Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 31 :

The function $f (x) = \tan x$ is discontinuous on the set:

${x = n π : n \in Z}$
${x = 2 n π : n \in Z}$
${x = (2 n + 1) \frac{π}{2} : n \in Z}$
${x = \frac{n π}{2} : n \in Z}$

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Ans. ${x = (2 n + 1) \frac{π}{2} : n \in Z}$

Explanation:

When $\tan ((2 n + 1) \frac{π}{2}) = \tan (n π + \frac{π}{2}) = - \cot n π .$

It is not defined at the integral points $(n \in Z)$ .

Hence, $f (x) = \tan x$ is discontinuous at $(2 n + 1) \frac{π}{2}$

Q 32 :

If $f (x) = x \sin \frac{1}{x}$ , where $x \neq 0$ , then the value of the function $f$ at $x = 0$ , so that the function is continuous at $x = 0$ , is:

0
- 1
1
None of these

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Ans. 0

Explanation:

We have $f (x) = x \sin \frac{1}{x},$ where $x \neq 0 .$

Since, $f (x)$ is continuous at $x = 0$ ,

we must have $\lim_{x \to 0} x \sin \frac{1}{x} = f (0)$

$∴$ $f (0) = 0 \times$ [an oscillating value between - 1 and 1]

$∴$ $f (0) = 0$

Q 33 :

If $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix},$ is continuous at $x = \frac{π}{2}$ , then

$m = 1, n = 0$
$m = \frac{n π}{2} + 1$
$n = \frac{m π}{2}$
$m = n = \frac{π}{2}$

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Ans. $n = \frac{m π}{2}$

Explanation:

We have, $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$

$∴$ $L.H.L. = \lim_{x \to {\frac{π}{2}}^{-}} (m x + 1) = \lim_{h \to 0} [m (\frac{π}{2} - h) + 1] = \frac{m π}{2} + 1$

and $R.H.L. = \lim_{x \to {\frac{π}{2}}^{+}} (\sin x + n) = \lim_{h \to 0} [\sin (\frac{π}{2} + h) + n]$

$= \lim_{h \to 0} (\cos h + n) = 1 + n$

We must have $L.H.L. = R.H.L.$

$\Rightarrow \frac{m π}{2} + 1 = n + 1$

$∴$ $n = \frac{m π}{2}$

Q 34 :

The set of points, where the function $f$ given by $f (x) =$ $| 2 x - 1 |$ $\sin x$ is differentiable is :

$R$
$R - {\frac{1}{2}}$
$(0, \infty)$
None of these

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Ans. $R - {\frac{1}{2}}$

Explanation:

Let $u (x) = \sin x$

$v (x) =$ $| x |$

$f (x) = v o u (x) = v [u (x)]$

$= v (\sin x)$

$=$ $| \sin x |$

$∵$ $u (x) = \sin x$ is a continuous function and $v (x) =$ $| x |$ is a continuous function.

$∴$ $f (x) = v \circ u (x)$ is also continuous everywhere but $v (x)$ is not differentiable at $x = 0$ .

$\Rightarrow f (x)$ is not differentiable where $\sin x = 0$

$x = n π, n \in Z$

Hence, $f (x)$ is continuous everywhere but not differentiable at $x = n π, n \in Z .$

$∵$ At $x = \frac{1}{2}$ , curve has two tangents at that point, therefore, from the graph it is clear that $y =$ $| 2 x - 1 |$ is not differentiable at $x = \frac{1}{2} .$

$∴$ $f (x) is differentiable in R - {\frac{1}{2}} .$

Q 35 :

Let $f (x) =$ ${\begin{matrix} (x - 1) \sin (\frac{1}{x - 1}), & if x \neq 1 \\ 0, & if x = 1 \end{matrix} .$ Then, which of the following is true?

$f$ is differentiable at $x = 1$ but not at $x = 0$
$f$ is neither differentiable at $x = 0$ nor at $x = 1$
$f$ is differentiable at $x = 0$ and $x = 1$
$f$ is differentiable at $x = 0$ but not $x = 1$

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Ans. $f$ is differentiable at $x = 0$ but not $x = 1$

Explanation:

We observe that,

$\lim_{x \to 1} \frac{f (x) - f (1)}{x - 1} = \lim_{x \to 1} \frac{(x - 1) \sin (\frac{1}{x - 1})}{x - 1}$

$= \lim_{x \to 1} \sin (\frac{1}{x - 1})$

= An oscillating number between -1 and 1

$∴$ $\lim_{x \to 1} \frac{f (x) - f (1)}{x - 1}$ does not exist.

$\Rightarrow$ $f (x)$ is not differentiable at $x = 1$ .

