , where k is a constant.

Step 7: Differentiate the equation of the circle with respect to x: $2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x - a}{y - b}$ .

Step 8: The slope of the tangent to the circle is the same as the slope of the orthogonal trajectory. Therefore, the orthogonal trajectory of the straight lines passing through (a, b) is a circle centered at (a, b).

Why other options are wrong:
- **Option A ( $x-a=k(y-b)$ ): This represents another family of straight lines, not an orthogonal trajectory.
- Option B ( $(x-a)(y-b)=k$ ): This represents a hyperbola, not an orthogonal trajectory to the given lines.
- Option C ( $(x-a)^{2}=k(y-b)$ ): This represents a parabola, not an orthogonal trajectory to the given lines.

Key Concepts:
- Orthogonal trajectories
- Differentiation
- Equation of a straight line
- Equation of a circle

Memory Aids:
- Orthogonal means perpendicular.

Quick Tips:**
- Remember that the slopes of orthogonal curves are negative reciprocals of each other.

79 Medium

79 of 97

Application +4 | -1

The point of inflexion for the curve

y = {x^{5/2}}

is

View answer and explanation

Correct option: B

(0,0)

Option B ((0,0)) is correct because it's the point where the second derivative changes sign, indicating a point of inflection.

Solution approach

Correct Answer + Core Reason: (0,0). The point of inflection occurs where the second derivative is zero and changes sign.

Step-by-Step Solution:

Step 1: Find the first derivative of $y = x^{5/2}$ : $\frac{{dy}}{{dx}} = \frac{5}{2}x^{3/2}$ .

Step 2: Find the second derivative of $y = x^{5/2}$ : $\frac{{d^2y}}{{dx^2}} = \frac{15}{4}x^{1/2}$ .

Step 3: Set the second derivative equal to zero to find potential points of inflection: $\frac{15}{4}x^{1/2} = 0 \Rightarrow x = 0$ .

Step 4: Check if the second derivative changes sign at x = 0. For $x < 0$ , the function is not defined. For $x > 0$ , the second derivative is positive. However, since the function is only defined for $x \ge 0$ , we consider the point (0,0).

Step 5: When x=0, y = 0. Therefore, the point is (0,0).

Step 6: Analyze the behavior of the second derivative around x=0. Since the second derivative is 0 at x=0 and positive for x>0, (0,0) is a point of inflection.

Why other options are wrong:
- Option A (1,1): At (1,1), the second derivative is $\frac{15}{4}$ , which is not zero, so it's not a point of inflection.
- Option C (1,0): This point does not lie on the curve $y = x^{5/2}$ since $0 \neq 1^{5/2}$ .
- Option D (0,1): This point does not lie on the curve $y = x^{5/2}$ since $1 \neq 0^{5/2}$ .

Key Concepts:
- Points of inflection
- Second derivative test

Memory Aids:
- A point of inflection occurs where the concavity of the curve changes.

Quick Tips:
- Find the second derivative and set it equal to zero to find potential points of inflection.
- Check if the second derivative changes sign at the potential points of inflection.

80 Medium

80 of 97

Analyzing +4 | -1

The point in the interval

\;\left[ {0\;,2\pi } \right]

, where

f\left( x \right) = {e^x}\;\sin x\;

has maximum slope, is

View answer and explanation

Correct option: B

\frac{\pi }{2}

Option B (

\frac{\pi}{2}

) is correct because the second derivative of the function is zero at this point, and the third derivative is negative, indicating a maximum slope.

Solution approach

Correct Answer + Core Reason: $\frac{\pi}{2}$ . The maximum slope occurs where the second derivative is zero and the third derivative is negative.

Step-by-Step Solution:

Step 1: Find the first derivative of the function $f(x) = e^x \sin x$ .
$f'(x) = e^x \cos x + e^x \sin x = e^x (\cos x + \sin x)$

Step 2: Find the second derivative of the function.
$f''(x) = e^x (\cos x + \sin x) + e^x (-\sin x + \cos x) = 2e^x \cos x$

Step 3: Find the third derivative of the function.
$f'''(x) = 2e^x \cos x + 2e^x(-\sin x) = 2e^x (\cos x - \sin x)$

Step 4: Set the second derivative equal to zero to find potential points of maximum or minimum slope.
$f''(x) = 0 \Rightarrow 2e^x \cos x = 0 \Rightarrow \cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$ in the interval $[0, 2\pi]$ .

