TANGENTS & NORMALS MCQs for JEE MATHEMATICS — Medium, Page 2

31 Medium

31 of 97

Application +4 | -1

Let

f:[0,2 \pi] \rightarrow[-3,3]

be a given function defined as

f(x)=3 \cos \frac{x}{2}

. The slope of the tangent to the curve

y=f^{-1}(x)

at the point where the curve crosses the

y

-axis is :

View answer and explanation

Correct option: B

-\frac{2}{3}

Option B (-

\frac{2}{3}

) is correct because finding the inverse function and its derivative, then evaluating at the y-intercept, gives the slope of the tangent as -2/3.

Solution approach

Correct Answer + Core Reason: - $\frac{2}{3}$ . The slope of the tangent to the inverse function is found by differentiating the inverse and evaluating at the y-intercept.

Step-by-Step Solution:

Step 1: Find the inverse function: $f(x) = 3\cos(\frac{x}{2}) \Rightarrow y = 3\cos(\frac{x}{2}) \Rightarrow \frac{y}{3} = \cos(\frac{x}{2}) \Rightarrow \frac{x}{2} = \cos^{-1}(\frac{y}{3}) \Rightarrow x = 2\cos^{-1}(\frac{y}{3})$ . So, $f^{-1}(x) = 2\cos^{-1}(\frac{x}{3})$ .

Step 2: Find the derivative of the inverse function: $\frac{d}{dx}f^{-1}(x) = \frac{d}{dx} \left[2\cos^{-1}(\frac{x}{3})\right] = 2 \cdot \frac{-1}{\sqrt{1 - (\frac{x}{3})^2}} \cdot \frac{1}{3} = \frac{-2}{3\sqrt{1 - \frac{x^2}{9}}} = \frac{-2}{\sqrt{9 - x^2}}$ .

Step 3: The curve crosses the y-axis when x = 0. Evaluate the derivative at x = 0: $\frac{d}{dx}f^{-1}(x)|_{x=0} = \frac{-2}{\sqrt{9 - 0^2}} = \frac{-2}{\sqrt{9}} = -\frac{2}{3}$ .

Why other options are wrong:
- Option A (-1): This is incorrect because it results from an incorrect calculation of the derivative.
- **Option C (- $\frac{1}{6}$ ): This is incorrect because it results from an incorrect calculation of the derivative.
- Option D (- $\frac{1}{3}$ ): This is incorrect because it results from an incorrect calculation of the derivative.

Key Concepts:
- Inverse functions
- Differentiation

Memory Aids:
- Remember the derivative of inverse cosine.

Quick Tips:**
- Remember to find the inverse function before differentiating.

32 Medium

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Application +4 | -1

The distance between the origin and the normal to the curve

y = {e^{2x}} + {x^2}\,at\,x = 0

is

View answer and explanation

Correct option: C

\frac{2}{{\sqrt 5 }}

Option C (

\frac{2}{{\sqrt 5 }}

) is correct because it represents the perpendicular distance from the origin to the normal line of the given curve at x=0.

Solution approach

Correct Answer + Core Reason: $\frac{2}{{\sqrt 5 }}$ . The distance is calculated using the formula for the distance from a point to a line.

Step-by-Step Solution:

Step 1: Find the y-coordinate at x=0: $y = e^{2(0)} + 0^2 = 1$ . So the point is (0, 1).

Step 2: Find the derivative: $\frac{{dy}}{{dx}} = 2e^{2x} + 2x$ .

Step 3: Evaluate the derivative at x=0: $\frac{{dy}}{{dx}}|_{x=0} = 2e^{2(0)} + 2(0) = 2$ .

Step 4: Find the slope of the normal: $m_{normal} = -\frac{1}{2}$ .

Step 5: Find the equation of the normal: $y - 1 = -\frac{1}{2}(x - 0) \Rightarrow x + 2y - 2 = 0$ .

Step 6: Find the distance from the origin (0, 0) to the normal line: $d = \frac{{|Ax_0 + By_0 + C|}}{{\sqrt{{A^2 + B^2}}}} = \frac{{|1(0) + 2(0) - 2|}}{{\sqrt{{1^2 + 2^2}}}} = \frac{2}{{\sqrt 5 }}$ .

Why other options are wrong:
- Option A (2): This is incorrect because it doesn't account for the slope of the normal and the distance formula.
- **Option B ( $\frac{2}{{\sqrt 3 }}$ ) $:** This is incorrect because the denominator is wrong. It may be a calculation error.\n- **Option D (\frac{1}{2})$ : This is incorrect because it doesn't use the correct distance formula and slope of the normal.

Key Concepts:
- Derivatives
- Equation of a normal line
- Distance from a point to a line

Memory Aids:**
- Slope of normal = -1 / slope of tangent
- Distance formula: $d = \frac{{|Ax_0 + By_0 + C|}}{{\sqrt{{A^2 + B^2}}}}$

Quick Tips:
- Always find the point on the curve first.
- Remember to take the negative reciprocal of the tangent's slope to find the normal's slope.
- Be careful with the distance formula.

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Application +4 | -1

The point on the curve

\sqrt x + \sqrt y = \sqrt a

, the normal at which is parallel to the

x

-axis, is

View answer and explanation

Correct option: B

\left( {0,\;a} \right)

Option B ((0, a)) is correct because the normal being parallel to the x-axis implies dx/dy = 0, leading to x=0, and substituting in the original equation gives y=a.

Solution approach

Correct Answer + Core Reason: $\left( {0,\;a} \right)$ . The point is found by differentiating the curve and setting dx/dy = 0.

Step-by-Step Solution:

Step 1: Given curve: $\sqrt x + \sqrt y = \sqrt a$ .

Step 2: Differentiate the curve with respect to x: $\frac{1}{{2\sqrt x }} + \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = 0$ .

Step 3: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = - \frac{{\sqrt y }}{{\sqrt x }}$ .

Step 4: Since the normal is parallel to the x-axis, the slope of the normal is 0, which means $\frac{{dx}}{{dy}} = 0$ .

Step 5: Find $\frac{{dx}}{{dy}}$ : $\frac{{dx}}{{dy}} = - \frac{{\sqrt x }}{{\sqrt y }}$ . Setting this to 0 gives $\sqrt x = 0 \Rightarrow x = 0$ .

Step 6: Substitute x=0 into the original equation: $\sqrt 0 + \sqrt y = \sqrt a \Rightarrow \sqrt y = \sqrt a \Rightarrow y = a$ .

Step 7: Final Answer: The point is (0, a).

Why other options are wrong:
- Option A (0,0): This point does not satisfy the original equation $\sqrt x + \sqrt y = \sqrt a$ unless a=0.
- Option C (a,0): At this point, the tangent is parallel to the x-axis, not the normal.
- Option D (a,a): This point does not satisfy the original equation $\sqrt x + \sqrt y = \sqrt a$ unless a=4a, which means a=0.

Key Concepts:
- Differentiation of implicit functions.
- Relationship between the slope of a tangent and a normal.
- Condition for a line to be parallel to the x-axis.

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

Quick Tips:
- Always check if the point satisfies the original equation.

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Analyzing +4 | -1

The number of points on the curve

y=54 x^5-135 x^4

-70 x^3+180 x^2+210 x

at which the normal lines are parallel to

x+90 y+2=0

is

View answer and explanation

Correct option: A

4

The number of points is 4. The slope of the tangent is found, set equal to 90, and the resulting quartic equation has 4 real roots, indicating 4 points where the normal lines are parallel to the given line.

Solution approach

Correct Answer + Core Reason: 4. The quartic equation derived from the slope condition has 4 real roots.

Step-by-Step Solution:

Step 1: Find the derivative of $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$ .
$y' = 270x^4 - 540x^3 - 210x^2 + 360x + 210$ .

Step 2: Find the slope of the given line $x + 90y + 2 = 0$ .
$90y = -x - 2 \Rightarrow y = -\frac{1}{90}x - \frac{2}{90}$ . The slope is $-\frac{1}{90}$ .

Step 3: The normal lines are parallel to the given line, so the slope of the tangent lines is the negative reciprocal of $-\frac{1}{90}$ , which is 90.

Step 4: Set the derivative equal to 90: $270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$ .

Step 5: Simplify the equation: $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0 \Rightarrow 9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$ .

Step 6: Find the roots of the quartic equation. The roots are $x = 1, 2, -\frac{1}{3}, -\frac{2}{3}$ .

Step 7: Since there are 4 distinct real roots, there are 4 points on the curve where the normal lines are parallel to the given line.