$= \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0}$

$= \lim_{x \to 0} \frac{(x - 1) \sin (\frac{1}{x - 1}) - \sin 1}{x}$

$= \lim_{x \to 0} \frac{x \sin (\frac{1}{x - 1})}{x} - \lim_{x \to 0} \frac{\sin (\frac{1}{x - 1}) + \sin 1}{x}$

$= - \sin 1 - \lim_{x \to 0} \frac{2 \sin (\frac{x}{2 (x - 1)}) \cos {\frac{2 - x}{2 (x - 1)}}}{{\frac{x}{2 (x - 1)}} 2 (x - 1)}$

$= - \sin 1 + \cos 1$

$∴$ $f (x) is differentiable at x = 0$

Q 36 :

The function $f (x) = e^{| x |}$ is :

$f' (0) = 1$
$f' (0) = - 1$
$f' (0) = 0$
$f' (0)$ does not exist

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Ans. $f' (0)$ does not exist.

Explanation:

$f (x) = e^{| x |}$ $\Rightarrow$ $f (x) =$ ${\begin{matrix} e^{x}, & x \geq 0 \\ e^{- x}, & x < 0 \end{matrix}$

$RHD = \lim_{h \to 0} \frac{e^{h} - f (0)}{h}$

$\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$

$LHD = \lim_{h \to 0} \frac{e^{- h} - f (0)}{- h}$

$= \lim_{h \to 0} \frac{e^{- h} - 1}{- h} = - 1$

Since, $RHD \neq LHD$

that implies $f$ is not differentiable at $x = 0$ .

$f' (0)$ does not exist.

Q 37 :

Let $f (x) =$ $| \sin x |$ . Then

$f$ is everywhere differentiable
$f$ is everywhere continuous but not differentiable at $x = n π, n \in Z$
$f$ is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$
None of these

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Ans. $f$ is everywhere continuous but not differentiable at $x = n π, n \in Z$

Explanation:

We have, $f (x) =$ $| \sin x |$

We know that $| x |$ and $\sin x$ are continuous for all real $x$ .

So, $| \sin x |$ is also continuous for all real $x$ .

Since, $| x |$ is non-differentiable at $x = 0$

therefore, $| \sin x |$ is non-differentiable when $\sin x = 0$ or $x = n π, n \in Z$

Hence, $f (x)$ is continuous everywhere but not differentiable at $x = n π, n \in Z .$

Q 38 :

The function $f (x) = e^{| x |}$ is :

Continuous everywhere but not differentiable at $x = 0$
Continuous and differentiable everywhere
Not continuous at $x = 0$
None of the above

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Ans. Continuous everywhere but not differentiable at $x = 0$

Explanation:

$f (x) = e^{| x |}$ $=$ ${\begin{matrix} e^{x}, & x \geq 0 \\ e^{- x}, & x < 0 \end{matrix}$

$f' (x) =$ ${\begin{matrix} e^{x}, & x \geq 0 \\ - e^{- x}, & x < 0 \end{matrix}$

$f (0) = e^{0} = 1$

$RHL = \lim_{x \to 0^{+}} e^{x} = e^{0} = 1$

$LHL = \lim_{x \to 0^{-}} e^{- x} = e^{- 0} = 1$

If $x = 0$ ,

$f (0) = LHL = RHL$

$∴$ $f (x)$ is continuous at $x = 0$

$LHD = \lim_{x \to 0^{-}} f' (x) = \lim_{x \to 0^{-}} (- e^{- x}) = - e^{- 0} = - 1$

$RHD = \lim_{x \to 0^{+}} e^{x} = e^{0} = 1$

$LHD \neq RHD$

$∴$ $f (x)$ is not differentiable at $x = 0$

Q 39 :

The differential coefficient of $\sin (\cos (x^{2}))$ with respect to $x$ is :

$- 2 x \sin (x^{2}) \cos (\cos (x^{2}))$
$2 x \sin (x^{2}) \cos (x^{2})$
$2 x \sin (x^{2}) \cos (x^{2}) \cos x$
None of the above

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Ans. $- 2 x s i n (x^{2}) c o s (c o s (x^{2}))$

Explanation:

We have, $y = \sin (\cos (x^{2}))$

Therefore, $\frac{d y}{d x} = \frac{d}{d x} [\sin (\cos (x^{2}))]$

$= \cos (\cos (x^{2})) \cdot \frac{d}{d x} (\cos (x^{2}))$

$= \cos (\cos (x^{2})) (- \sin (x^{2})) \cdot \frac{d}{d x} (x^{2})$

$= - \sin (x^{2}) \cos (\cos (x^{2})) (2 x)$

$= - 2 x \sin (x^{2}) \cos (\cos (x^{2}))$

Q 40 :

If $y = \sqrt{3 x + 2} + \frac{1}{\sqrt{2 x^{2} + 4}}$ , then $\frac{d y}{d x}$ is equal to:

$\frac{3}{2 \sqrt{3 x + 2}} - \frac{2 x}{{(2 x^{2} + 4)}^{3 / 2}}$
$\frac{3}{2 \sqrt{3 x + 2}} - \frac{2 x}{{(2 x^{2} + 4)}^{3 / 2}}$
$\frac{3}{2 \sqrt{3 x + 2}} - \frac{2 x}{{(2 x^{2} + 4)}^{3 / 2}}$
None of the above

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Ans. $\frac{3}{2 \sqrt{3 x + 2}} - \frac{2 x}{{(2 x^{2} + 4)}^{3 / 2}}$

Explanation:

Let $y = \sqrt{3 x + 2} + \frac{1}{\sqrt{2 x^{2} + 4}}$

$= {(3 x + 2)}^{\frac{1}{2}} + {(2 x^{2} + 4)}^{- \frac{1}{2}}$

$∴$ $\frac{d y}{d x} = \frac{1}{2} {(3 x + 2)}^{\frac{1}{2} - 1} \cdot \frac{d}{d x} (3 x + 2) + (- \frac{1}{2}) {(2 x^{2} + 4)}^{- \frac{1}{2} - 1} \cdot \frac{d}{d x} (2 x^{2} + 4)$

$= \frac{1}{2} {(3 x + 2)}^{- \frac{1}{2}} . (3) - (\frac{1}{2}) {(2 x^{2} + 4)}^{- \frac{3}{2}} (4 x)$

$= \frac{3}{2 \sqrt{3 x + 2}} - \frac{2 x}{{(2 x^{2} + 4)}^{3 / 2}}$

Q 41 :

The derivative of $2 x + 3 y = \sin y$ is:

$\frac{2}{\cos y}$
$\frac{2}{\cos y + 3}$
$\frac{2}{\cos y - 3}$
None of these

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Ans. $\frac{2}{c o s y - 3}$

Explanation:

Given, $2 x + 3 y = \sin y$

On differentiating both sides w.r.t. $x$ , we get

$\frac{d}{d x} (2 x + 3 y) = \frac{d}{d x} (\sin y)$

$\Rightarrow$ $2 + 3 \frac{d y}{d x} = \cos y \frac{d y}{d x}$

$\Rightarrow$ $3 \frac{d y}{d x} - \cos y \frac{d y}{d x} = - 2$

$\Rightarrow$ $(3 - \cos y) \frac{d y}{d x} = - 2$

$\Rightarrow$ $\frac{d y}{d x} = \frac{2}{\cos y - 3}$

Q 42 :

If $x + \sin y = \log x$ , then $\frac{d y}{d x}$ is equal to:

$\frac{1 - x}{x \sin y}$
$\frac{1 - x}{x \cos y}$
$\frac{1 + x}{x \cos y}$
None of these

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Ans. $\frac{1 - x}{x c o s y}$

Explanation:

$\sin y + x = \log x$

On differentiating with respect to $x$ , we get

$\cos y \frac{d y}{d x} + 1 = \frac{1}{x}$

$\Rightarrow$ $\cos y \frac{d y}{d x} = \frac{1}{x} - 1$

$\Rightarrow$ $\cos y \frac{d y}{d x} = \frac{1 - x}{x}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{1 - x}{x \cos y}$

Q 43 :

If $2 x + 3 y = \sin x$ , then $\frac{d y}{d x}$ is equal to:

$\frac{\cos x + 2}{3}$
$\frac{\cos x - 2}{3}$
$\cos x + 2$
None of these

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Ans. $\frac{c o s x - 2}{3}$

Explanation:

Given, $2 x + 3 y = \sin x$

On differentiating both sides w.r.t. $x$ , we get

$\frac{d}{d x} (2 x + 3 y) = \frac{d}{d x} (\sin x)$

$\Rightarrow$ $2 + 3 \frac{d y}{d x} = \cos x$

$\Rightarrow$ $3 \frac{d y}{d x} = \cos x - 2$

$\Rightarrow$ $\frac{d y}{d x} = \frac{\cos x - 2}{3}$

Q 44 :

If $y = \sqrt{\sin x + y}$ , then $\frac{d y}{d x}$ is equal to:

$\frac{\cos x}{2 y - 1}$
$\frac{\cos x}{1 - 2 y}$
$\frac{\sin x}{1 - 2 y}$
$\frac{\sin x}{2 y - 1}$

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Ans. $\frac{c o s x}{2 y - 1}$

Explanation:

$∵$ $y = {(\sin x + y)}^{^{\frac{1}{2}}}$

$∴$ $\frac{d y}{d x} = \frac{1}{2} {(\sin x + y)}^{- \frac{1}{2}} \cdot \frac{d}{d x} (\sin x + y)$ [By chain rule derivative]

$\Rightarrow$ $\frac{d y}{d x} = \frac{1}{2} \cdot \frac{1}{{(\sin x + y)}^{1 / 2}} (\cos x + \frac{d y}{d x})$