Step 5: Evaluate the third derivative at the potential points.\nAt $x = \frac{\pi}{2}$ , $f'''(x) = 2e^{\frac{\pi}{2}} (\cos \frac{\pi}{2} - \sin \frac{\pi}{2}) = 2e^{\frac{\pi}{2}} (0 - 1) = -2e^{\frac{\pi}{2}} < 0$ , which indicates a maximum slope.\nAt $x = \frac{3\pi}{2}$ , $f'''(x) = 2e^{\frac{3\pi}{2}} (\cos \frac{3\pi}{2} - \sin \frac{3\pi}{2}) = 2e^{\frac{3\pi}{2}} (0 - (-1)) = 2e^{\frac{3\pi}{2}} > 0$ , which indicates a minimum slope.

Step 6: Conclude that the maximum slope occurs at $x = \frac{\pi}{2}$ .

Why other options are wrong:
- Option A: $\frac{\pi}{4}$ is incorrect because the second derivative is not zero at this point.
- Option C: $\pi$ is incorrect because the second derivative is not zero at this point.
- Option D: $\frac{3\pi}{2}$ is incorrect because the third derivative is positive at this point, indicating a minimum slope.

Key Concepts:
- Finding points of maximum/minimum slope using derivatives.
- Second derivative test for concavity.

Memory Aids:
- Maximum slope occurs when the second derivative is zero and the third derivative is negative.

Quick Tips:
- Remember to find both the second and third derivatives to determine whether a point is a maximum or minimum slope.

81 Medium

81 of 97

Application +4 | -1

The intercepts made by the tangent to the curve

y = \int \limits_0^x \left| t \right|dt

, which is parallel to the line

y = 2x

, on

y

-axis are equal to

View answer and explanation

Correct option: B

- 2,2

Option B ( - 2,2) is correct because these are the y-intercepts of the tangent lines to the curve that are parallel to the line y = 2x, found by equating the derivatives and solving for the points of tangency.

Solution approach

Correct Answer + Core Reason: $- 2,2$ . The y-intercepts are found by equating the derivatives and solving for the points of tangency.

Step-by-Step Solution:

Step 1: Given $y = \int_0^x |t| dt$ , find the derivative $\frac{dy}{dx}$ . Since $\frac{d}{dx} \int_0^x f(t) dt = f(x)$ , we have $\frac{dy}{dx} = |x|$ .

Step 2: The tangent is parallel to the line $y = 2x$ , so the slope of the tangent is 2. Therefore, $|x| = 2$ , which means $x = 2$ or $x = -2$ .

Step 3: Find the corresponding $y$ values for $x = 2$ and $x = -2$ .\nFor $x = 2$ , $y = \int_0^2 |t| dt = \int_0^2 t dt = [\frac{t^2}{2}]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2$ .\nFor $x = -2$ , $y = \int_0^{-2} |t| dt = -\int_{-2}^0 t dt = -[\frac{t^2}{2}]_{-2}^0 = -(\frac{0^2}{2} - \frac{(-2)^2}{2}) = -(-2) = 2$ . However, since x is negative, we need to consider the integral from 0 to x as -integral from x to 0. So, y = - ∫_x⁰ |t| dt = - ∫_{-2}^0 -t dt = ∫_{-2}^0 t dt = [t²/2]_{-2}^0 = 0 - 4/2 = -2. So y = 2 for x = 2 and y = 2 for x = -2.\nThere is an error in the original solution. For x = -2, y = ∫_0^-2 |t| dt = ∫_{-2}^0 -t dt = [-t²/2]_{-2}^0 = 0 - (-4/2) = 2.

Step 4: Find the equation of the tangent line for each point.\nFor (2, 2), the equation is $y - 2 = 2(x - 2) \Rightarrow y = 2x - 2$ . The y-intercept is -2.\nFor (-2, 2), the equation is $y - 2 = 2(x + 2) \Rightarrow y = 2x + 6$ . There is an error here. Since y = ∫_0^x |t| dt, when x < 0, y = ∫_0^x -t dt. So, when x = -2, y = ∫_0^-2 -t dt = [-t²/2]

82 Medium

82 of 97

Application +4 | -1

The point at which the tangent to the curve

y = 2{x^2} - x + 1

is parallel to

y = 3x + 9

, will be

View answer and explanation

Correct option: B

(1,2)

Option B (1,2) is correct because the tangent to the curve is parallel to the line when their slopes are equal, which occurs at the point (1, 2).