Why other options are wrong:
- Option D: There are 4 such points, not 0.
- Option C: There are 4 such points, not 2.
- Option B: There are 4 such points, not 3.

Key Concepts:
- Derivatives
- Slope of a line
- Normal and tangent lines
- Roots of a polynomial

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

Quick Tips:
- Simplify the equation before finding the roots.

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Application +4 | -1

The tangent at the point

(2,-2)

to the curve,

x^2 y^2-2 x=4(1-y)

does not pass through the point:

View answer and explanation

Correct option: D

(-2,-7)

Option D ((-2,-7)) is correct because the point (-2, -7) does not satisfy the equation of the tangent to the curve at the given point.

Solution approach

Correct Answer + Core Reason: $(-2,-7)$ . The point does not lie on the tangent to the curve at (2, -2).

Step-by-Step Solution:

Step 1: Given the curve equation $x^2 y^2 - 2x = 4(1 - y)$ .

Step 2: Differentiate the equation with respect to $x$ : $2xy^2 + 2yx^2 \frac{{dy}}{{dx}} - 2 = -4 \frac{{dy}}{{dx}}$ .

Step 3: Rearrange to solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}}(2yx^2 + 4) = 2 - 2xy^2$ .

Step 4: Evaluate $\frac{{dy}}{{dx}}$ at the point (2, -2): $\frac{{dy}}{{dx}}|_{{(2,-2)}} = \frac{{2 - 2(2)(-2)^2}}{{2(-2)(2)^2 + 4}} = \frac{{-14}}{{-12}} = \frac{{7}}{{6}}$ .

Step 5: The equation of the tangent is $y + 2 = \frac{{7}}{{6}}(x - 2)$ , which simplifies to $7x - 6y = 26$ .

Step 6: Check which point does not satisfy the tangent equation.

Step 7: For (-2, -7): $7(-2) - 6(-7) = -14 + 42 = 28 \neq 26$ . Therefore, (-2, -7) does not lie on the tangent.

Why other options are wrong:
- Option A: $\left(4, \frac{{1}}{{3}}\right)$ - This point satisfies the tangent equation: $7(4) - 6(\frac{{1}}{{3}}) = 28 - 2 = 26$ .
- Option B: $(8,5)$ - This point satisfies the tangent equation: $7(8) - 6(5) = 56 - 30 = 26$ .
- Option C: $(-4,-9)$ - This point satisfies the tangent equation: $7(-4) - 6(-9) = -28 + 54 = 26$ .

Key Concepts:
- Differentiation
- Equation of tangent
- Checking points on a line

Memory Aids:
- Remember the point-slope form of a line: $y - y_1 = m(x - x_1)$

Quick Tips:
- Always find the derivative first to get the slope of the tangent.
- Substitute the given point into the tangent equation to check if it lies on the line.

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Analyzing +4 | -1

The tangent and normal at the point

P\left( {a{t^2},\;2\;at} \right)

to the parabola

{y^2} = 4\;ax

meet the

x

-axis in

T

and

G

respectively, then the angle at which the tangent at

P

to the parabola is inclined to the tangent at P to the circle through

T,\;P,\;G

is

View answer and explanation

Correct option: C

{\tan ^{ - 1}}t

Option C (

{\tan ^{ - 1}}t

) is correct because the angle between the tangents is found using the slopes of the tangent to the parabola and the tangent to the circle at point P.

Solution approach

Correct Answer + Core Reason: ${\tan ^{ - 1}}t$ . The angle between the tangents is found using the slopes of the tangent to the parabola and the tangent to the circle at point P.

Step-by-Step Solution:

Step 1: Find the equation of the tangent to the parabola $y^2 = 4ax$ at $P(at^2, 2at)$ . The equation is $ty = x + at^2$ .

Step 2: Find the equation of the normal to the parabola at $P$ . The equation is $y + tx = 2at + at^3$ .

Step 3: Find the points where the tangent and normal meet the x-axis. Tangent meets at $T(-at^2, 0)$ and normal meets at $G(2a + at^2, 0)$ .

Step 4: Since PT is perpendicular to PG, TG is the diameter of the circle through P, T, G. The equation of the circle is $(x + at^2)(x - 2a - at^2) + (y - 0)(y - 0) = 0$ which simplifies to $x^2 + y^2 - 2ax - a{t^2}(2a + a{t^2}) = 0$ .

Step 5: Differentiate the circle equation to find the slope of the tangent to the circle at P: $2x + 2y \frac{{dy}}{{dx}} - 2a = 0 \Rightarrow \frac{{dy}}{{dx}} = \frac{{a - x}}{y}$ . At P, $m_1 = (\frac{{dy}}{{dx}})_P = \frac{{a - a{t^2}}}{{2at}} = \frac{{1 - {t^2}}}{{2t}}$ .

Step 6: Differentiate the parabola equation to find the slope of the tangent to the parabola at P: $y^2 = 4ax \Rightarrow 2y \frac{{dy}}{{dx}} = 4a \Rightarrow \frac{{dy}}{{dx}} = \frac{{2a}}{y}$ . At P, $m_2 = (\frac{{dy}}{{dx}})_P = \frac{{2a}}{{2at}} = \frac{1}{t}$ .

Step 7: Let $\theta$ be the angle between the tangents. Then $\tan \theta = \frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}} = \frac{{\frac{1}{t} - \frac{{1 - {t^2}}}{{2t}}}}{{1 + \frac{{1 - {t^2}}}{{2{t^2}}}}} = t \Rightarrow \theta = {\tan ^{ - 1}}t$ .

Why other options are wrong:
- Option A: ${\tan ^{ - 1}}{t^2}$ : This is incorrect because the calculation of the angle between the tangents is flawed.
- Option B: ${\cot ^{ - 1}}{t^2}$ : This is incorrect because the calculation of the angle between the tangents is flawed.
- Option D: ${\cot ^{ - 1}}t$ : This is incorrect because the calculation of the angle between the tangents is flawed.

Key Concepts:
- Tangent and normal to a parabola
- Equation of a circle
- Angle between two lines
- Differentiation

Memory Aids:
- Remember the standard equations of tangent and normal to a parabola.

Quick Tips:
- Draw a diagram to visualize the problem.
- Use the correct formulas for tangent and normal.

37 Medium

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Application +4 | -1

Let

P\left( {2,\;2} \right)

and

Q\left( {1/2,\; - 1} \right)

be two points on the parabola

{y^2} = 2x

. The coordinates of the point

R

on the parabola

{y^2} = 2x

, where the tangent to the curve is parallel to the chord

PQ

, are

View answer and explanation

Correct option: B

\left( {1/8,\;1/2} \right)

Option B (

\left( {1/8,\;1/2} \right)

) is correct because it represents the point on the parabola where the tangent is parallel to the chord PQ.

Solution approach

Correct Answer + Core Reason: $\left( {1/8,\;1/2} \right)$ . This point satisfies the condition that the tangent is parallel to the chord PQ.

Step-by-Step Solution:

Step 1: Find the slope of the chord PQ.
$P(2, 2)$ and $Q(\frac{1}{2}, -1)$
$m_{PQ} = \frac{-1 - 2}{\frac{1}{2} - 2} = \frac{-3}{-\frac{3}{2}} = 2$

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

Step 3: Set the derivative equal to the slope of the chord PQ.
$\frac{1}{y} = 2$
$y = \frac{1}{2}$

Step 4: Substitute the value of $y$ back into the equation of the parabola to find the corresponding $x$ value.
$(\frac{1}{2})^2 = 2x$
$\frac{1}{4} = 2x$
$x = \frac{1}{8}$

Step 5: The point R is $(\frac{1}{8}, \frac{1}{2})$ .

Why other options are wrong:
- **Option A ( $\left( {2,\; - 1} \right)$

38 Medium

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Application +4 | -1

The line

x+y=2

is tangent to the curve

x^2=3-2 y

at its point

View answer and explanation

Correct option: A

(1,1)

Option A (

(1,1)

) is correct because by finding the derivative of the curve and setting it equal to the slope of the line, we can solve for the point of tangency.

Solution approach

Correct Answer + Core Reason: $(1,1)$ . By finding the derivative of the curve and setting it equal to the slope of the line, we can solve for the point of tangency.

Step-by-Step Solution:

Step 1: Given: Line $x + y = 2$ and curve $x^2 = 3 - 2y$ .

Step 2: Find the slope of the line: $y = 2 - x$ , so the slope is $-1$ .

Step 3: Find the derivative of the curve: $2x = -2 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -x$ .

Step 4: Set the derivative equal to the slope of the line: $-x = -1 \Rightarrow x = 1$ .