$\Rightarrow$ $\frac{d y}{d x} = \frac{1}{2 y} (\cos x + \frac{d y}{d x})$ $[∵ {(\sin x + y)}^{1 / 2} = y]$

$\Rightarrow$ $\frac{d y}{d x} (1 - \frac{1}{2 y}) = \frac{\cos x}{2 y}$

$∴$ $\frac{d y}{d x} = \frac{\cos x}{2 y} \cdot \frac{2 y}{2 y - 1} = \frac{\cos x}{2 y - 1}$

Q 45 :

If $\cos y = x \cos (a + y)$ with $\cos a = 1$ , then $\frac{d y}{d x}$ is equal to:

$\frac{\sin^{2} (a + y)}{\sin a}$
$\frac{\cos^{2} (a + y)}{\sin a}$
$\sin^{2} (a + y) \sin a$
None of these

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Ans. $\frac{\cos^{2} (a + y)}{\sin a}$

Explanation:

Given, $\cos y = x \cos (a + y)$

$\Rightarrow$ $x = \frac{\cos y}{\cos (a + y)}$

$∴$ $\frac{d y}{d x} = \frac{d}{d y} [\frac{\cos y}{\cos (a + y)}]$

$= \frac{\cos (a + y) (- \sin y) - \cos y [- \sin (a + y)]}{\cos^{2} (a + y)}$

$= \frac{\cos y \sin (a + y) - \sin y \cos (a + y)}{\cos^{2} (a + y)}$

$= \frac{\sin ((a + y) - y)}{\cos^{2} (a + y)} = \frac{\sin a}{\cos^{2} (a + y)}$ $[∵ \sin (A - B) = \sin A \cos B - \cos A \sin B]$

Hence, $\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{\cos^{2} (a + y)}{\sin a}$

Q 46 :

If $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}),$ then $\frac{d y}{d x}$ is equal to:

$\frac{1}{1 + x^{2}}$
$\frac{2}{1 + x^{2}}$
$\frac{2}{1 - x^{2}}$
$\frac{- 2}{1 + x^{2}}$

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Ans. $\frac{2}{1 + x^{2}}$

Explanation:

Let $x = \tan θ$

Then $y = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ})$

Using the identity $\frac{2 \tan θ}{1 + \tan^{2} θ} = \sin 2 θ,$

we get $y = \sin^{- 1} (\sin 2 θ)$

Since $2 θ$ lies in the principal value range, $y = 2 θ = 2 \tan^{- 1} x$

Differentiating,

$\frac{d y}{d x} = 2 \cdot \frac{1}{1 + x^{2}} = \frac{2}{1 + x^{2}}$

Q 47 :

If $y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ , $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ , then $\frac{d y}{d x}$ is:

$\frac{3}{1 + x^{2}}$
$\frac{1}{1 + x^{2}}$
$\frac{- 3}{1 + x^{2}}$
$\frac{3}{1 - x^{2}}$

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Ans. $\frac{3}{1 + x^{2}}$

Explanation:

$y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) = 3 \tan^{- 1} x$

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

Q 48 :

If $y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}$ , $- 1 \leq x < 1$ , then $\frac{d y}{d x}$ is equal to:

0
1
2
3

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Ans. 0

Explanation:

Put $\sin^{- 1} \sqrt{1 - x^{2}} = \cos^{- 1} x$

$∴$ $y = \sin^{- 1} x + \cos^{- 1} x$

$= \frac{π}{2}$

$∴$ $\frac{d y}{d x} = 0$

Q 49 :

Derivative of $\cot^{- 1} [\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}], 0 < x < \frac{π}{2}$ is:

$\frac{1}{2}$
$1$
$2$
None of these

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Ans. $\frac{1}{2}$

Explanation:

$Let y = \cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}})$

$y = \cot^{- 1} [\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}]$

$\Rightarrow$ $y = \cot^{- 1} (\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}) = \cot^{- 1} (\cot \frac{x}{2}) = \frac{x}{2}$

$∴$ $\frac{d y}{d x} = \frac{1}{2}$

Q 50 :

If $y^{x} = e^{y - x}$ , then $\frac{d y}{d x}$ is equal to:

$\frac{1 + \log y}{y \log y}$
$\frac{{(1 + \log y)}^{2}}{y \log y}$
$\frac{1 + \log y}{{(\log y)}^{2}}$
$\frac{{(1 + \log y)}^{2}}{\log y}$

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Ans. $\frac{{(1 + l o g y)}^{2}}{l o g y}$

Explanation:

We have, $y^{x} = e^{y - x}$

Taking log both sides, we get

$x \log y = y - x$

$\Rightarrow$ $x (\log y + 1) = y$

$\Rightarrow$ $\frac{y}{1 + \log y} = x$

On differentiating with respect to $x$ , we get

$(\frac{(1 + \log y) - (\frac{1}{y}) y}{{(1 + \log y)}^{2}}) \frac{d y}{d x} = 1$