Solution approach

Correct Answer + Core Reason: (1,2). The point is found by equating the derivative of the curve to the slope of the line.

Step-by-Step Solution:

Step 1: Find the derivative of the curve $y = 2x^2 - x + 1$ : $\frac{dy}{dx} = 4x - 1$ .

Step 2: Find the slope of the line $y = 3x + 9$ : The slope is 3.

Step 3: Set the derivative equal to the slope: $4x - 1 = 3$ .

Step 4: Solve for x: $4x = 4$ , so $x = 1$ .

Step 5: Substitute x = 1 into the equation of the curve to find the corresponding y-value: $y = 2(1)^2 - 1 + 1 = 2$ .

Step 6: The point is (1, 2).

Why other options are wrong:
- Option A: This is incorrect. The point (2, 1) does not lie on the curve and the tangent at x=2 is not parallel to the given line.
- Option C: This is incorrect. The point (3, 9) does not lie on the curve and the tangent at x=3 is not parallel to the given line.
- Option D: This is incorrect. The point (-2, 1) does not lie on the curve and the tangent at x=-2 is not parallel to the given line.

Key Concepts:
- Derivative as slope of tangent
- Parallel lines have equal slopes

Memory Aids:
- The derivative of a function at a point gives the slope of the tangent line at that point.

Quick Tips:
- To find where a tangent line is parallel to another line, equate the derivative of the function to the slope of the line.

83 Medium

83 of 97

Application +4 | -1

Let

l

be the line through

(0,0)

and tangent to the curve

y=x^{3}+x+16

. Then the slope of

l

equal to :

View answer and explanation

Correct option: D

13

Option D (13) is correct because finding the tangent line equation and using the point of tangency allows us to solve for the slope 'm', which equals 13.

Solution approach

Correct Answer + Core Reason: 13. The slope of the line through (0,0) and tangent to the curve is found by equating the line's equation to the curve and using the tangency condition.

Step-by-Step Solution:

Step 1: Let the equation of the line be $y = mx$ .

Step 2: Let the point of tangency be $(x_1, y_1)$ . Then $y_1 = x_1^3 + x_1 + 16$ and $y_1 = mx_1$ .

Step 3: Differentiate the curve equation: $\frac{dy}{dx} = 3x^2 + 1$ .

Step 4: The slope of the tangent at $(x_1, y_1)$ is $m = 3x_1^2 + 1$ .

Step 5: Substitute $y_1 = mx_1$ into the curve equation: $mx_1 = x_1^3 + x_1 + 16$ .

Step 6: Substitute $m = 3x_1^2 + 1$ into the equation from Step 5: $(3x_1^2 + 1)x_1 = x_1^3 + x_1 + 16 \Rightarrow 3x_1^3 + x_1 = x_1^3 + x_1 + 16 \Rightarrow 2x_1^3 = 16 \Rightarrow x_1^3 = 8 \Rightarrow x_1 = 2$ .

Step 7: Substitute $x_1 = 2$ into the equation for m: $m = 3(2^2) + 1 = 3(4) + 1 = 12 + 1 = 13$ .

Why other options are wrong:
- Option A (10): This is incorrect because it results from an incorrect calculation of the slope.
- Option B (11): This is incorrect because it results from an incorrect calculation of the slope.
- Option C (17): This is incorrect because it results from an incorrect calculation of the slope.

Key Concepts:
- Equation of a tangent line
- Differentiation

Memory Aids:
- The slope of the tangent line is the derivative evaluated at the point of tangency.

Quick Tips:
- Remember to find the derivative before finding the tangent line equation.

84 Medium

84 of 97

Analyzing +4 | -1

If the curves

\frac{x^2}{a^2}+\frac{y^2}{12}=1

and

y^3=8 x

intersect at right angles, then the value of

a^2

is equal to

View answer and explanation

Correct option: D

4

Option D (4) is correct because it is the value of

a^2

that satisfies the condition for the curves to intersect at right angles.