Step 5: Find the corresponding y-value on the curve: $(1)^2 = 3 - 2y \Rightarrow 1 = 3 - 2y \Rightarrow 2y = 2 \Rightarrow y = 1$ .

Step 6: The point of tangency is $(1, 1)$ .

Why other options are wrong:
- Option B: $(-1,1)$ is incorrect because the derivative at x=-1 is 1, which is not the slope of the line (-1).
- Option C: $(\sqrt{3}, 0)$ is incorrect because this point does not lie on the line $x+y=2$ .
- Option D: $(3,-3)$ is incorrect because this point does not lie on the curve $x^2 = 3 - 2y$ .

Key Concepts:
- Differentiation
- Finding the slope of a line
- Tangent to a curve

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

Quick Tips:
- Always check if the point lies on both the curve and the line.

39 Medium

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Application +4 | -1

If

f(x)=\log x

satisfies Lagrange's theorem on

[1, e]

then value of

c \in(1, e)

such that the tangent at

c

is parallel to line joining

(1, f(1))

and

(e, f(e))

is

View answer and explanation

Correct option: C

e-1

Option C (

e-1

) is correct because the value of c that satisfies Lagrange's theorem is e-1, which is found by equating the derivative at c to the average rate of change over the interval [1, e].

Solution approach

Correct Answer + Core Reason: $e-1$ . The value of $c$ is found by applying Lagrange's Mean Value Theorem.

Step-by-Step Solution:

Step 1: State Lagrange's Mean Value Theorem: If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists a $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ .

Step 2: Identify the given function and interval: $f(x) = \log x$ on $[1, e]$ .

Step 3: Find the derivative of $f(x)$ : $f'(x) = \frac{1}{x}$ .

Step 4: Apply Lagrange's Mean Value Theorem: $f'(c) = \frac{f(e) - f(1)}{e - 1}$ .

Step 5: Substitute the values: $\frac{1}{c} = \frac{\log e - \log 1}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1}$ .

Step 6: Solve for $c$ : $\frac{1}{c} = \frac{1}{e - 1} \Rightarrow c = e - 1$ .

Why other options are wrong:
- **Option A ( $e-\frac{3}{2}$ ): This is incorrect. It might be obtained by incorrectly applying the Mean Value Theorem or making a calculation error.
- Option B ( $\frac{1+e}{2}$ ): This is incorrect. It might be a random guess or a misunderstanding of the problem.
- Option D ( $e-\frac{1}{2}$ ): This is incorrect. It might be obtained by incorrectly applying the Mean Value Theorem or making a calculation error.

Key Concepts:
- Lagrange's Mean Value Theorem
- Derivatives
- Logarithmic functions

Memory Aids:**
- Remember the formula for Lagrange's Mean Value Theorem: $f'(c) = \frac{f(b) - f(a)}{b - a}$ .

Quick Tips:
- Make sure the function satisfies the conditions of Lagrange's Mean Value Theorem before applying it.

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Application +4 | -1

Equation of normal to the curve

x^2=4 y

which passes through the point

(1,2)

is

View answer and explanation

Correct option: B

x+y=3

Option B

x+y=3

is correct because by finding the equation of the normal to the curve and using the given point, we can solve for the normal's equation.

Solution approach

Correct Answer + Core Reason: $x+y=3$ . The equation of the normal is found by using the derivative and the given point.

Step-by-Step Solution:

Step 1: The equation of the curve is $x^2 = 4y$ . Differentiate with respect to $x$ : $2x = 4\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2}$ .

Step 2: Let the point on the curve be $(2t, t^2)$ . The slope of the tangent at this point is $\frac{dy}{dx} = \frac{2t}{2} = t$ .

Step 3: The slope of the normal at this point is $-\frac{1}{t}$ .

Step 4: The equation of the normal is $y - t^2 = -\frac{1}{t}(x - 2t)$ .

Step 5: The normal passes through the point $(1, 2)$ . Substitute this point into the equation of the normal: $2 - t^2 = -\frac{1}{t}(1 - 2t)$ .

Step 6: Simplify the equation: $2t - t^3 = -1 + 2t \Rightarrow t^3 = 1 \Rightarrow t = 1$ .

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

Step 8: Final Answer: The equation of the normal is $x + y = 3$ .

Why other options are wrong:
- Option A: $x+y=-3$ : This is incorrect as it does not satisfy the condition of being normal to the curve and passing through the given point.
- Option C: $2x=y$ : This is incorrect as it does not satisfy the condition of being normal to the curve and passing through the given point.
- Option D: $x-y=7$ : This is incorrect as it does not satisfy the condition of being normal to the curve and passing through the given point.

Key Concepts:
- Differential Calculus
- Equation of a Normal to a Curve

Memory Aids:
- Remember that 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, then find the negative reciprocal for the slope of the normal.

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Analyzing +4 | -1

The angle of intersection of the curves

y = {x^2}\,and\,x = {y^2}

is

View answer and explanation

Correct option: D

{\tan ^{ - 1}}\left( {\frac{3}{4}} \right)

Option D (

{\tan ^{ - 1}}(\frac{3}{4})

) is correct because it is the angle calculated using the slopes of the two curves at their point of intersection.

Solution approach

Correct Answer + Core Reason: ${\tan ^{ - 1}}(\frac{3}{4})$ . The angle of intersection is found using the slopes of the tangents at the point of intersection.

Step-by-Step Solution:

Step 1: Find the point(s) of intersection of the curves $y = x^2$ and $x = y^2$ :\nSubstitute $y = x^2$ into $x = y^2$ :
$x = (x^2)^2$
$x = x^4$
$x^4 - x = 0$
$x(x^3 - 1) = 0$
$x = 0$ or $x = 1$ \nIf $x = 0$ , then $y = 0^2 = 0$ . If $x = 1$ , then $y = 1^2 = 1$ .\nSo the points of intersection are $(0, 0)$ and $(1, 1)$ .

Step 2: Find the derivatives of the curves:\nFor $y = x^2$ , $\frac{dy}{dx} = 2x$ .\nFor $x = y^2$ , $1 = 2y \frac{dy}{dx}$ , so $\frac{dy}{dx} = \frac{1}{2y}$ .

Step 3: Evaluate the derivatives at the point of intersection $(1, 1)$ :\nFor $y = x^2$ , $m_1 = \frac{dy}{dx}|_{(1, 1)} = 2(1) = 2$ .\nFor $x = y^2$ , $m_2 = \frac{dy}{dx}|_{(1, 1)} = \frac{1}{2(1)} = \frac{1}{2}$ .

Step 4: Calculate the angle of intersection using the formula:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2 - \frac{1}{2}}{1 + 2(\frac{1}{2})} \right| = \left| \frac{\frac{3}{2}}{2} \right| = \frac{3}{4}$
$\theta = {\tan ^{ - 1}}(\frac{3}{4})$

Why other options are wrong:
- **Option A ( ${\tan ^{ - 1}}(\frac{4}{3})$ ): This is the inverse of the correct tangent value, indicating the slopes were subtracted in the wrong order or the formula was misapplied.
- Option B ( ${\tan ^{ - 1}}(1)$ ): This would imply the slopes are equal, which is not the case here.
- Option C ( ${90^0}$ ): This would imply the tangents are perpendicular, which is not the case here.

Key Concepts:
- Finding points of intersection of curves.
- Finding the derivative of a function.
- Calculating the angle between two curves using their slopes.

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

Quick Tips:
- Always find the points of intersection before calculating the angle between the curves.
- Remember to use the correct formula for the angle between two lines.

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Application +4 | -1

The curve

y - {e^{xy}} + x = 0

has a vertical tangent at the point

View answer and explanation

Correct option: A

(1, 0)

Option A ((1, 0)) is correct because it satisfies the condition for a vertical tangent, where the derivative approaches infinity.

Solution approach

Correct Answer + Core Reason: (1, 0). This point satisfies the condition for a vertical tangent.

Step-by-Step Solution:

Step 1: Given equation is $y - {e^{xy}} + x = 0$ .

Step 2: Rewrite the equation: ${e^{xy}} = x + y$ .

Step 3: Take the natural logarithm of both sides: $\log \left( {x + y} \right) = xy$ .

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

Step 5: Solve for $\frac{{dy}}{{dx}}$ :
$\frac{{dy}}{{dx}}\left( {\frac{1}{{y + x}} - x} \right) = y - \frac{1}{{y + x}} \Rightarrow \frac{{dy}}{{dx}} = \frac{{y\left( {y + x} \right) - 1}}{{1 - x\left( {y + x} \right)}}$ .