$\Rightarrow$ $\frac{\log y}{{(1 + \log y)}^{2}} \cdot \frac{d y}{d x} = 1$

$\Rightarrow$ $\frac{d y}{d x} = \frac{{(1 + \log y)}^{2}}{\log y}$

Q 51 :

If $x = a t^{2}$ and $y = 2 a t$ , then $\frac{d y}{d x}$ is equal to:

$t$
$\frac{1}{t}$
$\frac{- 1}{t^{2}}$
None of these

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Ans. $\frac{1}{t}$

Explanation:

Given that, $x = a t^{2}, y = 2 a t$

So $\frac{d x}{d t} = 2 a t$

and $\frac{d y}{d t} = 2 a$

$∴$ $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t}$

Q 52 :

If $x = a (\cos θ + θ \sin θ)$ and $y = a (\sin θ - θ \cos θ)$ , then $\frac{d y}{d x}$ is equal to:

$\tan θ$
$\cos θ$
$\sin θ$
$\cot θ$

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Ans. $t a n θ$

Explanation:

Given, $x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ)$

On differentiating with respect to $θ$ , we get

$\frac{d x}{d θ} = a \frac{d}{d θ} (\cos θ + θ \sin θ)$

$= a {\frac{d}{d θ} (\cos θ) + \frac{d}{d θ} (θ \sin θ)}$

$= a {- \sin θ + (θ \cos θ + \sin θ)}$

$= a θ \cos θ$

[using product rule in $\frac{d}{d θ} (θ \sin θ)$ ]

and $\frac{d y}{d θ} = a \frac{d}{d θ} (\sin θ - θ \cos θ)$

$= a {\frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (θ \cos θ)}$

$= a {\cos θ - (θ (- \sin θ) + \cos θ)}$

$= a θ \sin θ$

[using product rule in $\frac{d}{d θ} (θ \cos θ)$ ]

$∴$ $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ$

Q 53 :

The derivative of $\cos^{- 1} (2 x^{2} - 1)$ w.r.t. $\cos^{- 1} x$ is:

$2$
$\frac{- 1}{2 \sqrt{1 - x^{2}}}$
$\frac{2}{x}$
$1 - x^{2}$

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Ans. 2

Explanation:

$Let u = \cos^{- 1} (2 x^{2} - 1) and v = \cos^{- 1} x$

$∴$ $\frac{d u}{d x} = \frac{- 1}{\sqrt{1 - {(2 x^{2} - 1)}^{2}}} \cdot 4 x = \frac{- 4 x}{\sqrt{1 - (4 x^{4} + 1 - 4 x^{2})}}$

$= \frac{- 4 x}{\sqrt{- 4 x^{4} + 4 x^{2}}} = \frac{- 4 x}{\sqrt{4 x^{2} (1 - x^{2})}} = \frac{- 2}{\sqrt{1 - x^{2}}}$

$and$ $\frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}}$

$∴$ $\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 2 / \sqrt{1 - x^{2}}}{- 1 / \sqrt{1 - x^{2}}} = 2$

Q 54 :

The derivative of $\sin^{2} x$ with respect to $e^{\cos x}$ is:

$\frac{2 \cos x}{e^{\cos x}}$
$- \frac{2 \cos x}{e^{\cos x}}$
$\frac{2}{e^{\cos x}}$
None of these

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Ans. $- \frac{2 \cos x}{e^{\cos x}}$

Explanation:

Let $u (x) = \sin^{2} x$ and $v (x) = e^{\cos x} .$ We want to find $\frac{d u}{d v} .$

Thus, $\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} .$ Clearly, $\frac{d u}{d x} = 2 \sin x \cos x$

and $\frac{d v}{d x} = e^{\cos x} (- \sin x) = - (\sin x) e^{\cos x}$

Therefore, $\frac{d u}{d v} = \frac{2 \sin x \cos x}{- \sin x e^{\cos x}} = - \frac{2 \cos x}{e^{\cos x}}$

Q 55 :

If $y = \cos^{- 1} x$ , then the value of $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone is:

$- \cot y {cosec}^{2} y$
$cosec y \cot^{2} y$
$- \cot y cosec y$
None of the above

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Ans. $- c o t y c o s e c^{2} y$

Explanation:

Given, $y = \cos^{- 1} x$

$\Rightarrow$ $x = \cos y$

On differentiating with respect to $y$ , we get

$\frac{d x}{d y} = - \sin y$

$\Rightarrow$ $\frac{d y}{d x} = - cosec y$ $\dots (i)$

Again, differentiating with respect to $x$ , we get

$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (- cosec y) = - (- cosec y \cot y) \frac{d y}{d x}$

$= cosec y \cot y (- cosec y)$

$= - \cot y {cosec}^{2} y$ $[from Eq. (i)]$

Q 56 :