Solution approach

Correct Answer + Core Reason: 4. The value of $a^2$ is found by using the condition that the product of the slopes of the tangents at the point of intersection is -1.

Step-by-Step Solution:

Step 1: Find the derivative of the first curve $\frac{x^2}{a^2} + \frac{y^2}{12} = 1$ .\nDifferentiate with respect to x:
$\frac{2x}{a^2} + \frac{2y}{12} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{12x}{a^2y}$

Step 2: Find the derivative of the second curve $y^3 = 8x$ .\nDifferentiate with respect to x:
$3y^2 \frac{dy}{dx} = 8$
$\frac{dy}{dx} = \frac{8}{3y^2}$

Step 3: Since the curves intersect at right angles, the product of their slopes is -1.
$\left(-\frac{12x}{a^2y}\right) \left(\frac{8}{3y^2}\right) = -1$
$\frac{-96x}{3a^2y^3} = -1$
$\frac{32x}{a^2y^3} = 1$

Step 4: Substitute $y^3 = 8x$ into the equation.
$\frac{32x}{a^2(8x)} = 1$
$\frac{4}{a^2} = 1$
$a^2 = 4$

Why other options are wrong:
- Option A (16): If $a^2 = 16$ , then $\frac{32x}{16(8x)} = \frac{1}{4} \neq 1$ , so the curves do not intersect at right angles.
- Option B (12): If $a^2 = 12$ , then $\frac{32x}{12(8x)} = \frac{1}{3} \neq 1$ , so the curves do not intersect at right angles.
- Option C (8): If $a^2 = 8$ , then $\frac{32x}{8(8x)} = \frac{1}{2} \neq 1$ , so the curves do not intersect at right angles.

Key Concepts:
- Implicit differentiation
- Orthogonal intersection of curves
- Product of slopes for perpendicular lines

Memory Aids:
- Two curves intersect at right angles if the product of their slopes at the point of intersection is -1.

Quick Tips:
- Remember to use implicit differentiation when the function is not explicitly defined as y = f(x).

85 Medium

85 of 97

Analyzing +4 | -1

The maximum slope of the curve

y = - {x^3} + 3{x^2} + 2x - 27

is

View answer and explanation

Correct option: A

5

Option A (5) is correct because it is the maximum value of the derivative of the given function, found by setting the second derivative to zero.

Solution approach

Correct Answer + Core Reason: 5. The maximum slope is found by maximizing the derivative of the function.

Step-by-Step Solution:

Step 1: Find the first derivative of the function $y = -x^3 + 3x^2 + 2x - 27$ .
$\frac{{dy}}{{dx}} = -3x^2 + 6x + 2$

Step 2: Let $z = \frac{{dy}}{{dx}} = -3x^2 + 6x + 2$ . Find the derivative of $z$ with respect to $x$ .
$\frac{{dz}}{{dx}} = -6x + 6$

Step 3: To find the maximum or minimum, set $\frac{{dz}}{{dx}} = 0$ .
$-6x + 6 = 0$
$x = 1$

Step 4: Find the second derivative of $z$ with respect to $x$ .
$\frac{{d^2z}}{{dx^2}} = -6$ \nSince $\frac{{d^2z}}{{dx^2}} < 0$ , the value is maximum at $x = 1$ .

Step 5: Substitute $x = 1$ into the equation for $z$ .
$z = -3(1)^2 + 6(1) + 2 = -3 + 6 + 2 = 5$

Step 6: The maximum slope is 5.

Why other options are wrong:
- Option B (-5): This is the negative of the correct answer. It might be obtained by incorrectly calculating the derivative or not considering the sign.
- Option C (1/5): This is the reciprocal of the correct answer. There's no direct mathematical operation that would lead to this value.
- Option D (None of these): This is incorrect because the maximum slope is indeed 5, which is a valid option.

Key Concepts:
- Differentiation
- Finding maxima and minima
- Second derivative test

Memory Aids:
- To find the maximum or minimum of a function, set its derivative equal to zero.
- If the second derivative is negative, the function has a maximum at that point.

Quick Tips:
- Always find the first derivative and set it to zero to find critical points.
- Use the second derivative test to determine whether the critical point is a maximum or minimum.

86 Medium

86 of 97

Analyzing +4 | -1

The shortest distance between the line

y - x = 1

and the curve

x = {y^2}

is

View answer and explanation

Correct option: A

\frac{{3\sqrt 2 }}{8}

Option A (

\frac{{3\sqrt 2 }}{8}

) is correct because it represents the shortest distance between the line and the curve, calculated using the formula for the distance between a point and a line.

Solution approach

Correct Answer + Core Reason: $\frac{{3\sqrt 2 }}{8}$ . The shortest distance is found using the formula for the distance between a point and a line.

Step-by-Step Solution:\nStep 1: Write the equation of the line: $y - x = 1 \Rightarrow x - y + 1 = 0$ \nStep 2: Write the equation of the curve: $x = {y^2}$ \nStep 3: Find the point on the curve where the tangent is parallel to the line. The slope of the line is 1.\nStep 4: Differentiate the equation of the curve with respect to x: $1 = 2y\frac{{dy}}{{dx}} \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2y}}$ \nStep 5: Set the derivative equal to the slope of the line: $\frac{1}{{2y}} = 1 \Rightarrow y = \frac{1}{2}$ \nStep 6: Find the corresponding x-coordinate: $x = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$ \nStep 7: The point on the curve is $\left( {\frac{1}{4},\frac{1}{2}} \right)$ .\nStep 8: Use the formula for the distance between a point and a line: $d = \frac{{\left| {Ax + By + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$ \nStep 9: Substitute the values: $d = \frac{{\left| {1\left( {\frac{1}{4}} \right) - 1\left( {\frac{1}{2}} \right) + 1} \right|}}{{\sqrt {{1^2} + {{( - 1)}^2}} }} = \frac{{\left| {\frac{1}{4} - \frac{1}{2} + 1} \right|}}{{\sqrt 2 }} = \frac{{\left| {\frac{3}{4}} \right|}}{{\sqrt 2 }} = \frac{3}{{4\sqrt 2 }}$ \nStep 10: Rationalize the denominator: $d = \frac{3}{{4\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{3\sqrt 2 }}{8}$

Why other options are wrong:
- **Option B ( $\frac{{2\sqrt 3 }}{8}$ ): Incorrect because the distance is calculated incorrectly.
- Option C ( $\frac{{3\sqrt 2 }}{5}$ ): Incorrect because the distance is calculated incorrectly.
- Option D ( $\frac{{\sqrt 3 }}{4}$ ): Incorrect because the distance is calculated incorrectly.

Key Concepts:
- Derivatives
- Tangent to a Curve
- Distance between a Point and a Line

Memory Aids:
- The shortest distance between a line and a curve occurs where the tangent to the curve is parallel to the line.

Quick Tips:**
- Remember the formula for the distance between a point and a line.

87 Medium

87 of 97

Application +4 | -1

The equation of the tangent to the curve

y=\sqrt{9-2 x^2}

at the point where the ordinate & the abscissa are equal is

View answer and explanation

Correct option: D

2 x+y-3 \sqrt{3}=0

Option D (

2 x+y-3 \sqrt{{3}}=0

) is correct because it is the equation of the tangent calculated at the point where the ordinate and abscissa are equal on the curve.

Solution approach

Correct Answer + Core Reason: $2 x+y-3 \sqrt{{3}}=0$ . This is the equation of the tangent at the specified point.

Step-by-Step Solution:

Step 1: Find the point on the curve where the ordinate and abscissa are equal. Let the point be $(x_1, y_1)$ . Given $x_1 = y_1$ .

Step 2: Substitute $x_1$ for $y$ in the equation $y = \sqrt{{9 - 2x^2}}$ :\nx_1 = $\sqrt{{9 - 2x_1^2}}$ $\n$ x_1² = 9 - 2x_1² $\n$ 3x_1² = 9 $\n$ x_1² = 3 $\n$ x_1 = ± $\sqrt{{3}}$ $\nSince$ y > 0 $, the point is$ ( $\sqrt{{3}}$ , $\sqrt{{3}}$ ) $.\n\n**Step 3:** Find the derivative$ $\frac{{dy}}{{dx}}$ $of the curve$ y = $\sqrt{{9 - 2x^2}}$ $:\n$ y² = 9 - 2x² $\n$ 2y $\frac{{dy}}{{dx}}$ = -4x $\n$ $\frac{{dy}}{{dx}}$ = $\frac{{-2x}}{{y}}$ $\n\n**Step 4:** Evaluate the derivative at the point$ ( $\sqrt{{3}}$ , $\sqrt{{3}}$ ) $:\n$ \left( $\frac{{dy}}{{dx}}$ \right)_{{( $\sqrt{{3}}$ , $\sqrt{{3}}$ )}} = \frac{{-2 $\sqrt{{3}}$ }}{{ $\sqrt{{3}}$ }} = -2 $\n\n**Step 5:** Find the equation of the tangent using the point-slope form:$ y - y_1 = m(x - x_1) $, where$ m = -2 $and$ (x_1, y_1) = ( $\sqrt{{3}}$ , $\sqrt{{3}}$ ) $:\ny - \sqrt{{3}} = -2(x - \sqrt{{3}})$ \ny - $\sqrt{{3}}$ = -2x + 2 $\sqrt{{3}}$ $\n$ 2x + y - 3 $\sqrt{{3}}$ = 0 $\n\n**Why other options are wrong:**\n- **Option A ($ 2 x+y- $\sqrt{{3}}$ =0 $):** This equation does not pass through the point$ ( $\sqrt{{3}}$ , $\sqrt{{3}}$ ) $.\n- **Option B ($ 2 x+y-3=0 $):** This equation does not pass through the point$ ( $\sqrt{{3}}$ , $\sqrt{{3}}$ ) $.\n- **Option C ($ 2 x-y-3 $\sqrt{{3}}$ =0 $):** The slope of this line is incorrect.\n\n**Key Concepts:**\n- Finding the derivative of a function.\n- Equation of a tangent to a curve.\n\n**Memory Aids:**\n- Remember the point-slope form of a line:$ y - y_1 = m(x - x_1) $.\n\n**Quick Tips:**\n- Always find the point of tangency before finding the equation of the tangent.$

88 Medium

88 of 97

Application +4 | -1

The length of subtangent to the curve

{x^2}{y^2} = {a^4}

at the point

\left( { - a,\;a} \right)

is

View answer and explanation

Correct option: C

a

Option C (

a

) is correct because the length of the subtangent is calculated by finding the derivative and using the formula for subtangent length.

Solution approach

Correct Answer + Core Reason: $a$ . The length of the subtangent is found by calculating the derivative and applying the subtangent formula.

Step-by-Step Solution:

Step 1: Given the equation of the curve: ${x^2}{y^2} = {a^4}$ .

Step 2: Differentiate the equation with respect to $x$ :
$2x^2y \frac{{dy}}{{dx}} + 2xy^2 = 0$

Step 3: Solve for $\frac{{dy}}{{dx}}$ :
$\frac{{dy}}{{dx}} = - \frac{{2xy^2}}{{2x^2y}} = - \frac{y}{x}$

Step 4: Evaluate $\frac{{dy}}{{dx}}$ at the point $(-a, a)$ :
$\left( \frac{{dy}}{{dx}} \right)_{(-a, a)} = - \frac{a}{{-a}} = 1$

Step 5: Calculate the length of the subtangent using the formula: Subtangent = $\frac{y}{{\frac{{dy}}{{dx}}}}$ \nSubtangent = $\frac{a}{1} = a$

Why other options are wrong:
- Option A: $3a$ is incorrect. This might be obtained by an incorrect differentiation or an error in applying the subtangent formula.
- Option B: $2a$ is incorrect. This might arise from a mistake in simplifying the derivative or applying the subtangent formula.
- Option D: $4a$ is incorrect. This is likely due to a significant error in the differentiation or the subtangent calculation.

Key Concepts:
- Differentiation of implicit functions
- Length of subtangent formula: $\frac{y}{{\frac{{dy}}{{dx}}}}$

Memory Aids:
- Remember the formula for the length of the subtangent.
- Implicit differentiation involves differentiating each term with respect to $x$ and using the chain rule where necessary.

Quick Tips:
- Always double-check the differentiation and simplification steps.
- Ensure the correct formula for the subtangent is used.

89 Medium

89 of 97

Application +4 | -1

The equation(s) of the tangent(s) to the curve

y = {x^4}

from the point (2, 0) not on the curve is given by

View answer and explanation

Correct option: C

y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)

Option C (

y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)

) is correct because it represents the tangent to the curve

y = {x^4}

from the point (2, 0) not on the curve.

Solution approach

Correct Answer + Core Reason: $y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)$ . This is the equation of the tangent line to the curve $y=x^4$ passing through (2,0).

Step-by-Step Solution:

Step 1: Let the point of tangency be $(x_1, y_1)$ on the curve $y = x^4$ . Then $y_1 = x_1^4$ .

Step 2: Find the derivative of $y$ with respect to $x$ : $\frac{dy}{dx} = 4x^3$ .

Step 3: The slope of the tangent at $(x_1, y_1)$ is $4x_1^3$ .

Step 4: The equation of the tangent line at $(x_1, y_1)$ is $y - y_1 = 4x_1^3(x - x_1)$ .

Step 5: Since the tangent passes through (2, 0), substitute $x = 2$ and $y = 0$ into the tangent equation: $0 - x_1^4 = 4x_1^3(2 - x_1)$ .

Step 6: Simplify the equation: $-x_1^4 = 8x_1^3 - 4x_1^4 \Rightarrow 3x_1^4 - 8x_1^3 = 0 \Rightarrow x_1^3(3x_1 - 8) = 0$ .

Step 7: Solve for $x_1$ : $x_1 = 0$ or $x_1 = \frac{8}{3}$ .

Step 8: If $x_1 = 0$ , then $y_1 = 0^4 = 0$ . The tangent equation is $y - 0 = 4(0)^3(x - 0) \Rightarrow y = 0$ .

Step 9: If $x_1 = \frac{8}{3}$ , then $y_1 = (\frac{8}{3})^4 = \frac{4096}{81}$ . The slope is $4(\frac{8}{3})^3 = 4(\frac{512}{27}) = \frac{2048}{27}$ .

Step 10: The tangent equation is $y - \frac{4096}{81} = \frac{2048}{27}(x - \frac{8}{3})$ .

Step 11: Final Answer: $y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)$ .

Why other options are wrong:
- **Option A ( $y = \frac{{4098}}{{81}}$ ): This is a horizontal line and not the correct tangent.
- Option B ( $y - 1 = 5\left( {x - 1} \right)$ ): This is a line, but it's not tangent to the curve at the correct point.
- Option D ( $y - \frac{{32}}{{243}} = \frac{{80}}{{81}}\left( {x - \frac{2}{3}} \right)$ ): This is also a line, but it's not tangent to the curve at the correct point.

Key Concepts:
- Differentiation
- Tangent line equation
- Finding points of tangency

Memory Aids:
- The derivative gives the slope of the tangent line.

Quick Tips:**
- Always find the derivative first.
- Use the point-slope form of a line to find the tangent equation.

90 Medium

90 of 97

Application +4 | -1

The point on the curve

3 y=6 x-5 x^3

the normal at which passes through the origin, is

View answer and explanation

Correct option: A

(1,1 / 3)

Option A ((1, 1/3)) is correct because the normal to the curve at this point passes through the origin, satisfying both the curve's equation and the normal's equation.

Solution approach

Correct Answer + Core Reason: (1, 1/3). The normal at this point passes through the origin and the point lies on the curve.

Step-by-Step Solution:

Step 1: Let the point on the curve be $(x_1, y_1)$ . Since the point lies on the curve, $3y_1 = 6x_1 - 5x_1^3$ .

Step 2: Find the derivative of the curve $3y = 6x - 5x^3$ with respect to $x$ . $3\frac{{dy}}{{dx}} = 6 - 15x^2 \Rightarrow \frac{{dy}}{{dx}} = 2 - 5x^2$ .

Step 3: Evaluate the derivative at $(x_1, y_1)$ to find the slope of the tangent at that point. $\frac{{dy}}{{dx}}|_{(x_1, y_1)} = 2 - 5x_1^2$ .

Step 4: Find the slope of the normal at $(x_1, y_1)$ . The slope of the normal is the negative reciprocal of the slope of the tangent: $m_{{normal}} = \frac{{-1}}{{2 - 5x_1^2}}$ .

Step 5: Write the equation of the normal using the point-slope form: $y - y_1 = m_{{normal}}(x - x_1) \Rightarrow y - y_1 = \frac{{-1}}{{2 - 5x_1^2}}(x - x_1)$ .

Step 6: Since the normal passes through the origin $(0, 0)$ , substitute $x = 0$ and $y = 0$ into the equation of the normal: $0 - y_1 = \frac{{-1}}{{2 - 5x_1^2}}(0 - x_1) \Rightarrow -y_1 = \frac{{x_1}}{{2 - 5x_1^2}} \Rightarrow y_1 = \frac{{x_1}}{{5x_1^2 - 2}}$ .

Step 7: Substitute this expression for $y_1$ into the equation of the curve: $3y_1 = 6x_1 - 5x_1^3 \Rightarrow 3(\frac{{x_1}}{{5x_1^2 - 2}}) = 6x_1 - 5x_1^3$ .

Step 8: Simplify and solve for $x_1$ : $3x_1 = (6x_1 - 5x_1^3)(5x_1^2 - 2) \Rightarrow 3x_1 = 30x_1^3 - 12x_1 - 25x_1^5 + 10x_1^3 \Rightarrow 25x_1^5 - 40x_1^3 + 15x_1 = 0 \Rightarrow 5x_1(5x_1^4 - 8x_1^2 + 3) = 0 \Rightarrow 5x_1(5x_1^2 - 3)(x_1^2 - 1) = 0$ .

Step 9: The possible values for $x_1$ are $0, \pm 1, \pm \sqrt{{\frac{{3}}{{5}}}}$ .\nIf $x_1 = 0$ , then $y_1 = 0$ . But the normal at (0,0) is not defined.\nIf $x_1 = 1$ , then $y_1 = \frac{{1}}{{5(1)^2 - 2}} = \frac{{1}}{{3}}$ . So, the point is $(1, \frac{{1}}{{3}})$ .\nIf $x_1 = -1$ , then $y_1 = \frac{{-1}}{{5(-1)^2 - 2}} = -\frac{{1}}{{3}}$ . So, the point is $(-1, -\frac{{1}}{{3}})$ .\nIf $x_1 = \sqrt{{\frac{{3}}{{5}}}}$ , then $y_1 = \frac{{\sqrt{{\frac{{3}}{{5}}}}}}{{5(\frac{{3}}{{5}}) - 2}} = \frac{{\sqrt{{\frac{{3}}{{5}}}}}}{{1}} = \sqrt{{\frac{{3}}{{5}}}}$ .\nIf $x_1 = -\sqrt{{\frac{{3}}{{5}}}}$ , then $y_1 = -\sqrt{{\frac{{3}}{{5}}}}$ .

Step 10: Check which of the points satisfy the equation of the curve: $3y = 6x - 5x^3$ .\nFor $(1, \frac{{1}}{{3}})$ : $3(\frac{{1}}{{3}}) = 1$ and $6(1) - 5(1)^3 = 6 - 5 = 1$ . This point satisfies the equation.\nFor $(-1, -\frac{{1}}{{3}})$ : $3(-\frac{{1}}{{3}}) = -1$ and $6(-1) - 5(-1)^3 = -6 + 5 = -1$ . This point satisfies the equation.

Step 11: From the given options, only $(1, \frac{{1}}{{3}})$ is present.

Why other options are wrong:
- Option B: The point $(\frac{{1}}{{3}}, 1)$ does not lie on the curve $3y = 6x - 5x^3$ .
- Option C: The point $(2, -\frac{{28}}{{3}})$ does not satisfy the condition that the normal at that point passes through the origin.
- Option D: While the point $(-1, -\frac{{1}}{{3}})$ does satisfy the condition, it's not the only solution, and option A is the only one present in the choices.

Key Concepts:
- Differentiation to find the slope of a tangent
- Normal to a curve
- Equation of a line

Memory Aids:
- The slope of the normal is the negative reciprocal of the slope of the tangent.

Quick Tips:
- Always check if the point lies on the curve.
- Remember that the normal is perpendicular to the tangent.

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