Step 6: For a vertical tangent, $\frac{{dy}}{{dx}} = \infty$ , which means the denominator must be zero:
$1 - x\left( {x + y} \right) = 0$ .

Step 7: Check the given points to see which one satisfies the condition $1 - x\left( {x + y} \right) = 0$ .\nFor (1, 0): $1 - 1\left( {1 + 0} \right) = 1 - 1 = 0$ . Therefore, (1, 0) is the point where the curve has a vertical tangent.

Why other options are wrong:
- Option B: at no point - The point (1,0) satisfies the condition.
- Option C: (0, 1) - For (0, 1): $1 - 0\left( {0 + 1} \right) = 1 \ne 0$ .
- Option D: (0, 0) - For (0, 0): $1 - 0\left( {0 + 0} \right) = 1 \ne 0$ .

Key Concepts:
- Implicit differentiation
- Vertical tangents

Memory Aids:
- A vertical tangent occurs when the derivative is undefined (denominator is zero).

Quick Tips:
- Differentiate implicitly, solve for dy/dx, and set the denominator equal to zero.

43 Medium

43 of 97

Application +4 | -1

The length of the normal to the curve

x = a\left( {\theta + \sin \theta } \right),y = a\left( {1 - \cos \theta } \right)at\;\theta = \frac{\pi }{2}

is

View answer and explanation

Correct option: D

\sqrt {2} a

Option D (

\sqrt {2} a

) is correct because it is the calculated length of the normal to the curve at the given parameter value.

Solution approach

Correct Answer + Core Reason: $\sqrt {2} a$ . The length of the normal is calculated using the derivative and the y-coordinate at $\theta = \frac{\pi}{2}$ .

Step-by-Step Solution:

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

Step 2: Evaluate $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$ .
$\frac{dy}{dx}|_{\theta = \frac{\pi}{2}} = \frac{\sin(\frac{\pi}{2})}{1 + \cos(\frac{\pi}{2})} = \frac{1}{1 + 0} = 1$

Step 3: Find the y-coordinate at $\theta = \frac{\pi}{2}$ .
$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$

Step 4: Calculate the length of the normal.
Length of normal $= y\sqrt{1 + (\frac{dy}{dx})^2} = a\sqrt{1 + 1^2} = a\sqrt{2} = \sqrt{2}a$

Why other options are wrong:
- **Option A ( $2a$ ): This is incorrect because it doesn't account for the correct formula for the length of the normal.
- Option B ( $\frac{a}{2}$ ): This is incorrect and likely a result of miscalculation or using the wrong formula.
- Option C ( $\frac{a}{{\sqrt 2 }}$ ): This is incorrect and likely a result of miscalculation or using the wrong formula.

Key Concepts:
- Parametric differentiation.
- Length of the normal to a curve.

Memory Aids:**
- Length of normal $= y\sqrt{1 + (\frac{dy}{dx})^2}$

Quick Tips:
- Remember to find both $\frac{dy}{dx}$ and the y-coordinate at the given point.

44 Medium

44 of 97

Application +4 | -1

The value of

\mathrm{n}

for which the length of subnormal of the curve

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

is constant, is

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because it makes the length of the subnormal constant, independent of x and y.

Solution approach

Correct Answer + Core Reason: -2. The value of n that makes the subnormal length constant is found by setting the exponent of y to zero.

Step-by-Step Solution:

Step 1: Take the natural logarithm of both sides of the equation $xy^n = a^{n+1}$ .
$\ln(xy^n) = \ln(a^{n+1})$
$\ln x + n \ln y = (n+1) \ln a$

Step 2: Differentiate both sides with respect to x.
$\frac{1}{x} + n \frac{1}{y} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{y}{nx}$

Step 3: Find the length of the subnormal.\nLength of subnormal = $y \left| \frac{dy}{dx} \right| = y \left| -\frac{y}{nx} \right| = \frac{y^2}{n|x|}$

Step 4: Express x in terms of y using the original equation $xy^n = a^{n+1}$ , so $x = \frac{a^{n+1}}{y^n}$ .\nSubstitute this into the subnormal length expression:\nSubnormal length = $\frac{y^2}{n \left| \frac{a^{n+1}}{y^n} \right|} = \frac{y^{n+2}}{n |a^{n+1}|}$

Step 5: For the subnormal length to be constant, the exponent of y must be zero.\nTherefore, $n+2 = 0$ , which implies $n = -2$ .

Why other options are wrong:
- Option A (-4): If n = -4, the subnormal length would be $\frac{y^{-2}}{-4a^{-3}}$ , which is not constant.
- Option B (6): If n = 6, the subnormal length would be $\frac{y^{8}}{6a^{7}}$ , which is not constant.
- Option C (4): If n = 4, the subnormal length would be $\frac{y^{6}}{4a^{5}}$ , which is not constant.

Key Concepts:
- Implicit differentiation
- Subnormal length
- Logarithmic differentiation

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

Quick Tips:
- Use logarithmic differentiation to simplify the process.
- Substitute back into the expression to check for constancy.

45 Medium

45 of 97

Application +4 | -1

The angle between the curves

y = {x^2}\,and\,{y^2} - x = 0

at the point (1,1), is

View answer and explanation

Correct option: D

{\tan ^{ - 1}}\frac{3}{4}

Option D (

{\tan ^{ - 1}}\frac{3}{4}

) is correct because it represents the angle between the two curves at their point of intersection (1,1), calculated using the slopes of their tangents.

Solution approach

Correct Answer + Core Reason: ${\tan ^{ - 1}}\frac{3}{4}$ . The angle between the curves is found using the slopes of the tangents at the point of intersection.

Step-by-Step Solution:

Step 1: Find the derivative of the first curve, $y = x^2$ : $\frac{{dy}}{{dx}} = 2x$ . At (1,1), $m_1 = 2(1) = 2$ .

Step 2: Find the derivative of the second curve, $y^2 = x$ : $2y\frac{{dy}}{{dx}} = 1 \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2y}}$ . At (1,1), $m_2 = \frac{1}{{2(1)}} = \frac{1}{2}$ .

Step 3: Use the formula for the angle between two lines: $\tan \theta = \left| {\frac{{m_1 - m_2}}{{1 + m_1 m_2}}} \right| = \left| {\frac{{2 - \frac{1}{2}}}{{1 + 2 \times \frac{1}{2}}}} \right| = \frac{3}{4}$ .

Step 4: Find the angle: $\theta = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right)$ .

Why other options are wrong:
- Option A: This answer is incorrect. It may be a result of incorrect calculation of the slopes or using the wrong formula for the angle between two lines.
- 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 using the wrong formula.

Key Concepts:
- Derivatives as slopes of tangents.
- Angle between two lines.

Memory Aids:
- The formula for the angle between two lines is $\tan \theta = \left| {\frac{{m_1 - m_2}}{{1 + m_1 m_2}}} \right|$ .

Quick Tips:
- Always find the derivatives of both curves before finding the angle.

46 Medium

46 of 97

Application +4 | -1

For the parabola

{y^2} = 4ax

, the ratio of the subtangent to the abscissa is

View answer and explanation

Correct option: B

2:1

Option B (

x + 2\\;y = \frac{\pi }{2}

) is correct because it is the equation of the tangent to the curve that is parallel to the given line.

Solution approach

Correct Answer + Core Reason: $x + 2\\;y = \frac{\pi }{2}$ . The equation of the tangent is found by finding the points where the derivative equals the slope of the parallel line.

Step-by-Step Solution:

Step 1: Given $y = \cos(x+y)$ .

Step 2: Differentiate with respect to $x$ : $\frac{{dy}}{{dx}} = -\sin(x+y)(1 + \frac{{dy}}{{dx}})$ .

Step 3: The line $x + 2y = 0$ has a slope of $-\frac{1}{2}$ . Since the tangent is parallel to this line, $\frac{{dy}}{{dx}} = -\frac{1}{2}$ .

Step 4: Substitute $\frac{{dy}}{{dx}} = -\frac{1}{2}$ into the derivative equation: $-\frac{1}{2} = -\sin(x+y)(1 - \frac{1}{2}) \Rightarrow -\frac{1}{2} = -\sin(x+y)(\frac{1}{2}) \Rightarrow \sin(x+y) = 1$ .

Step 5: Since $\sin(x+y) = 1$ , $x+y = \frac{\pi}{2} + 2n\pi$ , where $n$ is an integer.

Step 6: Substitute $\sin(x+y) = 1$ into $y = \cos(x+y)$ . Since $\sin^2(x+y) + \cos^2(x+y) = 1$ , we have $1 + y^2 = 1 \Rightarrow y = 0$ .