If $y = {(\frac{1}{x})}^{x},$ then the value of $e^{e} {(\frac{d^{2} y}{d x^{2}})}_{x = e}$ is:

$2 - \frac{1}{e}$
$4 - \frac{1}{e}$
$\frac{1}{e}$
$1 - \frac{1}{e}$

1

2

3

4

(2)

Ans. $4 - \frac{1}{e}$

Explanation:

$y = {(\frac{1}{x})}^{x}$

Taking logarithm on both sides,

$\log y = x \log (\frac{1}{x}) = - x \log x$

Differentiating with respect to $x$ ,

$\frac{1}{y} \frac{d y}{d x} = - [x \cdot \frac{1}{x} + \log x]$

$\frac{d y}{d x} = - y (1 + \log x)$

$\frac{d^{2} y}{d x^{2}} = - [y \cdot \frac{1}{x} + (1 + \log x) \frac{d y}{d x}]$

$= - [\frac{y}{x} + (1 + \log x) {- y (1 + \log x)}]$

$= - {(\frac{1}{x})}^{x + 1} + {(1 + \log x)}^{2} {(\frac{1}{x})}^{x}$

When $x = e$ ,

$e^{e} {[\frac{d^{2} y}{d x^{2}}]}_{x = e} = [- {(\frac{1}{e})}^{e + 1} + 4 {(\frac{1}{e})}^{e}] e^{e}$

$= - e^{- e - 1 + e} + 4 e^{- e + e}$

$= 4 - \frac{1}{e}$

Q 57 :

If $y = e^{n x},$ then $n^{th}$ derivative of $y$ is:

$e^{n x}$
$n^{2} e^{n x}$
$n y$
$n^{n} y$

1

2

3

4

(4)

Ans. $n^{n} y$

Explanation:

$y = e^{n x}$

$y = n e^{n x}$

$y_{1} = n e^{n x}$

$y_{2} = n^{2} e^{n x}$

...............................

$y_{n}^{''} = n^{n} y$

Q 58 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 - \cos 2 x}}{x \sqrt{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix},$ then the value of $k$ which makes the function $f$ continuous at $x = 0$ is:

1
- 1
0
No value

1

2

3

4

(1)

Ans. 1

Explanation:

Given that the function $f$ is continuous at $x = 0$ ,

Then, $\lim_{x \to 0} f (x) = k$

$\Rightarrow$ $\lim_{x \to 0} \frac{\sqrt{1 - \cos 2 x}}{x \sqrt{2}} = k$

$\Rightarrow$ $\lim_{x \to 0} \frac{\sqrt{2 \sin^{2} x}}{x \sqrt{2}} = k$

$\Rightarrow$ $\lim_{x \to 0} \frac{\sqrt{2} \sin x}{x \sqrt{2}} = k$

$\Rightarrow$ $\lim_{x \to 0} \frac{\sin x}{x} = k$ $[∴ \lim_{x \to 0} \frac{\sin x}{x} = 1]$

$\Rightarrow$ $1 = k$

Q 59 :

If $y = \log (\sec e^{x^{2}}),$ then $\frac{d y}{d x} =$

$x^{2} e^{x^{2}} \tan e^{x^{2}}$
$e^{x^{2}} \tan e^{x^{2}}$
$2 x e^{x^{2}} \tan e^{x^{2}}$
$x e^{x^{2}} \tan e^{x^{2}}$

1

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4

(3)

Ans. $2 x e^{x^{2}} t a n e^{x^{2}}$

Explanation:

$y = \log (se c e^{x^{2}})$

$\frac{d y}{d x} = \frac{d \log (\sec e^{x^{2}})}{d (\sec e^{x^{2}})} \times \frac{d (\sec e^{x^{2}})}{d (e^{x^{2}})} \times \frac{d (e^{x^{2}})}{d (x^{2})} \times \frac{d (x^{2})}{d x}$

$= \frac{1}{\sec e^{x^{2}}} \times \sec e^{x^{2}} \tan e^{x^{2}} \times e^{x^{2}} \times 2 x$

$∴$ $\frac{d y}{d x} = 2 x e^{x^{2}} \tan e^{x^{2}}$

Q 60 :

If $y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}, 0 < x < 1,$ then:

$\frac{d y}{d x} = 0$
$\frac{d y}{d x} = \frac{π}{2}$
$\frac{d y}{d x} = \frac{π}{3}$
does not exist

1

2

3

4

(1)

Ans. $\frac{d y}{d x} = 0$

Explanation:

$y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}$

$y = \sin^{- 1} x + \cos^{- 1} x$

$y = 1$

$\frac{d y}{d x} = 0$

Topic Question Set

Q 31 :

The function $f (x) = \tan x$ is discontinuous on the set:

Q 32 :

If $f (x) = x \sin \frac{1}{x}$ , where $x \neq 0$ , then the value of the function $f$ at $x = 0$ , so that the function is continuous at $x = 0$ , is:

Q 33 :

If $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix},$ is continuous at $x = \frac{π}{2}$ , then

Q 34 :

The set of points, where the function $f$ given by $f (x) =$ $| 2 x - 1 |$ $\sin x$ is differentiable is :

Q 35 :

Let $f (x) =$ ${\begin{matrix} (x - 1) \sin (\frac{1}{x - 1}), & if x \neq 1 \\ 0, & if x = 1 \end{matrix} .$ Then, which of the following is true?

