TANGENTS & NORMALS MCQs for JEE MATHEMATICS — Medium

1 Medium

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Analyzing +4 | -1

The angle between the tangents at those points on the curve

x = {t^2} + 1

and

y = {t^2} - t - 6

where it meets

x

-axis is

View answer and explanation

Correct option: C

\pm {\tan ^{ - 1}}\left( {\frac{{10}}{{49}}} \right)

Option C (

\pm {\tan ^{ - 1}}\left( {\frac{{10}}{{49}}} \right)

) is correct because it represents the angle between the tangents at the points where the curve intersects the x-axis.

Solution approach

Correct Answer + Core Reason: $\pm {\tan ^{ - 1}}\left( {\frac{{10}}{{49}}} \right)$ . This is the angle between the tangents at the points where the curve intersects the x-axis.

Step-by-Step Solution:

Step 1: Given $x = t^2 + 1$ and $y = t^2 - t - 6$ .

Step 2: Find $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2t - 1$ .

Step 3: Find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 1}{2t}$ .

Step 4: The curve meets the x-axis when $y = 0$ , so $t^2 - t - 6 = 0 \Rightarrow (t - 3)(t + 2) = 0$ . Thus, $t = 3$ and $t = -2$ .

Step 5: Find the slopes at $t = 3$ and $t = -2$ :
- At $t = 3$ , $m_1 = \frac{2(3) - 1}{2(3)} = \frac{5}{6}$ .
- At $t = -2$ , $m_2 = \frac{2(-2) - 1}{2(-2)} = \frac{-5}{-4} = \frac{5}{4}$ .

Step 6: Find the angle between the tangents using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$ :
- $\tan \theta = \left| \frac{\frac{5}{6} - \frac{5}{4}}{1 + \frac{5}{6}\cdot\frac{5}{4}} \right| = \left| \frac{\frac{10 - 15}{12}}{\frac{24 + 25}{24}} \right| = \left| \frac{-5}{12} \cdot \frac{24}{49} \right| = \frac{10}{49}$ .

Step 7: Therefore, $\theta = \pm {\tan ^{ - 1}}\left( {\frac{10}{49}} \right)$ .

Step 8: Final Answer: $\pm {\tan ^{ - 1}}\left( {\frac{{10}}{{49}}} \right)$ .

Why other options are wrong:
- Option A: $\pm {\tan ^{ - 1}}\left( {\frac{4}{{29}}} \right)$ is incorrect because it does not represent the correct angle between the tangents.
- Option B: $\pm {\tan ^{ - 1}}\left( {\frac{5}{{49}}} \right)$ is incorrect because it does not represent the correct angle between the tangents.
- Option D: $\pm {\tan ^{ - 1}}\left( {\frac{8}{{29}}} \right)$ is incorrect because it does not represent the correct angle between the tangents.

Key Concepts:
- Parametric differentiation
- Finding the angle between two lines

Memory Aids:
- The angle between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$ .

Quick Tips:
- Remember to find the slopes at the correct values of the parameter $t$ .

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Application +4 | -1

The point

(0,5)

is closest to the curve

x^{2}=2 y

at

View answer and explanation

Correct option: D

(2 \sqrt{2}, 4)

Option D (

(2 \sqrt{{2}}, 4)

) is correct because by minimizing the distance between the point (0,5) and a general point on the curve, we find the closest point.

Solution approach

Correct Answer + Core Reason: $(2 \sqrt{{2}}, 4)$ . By minimizing the distance between the point and the curve, we find this point to be the closest.

Step-by-Step Solution:

Step 1: Let $P(t, \frac{{t^2}}{{2}})$ be a point on the curve $x^2 = 2y$ . Let $A = (0, 5)$ .

Step 2: The distance between $A$ and $P$ is $d = \sqrt{{(t - 0)^2 + (\frac{{t^2}}{{2}} - 5)^2}}$ .

Step 3: To minimize the distance, we can minimize the square of the distance: $z = d^2 = t^2 + (\frac{{t^2}}{{2}} - 5)^2 = t^2 + \frac{{t^4}}{{4}} - 5t^2 + 25 = \frac{{t^4}}{{4}} - 4t^2 + 25$ .

Step 4: Differentiate $z$ with respect to $t$ : $\frac{{dz}}{{dt}} = t^3 - 8t = t(t^2 - 8)$ .

Step 5: Set $\frac{{dz}}{{dt}} = 0$ to find critical points: $t(t^2 - 8) = 0$ , so $t = 0$ or $t = \pm 2\sqrt{{2}}$ .

Step 6: Find the second derivative: $\frac{{d^2z}}{{dt^2}} = 3t^2 - 8$ .

Step 7: Evaluate the second derivative at the critical points:
- At $t = 0$ , $\frac{{d^2z}}{{dt^2}} = -8 < 0$ , so this is a local maximum.
- At $t = \pm 2\sqrt{{2}}$ , $\frac{{d^2z}}{{dt^2}} = 3(8) - 8 = 16 > 0$ , so these are local minima.

Step 8: The points on the curve are $(2\sqrt{{2}}, 4)$ and $(-2\sqrt{{2}}, 4)$ .

Step 9: Since we are looking for the closest point, we choose either $(2\sqrt{{2}}, 4)$ or $(-2\sqrt{{2}}, 4)$ . The problem only lists $(2\sqrt{{2}}, 4)$ as an option.

Why other options are wrong:
- Option A: Incorrect because the point $(2 \sqrt{{2}}, 0)$ is not on the curve $x^2 = 2y$ .
- Option B: Incorrect because the point $(2, 2)$ is on the curve, but it is not the closest point to $(0, 5)$ .
- Option C: Incorrect because the point $(-2 \sqrt{{2}}, 0)$ is not on the curve $x^2 = 2y$ .

Key Concepts:
- Distance formula
- Optimization using calculus

Memory Aids:
- To minimize distance, minimize the square of the distance.

Quick Tips:
- Always check the second derivative to determine if a critical point is a minimum or maximum.

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Application +4 | -1

An equation of the tangent to the curve

y=x^4

from the point

(2,0)

not on the curve is

View answer and explanation

Correct option: A

y=0

Option A (

y=0

) is the correct answer because it represents a tangent line to the curve

y=x^4

passing through the point (2,0).

Solution approach

Correct Answer + Core Reason: $y=0$ . The tangent to the curve passes through the point (2,0).

Step-by-Step Solution:

Step 1: Let the point of contact be $(h, k)$ , where $k=h^4$ .

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

Step 3: Write the equation of the tangent at $(h, k)$ . $y-k=4 h^3(x-h)$ .

Step 4: Substitute $k=h^4$ into the tangent equation. $y-h^4=4 h^3(x-h)$ .

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

Step 6: Solve for $h$ . $-h^4=8h^3 - 4h^4 \Rightarrow 3h^4 - 8h^3 = 0 \Rightarrow h^3(3h-8) = 0$ . So $h=0$ or $h=\frac{8}{3}$ .

Step 7: Find the corresponding $k$ values. If $h=0$ , $k=0$ . If $h=\frac{8}{3}$ , $k=\left(\frac{8}{3}\right)^4$ .

Step 8: Find the tangent equations. If $(h,k)=(0,0)$ , the tangent is $y-0=4(0)^3(x-0) \Rightarrow y=0$ . If $(h,k)=\left(\frac{8}{3},\left(\frac{8}{3}\right)^4\right)$ , the tangent is $y-\left(\frac{8}{3}\right)^4=4\left(\frac{8}{3}\right)^3\left(x-\frac{8}{3}\right)$ .

Why other options are wrong:
- **Option B ( $x=0$ ): This is incorrect because it is a vertical line, and it is not tangent to the curve at (2,0).
- Option C ( $x+y=0$ ):** This is incorrect because it does not represent a tangent to the curve $y=x^4$ passing through (2,0).
- Option D (None of these): This is incorrect because $y=0$ is a valid tangent.

Key Concepts:
- Finding the equation of a tangent to a curve
- Differentiation

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

Quick Tips:
- Always find the derivative first to determine the slope of the tangent.

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Application +4 | -1

The equation of the normal to the curve

y = {e^{ - 2\left| x \right|}}

at the point where the curve cuts the line

x = 1/2

, is

View answer and explanation

Correct option: B

2e\left( {ex - 2y} \right) = {e^2} - 4

Option B (

2e(ex - 2y) = {e^2} - 4

) is correct because it is the equation of the normal to the curve at the specified point, found by calculating the derivative and using the point-slope form.

Solution approach

Correct Answer + Core Reason: $2e(ex - 2y) = {e^2} - 4$ . The equation of the normal is found by finding the derivative and using the point-slope form.

Step-by-Step Solution:

Step 1: Given the curve $y = e^{-2|x|}$ . Since we are considering the point where $x = \frac{1}{2}$ , we have $x > 0$ , so $y = e^{-2x}$ .

Step 2: The point where the curve cuts the line $x = \frac{1}{2}$ is $(\frac{1}{2}, e^{-1}) = (\frac{1}{2}, \frac{1}{e})$ .

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

Step 4: Evaluate the derivative at $x = \frac{1}{2}$ : $\frac{dy}{dx}|_{x=\frac{1}{2}} = -2e^{-2(\frac{1}{2})} = -2e^{-1} = -\frac{2}{e}$ .

Step 5: The slope of the normal is the negative reciprocal of the slope of the tangent: $m_{normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{-\frac{2}{e}} = \frac{e}{2}$ .

Step 6: Use the point-slope form of a line to find the equation of the normal: $y - y_1 = m(x - x_1)$ , where $(x_1, y_1) = (\frac{1}{2}, \frac{1}{e})$ .
$y - \frac{1}{e} = \frac{e}{2}(x - \frac{1}{2})$
$y - \frac{1}{e} = \frac{e}{2}x - \frac{e}{4}$
$4y - \frac{4}{e} = 2ex - e^2$
$4ey - 4 = 2e^2x - e^3$
$2e^2x - 4ey = e^3 - 4$
$2e(ex - 2y) = e^2 - 4$

Step 7: Final Answer: $2e(ex - 2y) = {e^2} - 4$

Why other options are wrong:
- Option A: $2e(ex + 2y) = {e^2} - 4$ This is incorrect because the sign is wrong in the equation.
- Option C: $2e(ey - 2x) = {e^2} - 4$ This is incorrect because the variables are mixed up in the equation.
- Option D: None of these This is incorrect because the correct equation is given in Option B.

Key Concepts:
- Derivatives and slopes of tangents and normals
- Point-slope form of a line

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

Quick Tips:
- Find the derivative to get the slope of the tangent.
- Take the negative reciprocal to find the slope of the normal.
- Use the point-slope form to find the equation of the line.

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Application +4 | -1

If the two curves

y=a^x

and

y=b^x

intersect at an angle

\alpha

, then tan

\alpha

equals -

View answer and explanation

Correct option: A

\frac{\log b-\log a}{1+\log a \log b}

Option A (

\frac{{\log b - \log a}}{{1 + \log a \log b}}

) is correct because the tangent of the angle between the two curves is found by calculating the derivatives of the curves at their point of intersection and then using the formula for the tangent of the angle between two lines.

Solution approach

Correct Answer + Core Reason: $\frac{{\log b - \log a}}{{1 + \log a \log b}}$ . The tangent of the angle is found using derivatives and the formula for the angle between two lines.

Step-by-Step Solution:

Step 1: Find the point of intersection. The curves $y = a^x$ and $y = b^x$ intersect when $a^x = b^x$ . This occurs when $x = 0$ , so $y = a^0 = b^0 = 1$ . The point of intersection is (0, 1).

Step 2: Find the derivatives of the curves. $\frac{{dy}}{{dx}} = a^x \log a$ and $\frac{{dy}}{{dx}} = b^x \log b$ .

Step 3: Evaluate the derivatives at the point of intersection (0, 1). $m_1 = (a^x \log a)_{(0,1)} = a^0 \log a = \log a$ and $m_2 = (b^x \log b)_{(0,1)} = b^0 \log b = \log b$ .

Step 4: Use the formula for the tangent of the angle between two lines: $\tan \alpha = \left| \frac{{m_2 - m_1}}{{1 + m_1 m_2}} \right| = \left| \frac{{\log b - \log a}}{{1 + \log a \log b}} \right|$ . Since the question doesn't specify the order, we can assume the positive value.

Step 5: Write the final expression: $\tan \alpha = \frac{{\log b - \log a}}{{1 + \log a \log b}}$ .

Why other options are wrong:
- **Option B ( $\frac{{\log a + \log b}}{{1 - \log a \log b}}$ ): This is incorrect because it uses the wrong sign in the numerator and denominator of the formula for the tangent of the angle between two lines.
- Option C ( $\frac{{\log a - \log b}}{{1 - \log a \log b}}$ ): This is incorrect because it uses the wrong sign in the numerator and denominator of the formula for the tangent of the angle between two lines.
- Option D (None of these): This is incorrect because option A is the correct answer.

Key Concepts:
- Derivatives of exponential functions
- Angle between two lines

Memory Aids:**
- Remember the derivative of $a^x$ is $a^x \log a$ .
- Remember the formula for the tangent of the angle between two lines: $\tan \alpha = \left| \frac{{m_2 - m_1}}{{1 + m_1 m_2}} \right|$ .

Quick Tips:
- Find the point of intersection first.
- Calculate the derivatives at the point of intersection.
- Apply the formula for the angle between two lines.

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Application +4 | -1

If the curve

y = a{x^3} + b{x^2} + cx

is inclined at

{45^0}

to

x

-axis at (0, 0) but touches

x

-axis at (1, 0), then

View answer and explanation

Correct option: A

a = 1,\;b = - 2,\;c = 1

Option A (

a = 1,\;b = - 2,\;c = 1

) is correct because these values satisfy the given conditions of the curve's inclination and tangency.

Solution approach

Correct Answer + Core Reason: $a = 1,\;b = - 2,\;c = 1$ . The values are found by using the given conditions for the curve.

Step-by-Step Solution:

Step 1: Given the curve $y = a{x^3} + b{x^2} + cx$ .

Step 2: Find the derivative $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = 3a{x^2} + 2bx + c$ .

Step 3: The curve is inclined at $45^0$ to the x-axis at (0, 0), so $\frac{{dy}}{{dx}}|_{x=0} = \tan 45^0 = 1$ . Thus, $3a(0)^2 + 2b(0) + c = 1 \Rightarrow c = 1$ .

Step 4: The curve touches the x-axis at (1, 0), so $y(1) = 0$ . Thus, $a(1)^3 + b(1)^2 + c(1) = 0 \Rightarrow a + b + c = 0$ .

Step 5: Also, at (1,0) the tangent touches the x-axis, so $\frac{{dy}}{{dx}}|_{x=1} = 0$ . Thus, $3a(1)^2 + 2b(1) + c = 0 \Rightarrow 3a + 2b + c = 0$ .

Step 6: Substitute $c = 1$ into the equations:
$a + b + 1 = 0 \Rightarrow a + b = -1$
$3a + 2b + 1 = 0 \Rightarrow 3a + 2b = -1$

Step 7: Solve the system of equations:
$a + b = -1$
$3a + 2b = -1$ \nMultiply the first equation by 2: $2a + 2b = -2$ .\nSubtract this from the second equation: $(3a + 2b) - (2a + 2b) = -1 - (-2) \Rightarrow a = 1$ .

Step 8: Substitute $a = 1$ into $a + b = -1$ : $1 + b = -1 \Rightarrow b = -2$ .

Step 9: Therefore, $a = 1$ , $b = -2$ , and $c = 1$ .

Why other options are wrong:
- Option B: $a = 1,\;b = 1,\;c = - 2$ - These values do not satisfy the given conditions.
- Option C: $a = - 2,\;b = 1,\;c = 1$ - These values do not satisfy the given conditions.
- Option D: $a = - 1,\;b = 2,\;c = 1$ - These values do not satisfy the given conditions.

Key Concepts:
- Finding the derivative of a function
- Using given conditions to solve for unknowns
- Tangent to a curve

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

Quick Tips:
- Always use all given conditions to form equations and solve for the unknowns.

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Application +4 | -1

If the tangent at a point on the ellipse

\frac{x^2}{27}+\frac{y^2}{3}=1

meets the coordinate axes at

A

and

B

and,

O

is the origin, the minimum area (in sq. units). of the triangle

O A B

, is

View answer and explanation

Correct option: A

9

Option A (9) is correct because the minimum area of the triangle OAB is found to be 9 square units when sin(2*theta) is maximized.

Solution approach

Correct Answer + Core Reason: 9. The minimum area of the triangle is 9 square units.

Step-by-Step Solution:

Step 1: Let $P(3\sqrt{3} \cos \theta, \sqrt{3} \sin \theta)$ be a point on the ellipse $\frac{x^2}{27} + \frac{y^2}{3} = 1$ .

Step 2: The equation of the tangent at $P$ is $\frac{x}{3\sqrt{3}} \cos \theta + \frac{y}{\sqrt{3}} \sin \theta = 1$ .

Step 3: This tangent meets the coordinate axes at $A(3\sqrt{3} \sec \theta, 0)$ and $B(0, \sqrt{3} \csc \theta)$ .

Step 4: Let $\Delta$ be the area of triangle $OAB$ . Then, $\Delta = \frac{1}{2}(OA \times OB) = \frac{1}{2} \times \frac{3\sqrt{3}}{\cos \theta} \times \frac{\sqrt{3}}{\sin \theta} = \frac{9}{\sin 2\theta}$ .

Step 5: The area $\Delta$ is minimum when $\sin 2\theta$ is maximum. The maximum value of $\sin 2\theta$ is 1.

Step 6: Hence, the minimum value of $\Delta$ is 9.

Why other options are wrong:
- Option B: This is incorrect as the minimum area is 9, not 9/2.
- Option C: This is incorrect as the minimum area is 9, not $9\sqrt{3}$ .
- Option D: This is incorrect as the minimum area is 9, not $3\sqrt{3}$ .

Key Concepts:
- Equation of tangent to an ellipse
- Area of a triangle
- Trigonometric functions

Memory Aids:
- Remember the parametric form of an ellipse: $(a\cos \theta, b\sin \theta)$ .

Quick Tips:
- To minimize the area, maximize the denominator.

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Analyzing +4 | -1

The equation of the tangent to curve

y\left( {2x - 1} \right){e^{2\left( {1 - x} \right)}}

at the points its maximum, is

View answer and explanation

Correct option: A

y - 1 = 0

Option A (

y - 1 = 0

) is correct because the tangent to the curve at its maximum point (1,1) is a horizontal line, hence y=1.

Solution approach

Correct Answer + Core Reason: $y - 1 = 0$ . The tangent at the maximum point (1,1) is a horizontal line.

Step-by-Step Solution:

Step 1: Find the first derivative of the function $y = (2x - 1)e^{2(1 - x)}$ .
$\frac{dy}{dx} = 2e^{2(1 - x)} - 2(2x - 1)e^{2(1 - x)} = 2e^{2(1 - x)}(2 - 2x) = 4e^{2(1 - x)}(1 - x)$

Step 2: Set the first derivative equal to zero to find critical points.
$\frac{dy}{dx} = 0 \Rightarrow 4e^{2(1 - x)}(1 - x) = 0 \Rightarrow x = 1$

Step 3: Find the second derivative to determine if the critical point is a maximum or minimum.
$\frac{d^2y}{dx^2} = -8e^{2(1 - x)}(1 - x) - 4e^{2(1 - x)}$

Step 4: Evaluate the second derivative at x=1.
$(\frac{d^2y}{dx^2})_{x = 1} = -4 < 0$ , which indicates that x=1 is a maximum.

Step 5: Find the y-coordinate of the maximum point.\nWhen $x = 1$ , $y = (2(1) - 1)e^{2(1 - 1)} = 1 \cdot e^0 = 1$ . So the maximum point is (1, 1).

Step 6: Find the equation of the tangent at the maximum point (1, 1).\nSince the derivative is 0 at x=1, the tangent is a horizontal line with equation $y = 1$ , or $y - 1 = 0$ .

Why other options are wrong:
- Option B: $x - 1 = 0$ is incorrect because it represents a vertical line at x=1, not the tangent at the maximum point.
- Option C: $x + y - 1 = 0$ is incorrect because it is not the tangent at the maximum point (1,1).
- Option D: $x - y + 1 = 0$ is incorrect because it is not the tangent at the maximum point (1,1).

Key Concepts:
- Finding critical points using the first derivative.
- Determining the nature of critical points using the second derivative.
- Equation of a tangent line.

Memory Aids:
- A horizontal tangent line has a slope of 0.

Quick Tips:
- Remember to check the second derivative to confirm whether a critical point is a maximum or minimum.

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Application +4 | -1

Let

f\left( x \right)\,and\,g\left( x \right)

be differentiable for

0 \le x \le 1,

, such that

f\left( 0 \right) = 2,g\left( 0 \right) = 0,f\left( 1 \right) = 6.

. Let there exist a real number

c

in [0,1] such that

f'\left( c \right) = 2g'\left( c \right),

then the value of

g\left( 1 \right)

must be

View answer and explanation

Correct option: B

2

Option B (2) is correct because by applying Rolle's theorem to the function

\phi(x) = f(x) - 2g(x)

, we find that g(1) = 2.

Solution approach

Correct Answer + Core Reason: 2. Applying Rolle's theorem to a constructed function yields g(1) = 2.

Step-by-Step Solution:

Step 1: Define a function $\phi(x) = f(x) - 2g(x)$ on [0, 1].

Step 2: Calculate $\phi(0) = f(0) - 2g(0) = 2 - 2(0) = 2$ .

Step 3: Calculate $\phi(1) = f(1) - 2g(1) = 6 - 2g(1)$ .

Step 4: Find the derivative of $\phi(x)$ : $\phi'(x) = f'(x) - 2g'(x)$ .

Step 5: Given that there exists a c in [0, 1] such that $f'(c) = 2g'(c)$ , we have $\phi'(c) = f'(c) - 2g'(c) = 0$ .

Step 6: Since $\phi'(c) = 0$ , we can apply Rolle's theorem, which implies $\phi(0) = \phi(1)$ .

Step 7: Therefore, $2 = 6 - 2g(1)$ . Solving for $g(1)$ , we get $2g(1) = 4$ , so $g(1) = 2$ .

Why other options are wrong:
- Option A: This is incorrect. The correct value is 2.
- Option C: This is incorrect. The correct value is 2.
- Option D: This is incorrect. The correct value is 2.

Key Concepts:
- Rolle's theorem
- Differentiable functions

Memory Aids:
- Rolle's theorem states that if a function is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists a c in (a, b) such that f'(c) = 0.

Quick Tips:
- Look for opportunities to apply Rolle's theorem or the Mean Value Theorem when given information about derivatives and function values.

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Application +4 | -1

The point of the curve

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

at which the normal is parallel to line

\;y - 2x + 1 = 0

View answer and explanation

Correct option: C

\left( {5,\; - 2} \right)

Option C ((5, -2)) is correct because it is found by equating the slope of the normal to the slope of the given line.

Solution approach

Correct Answer + Core Reason: $\left( {5,\; - 2} \right)$ . The point is found by finding the derivative and equating the slope of the normal to the slope of the given line.

Step-by-Step Solution:

Step 1: Given equation: $y^2 = 2(x - 3)$ .

Step 2: Differentiate both sides with respect to $x$ : $2y \frac{dy}{dx} = 2 \Rightarrow \frac{dy}{dx} = \frac{1}{y}$ .

Step 3: Find the slope of the normal: $m_{normal} = -\frac{1}{\frac{dy}{dx}} = -y$ .

Step 4: Given line: $y - 2x + 1 = 0 \Rightarrow y = 2x - 1$ . The slope of the line is 2.

Step 5: Since the normal is parallel to the line, equate their slopes: $-y = 2 \Rightarrow y = -2$ .

Step 6: Substitute $y = -2$ into the equation of the curve: $(-2)^2 = 2(x - 3) \Rightarrow 4 = 2x - 6 \Rightarrow 2x = 10 \Rightarrow x = 5$ .

Step 7: The point is (5, -2).

Step 8: Final Answer: (5, -2)

Why other options are wrong:
- Option A: (5, 2) is incorrect. This might be obtained by not considering the negative sign for the slope of the normal.
- Option B: $\left( { - \frac{1}{2},\; - 2} \right)$ is incorrect. This might be obtained by incorrect algebraic manipulation.
- Option D: $\left( {\frac{3}{2},\;2} \right)$ is incorrect. This might be obtained by incorrect algebraic manipulation.

Key Concepts:
- Derivatives and slopes
- Equation of normal
- Parallel lines

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

Quick Tips:
- Always find the derivative correctly and equate the slopes carefully.

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Analyzing +4 | -1

Let

P

be the point (other than the origin) of intersection of the curves

{y^2} = 4ax

and

a{y^2} = 4{x^3}

such that the normals to the two curves meet

x

-axis at

{G_1}\,and\,{G_2}

respectively. Then,

{G_1}{G_2} =

View answer and explanation

Correct option: B

4a

Option B (

4a

) is correct because it is the distance between the x-intercepts of the normals to the two curves at their point of intersection.

Solution approach

Correct Answer + Core Reason: $4a$ . The distance between the x-intercepts of the normals to the two curves at their point of intersection is $4a$ .

Step-by-Step Solution:

Step 1: Find the points of intersection of the curves ${y^2} = 4ax$ and $a{y^2} = 4{x^3}$ .

Step 2: Substitute ${y^2} = 4ax$ into $a{y^2} = 4{x^3}$ to get $a(4ax) = 4{x^3}$ , which simplifies to $4{a^2}x = 4{x^3}$ . Thus, ${x^3} - {a^2}x = 0$ , so $x(x^2 - a^2) = 0$ . The solutions are $x = 0$ and $x = \pm a$ .

Step 3: Since we are looking for the point other than the origin, consider $x = a$ . Then, ${y^2} = 4a(a) = 4a^2$ , so $y = \pm 2a$ . Let's take the point $P(a, 2a)$ .

Step 4: Find the slope of the tangent to ${y^2} = 4ax$ at $P(a, 2a)$ . Differentiating implicitly, $2y \frac{{dy}}{{dx}} = 4a$ , so $\frac{{dy}}{{dx}} = \frac{{2a}}{y}$ . At $P$ , $\frac{{dy}}{{dx}} = \frac{{2a}}{{2a}} = 1$ . The slope of the normal is $-1$ .

Step 5: Find the equation of the normal to ${y^2} = 4ax$ at $P(a, 2a)$ . Using the point-slope form, $y - 2a = -1(x - a)$ , which simplifies to $y = -x + 3a$ . The x-intercept is $G_1(3a, 0)$ .

Step 6: Find the slope of the tangent to $a{y^2} = 4{x^3}$ at $P(a, 2a)$ . Differentiating implicitly, $2ay \frac{{dy}}{{dx}} = 12{x^2}$ , so $\frac{{dy}}{{dx}} = \frac{{6{x^2}}}{{ay}}$ . At $P$ , $\frac{{dy}}{{dx}} = \frac{{6{a^2}}}{{a(2a)}} = 3$ . The slope of the normal is $-\frac{1}{3}$ .

Step 7: Find the equation of the normal to $a{y^2} = 4{x^3}$ at $P(a, 2a)$ . Using the point-slope form, $y - 2a = -\frac{1}{3}(x - a)$ , which simplifies to $y = -\frac{1}{3}x + \frac{7}{3}a$ . The x-intercept is $G_2(7a, 0)$ .

Step 8: Calculate the distance between $G_1$ and $G_2$ : ${G_1}{G_2} = |7a - 3a| = 4a$ .

Why other options are wrong:
- Option A: $2a$ . This is incorrect as the distance is calculated to be $4a$ .
- Option C: $a$ . This is incorrect as the distance is calculated to be $4a$ .
- Option D: None of these. This is incorrect because option B is correct.

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

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

Quick Tips:
- Always find the derivatives correctly using implicit differentiation.
- Be careful with algebraic manipulations.

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12 of 97

Application +4 | -1

The line which is parallel to

x

-axis and crosses the curve

y = \sqrt x

an angle of

{45^0}

, is

View answer and explanation

Correct option: B

y = \frac{1}{2}

Option B (

y = \frac{1}{2}

) is correct because it represents the line parallel to the x-axis that intersects the curve at a 45-degree angle, determined by equating the derivative to tan(45).

Solution approach

Correct Answer + Core Reason: $y = \frac{1}{2}$ . The line parallel to the x-axis intersects the curve at 45 degrees, meaning the derivative equals tan(45).

Step-by-Step Solution:

Step 1: Let $y = k$ be the line parallel to the x-axis.

Step 2: Find the intersection point of $y = k$ and $y = \sqrt{x}$ . Set $k = \sqrt{x}$ , so $x = k^2$ . The point of intersection is $(k^2, k)$ .

Step 3: Find the derivative of $y = \sqrt{x}$ with respect to $x$ : $\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt{x}}}$ .

Step 4: Evaluate the derivative at the point $(k^2, k)$ : $\left( {\frac{{dy}}{{dx}}} \right)_{(k^2, k)} = \frac{1}{{2\sqrt{k^2}}} = \frac{1}{{2k}}$ .

Step 5: The angle of intersection is 45 degrees, so the slope of the tangent line is $\tan(45^\circ) = 1$ .

Step 6: Set the derivative equal to 1: $\frac{1}{{2k}} = 1$ .

Step 7: Solve for $k$ : $2k = 1 \Rightarrow k = \frac{1}{2}$ .

Step 8: The equation of the line is $y = \frac{1}{2}$ .

Why other options are wrong:
- **Option A ( $y = \frac{1}{4}$ ): This is incorrect because it would result in an intersection angle different from 45 degrees.
- Option C ( $y = 1$ ): This is incorrect because it would result in an intersection angle different from 45 degrees.
- Option D ( $y = 4$ ): This is incorrect because it would result in an intersection angle different from 45 degrees.

Key Concepts:
- Derivatives
- Tangent lines
- Angle of intersection

Memory Aids:**
- $\tan(45^\circ) = 1$

Quick Tips:
- Remember that a line parallel to the x-axis has the form $y = k$ .
- The derivative gives the slope of the tangent line.

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13 of 97

Analyzing +4 | -1

Tangent of the angle at which the curves

y = {a^x}

and

y = {b^x}\;(a \ne b > 0)

intersect, is given by

View answer and explanation

Correct option: B

\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}

Option B (

\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}

) is correct because it is derived from the formula for the tangent of the angle between two curves using their derivatives at the point of intersection.

Solution approach

Correct Answer + Core Reason: $\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}$ . This is the tangent of the angle between the curves, found using derivatives.

Step-by-Step Solution:

Step 1: Find the point of intersection of the two curves $y = a^x$ and $y = b^x$ .

Step 2: Set the two equations equal to each other: $a^x = b^x$ .

Step 3: Divide both sides by $b^x$ : $\left(\frac{a}{b}\right)^x = 1$ .

Step 4: Since any number raised to the power of 0 is 1, we have $x = 0$ .

Step 5: Substitute $x = 0$ into either equation to find the y-coordinate: $y = a^0 = 1$ or $y = b^0 = 1$ . So, the point of intersection is $(0, 1)$ .

Step 6: Find the derivatives of both curves with respect to $x$ .

Step 7: For $y = a^x$ , $\frac{dy}{dx} = a^x \ln(a)$ .

Step 8: For $y = b^x$ , $\frac{dy}{dx} = b^x \ln(b)$ .

Step 9: Evaluate the derivatives at the point of intersection $(0, 1)$ .

Step 10: $m_1 = a^0 \ln(a) = \ln(a)$ .

Step 11: $m_2 = b^0 \ln(b) = \ln(b)$ .

Step 12: Use the formula for the tangent of the angle between two lines: $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ .

Step 13: Substitute the values of $m_1$ and $m_2$ : $\tan(\theta) = \left| \frac{\ln(a) - \ln(b)}{1 + \ln(a) \ln(b)} \right|$ .

Step 14: Use the logarithm property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$ : $\tan(\theta) = \left| \frac{\ln(\frac{a}{b})}{1 + \ln(a) \ln(b)} \right|$ . Since the question asks for the tangent of the angle, we can rewrite the natural logs as base 10 logs: $\tan(\theta) = \left| \frac{\log(\frac{a}{b})}{\log(e) (1 + \log(a) \log(b))} \right|$ . However, the given options use natural logs, so we keep the expression as $\tan(\theta) = \left| \frac{\ln(\frac{a}{b})}{1 + \ln(a) \ln(b)} \right|$ .

Step 15: The correct answer is $\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}$ .

Why other options are wrong:
- Option A: $\frac{{\log ab}}{{1 + \log ab}}$ : This is incorrect because it doesn't follow from the formula for the angle between two curves.
- Option C: $\frac{{\log ab}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}$ : This is incorrect because it doesn't follow from the formula for the angle between two curves.
- Option D: None of these: This is incorrect because option B is the correct answer.

Key Concepts:
- Derivatives of exponential functions
- Angle between two curves
- Logarithm properties

Memory Aids:
- $\frac{d}{dx}(a^x) = a^x \ln(a)$
- $\ln(a) - \ln(b) = \ln(\frac{a}{b})$
- $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Quick Tips:
- Remember to find the point of intersection first.
- Use the correct formula for the angle between two curves.
- Simplify the expression using logarithm properties.

14 Medium

14 of 97

Application +4 | -1

Let

f(x)=x^3+x+1

and let

g(x)

be it's inverse function, then equation of the tangent to

y=g(x)

at

x=3

is

View answer and explanation

Correct option: A

x-4 y+1=0

Option A (

x-4 y+1=0

) is correct because it is the equation of the tangent line to the inverse function at the given point, found using the derivative of the inverse function.

Solution approach

Correct Answer + Core Reason: $x-4 y+1=0$ . The equation of the tangent is found using the derivative of the inverse function.

Step-by-Step Solution:

Step 1: Find a point on $g(x)$ at $x=3$ . Since $g(x)$ is the inverse of $f(x)$ , we need to find $y$ such that $f(y) = 3$ . So, $y^3 + y + 1 = 3 \Rightarrow y^3 + y - 2 = 0$ . By inspection, $y = 1$ is a solution. Thus, $f(1) = 3$ , so $g(3) = 1$ . The point is $(3, 1)$ .

Step 2: Find the derivative of $f(x)$ : $f'(x) = 3x^2 + 1$

Step 3: Find the derivative of $g(x)$ at $x=3$ : $g'(3) = \frac{{1}}{{f'(g(3))}} = \frac{{1}}{{f'(1)}} = \frac{{1}}{{3(1)^2 + 1}} = \frac{{1}}{{4}}$

Step 4: Find the equation of the tangent line to $y = g(x)$ at $(3, 1)$ : $y - 1 = \frac{{1}}{{4}}(x - 3) \Rightarrow 4(y - 1) = x - 3 \Rightarrow 4y - 4 = x - 3 \Rightarrow x - 4y + 1 = 0$

Why other options are wrong:
- Option B: Incorrect sign in the equation.
- Option C: Incorrect slope of the tangent line.
- Option D: Incorrect slope and sign in the equation.

Key Concepts:
- Inverse functions
- Derivative of an inverse function
- Equation of a tangent line

Memory Aids:
- Remember that $g'(x) = \frac{{1}}{{f'(g(x))}}$

Quick Tips:
- Find a point on the inverse function first, then use the derivative of the inverse function to find the equation of the tangent line.

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15 of 97

Application +4 | -1

If

x=t^2

and

y=2 t

, then the equation of the normal at

t=1

is

View answer and explanation

Correct option: A

x+y-3=0

Option A (

x+y-3=0

) is correct because by finding the derivative and the slope of the normal, we can determine the equation of the normal at the given point.

Solution approach

Correct Answer + Core Reason: $x+y-3=0$ . By finding the derivative and the slope of the normal, we can determine the equation of the normal at the given point.

Step-by-Step Solution:

Step 1: Given: $x = t^2$ and $y = 2t$ .

Step 2: Find the derivatives with respect to t: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$ .

Step 3: Find $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} = \frac{1}{t}$ .

Step 4: Evaluate $\frac{dy}{dx}$ at $t = 1$ : $\frac{dy}{dx}|_{t=1} = \frac{1}{1} = 1$ .

Step 5: Find the slope of the normal: $m_{normal} = -\frac{1}{\frac{dy}{dx}} = -1$ .

Step 6: Find the coordinates at $t = 1$ : $x = (1)^2 = 1$ and $y = 2(1) = 2$ . So the point is $(1, 2)$ .

Step 7: Find the equation of the normal: $y - 2 = -1(x - 1) \Rightarrow y - 2 = -x + 1 \Rightarrow x + y - 3 = 0$ .

Why other options are wrong:
- Option B: $x+y-1=0$ is incorrect because it does not satisfy the point (1,2).
- Option C: $x+y+1=0$ is incorrect because the slope is wrong and it does not satisfy the point (1,2).
- Option D: $x+y+3=0$ is incorrect because the slope is wrong and it does not satisfy the point (1,2).

Key Concepts:
- Parametric differentiation
- Equation of a normal

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

Quick Tips:
- Always find the slope of the normal before finding the equation.

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16 of 97

Application +4 | -1

If the subnormal at any point on

y = {a^{1 - n}}{x^n}

is of constant length, then the value of

y = {a^{1 - n}}{x^n}

, is

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because it makes the length of the subnormal constant, independent of x.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . The value of n is found by setting the exponent of x in the subnormal length expression to zero.

Step-by-Step Solution:

Step 1: Given the curve equation $y = {a^{1 - n}}{x^n}$ .

Step 2: The length of the subnormal is given by $y\frac{{dy}}{{dx}}$ .

Step 3: Differentiate the curve equation with respect to x: $\frac{{dy}}{{dx}} = n{a^{1 - n}}{x^{n - 1}}$ .

Step 4: Calculate the length of the subnormal: $y\frac{{dy}}{{dx}} = ({a^{1 - n}}{x^n})(n{a^{1 - n}}{x^{n - 1}}) = n{({a^{1 - n}})^2}{x^{2n - 1}}$ .

Step 5: For the subnormal to be of constant length, the exponent of x must be zero: $2n - 1 = 0$ .

Step 6: Solve for n: $2n = 1 \Rightarrow n = \frac{1}{2}$ .

Why other options are wrong:
- Option A (1): If n=1, the subnormal length is not constant.
- Option C (2): If n=2, the subnormal length is not constant.
- **Option D ( $-2$ ): If n=-2, the subnormal length is not constant.

Key Concepts:
- Differentiation
- Subnormal Length

Memory Aids:**
- Remember the formula for the length of the subnormal: $y\frac{{dy}}{{dx}}$ .

Quick Tips:
- For the subnormal to be constant, the expression should not contain any variable.

17 Medium

17 of 97

Application +4 | -1

If

\int\limits_{{\pi / 2}}^x \sqrt{\left(3-2 \sin ^2 t\right)}+\int\limits_{0}^y \cos t d t=0

, then

\left(\frac{d y}{d x}\right)_{\pi, \pi}

is equal to

View answer and explanation

Correct option: C

\sqrt{3}

Option C (

\sqrt{3}

) is correct because after differentiating the given equation and substituting the given point, we find the derivative to be

\sqrt{3}

.

Solution approach

Correct Answer + Core Reason: $\sqrt{3}$ . The derivative is found by implicit differentiation and substitution.

Step-by-Step Solution:

Step 1: Differentiate both sides of the equation with respect to $x$ .
$\int\limits_{{\pi / 2}}^x \sqrt{\left(3-2 \sin ^2 t\right)}dt+\int\limits_{0}^y \cos t dt=0$

Step 2: Apply the Fundamental Theorem of Calculus and the chain rule.
$\sqrt{3-2\sin^2(x)} + \cos(y) \frac{dy}{dx} = 0$

Step 3: Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = -\frac{\sqrt{3-2\sin^2(x)}}{\cos(y)}$

Step 4: Substitute $x = \pi$ and $y = \pi$ into the expression for $\frac{dy}{dx}$ .
$\left.\frac{dy}{dx}\right|_{(\pi, \pi)} = -\frac{\sqrt{3-2\sin^2(\pi)}}{\cos(\pi)} = -\frac{\sqrt{3-0}}{-1} = \sqrt{3}$

Why other options are wrong:
- Option A (-3): This is incorrect because it doesn't correctly account for the signs after substitution.
- Option B (0): This is incorrect because the derivative is not zero at the given point.
- Option D (None of these): This is incorrect because $\sqrt{3}$ is the correct answer.

Key Concepts:
- Fundamental Theorem of Calculus
- Implicit Differentiation
- Chain Rule

Memory Aids:
- Remember the chain rule when differentiating composite functions.

Quick Tips:
- Always remember to apply the chain rule when differentiating implicitly.

18 Medium

18 of 97

Application +4 | -1

The tangent to the curve

y=e^{k x}

at a point

(0,1)

meets the

x

-axis at

(a, 0)

where

a \in[-2,-1]

, then

k \in

:

View answer and explanation

Correct option: D

\left[\frac{1}{2}, 1\right]

Option D ([

\frac{1}{2}, 1

]) is correct because finding the tangent equation at (0,1) and using the given range for the x-intercept 'a' allows us to determine the range for k.

Solution approach

Correct Answer + Core Reason: $\left[\frac{1}{2}, 1\right]$ . The tangent to the curve at (0,1) meets the x-axis at (a,0), and the given range of 'a' determines the range of k.

Step-by-Step Solution:

Step 1: Find the derivative of the curve $y = e^{kx}$ : $\frac{dy}{dx} = ke^{kx}$ .

Step 2: Evaluate the derivative at the point (0,1): $\frac{dy}{dx}|_{(0,1)} = ke^{k(0)} = k$ . This is the slope of the tangent at (0,1).

Step 3: Write the equation of the tangent line at (0,1): $y - 1 = k(x - 0) \Rightarrow y = kx + 1$ .

Step 4: The tangent line meets the x-axis at (a,0). Substitute this point into the tangent equation: $0 = ka + 1 \Rightarrow a = -\frac{1}{k}$ .

Step 5: Given that $a \in [-2, -1]$ , we have $-2 \leq -\frac{1}{k} \leq -1$ .

Step 6: Multiply by -1: $1 \leq \frac{1}{k} \leq 2$ .

Step 7: Take the reciprocal: $\frac{1}{2} \leq k \leq 1$ . Therefore, $k \in [\frac{1}{2}, 1]$ .

Why other options are wrong:
- **Option A ( $\left[-\frac{1}{2}, 0\right]$ ): This is incorrect because the range of 'k' is positive, not negative.
- Option B ( $\left[-1, -\frac{1}{2}\right]$ ): This is incorrect because the range of 'k' is positive, not negative.
- Option C ([0,1]): This is incorrect because the lower bound of 'k' is 1/2, not 0.

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.

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19 of 97

Application +4 | -1

The equations of those tangents to

4 x^2-9 y^2=36

which are perpendicular to the straight line

5 x+2 y-10=0

, are

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculation shows that there are no real points on the curve where the tangent is perpendicular to the given line.

Solution approach

Correct Answer + Core Reason: None of these. There are no real points on the curve where the tangent is perpendicular to the given line.

Step-by-Step Solution:

Step 1: Find the derivative of the given equation $4x^2 - 9y^2 = 36$ with respect to $x$ .

Step 2: Differentiate both sides: $8x - 18y \frac{{dy}}{{dx}} = 0$ .

Step 3: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = \frac{{4x}}{{9y}}$ . This represents the slope of the tangent to the curve.

Step 4: Find the slope of the line $5x + 2y - 10 = 0$ . Rearrange to get $y = -\frac{{5}}{{2}}x + 5$ , so the slope is $-\frac{{5}}{{2}}$ .

Step 5: Since the tangent is perpendicular to the line, the product of their slopes is -1. Therefore, $(\frac{{4x}}{{9y}})(-\frac{{5}}{{2}}) = -1$ .

Step 6: Solve for $y$ : $y = \frac{{10x}}{{9}}$ .

Step 7: Substitute this value of $y$ into the original equation $4x^2 - 9y^2 = 36$ : $4x^2 - 9(\frac{{10x}}{{9}})^2 = 36$ .

Step 8: Simplify: $4x^2 - \frac{{100x^2}}{{9}} = 36 \Rightarrow 36x^2 - 100x^2 = 324 \Rightarrow -64x^2 = 324$ .

Step 9: Solve for $x^2$ : $x^2 = -\frac{{324}}{{64}} = -\frac{{81}}{{16}}$ . Since $x^2$ cannot be negative, there are no real solutions for $x$ and $y$ .

Step 10: Conclusion: Since there are no real points on the curve where the tangent is perpendicular to the given line, none of the provided options are correct.

Why other options are wrong:
- Option A: This option assumes a real solution exists and attempts to construct a tangent line, but the initial assumption is incorrect.
- Option B: This option also assumes a real solution exists and provides a tangent line equation, but it's based on a non-existent point.
- Option C: Similar to options A and B, this option provides a tangent line equation based on a false premise.

Key Concepts:
- Differentiation to find the slope of a tangent
- Perpendicular lines and their slopes
- Solving simultaneous equations

Memory Aids:
- The product of slopes of perpendicular lines is -1.

Quick Tips:
- Always check for real solutions before constructing tangent lines.
- Remember that $x^2$ cannot be negative for real values of $x$ .

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20 of 97

Analyzing +4 | -1

All points on the curve

{y^2} = 4a\left( {x + a\sin \frac{x}{a}} \right)

at which the tangents are parallel to the axis of

x

lie on a

View answer and explanation

Correct option: B

Parabola

Option B (Parabola) is correct because the points where the tangents are parallel to the x-axis lie on the parabola

{y^2} = 4ax

.

Solution approach

Correct Answer + Core Reason: Parabola. The points where the tangents are parallel to the x-axis lie on a parabola.

Step-by-Step Solution:

Step 1: Given ${y^2} = 4a(x + a\sin(\frac{x}{a}))$ .

Step 2: Differentiate both sides with respect to x: $2y \frac{{dy}}{{dx}} = 4a(1 + \cos(\frac{x}{a})) \frac{1}{a} = 4a(1 + \cos(\frac{x}{a}))$ .

Step 3: For tangents parallel to the x-axis, $\frac{{dy}}{{dx}} = 0$ . Thus, $4a(1 + \cos(\frac{x}{a})) = 0$ , which implies $\cos(\frac{x}{a}) = -1$ .

Step 4: Solve for x: $\frac{x}{a} = (2n + 1)\pi$ , where n is an integer. Thus, $x = a(2n + 1)\pi$ .

Step 5: Since $\cos(\frac{x}{a}) = -1$ , then $\sin(\frac{x}{a}) = 0$ .

Step 6: Substitute $\sin(\frac{x}{a}) = 0$ into the original equation: ${y^2} = 4a(x + a(0)) = 4ax$ .

Step 7: The equation ${y^2} = 4ax$ represents a parabola.

Why other options are wrong:
- Option A: This is incorrect because the points lie on a parabola, not a circle.
- Option C: This is incorrect because the points lie on a parabola, not a line.
- Option D: This is incorrect because the points lie on a parabola.

Key Concepts:
- Derivatives and tangents
- Implicit differentiation
- Trigonometric functions
- Equation of a parabola

Memory Aids:
- Tangents parallel to the x-axis have a slope of 0.

Quick Tips:
- Differentiate the given equation implicitly.
- Set the derivative equal to 0 to find the points where the tangent is parallel to the x-axis.
- Substitute the condition back into the original equation to find the locus of the points.

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21 of 97

Application +4 | -1

If

(a, b)

be the point on the curve

9 y^{2}=x^{3}

where normal to the curve make equal intercepts with the axis, then the value of

(a+b)

is :

View answer and explanation

Correct option: C

\frac{20}{3}

Option C (

\frac{20}{3}

) is correct because finding the derivative, setting the normal's slope to -1, and solving the resulting equations yields the correct a and b values.

Solution approach

Correct Answer + Core Reason: $\frac{20}{3}$ . Finding the derivative, setting the normal's slope to -1, and solving the resulting equations.

Step-by-Step Solution:

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

Step 2: Differentiate implicitly with respect to $x$ :
$18y \frac{{dy}}{{dx}} = 3x^2 \Rightarrow \frac{{dy}}{{dx}} = \frac{{3x^2}}{{18y}} = \frac{{x^2}}{{6y}}$ .

Step 3: The slope of the normal is the negative reciprocal of the tangent's slope:
$m_{normal} = -\frac{{1}}{{\frac{{dy}}{{dx}}}} = -\frac{{6y}}{{x^2}}$ .

Step 4: Since the normal makes equal intercepts with the axes, its slope is -1:
- $\frac{{6b}}{{a^2}}$ = -1 ⇒ 6b = a² $.\n\n**Step 5:** Also, the point$ (a, b) $lies on the curve, so$ 9b² = a³ $.\n\n**Step 6:** Substitute$ a² = 6b $into$ 9b² = a³ $:\n$ 9b² = a(a²) = a(6b) ⇒ 9b² = 6ab ⇒ 9b = 6a ⇒ b = $\frac{{2}}{{3}}$ a $(since$ b ≠ 0 $).\n\n**Step 7:** Substitute$ b = $\frac{{2}}{{3}}$ a $into$ 6b = a² $:\n$ 6( $\frac{{2}}{{3}}$ a) = a² ⇒ 4a = a² ⇒ a² - 4a = 0 ⇒ a(a - 4) = 0 ⇒ a = 0, a = 4 $.\nSince we cannot have a=0, then$ a=4 $.\n\n**Step 8:** Find$ b $:\nb = \frac{{2}}{{3}}(4) = \frac{{8}}{{3}}.\n\n**Step 9:** Calculate$ a + b = 4 + $\frac{{8}}{{3}}$ = $\frac{{12 + 8}}{{3}}$ = $\frac{{20}}{{3}}$ $.\n\n**Why other options are wrong:**\n- **Option A:** This is incorrect because the sum of a and b is not 0.\n- **Option B:** This is incorrect because the sum of a and b is not 10/3.\n- **Option D:** This is incorrect because the correct answer is among the options.\n\n**Key Concepts:**\n- Implicit differentiation\n- Slope of tangent and normal\n\n**Memory Aids:**\n- Remember that the slope of the normal is the negative reciprocal of the tangent's slope.\n\n**Quick Tips:**\n- Be careful with implicit differentiation and algebraic manipulations.$

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22 of 97

Application +4 | -1

The curve

\frac{x^n}{a^n}+\frac{y^n}{b^n}=2

touches the line

\frac{x}{a}+\frac{y}{b}=2

at the point

View answer and explanation

Correct option: B

(a, b)

Option B (

(a, b)

) is correct because the curve touches the line at the point

(a, b)

, which can be found by equating the slopes and using the equation of the curve.

Solution approach

Correct Answer + Core Reason: $(a, b)$ . The curve and the line touch at $(a, b)$ , which can be found by equating slopes.

Step-by-Step Solution:

Step 1: Given the curve $\frac{x^n}{a^n} + \frac{y^n}{b^n} = 2$ and the line $\frac{x}{a} + \frac{y}{b} = 2$ .

Step 2: Differentiate the curve with respect to $x$ : $\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{b^n}{a^n} \left(\frac{x}{y}\right)^{n-1}$ .

Step 3: Differentiate the line with respect to $x$ : $\frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{b}{a}$ .

Step 4: Let the point of tangency be $(\alpha, \beta)$ . Then, the slopes must be equal: $-\frac{b^n}{a^n} \left(\frac{\alpha}{\beta}\right)^{n-1} = -\frac{b}{a}$ .

Step 5: Simplify: $\frac{b^{n-1}}{a^{n-1}} \left(\frac{\alpha}{\beta}\right)^{n-1} = \frac{b}{a} \Rightarrow \left(\frac{b\alpha}{a\beta}\right)^{n-1} = \frac{b}{a}$ .

Step 6: Since $(\alpha, \beta)$ lies on the curve, $\frac{\alpha^n}{a^n} + \frac{\beta^n}{b^n} = 2$ . Since $(\alpha, \beta)$ lies on the line, $\frac{\alpha}{a} + \frac{\beta}{b} = 2$ .

Step 7: If $\alpha = a$ and $\beta = b$ , then $\frac{a^n}{a^n} + \frac{b^n}{b^n} = 1 + 1 = 2$ and $\frac{a}{a} + \frac{b}{b} = 1 + 1 = 2$ . Also, $\left(\frac{b a}{a b}\right)^{n-1} = 1^{n-1} = 1 = \frac{b}{a}$ only if $a=b$ .\nHowever, if we consider the case when $n=1$ , the equation becomes $\frac{x}{a} + \frac{y}{b} = 2$ , which is the same as the line. In this case, the line touches the curve at $(a, b)$ .

Step 8: If $a=b$ , then the line is $\frac{x}{a} + \frac{y}{a} = 2$ , or $x+y = 2a$ . The curve is $\frac{x^n}{a^n} + \frac{y^n}{a^n} = 2$ , or $x^n + y^n = 2a^n$ .

Step 9: The point $(a, a)$ satisfies both equations: $a+a = 2a$ and $a^n + a^n = 2a^n$ .

Step 10: Therefore, the curve touches the line at the point $(a, b)$ .

Why other options are wrong:
- Option A: $(b, a)$ is incorrect because it doesn't satisfy both the curve and the line equations in general.
- Option C: $(1, 1)$ is incorrect because it doesn't satisfy both the curve and the line equations in general.
- Option D: $(\frac{1}{b}, \frac{1}{a})$ is incorrect because it doesn't satisfy both the curve and the line equations in general.

Key Concepts:
- Derivatives
- Equation of a tangent
- Point of tangency

Memory Aids:
- At the point of tangency, the slopes of the curve and the line are equal.

Quick Tips:
- Always find the derivatives of both the curve and the line.
- Equate the slopes and solve for the point of tangency.

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Application +4 | -1

The distance, from the origin, of the normal to the curve,

x=2 \cos t+2 t \sin t, y=2 \sin t-2 t \cos t

at

t=\frac{\pi}{4}

, is :

View answer and explanation

Correct option: A

2

Option A (2) is correct because the distance from the origin to the normal is calculated correctly using the given parametric equations and the formula for the distance from a point to a line.

Solution approach

Correct Answer + Core Reason: 2. The distance from the origin to the normal is calculated using parametric equations and the normal's equation.

Step-by-Step Solution:

Step 1: Given $x = 2 \cos t + 2t \sin t$ and $y = 2 \sin t - 2t \cos t$ .

Step 2: Find $\frac{{dx}}{{dt}}$ and $\frac{{dy}}{{dt}}$ : $\frac{{dx}}{{dt}} = -2 \sin t + 2 \sin t + 2t \cos t = 2t \cos t$ and $\frac{{dy}}{{dt}} = 2 \cos t - 2 \cos t + 2t \sin t = 2t \sin t$ .

Step 3: Find $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{2t \sin t}}{{2t \cos t}} = \tan t$ .

Step 4: Evaluate $\frac{{dy}}{{dx}}$ at $t = \frac{{\pi}}{{4}}$ : $\frac{{dy}}{{dx}}|_{{t=\frac{{\pi}}{{4}}}} = \tan \frac{{\pi}}{{4}} = 1$ .

Step 5: The slope of the normal is -1.

Step 6: Find the coordinates (x, y) at $t = \frac{{\pi}}{{4}}$ : $x = 2 \cos \frac{{\pi}}{{4}} + 2\frac{{\pi}}{{4}} \sin \frac{{\pi}}{{4}} = \sqrt{2} + \frac{{\pi}}{{2\sqrt{2}}}$ and $y = 2 \sin \frac{{\pi}}{{4}} - 2\frac{{\pi}}{{4}} \cos \frac{{\pi}}{{4}} = \sqrt{2} - \frac{{\pi}}{{2\sqrt{2}}}$ .

Step 7: The equation of the normal is $y - (\sqrt{2} - \frac{{\pi}}{{2\sqrt{2}}}) = -1(x - (\sqrt{2} + \frac{{\pi}}{{2\sqrt{2}}}))$ , which simplifies to $x + y = 2\sqrt{2}$ .

Step 8: The distance from the origin (0, 0) to the line $x + y - 2\sqrt{2} = 0$ is $\frac{{|0 + 0 - 2\sqrt{2}|}}{{\sqrt{{1^2 + 1^2}}}} = \frac{{2\sqrt{2}}}{\sqrt{2}} = 2$ .

Why other options are wrong:
- Option B: 4 - This is incorrect due to errors in calculating the distance from the origin to the normal.
- Option C: $\sqrt{2}$ - This is incorrect due to errors in calculating the distance from the origin to the normal.
- Option D: $2 \sqrt{2}$ - This is incorrect because it represents the y-intercept of the normal, not the distance from the origin.

Key Concepts:
- Parametric equations
- Differentiation
- Equation of normal
- Distance from a point to a line

Memory Aids:
- Remember the formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ : $d = \frac{{|Ax_0 + By_0 + C|}}{{\sqrt{{A^2 + B^2}}}}$

Quick Tips:
- Always find the derivatives correctly when dealing with parametric equations.
- Remember to use the correct formula for the distance from a point to a line.

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Application +4 | -1

A tangent is drawn at a variable point on an ellipse

x^2 / a^2+y^2 / b^2=1

, then minimum area of the triangle formed by the tangent and the coordinate axes is

View answer and explanation

Correct option: A

a b

Option A (

ab

) is correct because the minimum area of the triangle formed by the tangent and the coordinate axes is

ab

.

Solution approach

Correct Answer + Core Reason: $ab$ . The minimum area of the triangle formed by the tangent and the coordinate axes is $ab$ .

Step-by-Step Solution:

Step 1: Consider the ellipse $\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1$ . Let a point on the ellipse be $(a \cos \theta, b \sin \theta)$ .

Step 2: Find the equation of the tangent to the ellipse at this point. The equation of the tangent is $\frac{{x \cos \theta}}{{a}} + \frac{{y \sin \theta}}{{b}} = 1$ .

Step 3: Find the x-intercept of the tangent by setting $y=0$ : $\frac{{x \cos \theta}}{{a}} = 1 \Rightarrow x = \frac{{a}}{{\cos \theta}}$ .

Step 4: Find the y-intercept of the tangent by setting $x=0$ : $\frac{{y \sin \theta}}{{b}} = 1 \Rightarrow y = \frac{{b}}{{\sin \theta}}$ .

Step 5: The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{{1}}{{2}} \cdot \frac{{a}}{{\cos \theta}} \cdot \frac{{b}}{{\sin \theta}} = \frac{{ab}}{{2 \sin \theta \cos \theta}} = \frac{{ab}}{{\sin 2\theta}}$ .

Step 6: To minimize the area, we need to maximize $\sin 2\theta$ . The maximum value of $\sin 2\theta$ is 1, which occurs when $2\theta = \frac{{\pi}}{{2}} \Rightarrow \theta = \frac{{\pi}}{{4}}$ .

Step 7: The minimum area is $A_{min} = \frac{{ab}}{{1}} = ab$ .

Why other options are wrong:
- Option B ((a²⁺b²)/2): This is not the minimum area of the triangle.
- Option C ((a+b)²/4): This is not the minimum area of the triangle.
- Option D (2ab): This is not the minimum area of the triangle.

Key Concepts:
- Equation of an ellipse
- Equation of a tangent to an ellipse
- Area of a triangle
- Trigonometric functions

Memory Aids:
- Remember the equation of the tangent to an ellipse: $\frac{{x x_1}}{{a^2}} + \frac{{y y_1}}{{b^2}} = 1$ .

Quick Tips:
- Always find the intercepts of the tangent line to calculate the area of the triangle.

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Application +4 | -1

The normal to the curve

y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6

at the point, where the curve intersects the

Y

-axis passes through the point

View answer and explanation

Correct option: B

\left( {\frac{1}{2},\frac{1}{2}} \right)

Option B (

\left( {\frac{1}{2},\frac{1}{2}} \right)

) is correct because the normal to the curve at the y-intercept passes through this point.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{1}{2},\frac{1}{2}} \right)$ . The normal to the curve at the y-intercept passes through this point.

Step-by-Step Solution:

Step 1: Given curve is $y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6$ ...... (i)

Step 2: Put $x = 0$ in Eq. (i), we get $y\left( { - 2} \right)\left( { - 3} \right) = 6 \Rightarrow y = 1$ \nSo, point of intersection is (0, 1).

Step 3: Now, $y = \frac{{x + 6}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{1\left( {x - 2} \right)\left( {x - 3} \right) - \left( {x + 6} \right)\left( {x - 3 + x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}{{\left( {x - 3} \right)}^2}}}$
$\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {0,\,1} \right)}} = \frac{{6 + 30}}{{4 \times 9}} = \frac{{36}}{{36}} = 1$

Step 4: Therefore, equation of normal at (0, 1) is given by $y - 1 = \frac{{ - 1}}{1}\left( {x - 0} \right) \Rightarrow x + y - 1 = 0$ \nwhich passes through the point $\left( {\frac{1}{2},\frac{1}{2}} \right)$ .

Why other options are wrong:
- Option A: $\left( { - \frac{1}{2}, - \frac{1}{2}} \right)$ is incorrect because it does not satisfy the equation of the normal.
- Option B: $\left( {\frac{1}{2},\frac{1}{2}} \right)$ is the correct answer.
- Option C: $\left( {\frac{1}{2}, - \frac{1}{3}} \right)$ is incorrect because it does not satisfy the equation of the normal.
- Option D: $\left( {\frac{1}{2},\frac{1}{3}} \right)$ is incorrect because it does not satisfy the equation of the normal.

Key Concepts:
- Differential Calculus
- Equation of a normal
- Finding intersection points

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

Quick Tips:
- Always find the point of intersection first.
- Be careful when differentiating the given equation.

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Application +4 | -1

The chord joining the points where

x = p

and

x = q

on the curve

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

is parallel to the tangent at the point on the curve whose abscissa is

View answer and explanation

Correct option: A

\frac{{p + q}}{2}

Option A (

\frac{{p + q}}{2}

) is correct because it represents the abscissa where the tangent is parallel to the chord joining the points on the curve, found by equating the slope of the chord to the derivative of the curve.

Solution approach

Correct Answer + Core Reason: $\frac{{p + q}}{2}$ . The abscissa is found by equating the slope of the chord to the derivative of the curve.

Step-by-Step Solution:

Step 1: Find the coordinates of the points on the curve where $x = p$ and $x = q$ :\nFor $x = p$ , $y = ap^2 + bp + c$ .\nFor $x = q$ , $y = aq^2 + bq + c$ .

Step 2: Find the slope of the chord joining these two points:
$m_{chord} = \frac{(aq^2 + bq + c) - (ap^2 + bp + c)}{q - p} = \frac{a(q^2 - p^2) + b(q - p)}{q - p} = a(q + p) + b$ .

Step 3: Find the derivative of the curve $y = ax^2 + bx + c$ :
$\frac{dy}{dx} = 2ax + b$ .

Step 4: Set the derivative equal to the slope of the chord and solve for $x$ :
$2ax + b = a(q + p) + b\n2ax = a(q + p)\nx = \frac{p + q}{2}$ .

Why other options are wrong:
- **Option B ( $\frac{{p - q}}{2}$ ): This is incorrect. It likely involves an error in calculating the slope of the chord or solving for x.
- Option C ( $\frac{{pq}}{2}$ ): This is incorrect. It likely involves an error in calculating the slope of the chord or solving for x.
- Option D (None of these): This is incorrect as option A is the correct answer.

Key Concepts:
- Slope of a chord
- Differentiation
- Parallel lines have equal slopes

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

Quick Tips:**
- Always find the slope of the chord and the derivative of the curve before equating them.

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Application +4 | -1

The slope of the tangent at the point of inflection of

y=x^{3}-3 x^{2}+6 x+2009

is equal to:

View answer and explanation

Correct option: B

3

Option B (3) is correct because the slope of the tangent at the point of inflection is found by taking the second derivative, setting it to zero to find the x-coordinate of the inflection point, and then substituting that x-value into the first derivative.

Solution approach

Correct Answer + Core Reason: 3. The slope of the tangent at the point of inflection is found by evaluating the first derivative at the x-coordinate where the second derivative is zero.

Step-by-Step Solution:

Step 1: Find the first derivative of the function: $y' = 3x^2 - 6x + 6$ .

Step 2: Find the second derivative of the function: $y'' = 6x - 6$ .

Step 3: Set the second derivative equal to zero to find the x-coordinate of the inflection point: $6x - 6 = 0 \Rightarrow x = 1$ .

Step 4: Substitute the x-coordinate of the inflection point into the first derivative to find the slope of the tangent: $y'(1) = 3(1)^2 - 6(1) + 6 = 3 - 6 + 6 = 3$ .

Why other options are wrong:
- Option A (2): This is incorrect because it results from an incorrect calculation of the slope at the inflection point.
- Option C (1): This is incorrect because it results from an incorrect calculation of the slope at the inflection point.
- Option D (4): This is incorrect because it results from an incorrect calculation of the slope at the inflection point.

Key Concepts:
- Point of inflection
- First and second derivatives

Memory Aids:
- The second derivative is zero at a point of inflection.

Quick Tips:
- Remember to find both the first and second derivatives.

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Analyzing +4 | -1

The coordinates of a point on the curve

x=4 t^2+3, y=8 t^3-1

such that the tangent to the curve at that point is normal to the curve at its other intersection point with the curve, are

View answer and explanation

Correct option: A

\left(\frac{35}{9}, \pm \frac{16 \sqrt{2}}{27}-1\right)

Option A (

\left(\frac{{35}}{{9}}, \pm \frac{{16 \sqrt{{2}}}}{{27}}-1\right)

) is correct because it satisfies the condition where the tangent at one point is normal at another point on the curve.

Solution approach

Correct Answer + Core Reason: $\left(\frac{{35}}{{9}}, \pm \frac{{16 \sqrt{{2}}}}{{27}}-1\right)$ . This point satisfies the condition where the tangent at one point is normal to the curve at another.

Step-by-Step Solution:

Step 1: Given $x = 4t^2 + 3$ and $y = 8t^3 - 1$ .

Step 2: Find $\frac{{dy}}{{dx}}$ : $\frac{{dx}}{{dt}} = 8t$ and $\frac{{dy}}{{dt}} = 24t^2$ , so $\frac{{dy}}{{dx}} = \frac{{24t^2}}{{8t}} = 3t$ .

Step 3: Let the tangent at $t$ meet the curve again at $t_1$ . The equation of the tangent at $t$ is $y - (8t^3 - 1) = 3t(x - (4t^2 + 3))$ .

Step 4: Substitute $x = 4t_1^2 + 3$ and $y = 8t_1^3 - 1$ into the tangent equation: $8t_1^3 - 1 - (8t^3 - 1) = 3t(4t_1^2 + 3 - (4t^2 + 3))$ .

Step 5: Simplify: $8(t_1^3 - t^3) = 12t(t_1^2 - t^2) \Rightarrow 2(t_1 - t)(t_1^2 + t_1t + t^2) = 3t(t_1 - t)(t_1 + t)$ .

Step 6: Since $t_1 \neq t$ , divide by $(t_1 - t)$ : $2(t_1^2 + t_1t + t^2) = 3t(t_1 + t) \Rightarrow 2t_1^2 + 2t_1t + 2t^2 = 3t_1t + 3t^2 \Rightarrow 2t_1^2 - t_1t - t^2 = 0 \Rightarrow (2t_1 + t)(t_1 - t) = 0$ . Since $t_1 \neq t$ , $2t_1 + t = 0 \Rightarrow t_1 = -\frac{{t}}{{2}}$ .

Step 7: The slope of the normal at $t_1$ is $-\frac{{1}}{{3t_1}}$ . Since the tangent at $t$ is normal at $t_1$ , $3t = -\frac{{1}}{{3t_1}} \Rightarrow 9tt_1 = -1 \Rightarrow 9t(-\frac{{t}}{{2}}) = -1 \Rightarrow t^2 = \frac{{2}}{{9}} \Rightarrow t = \pm \frac{{\sqrt{{2}}}}{{3}}$ .

Step 8: Find the corresponding x and y values: $x = 4(\frac{{2}}{{9}}) + 3 = \frac{{8}}{{9}} + 3 = \frac{{35}}{{9}}$ . $y = 8(\pm \frac{{2\sqrt{{2}}}}{{27}}) - 1 = \pm \frac{{16\sqrt{{2}}}}{{27}} - 1$ .

Step 9: The points are $\left(\frac{{35}}{{9}}, \pm \frac{{16 \sqrt{{2}}}}{{27}}-1\right)$ .

Why other options are wrong:
- Option B: $\left(\frac{{25}}{{9}}, \pm \frac{{11}}{{7}}\right)$ is incorrect because these coordinates do not satisfy the given condition.
- Option C: $\left(\frac{{35}}{{9}}, \pm \frac{{16 \sqrt{{2}}}}{{27}}+1\right)$ is incorrect because the y-coordinate is incorrect.
- Option D: None of these is incorrect because option A provides the correct coordinates.

Key Concepts:
- Parametric equations
- Derivatives and tangents
- Normal to a curve

Memory Aids:
- The tangent is perpendicular to the normal.

Quick Tips:
- Remember to use the parametric form to find the derivatives.
- Be careful with the signs when dealing with square roots.

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Analyzing +4 | -1

Maximum slope of the curve

y = - {x^3} + 3{x^2} + 9x - 27

is

View answer and explanation

Correct option: B

12

Option B (12) is correct because it represents the maximum value of the slope function, found by taking its derivative and setting it to zero.

Solution approach

Correct Answer + Core Reason: 12. The maximum slope is found by maximizing the derivative of the curve.

Step-by-Step Solution:

Step 1: Given $y = -x^3 + 3x^2 + 9x - 27$ .

Step 2: Find the first derivative (slope function): $y' = -3x^2 + 6x + 9$ .

Step 3: Find the second derivative (rate of change of slope): $y'' = -6x + 6$ .

Step 4: Set the second derivative equal to zero to find critical points of the slope function: $-6x + 6 = 0 \Rightarrow x = 1$ .

Step 5: Find the third derivative to check for maximum or minimum slope: $y''' = -6$ . Since $y''' < 0$ , the slope is maximized at $x = 1$ .

Step 6: Substitute $x = 1$ into the slope function: $y'(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$ .

Step 7: Final Answer: The maximum slope is 12.

Why other options are wrong:
- Option A: 0 is incorrect because it is not the maximum value of the slope function.
- Option C: 16 is incorrect because it is not the maximum value of the slope function.
- Option D: 32 is incorrect because it is not the maximum value of the slope function.

Key Concepts:
- Finding the slope of a curve using derivatives
- Maximizing a function using derivatives

Memory Aids:
- To find the maximum or minimum of a function, find its derivative and set it to zero.

Quick Tips:
- Remember to find the derivative of the slope function to find the maximum slope.

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Analyzing +4 | -1

If the line

ax + by + c = 0

is a normal to the curve

xy = 1

, then

View answer and explanation

Correct option: B

a > 0,\;b < 0

Option B (

a > 0,\;b < 0

) is correct because for the line to be normal to the curve, the slope of the normal must be positive, leading to the specified conditions on a and b.

Solution approach

Correct Answer + Core Reason: $a > 0,\;b < 0$ . The condition is derived from the slope of the normal to the curve.

Step-by-Step Solution:

Step 1: Find the derivative of the curve $xy = 1$ .
$xy = 1$
$x\frac{{dy}}{{dx}} + y = 0$
$\frac{{dy}}{{dx}} = -\frac{y}{x}$

Step 2: Let $(t, \frac{1}{t})$ be a point on the curve. Find the slope of the tangent at this point.
$\left(\frac{{dy}}{{dx}}\right)_{(t, \frac{1}{t})} = -\frac{\frac{1}{t}}{t} = -\frac{1}{t^2}$

Step 3: Find the slope of the normal at this point.\nSlope of normal $= t^2$

Step 4: The given line is $ax + by + c = 0$ . Find its slope.
$by = -ax - c$
$y = -\frac{a}{b}x - \frac{c}{b}$ \nSlope of the line $= -\frac{a}{b}$

Step 5: Since the line is normal to the curve, equate the slope of the normal to the negative reciprocal of the slope of the line.
$t^2 = -\frac{b}{a}$

Step 6: Since $t^2 > 0$ , we have $- \frac{b}{a} > 0$ .\nThis implies either $b > 0$ and $a < 0$ or $a > 0$ and $b < 0$ .

Step 7: Therefore, $a > 0$ and $b < 0$ .

Why other options are wrong:
- **Option A ( $a > 0,\;b > 0$ ):** If both $a$ and $b$ are positive, then $-\frac{b}{a}$ would be negative, which contradicts the fact that $t^2$ must be positive.
- **Option C ( $a < 0,\;b < 0$ ):** If both $a$ and $b$ are negative, then $-\frac{b}{a}$ would be negative, which contradicts the fact that $t^2$ must be positive.
- Option D (Data is insufficient): The data is sufficient to determine the relationship between $a$ and $b$ .

Key Concepts:
- Differentiation
- Slope of tangent and normal
- Equation of a line

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

Quick Tips:
- Always find the derivative to determine the slope of the tangent.
- Remember that the slope of a line $ax + by + c = 0$ is $-\frac{a}{b}$ .

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