Step 7: If $y = 0$ , then $\sin(x) = 1$ , so $x = \frac{\pi}{2} + 2n\pi$ .

Step 8: The point is $(\frac{\pi}{2}, 0)$ .

Step 9: The equation of the tangent is $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \Rightarrow y = -\frac{1}{2}x + \frac{\pi}{4} \Rightarrow x + 2y = \frac{\pi}{2}$ .

Why other options are wrong:
- **Option A ( $x + 2\\;y = 1$ ): Incorrect equation of the tangent.
- Option C ( $x + 2\\;y = \frac{\pi }{4}$ ): Incorrect equation of the tangent.
- Option D (None of these):** The correct equation is $x + 2y = \frac{\pi}{2}$ .

Key Concepts:
- Derivatives
- Equation of a tangent line
- Trigonometric functions

Memory Aids:
- Remember the relationship between parallel lines and their slopes.

Quick Tips:
- Differentiate implicitly.
- Use trigonometric identities to simplify.

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47 of 97

Understanding +4 | -1

The product of the lengths of subtangent and subnormal at any point of a curve is

View answer and explanation

Correct option: B

Square of the ordinate

Option B (Square of the ordinate) is correct because the product of the lengths of the subtangent and subnormal is equal to y², where y is the ordinate.

Solution approach

Correct Answer + Core Reason: Square of the ordinate. The product of subtangent and subnormal lengths is y².

Step-by-Step Solution:

Step 1: Recall the formula for the length of the subtangent: Subtangent = $y \frac{{dx}}{{dy}}$ .

Step 2: Recall the formula for the length of the subnormal: Subnormal = $y \frac{{dy}}{{dx}}$ .

Step 3: Calculate the product of the lengths of the subtangent and subnormal: Product = $(y \frac{{dx}}{{dy}}) (y \frac{{dy}}{{dx}}) = y^2 \frac{{dx}}{{dy}} \frac{{dy}}{{dx}} = y^2$ .

Step 4: Recognize that $y$ is the ordinate.

Step 5: Therefore, the product is the square of the ordinate.

Why other options are wrong:
- Option A (Square of the abscissa): The product is related to the ordinate, not the abscissa.
- Option C (Constant): The product is not necessarily constant; it depends on the ordinate $y$ .
- Option D (None of these): This is incorrect because the square of the ordinate is the correct answer.

Key Concepts:
- Subtangent
- Subnormal
- Derivatives

Memory Aids:
- Remember the formulas for subtangent and subnormal.

Quick Tips:
- The subtangent involves $\frac{{dx}}{{dy}}$ , and the subnormal involves $\frac{{dy}}{{dx}}$ .

48 Medium

48 of 97

Application +4 | -1

If the tangent to the curve

y = x + \sin y

at a point

\left( {a,b} \right)

is parallel to the line joining

\left( {0,\frac{3}{2}} \right)

and

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

then

View answer and explanation

Correct option: A

\left| {b - a} \right| = 1

The correct answer is A (

\left| {b - a} \right| = 1

) because it is found by using the parallel line condition and the curve equation to find the relationship between a and b.

Solution approach

Correct Answer + Core Reason: $\left| {b - a} \right| = 1$ . The relationship between a and b is found using the parallel line condition and the curve equation.

Step-by-Step Solution:

Step 1: Given curve: $y = x + \sin y$

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

Step 3: The tangent at point $(a, b)$ is parallel to the line joining $\left( {0,\frac{3}{2}} \right)$ and $\left( {\frac{1}{2},2} \right)$ . So, the slope is $\frac{{2 - \frac{3}{2}}}{{\frac{1}{2} - 0}} = 1$

Step 4: Therefore, $\left. {\frac{{dy}}{{dx}}} \right|_{(a,b)} = 1 \Rightarrow \frac{1}{{1 - \cos b}} = 1 \Rightarrow 1 = 1 - \cos b \Rightarrow \cos b = 0 \Rightarrow \sin b = \pm 1$

Step 5: Since the point $(a, b)$ lies on the curve, $b = a + \sin b \Rightarrow b - a = \sin b$

Step 6: Therefore, $\left| {b - a} \right| = \left| {\sin b} \right| = 1$

Why other options are wrong:
- Option B: $\left| {a + b} \right| = 1$ is incorrect. It likely arises from an error in the algebraic manipulation or incorrect substitution.
- Option C: $b = a$ is incorrect. It likely arises from an error in the algebraic manipulation or incorrect substitution.
- Option D: $b = \frac{\pi }{2} + a$ is incorrect. It likely arises from an error in the algebraic manipulation or incorrect substitution.

Key Concepts:
- Tangent to a curve
- Slope of a line
- Parallel lines
- Differentiation

Memory Aids:
- Remember the relationship between the slope of a tangent and the derivative.

Quick Tips:
- Always double-check your differentiation and algebraic manipulations to avoid errors.

49 Medium

49 of 97

Application +4 | -1

The slope of the tangent to the curve

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

at the point where ordinate and abscissa are equal, is

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because it represents the slope of the tangent line at the point where the x and y coordinates are equal on the given curve.

Solution approach

Correct Answer + Core Reason: $- 1$ . The slope of the tangent is found by differentiating the curve equation and evaluating at the point where x=y.

Step-by-Step Solution:

Step 1: Given the curve $y = \sqrt {9 - {x^2}}$ .

Step 2: Find the derivative $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = \frac{{-2x}}{{2\sqrt {9 - {x^2}} }} = \frac{{-x}}{{\sqrt {9 - {x^2}} }}$ .

Step 3: Find the point where the ordinate and abscissa are equal, i.e., $y = x$ . So, $x = \sqrt {9 - {x^2}}$ .

Step 4: Square both sides: ${x^2} = 9 - {x^2} \Rightarrow 2{x^2} = 9 \Rightarrow {x^2} = \frac{9}{2} \Rightarrow x = \pm \frac{3}{{\sqrt 2 }}$ . Since $y = \sqrt {9 - {x^2}}$ is positive, $x$ must be positive, so $x = \frac{3}{{\sqrt 2 }}$ . Thus, the point is $(\frac{3}{{\sqrt 2 }}, \frac{3}{{\sqrt 2 }})$ .

Step 5: Evaluate the derivative at this point: $\frac{{dy}}{{dx}}|_{x = \frac{3}{{\sqrt 2 }}} = \frac{{-\frac{3}{{\sqrt 2 }}}}{{\sqrt {9 - {(\frac{3}{{\sqrt 2 }})}^2} }} = \frac{{-\frac{3}{{\sqrt 2 }}}}{{\sqrt {9 - \frac{9}{2}} }} = \frac{{-\frac{3}{{\sqrt 2 }}}}{{\sqrt {\frac{9}{2}} }} = \frac{{-\frac{3}{{\sqrt 2 }}}}{{\frac{3}{{\sqrt 2 }}}} = -1$ .

Step 6: The slope of the tangent is -1.

Why other options are wrong:
- Option A: 1 - This is the negative reciprocal of the correct slope, which would be the slope of the normal line.
- Option C: 0 - This would be the slope of a horizontal line, which is not the case here.
- Option D: None of these - The correct answer is -1, which is one of the options.

Key Concepts:
- Finding the derivative of a function
- Finding the slope of a tangent line
- Solving for the intersection of curves

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

Quick Tips:
- Always check the domain of the function and the sign of the derivative.

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50 of 97

Application +4 | -1

The angle of intersection of the curves

y = {x^2},\;y = 7 - {x^3}

at (1, 1) is

View answer and explanation

Correct option: C

\pi /2

Option C (

\pi /2

) is correct because the product of the slopes of the tangents to the two curves at the point of intersection is -1, indicating perpendicularity.

Solution approach

Correct Answer + Core Reason: $\pi /2$ . The curves intersect at right angles because the product of their slopes at the intersection point is -1.

Step-by-Step Solution:

Step 1: Verify that the curves intersect at (1, 1). For $y = x^2$ , when $x = 1$ , $y = 1^2 = 1$ . For $y = 7 - x^3$ , when $x = 1$ , $y = 7 - 1^3 = 6$ . There seems to be a typo in the question. It should be $y = 7 - x^3$ should be $y = 7/6 - x^3/6$ or $6y = 7 - x^3$ . Assuming $6y = 7 - x^3$ , when $x = 1$ , $6y = 7 - 1 \Rightarrow y = 1$ . So the point of intersection is (1, 1).

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

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

Step 4: Find the derivative of $6y = 7 - x^3$ . $6\frac{{dy}}{{dx}} = -3x^2 \Rightarrow \frac{{dy}}{{dx}} = -\frac{1}{2}x^2$ .

Step 5: Evaluate the derivative at $x = 1$ : $m_2 = \left( \frac{{dy}}{{dx}} \right)_{x=1} = -\frac{1}{2}(1)^2 = -\frac{1}{2}$ .

Step 6: Check if the curves are orthogonal (intersect at right angles). Multiply the slopes: $m_1 \cdot m_2 = 2 \cdot \left( -\frac{1}{2} \right) = -1$ . Since the product of the slopes is -1, the curves are orthogonal.

Step 7: The angle of intersection is $\pi /2$ .

Step 8: Final Answer: $\pi /2$ .

Why other options are wrong:
- **Option A ( $\pi /4$ ): This is incorrect because the product of the slopes is -1, indicating a right angle, not 45 degrees.
- Option B ( $\pi /3$ ): This is incorrect because the product of the slopes is -1, indicating a right angle, not 60 degrees.
- Option D (None of these):** This is incorrect because the curves do intersect at a specific angle, which is $\pi /2$ .

Key Concepts:
- Derivatives of functions
- Orthogonal curves
- Angle of intersection

Memory Aids:
- Two curves are orthogonal if the product of their slopes at the point of intersection is -1.

Quick Tips:
- Remember to find the derivatives of both curves and evaluate them at the point of intersection.
- Double-check your algebra when calculating the slopes.

51 Medium

51 of 97

Application +4 | -1

If the line

x \cos \alpha+y \sin \alpha=P

touches the curve

4 x^{3}=27 a y^{2}

, then

\frac{P}{a}=

:

View answer and explanation

Correct option: A

\cot ^{2} \alpha \cos \alpha

Option A (

\cot ^{2} \alpha \cos \alpha

) is correct because it is derived by finding the point of tangency and substituting it into the line equation.

Solution approach

Correct Answer + Core Reason: $\cot ^{2} \alpha \cos \alpha$ . The correct answer is found by parameterizing the curve, finding the derivative, and using the tangency condition.

Step-by-Step Solution:

Step 1: Parameterize the curve $4x^3 = 27ay^2$ . Let $x = \frac{3at}{2^{2/3}}$ , then $4(\frac{3at}{2^{2/3}})^3 = 27ay^2$ , which simplifies to $y^2 = a^2t^3$ , so $y = at^{3/2}$ .

Step 2: Find the derivative $\frac{dy}{dx}$ . We have $\frac{dx}{dt} = \frac{3a}{2^{2/3}}$ and $\frac{dy}{dt} = \frac{3}{2}at^{1/2}$ . Thus, $\frac{dy}{dx} = \frac{\frac{3}{2}at^{1/2}}{\frac{3a}{2^{2/3}}} = \frac{1}{2}2^{2/3}t^{1/2} = 2^{-1/3}t^{1/2}$ .

Step 3: The line $x \cos \alpha + y \sin \alpha = P$ touches the curve, so the slope of the tangent is $-\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha$ . Therefore, $2^{-1/3}t^{1/2} = -\cot \alpha$ . This implies that $t^{1/2} = -2^{1/3} \cot \alpha$ , so $t = 2^{2/3} \cot^2 \alpha$ .

Step 4: Substitute $x$ and $y$ into the line equation: $(\frac{3a}{2^{2/3}})(2^{2/3} \cot^2 \alpha) \cos \alpha + (a(2^{2/3} \cot^2 \alpha)^{3/2}) \sin \alpha = P$ .

Step 5: Simplify the equation: $3a \cot^2 \alpha \cos \alpha + a(2 \cot^3 \alpha) \sin \alpha = P$ . Since $t^{1/2} = -2^{1/3} \cot \alpha$ , we have $t^{3/2} = -2 \cot^3 \alpha$ .

Step 6: Rewrite the equation: $3a \cot^2 \alpha \cos \alpha - 2a \cot^3 \alpha \sin \alpha = P$ . This simplifies to $3a \cot^2 \alpha \cos \alpha - 2a \frac{\cos^3 \alpha}{\sin^2 \alpha} \sin \alpha = P$ , which further simplifies to $3a \cot^2 \alpha \cos \alpha - 2a \cot^2 \alpha \cos \alpha = P$ .

Step 7: Solve for $\frac{P}{a}$ : $\frac{P}{a} = \cot^2 \alpha \cos \alpha$ .

Why other options are wrong:
- Option B: $\cot ^{2} \alpha \sin \alpha$ is incorrect because it uses $\sin \alpha$ instead of $\cos \alpha$ in the final expression.
- Option C: $\tan ^{2} \alpha \cos \alpha$ is incorrect because it uses $\tan ^{2} \alpha$ instead of $\cot ^{2} \alpha$ .
- Option D: $\tan ^{2} \alpha \sin \alpha$ is incorrect because it uses $\tan ^{2} \alpha$ instead of $\cot ^{2} \alpha$ and $\sin \alpha$ instead of $\cos \alpha$ .

Key Concepts:
- Tangent to a curve
- Differentiation
- Parametric equations

Memory Aids:
- Remember the derivatives of trigonometric functions.
- Recall the relationship between slope and tangent.

Quick Tips:
- Parameterize the curve to simplify calculations.
- Use the tangency condition to relate the derivative to the line's slope.

52 Medium

52 of 97

Application +4 | -1

The angle between the tangents to the curve

{y^2} = 2ax

at the point where

x = \frac{a}{2}

, is

View answer and explanation

Correct option: D

\frac{\pi }{2}

Option D (

\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 tangents are perpendicular because the product of their slopes is -1.

Step-by-Step Solution:

Step 1: Given ${y^2} = 2ax$ . Differentiate with respect to $x$ : $2y\frac{{dy}}{{dx}} = 2a$ , so $\frac{{dy}}{{dx}} = \frac{a}{y}$ .

Step 2: When $x = \frac{a}{2}$ , substitute into the curve equation: ${y^2} = 2a\left( {\frac{a}{2}} \right) = {a^2}$ , so $y = \pm a$ .

Step 3: The two points on the curve are $P\left( {a/2,\;a} \right)$ and $Q\left( {a/2,\; - a} \right)$ .

Step 4: Find the slopes of the tangents at these points: ${\left( {\frac{{dy}}{{dx}}} \right)_P} = \frac{a}{a} = 1$ and ${\left( {\frac{{dy}}{{dx}}} \right)_Q} = \frac{a}{{ - a}} = - 1$ .

Step 5: The product of the slopes is ${\left( {\frac{{dy}}{{dx}}} \right)_P} \times {\left( {\frac{{dy}}{{dx}}} \right)_Q} = 1 \times - 1 = - 1$ .

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

Why other options are wrong:
- Option A: $\frac{\pi }{6}$ is incorrect. The tangents are perpendicular, not at an angle of $\frac{\pi }{6}$ .
- Option B: $\frac{\pi }{4}$ is incorrect. The tangents are perpendicular, not at an angle of $\frac{\pi }{4}$ .
- Option C: $\frac{\pi }{3}$ is incorrect. The tangents are perpendicular, not at an angle of $\frac{\pi }{3}$ .

Key Concepts:
- Differentiation
- Slope of a tangent
- Perpendicular lines

Memory Aids:
- Remember that if the product of the slopes of two lines is -1, the lines are perpendicular.

Quick Tips:
- Find the points on the curve first before finding the slopes of the tangents.

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53 of 97

Analyzing +4 | -1

Tangents are drawn from the origin to the curve

y = \cos \;x

. Their points of contact lie on

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because it represents the locus of the points of contact of tangents drawn from the origin to the curve

y = \cos x

.

Solution approach

Correct Answer + Core Reason: ${x^2}{y^2} = {x^2} - {y^2}$ . This equation represents the locus of the points of contact of tangents from the origin to the curve $y = \cos x$ .

Step-by-Step Solution:

Step 1: Let $(x_1, y_1)$ be a point of contact on the curve $y = \cos x$ .

Step 2: Find the derivative of $y = \cos x$ .
$\frac{dy}{dx} = -\sin x$

Step 3: The equation of the tangent at $(x_1, y_1)$ is:
$y - y_1 = -\sin x_1 (x - x_1)$

Step 4: Since the tangent passes through the origin (0, 0), substitute x = 0 and y = 0 into the tangent equation:
$0 - y_1 = -\sin x_1 (0 - x_1) \Rightarrow y_1 = x_1 \sin x_1$

Step 5: Since $(x_1, y_1)$ lies on the curve $y = \cos x$ , we have $y_1 = \cos x_1$ .

Step 6: Combine the equations $y_1 = x_1 \sin x_1$ and $y_1 = \cos x_1$ .
$\sin x_1 = \frac{y_1}{x_1}$

Step 7: Use the identity $\sin^2 x_1 + \cos^2 x_1 = 1$ .
$(\frac{y_1}{x_1})^2 + y_1^2 = 1 \Rightarrow \frac{y_1^2}{x_1^2} + y_1^2 = 1 \Rightarrow y_1^2 + x_1^2 y_1^2 = x_1^2 \Rightarrow x_1^2 = y_1^2 + x_1^2 y_1^2$

Step 8: Replace $(x_1, y_1)$ with $(x, y)$ to find the locus.
$x^2 = y^2 + x^2 y^2 \Rightarrow x^2 - y^2 = x^2 y^2 \Rightarrow x^2 y^2 = x^2 - y^2$

Why other options are wrong:
- Option A: Incorrect because the derivation does not lead to this equation.
- Option B: Incorrect because the derivation does not lead to this equation.
- Option D: Incorrect because the correct locus is found to be ${x^2}{y^2} = {x^2} - {y^2}$ .

Key Concepts:
- Tangents to curves
- Derivatives
- Locus of a point
- Trigonometric identities

Memory Aids:
- Remember the equation of a tangent: $y - y_1 = m(x - x_1)$
- $\sin^2 x + \cos^2 x = 1$

Quick Tips:
- Always find the derivative to get the slope of the tangent.
- Use the given conditions (e.g., passing through the origin) to find relationships between the coordinates.

54 Medium

54 of 97

Application +4 | -1

The tangent line at

(-1,15)

to the curve

y=2 x^{3}+13 x^{2}+5 x+9

passes through

View answer and explanation

Correct option: A

Origin

Option A (Origin) is correct because the equation of the tangent line at (-1, 15) is y + 15x = 0, which passes through the origin (0,0).

Solution approach

Correct Answer + Core Reason: Origin. The tangent line equation passes through the origin.

Step-by-Step Solution:

Step 1: Find the derivative of the curve: $y = 2x^3 + 13x^2 + 5x + 9$ . So, $\frac{{dy}}{{dx}} = 6x^2 + 26x + 5$ .

Step 2: Evaluate the derivative at x = -1: $\frac{{dy}}{{dx}}|_{-1} = 6(-1)^2 + 26(-1) + 5 = 6 - 26 + 5 = -15$ . This is the slope of the tangent line at (-1, 15).

Step 3: Find the equation of the tangent line using the point-slope form: $y - y_1 = m(x - x_1)$ . Substituting the point (-1, 15) and the slope -15, we get $y - 15 = -15(x + 1)$ .

Step 4: Simplify the equation: $y - 15 = -15x - 15$ , which simplifies to $y = -15x$ or $y + 15x = 0$ .

Step 5: Check which point satisfies the tangent line equation. For the origin (0, 0): $0 + 15(0) = 0$ , which is true.

Step 6: Final Answer: The tangent line passes through the origin.

Why other options are wrong:
- Option B ((0,9)): Substituting (0, 9) into the tangent line equation: $9 + 15(0) = 9 \neq 0$ . So, it doesn't lie on the tangent line.
- Option C ((2,3)): Substituting (2, 3) into the tangent line equation: $3 + 15(2) = 33 \neq 0$ . So, it doesn't lie on the tangent line.
- Option D ((1,4)): Substituting (1, 4) into the tangent line equation: $4 + 15(1) = 19 \neq 0$ . So, it doesn't lie on the tangent line.

Key Concepts:
- Derivatives
- Tangent Lines
- Point-Slope Form of a Line

Memory Aids:
- Remember the point-slope form: y - y1 = m(x - x1)
- The derivative at a point gives the slope of the tangent line at that point.

Quick Tips:
- Always find the derivative first to determine the slope of the tangent line.
- Substitute the given point and slope into the point-slope form to find the tangent line equation.
- Check if the given points satisfy the equation of the tangent line.

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55 of 97

Application +4 | -1

The angle between curves

y^2=4 x

and

x^2+y^2=5

at

(1,2)

is

View answer and explanation

Correct option: A

\tan ^{-1}(3)

Option A (

\tan^{-1}(3)

) is correct because by finding the derivatives of both curves at the point (1,2) and using the formula for the angle between two lines, we can find the angle between the curves.

Solution approach

Correct Answer + Core Reason: $\tan ^{-1}(3)$ . By finding the derivatives of both curves at the point (1,2) and using the formula for the angle between two lines, we can find the angle between the curves.

Step-by-Step Solution:

Step 1: Given: $y^2 = 4x$ and $x^2 + y^2 = 5$ .

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

Step 3: Evaluate $\frac{dy}{dx}$ at $(1, 2)$ : $\frac{dy}{dx}|_{(1,2)} = \frac{2}{2} = 1$ . Let $m_1 = 1$ .

Step 4: Find the derivative of $x^2 + y^2 = 5$ : $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$ .

Step 5: Evaluate $\frac{dy}{dx}$ at $(1, 2)$ : $\frac{dy}{dx}|_{(1,2)} = -\frac{1}{2}$ . Let $m_2 = -\frac{1}{2}$ .

Step 6: Find the angle between the curves: $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{1 - (-\frac{1}{2})}{1 + (1)(-\frac{1}{2})}| = |\frac{\frac{3}{2}}{\frac{1}{2}}| = 3$ .

Step 7: Therefore, $\theta = \tan^{-1}(3)$ .

Why other options are wrong:
- Option B: $\tan ^{-1}(2)$ is incorrect because the correct calculation leads to $\tan^{-1}(3)$ , not $\tan^{-1}(2)$ .
- Option C: $\frac{\pi}{2}$ is incorrect because the slopes are not negative reciprocals of each other.
- Option D:

56 Medium

56 of 97

Application +4 | -1

If the tangent to the curve

y = f\left( x \right) = x{\log _e}x,\left( {x > 0} \right)

at a point

\left( {c,f\left( c \right)} \right)

is parallel to the line segment joining to the points(1,0) and (e, e) then c=

View answer and explanation

Correct option: B

{e^{\left( {\frac{1}{{e - 1}}} \right)}}

Option B (e^(1/(e-1))) is correct because the derivative of the function at point c must equal the slope of the line segment joining the two points, which can be solved for c.

Solution approach

Correct Answer + Core Reason: ${e^{\left( {\frac{{1}}{{{e - 1}}}}\right)}}$ . The value of c is found by equating the derivative of the function at c to the slope of the line segment.

Step-by-Step Solution:

Step 1: Find the derivative of the function $y = f(x) = x \log_e x$ .
$\frac{dy}{dx} = \frac{d}{dx}(x \log_e x) = x \cdot \frac{1}{x} + \log_e x \cdot 1 = 1 + \log_e x$ .

Step 2: Evaluate the derivative at $x = c$ .
$\left(\frac{dy}{dx}\right)_{x=c} = 1 + \log_e c$ .

Step 3: Find the slope of the line segment joining the points (1,0) and (e,e).\nThe slope is $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$ .

Step 4: Equate the derivative at $x = c$ to the slope of the line segment.
$1 + \log_e c = \frac{e}{e - 1}$
$\log_e c = \frac{e}{e - 1} - 1 = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$

Step 5: Solve for c.\nc = e^{ $\frac{1}{e - 1}$ } $\n\n**Why other options are wrong:**\n- **Option A ($ $\frac{{e - 1}}{e}$ $):** This is incorrect because it likely involves an error in calculating the derivative or in solving for c.\n- **Option C ($ {e^{\left( {\frac{{1}}{{{1 - e}}}}\right)}} $):** This is incorrect because it likely involves an error in calculating the derivative or in solving for c.\n- **Option D ($ $\frac{e}{{e - 1}}$ $):** This is incorrect because it likely involves an error in calculating the derivative or in solving for c.\n\n**Key Concepts:**\n- Differentiation\n- Slope of a line\n- Logarithmic functions\n\n**Memory Aids:**\n- Remember the product rule for differentiation:$ $\frac{d}{dx}$ (uv) = u $\frac{dv}{dx}$ + v $\frac{du}{dx}$ $.\n\n**Quick Tips:**\n- Always double-check the derivative calculation.\n- Ensure the slope is calculated correctly.$

57 Medium

57 of 97

Analyzing +4 | -1

If the tangent at each point of the curve

y=\frac{2}{3} x^3-2 a x^2+2 x+5

makes an acute angle with the positive direction of

x

-axis, then

View answer and explanation

Correct option: B

-1 \leq a \leq 1

Option B (

-1 \leq a \leq 1

) is correct because the derivative of the curve must be non-negative for all x, leading to the given range for 'a'.

Solution approach

Correct Answer + Core Reason: $-1 \leq a \leq 1$ . The condition for an acute angle implies the derivative is non-negative, leading to this inequality.

Step-by-Step Solution:

Step 1: Find the derivative of the given curve: $y=\frac{{2}}{{3}} x^3-2 a x^2+2 x+5$ .

Step 2: Calculate $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = 2x^2 - 4ax + 2$ .

Step 3: Since the tangent makes an acute angle with the positive x-axis, $\frac{{dy}}{{dx}} \geq 0$ for all x.

Step 4: Thus, $2x^2 - 4ax + 2 \geq 0$ for all x, which simplifies to $x^2 - 2ax + 1 \geq 0$ for all x.

Step 5: For the quadratic to be non-negative, the discriminant must be non-positive: $D = (2a)^2 - 4(1)(1) \leq 0$ .

Step 6: Simplify: $4a^2 - 4 \leq 0 \Rightarrow a^2 - 1 \leq 0 \Rightarrow (a-1)(a+1) \leq 0$ .

Step 7: Solve the inequality: $-1 \leq a \leq 1$ .

Why other options are wrong:
- Option A: $a \geq 1$ is incorrect because it only considers one part of the solution where the discriminant is non-positive.
- Option C: $a \leq -1$ is incorrect because it only considers the other part of the solution where the discriminant is non-positive.
- Option D: None of these is incorrect because the correct range of 'a' is found to be $-1 \leq a \leq 1$ .

Key Concepts:
- Derivatives and tangents
- Quadratic inequalities
- Discriminant of a quadratic equation

Memory Aids:
- Acute angle implies non-negative derivative.
- For $ax^2 + bx + c \geq 0$ for all x, $D \leq 0$ .

Quick Tips:
- Remember to consider the discriminant when dealing with quadratic inequalities.
- The derivative represents the slope of the tangent.

58 Medium

58 of 97

Analyzing +4 | -1

The line

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

touches the curve

{\left( {\frac{x}{a}} \right)^n} + {\left( {\frac{y}{b}} \right)^n} = 2

at the point (a,b) for

View answer and explanation

Correct option: C

Any

n \in R

Option C (Any

n \in R

) is correct because the tangent to the curve at (a, b) is the same regardless of the value of n.

Solution approach

Correct Answer + Core Reason: Any $n \in R$ . The tangent to the curve at (a, b) is the same for all n.

Step-by-Step Solution:\nStep 1: Write the equation of the curve: ${\left( {\frac{x}{a}} \right)^n} + {\left( {\frac{y}{b}} \right)^n} = 2$ \nStep 2: Differentiate the equation with respect to x: $n{\left( {\frac{x}{a}} \right)^{n - 1}} \cdot \frac{1}{a} + n{\left( {\frac{y}{b}} \right)^{n - 1}} \cdot \frac{1}{b} \cdot \frac{{dy}}{{dx}} = 0$ \nStep 3: Solve for $\frac{{dy}}{{dx}}$ : $\frac{{dy}}{{dx}} = - \frac{{{{\left( {\frac{x}{a}} \right)}^{n - 1}} \cdot \frac{1}{a}}}{{{{\left( {\frac{y}{b}} \right)}^{n - 1}} \cdot \frac{1}{b}}} = - \frac{b}{a}{\left( {\frac{x}{a}} \right)^{n - 1}}{\left( {\frac{y}{b}} \right)^{1 - n}}$ \nStep 4: Evaluate $\frac{{dy}}{{dx}}$ at the point (a, b): ${\left. {\frac{{dy}}{{dx}}} \right|_{(a,b)}} = - \frac{b}{a}{\left( {\frac{a}{a}} \right)^{n - 1}}{\left( {\frac{b}{b}} \right)^{1 - n}} = - \frac{b}{a}$ \nStep 5: Write the equation of the tangent at (a, b) using the point-slope form: $y - b = - \frac{b}{a}(x - a)$ \nStep 6: Simplify the equation of the tangent: $ay - ab = - bx + ab \Rightarrow bx + ay = 2ab$ \nStep 7: Divide by ab: $\frac{x}{a} + \frac{y}{b} = 2$ \nStep 8: Since the equation of the tangent is independent of n, the line touches the curve at (a, b) for any $n \in R$ .

Why other options are wrong:
- **Option A ( $n = 2$ only): Incorrect because the tangent is the same for all values of n.
- Option B ( $n = - 3$ only): Incorrect because the tangent is the same for all values of n.
- Option D (None of these): Incorrect because the tangent is the same for all values of n.

Key Concepts:
- Differentiation
- Tangent to a Curve
- Implicit Differentiation

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

Quick Tips:**
- Remember to use implicit differentiation when the equation is not explicitly solved for y.

59 Medium

59 of 97

Analyzing +4 | -1

Angle between

{y^2} = x

and

{x^2} = y

at the origin is

View answer and explanation

Correct option: C

\frac{\pi }{2}

Option C (

\frac{\pi }{2}

) is correct because the tangents at the origin are the x and y axes, which are perpendicular.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{2}$ . The tangents at the origin are the x and y axes, which are perpendicular.

Step-by-Step Solution:

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

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

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

Step 4: Evaluate the slopes at the origin (0, 0):
- For $y^2 = x$ , as $y \to 0$ , $\frac{dy}{dx} \to \infty$ .
- For $x^2 = y$ , at $x = 0$ , $\frac{dy}{dx} = 0$ .

Step 5: The slopes are $\infty$ and 0, which means the tangents are the y-axis and x-axis, respectively.

Step 6: The angle between the x and y axes is $\frac{\pi}{2}$ .

Step 7: Final Answer: The angle between the curves at the origin is $\frac{\pi}{2}$ .

Why other options are wrong:
- Option A: $2{\tan ^{ - 1}}(\frac{3}{4})$ is incorrect because it does not represent the angle between the curves at the origin.
- Option B: ${\tan ^{ - 1}}(\frac{4}{3})$ is incorrect because it does not represent the angle between the curves at the origin.
- Option D: $\frac{\pi}{4}$ is incorrect because it does not represent the angle between the curves at the origin.

Key Concepts:
- Differentiation of implicit functions
- Finding the angle between two curves

Memory Aids:
- The angle between two curves is the angle between their tangents at the point of intersection.

Quick Tips:
- Remember that a slope of $\infty$ corresponds to a vertical tangent, and a slope of 0 corresponds to a horizontal tangent.

60 Medium

60 of 97

Application +4 | -1

If the straight line joining the points

(0,3)

and

(5,-2)

is tangent to the curve

y=\frac{c}{x+1}

; then the value of

c

is :

View answer and explanation

Correct option: C

4

Option C (4) is correct because the tangent condition and the equation of the line allow us to solve for 'c', which equals 4.

Solution approach

Correct Answer + Core Reason: 4. The straight line is tangent to the curve, allowing us to find the value of c using the derivative and the equation of the line.

Step-by-Step Solution:

Step 1: Find the slope of the line joining (0,3) and (5,-2): $m = \frac{-2 - 3}{5 - 0} = -1$ .

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

Step 3: Since the line is tangent to the curve, the slope of the tangent is equal to the slope of the line: $-\frac{c}{(x+1)^2} = -1 \Rightarrow (x+1)^2 = c$ .

Step 4: Let the point of tangency be $(x_0, y_0)$ . Then $y_0 = \frac{c}{x_0 + 1}$ .

Step 5: The point $(x_0, y_0)$ also lies on the line $y = -x + 3$ . So, $y_0 = -x_0 + 3$ .

Step 6: Equate the two expressions for $y_0$ : $\frac{c}{x_0 + 1} = -x_0 + 3$ .

Step 7: From Step 3, we have $x_0 + 1 = \sqrt{c}$ . Substitute this into the equation from Step 6: $\frac{c}{\sqrt{c}} = -x_0 + 3 \Rightarrow \sqrt{c} = -x_0 + 3 \Rightarrow x_0 = 3 - \sqrt{c}$ .

Step 8: Substitute $x_0 = 3 - \sqrt{c}$ into $x_0 + 1 = \sqrt{c}$ : $3 - \sqrt{c} + 1 = \sqrt{c} \Rightarrow 4 = 2\sqrt{c} \Rightarrow \sqrt{c} = 2 \Rightarrow c = 4$ .

Why other options are wrong:
- Option A (2): This is incorrect because it results from an incorrect calculation of c.
- Option B (3): This is incorrect because it results from an incorrect calculation of c.
- Option D (5): This is incorrect because it results from an incorrect calculation of c.

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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