Q 36 :

The function $f (x) = e^{| x |}$ is :

Q 37 :

Let $f (x) =$ $| \sin x |$ . Then

Q 38 :

The function $f (x) = e^{| x |}$ is :

Q 39 :

The differential coefficient of $\sin (\cos (x^{2}))$ with respect to $x$ is :

Q 40 :

If $y = \sqrt{3 x + 2} + \frac{1}{\sqrt{2 x^{2} + 4}}$ , then $\frac{d y}{d x}$ is equal to:

Q 41 :

The derivative of $2 x + 3 y = \sin y$ is:

Q 42 :

If $x + \sin y = \log x$ , then $\frac{d y}{d x}$ is equal to:

Q 43 :

If $2 x + 3 y = \sin x$ , then $\frac{d y}{d x}$ is equal to:

Q 44 :

If $y = \sqrt{\sin x + y}$ , then $\frac{d y}{d x}$ is equal to:

Q 45 :

If $\cos y = x \cos (a + y)$ with $\cos a = 1$ , then $\frac{d y}{d x}$ is equal to:

Q 46 :

If $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}),$ then $\frac{d y}{d x}$ is equal to:

Q 47 :

If $y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ , $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ , then $\frac{d y}{d x}$ is:

Q 48 :

If $y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}$ , $- 1 \leq x < 1$ , then $\frac{d y}{d x}$ is equal to:

Q 49 :

Derivative of $\cot^{- 1} [\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}], 0 < x < \frac{π}{2}$ is:

Q 50 :

If $y^{x} = e^{y - x}$ , then $\frac{d y}{d x}$ is equal to:

Q 51 :

If $x = a t^{2}$ and $y = 2 a t$ , then $\frac{d y}{d x}$ is equal to:

Q 52 :

If $x = a (\cos θ + θ \sin θ)$ and $y = a (\sin θ - θ \cos θ)$ , then $\frac{d y}{d x}$ is equal to:

Q 53 :

The derivative of $\cos^{- 1} (2 x^{2} - 1)$ w.r.t. $\cos^{- 1} x$ is:

Q 54 :

The derivative of $\sin^{2} x$ with respect to $e^{\cos x}$ is:

Q 55 :

If $y = \cos^{- 1} x$ , then the value of $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone is:

Q 56 :

If $y = {(\frac{1}{x})}^{x},$ then the value of $e^{e} {(\frac{d^{2} y}{d x^{2}})}_{x = e}$ is:

Q 57 :

If $y = e^{n x},$ then $n^{th}$ derivative of $y$ is:

Q 58 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 - \cos 2 x}}{x \sqrt{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix},$ then the value of $k$ which makes the function $f$ continuous at $x = 0$ is:

Q 59 :

If $y = \log (\sec e^{x^{2}}),$ then $\frac{d y}{d x} =$

Q 60 :

If $y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}, 0 < x < 1,$ then:

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Topic Question Set

Q 31 : The function f(x)=tanx is discontinuous on the set:

Q 32 : If f(x)=xsin1x, where x≠0, then the value of the function f at x=0, so that the function is continuous at x=0, is:

Q 33 : If f(x)={mx+1,if x≤π2sinx+n,if x>π2, is continuous at x=π2, then

Q 34 : The set of points, where the function f given by f(x)=|2x-1|sinx is differentiable is :

Q 35 : Let f(x)={(x-1)sin(1x-1),if x≠10,if x=1. Then, which of the following is true?

Q 36 : The function f(x)=e|x| is :

Q 37 : Let f(x)=|sinx|. Then

Q 38 : The function f(x)=e|x| is :

Q 39 : The differential coefficient of sin(cos(x2)) with respect to x is :

Q 40 : If y=3x+2+12x2+4, then dydx is equal to:

Q 41 : The derivative of 2x+3y=siny is:

Q 42 : If x+siny=logx, then dydx is equal to:

Q 43 : If 2x+3y=sinx, then dydx is equal to:

Q 44 : If y=sinx+y, then dydx is equal to:

Q 45 : If cosy=xcos(a+y) with cosa=1, then dydx is equal to:

Q 46 : If y=sin-1(2x1+x2), then dydx is equal to:

Q 47 : If y=tan-1(3x-x31-3x2), -13<x<13, then dydx is:

Q 48 : If y=sin-1x+sin-11-x2, -1≤x<1, then dydx is equal to:

Q 49 : Derivative of cot-1[1+sinx+1-sinx1+sinx-1-sinx], 0<x<π2 is:

Q 50 : If yx=e y-x, then dydx is equal to:

Q 51 : If x=at2 and y=2at, then dydx is equal to:

Q 52 : If x=a(cosθ+θsinθ) and y=a(sinθ-θcosθ), then dydx is equal to:

Q 53 : The derivative of cos-1(2x2-1) w.r.t. cos-1x is:

Q 54 : The derivative of sin2x with respect to ecosx is:

Q 55 : If y=cos-1x, then the value of d2ydx2 in terms of y alone is:

Q 56 : If y=(1x)x, then the value of ee(d2ydx2)x=e is:

Q 57 : If y=enx, then nth derivative of y is:

Q 58 : If f(x)={1-cos2xx2,x≠0k,x=0, then the value of k which makes the function f continuous at x=0 is:

Q 59 : If y=log(sec ex2), then dydx=

Q 60 : If y=elog(sin-1x)+elog(cos-1x), 0<x<1, then:

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Q 31 :

The function $f (x) = \tan x$ is discontinuous on the set:

Q 32 :

If $f (x) = x \sin \frac{1}{x}$ , where $x \neq 0$ , then the value of the function $f$ at $x = 0$ , so that the function is continuous at $x = 0$ , is:

Q 33 :

If $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix},$ is continuous at $x = \frac{π}{2}$ , then

Q 34 :

The set of points, where the function $f$ given by $f (x) =$ $| 2 x - 1 |$ $\sin x$ is differentiable is :

Q 35 :

Let $f (x) =$ ${\begin{matrix} (x - 1) \sin (\frac{1}{x - 1}), & if x \neq 1 \\ 0, & if x = 1 \end{matrix} .$ Then, which of the following is true?

Q 36 :

The function $f (x) = e^{| x |}$ is :

Q 37 :

Let $f (x) =$ $| \sin x |$ . Then

Q 38 :

The function $f (x) = e^{| x |}$ is :

Q 39 :

The differential coefficient of $\sin (\cos (x^{2}))$ with respect to $x$ is :

Q 40 :

If $y = \sqrt{3 x + 2} + \frac{1}{\sqrt{2 x^{2} + 4}}$ , then $\frac{d y}{d x}$ is equal to:

Q 41 :

The derivative of $2 x + 3 y = \sin y$ is:

Q 42 :

If $x + \sin y = \log x$ , then $\frac{d y}{d x}$ is equal to:

Q 43 :

If $2 x + 3 y = \sin x$ , then $\frac{d y}{d x}$ is equal to:

Q 44 :

If $y = \sqrt{\sin x + y}$ , then $\frac{d y}{d x}$ is equal to:

Q 45 :

If $\cos y = x \cos (a + y)$ with $\cos a = 1$ , then $\frac{d y}{d x}$ is equal to:

Q 46 :

If $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}),$ then $\frac{d y}{d x}$ is equal to:

Q 47 :

If $y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$ , $- \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ , then $\frac{d y}{d x}$ is:

Q 48 :

If $y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}$ , $- 1 \leq x < 1$ , then $\frac{d y}{d x}$ is equal to:

Q 49 :

Derivative of $\cot^{- 1} [\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}], 0 < x < \frac{π}{2}$ is:

Q 50 :

If $y^{x} = e^{y - x}$ , then $\frac{d y}{d x}$ is equal to:

Q 51 :

If $x = a t^{2}$ and $y = 2 a t$ , then $\frac{d y}{d x}$ is equal to:

Q 52 :

If $x = a (\cos θ + θ \sin θ)$ and $y = a (\sin θ - θ \cos θ)$ , then $\frac{d y}{d x}$ is equal to:

Q 53 :

The derivative of $\cos^{- 1} (2 x^{2} - 1)$ w.r.t. $\cos^{- 1} x$ is:

Q 54 :

The derivative of $\sin^{2} x$ with respect to $e^{\cos x}$ is:

Q 55 :

If $y = \cos^{- 1} x$ , then the value of $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone is:

Q 56 :

If $y = {(\frac{1}{x})}^{x},$ then the value of $e^{e} {(\frac{d^{2} y}{d x^{2}})}_{x = e}$ is:

Q 57 :

If $y = e^{n x},$ then $n^{th}$ derivative of $y$ is:

Q 58 :

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 - \cos 2 x}}{x \sqrt{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix},$ then the value of $k$ which makes the function $f$ continuous at $x = 0$ is:

Q 59 :

If $y = \log (\sec e^{x^{2}}),$ then $\frac{d y}{d x} =$

Q 60 :

If $y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}, 0 < x < 1,$ then: