Differentiability – JEE Advanced Previous Year Questions with Solutions
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Q. Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in R .$ If $f(x)$ is differentiable at $x=0,$ then (A) $f(x)$ is differentiable only in a finite interval containing zero (B) $f(x)$ is continuous $\forall x \in R$ (C) $f^{\prime}(x)$ is constant $\forall x \in R$ (D) $f(x)$ is differentiable except at finitely many points [JEE 2011, 4M]

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Sol. (B,C) $\mathrm{F}(\mathrm{x}+\mathrm{y})=\mathrm{F}(\mathrm{x})+\mathrm{F}(\mathrm{y})$ diff. both side $\mathrm{F}^{\prime}(\mathrm{x}+\mathrm{y})=\mathrm{F}^{\prime}(\mathrm{x})$ Put $\mathrm{x}=0, \mathrm{y}=\mathrm{x}$ $\Rightarrow \mathrm{F}^{\prime}(\mathrm{x})=\mathrm{F}^{\prime}(0)$ than $\Rightarrow \mathrm{F}(\mathrm{x})$ is linear functon.

Q. If (A) $f(x)$ is continuous at $x=-\frac{\pi}{2}$ (B) $f(x)$ is not differentiable at $x=0$ (C) ƒ(x) is differentiable at x = 1 (D) $f(x)$ is differentiable at $x=-\frac{3}{2}$ [JEE 2011, 4M]

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Sol. (A,B,C,D)

Q. Let $f(\mathrm{x})=$ $\left\{\begin{array}{cc}{\mathrm{x}^{2}\left|\cos \frac{\pi}{\mathrm{x}}\right|} & {\mathrm{x} \neq 0} \\ {0} & {, \quad \mathrm{x}=0}\end{array}, \mathrm{x} \in \mathbb{R}, \text { then } f \text { is }-\right.$ (A) differentiable both at x = 0 and at x = 2 (B) differentiable at x = 0 but not differentiable at x = 2 (C) not differentiable at x = 0 but differentiable at x = 2 (D) differentiable neither at x = 0 nor at x = 2 [JEE 2012, 3M, –1M]

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Sol. (B)

Q. Let $f_{1}: \square \rightarrow \square, f_{2}:[0, \infty) \rightarrow \square, f_{3}: \square \rightarrow \square$ and $f_{4}: \square \rightarrow[0, \infty)$ be defined by $f_{1}(x)=\left\{\begin{array}{ll}{|x|} & {\text { if } \quad x<0,} \\ {e^{x}} & {\text { if } \quad x \geq 0}\end{array} f_{2}(x)=x^{2}\right.$ [JEE(Advanced)-2014, 3(–1)]

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Sol. (D)

Q. Let $f: \square \rightarrow \square$ and $\mathrm{g}: \square \rightarrow \square$ be respectively given by $f(\mathrm{x})=|\mathrm{x}|+1$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}+1$ Define $\mathrm{h}: \square \rightarrow \square$ by $\quad \mathrm{h}(\mathrm{x})=\left\{\begin{array}{ll}{\max \{f(\mathrm{x}), \mathrm{g}(\mathrm{x})\}} & {\text { if } \quad \mathrm{x} \leq 0} \\ {\min \{f(\mathrm{x}), \mathrm{g}(\mathrm{x})\}} & {\text { if } \quad \mathrm{x}>0}\end{array}\right.$ The number of points at which h(x) is not differentiable is [JEE(Advanced)-2014, 3]

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Sol. 3

Q. Let $a, b \in \square$ and $f: \square \rightarrow \square$ be defined by $f(x)=\operatorname{acos}\left(\left|x^{3}-x\right|\right)+b|x| \sin \left(\left|x^{3}+x\right|\right)$ Then $f$ is $-$ (A) differentiable at x = 0 if a = 0 and b = 1 (B) differentiable at x = 1 if a = 1 and b = 0 (C) NOT differentiable at x = 0 if a = 1 and b = 0 (D) NOT differentiable at x = 1 if a = 1 and b = 1 [JEE(Advanced)-2016]

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Sol. (A,B)

Q. Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow \square$ and $\mathrm{g}:\left[-\frac{1}{2}, 2\right] \rightarrow \square$ be function defined by $f(\mathrm{x})=\left[\mathrm{x}^{2}-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x),$ where [y] denotes the greatest integer less than or equal to y for $\mathrm{y} \in \square .$ Then (A) $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$ (B) $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$ (C) $\mathrm{g}$ is NOT differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$ (D) $\mathrm{g}$ is NOT differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$ [JEE(Advanced)-2016]

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Sol. (B,C) $f(x)=\left[x^{2}\right]-3$ $g(x)=(|x|+|4 x-7|)\left(\left[x^{2}\right]-3\right)$ $\because f$ is discontinuous at $\mathrm{x}=1, \sqrt{2}, \sqrt{3}, 2$ in $\left[-\frac{1}{2}, 2\right]$ and $|\mathrm{x}|+|4 \mathrm{x}-7| \neq 0$ at $\mathrm{x}=1, \sqrt{2}, \sqrt{3}, 2$ $\Rightarrow \mathrm{g}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1, \sqrt{2}, \sqrt{3}$ in $\left(-\frac{1}{2}, 2\right)$ $\operatorname{In}(0-\delta, 0+\delta)$ $\mathrm{g}(\mathrm{x})=(|\mathrm{x}|+|4 \mathrm{x}-7|) \cdot(-3)$ $\Rightarrow \mathrm{g}^{\prime}$ is non derivable at $\mathrm{x}=0$ $\operatorname{In}\left(\frac{7}{4}-\delta, \frac{7}{4}+\delta\right)$ $g(x)=0$ as $f(x)=0$ $\Rightarrow$ Derivable at $x=\frac{7}{4}$ $\therefore \mathrm{g}^{\prime}$ is non-derivable at $0,1, \sqrt{2}, \frac{7}{4}$

Q. Let $f: \square \rightarrow \square$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f^{\prime}(\mathrm{y})+f^{\prime}(\mathrm{x}) f(\mathrm{y})$ for all $\mathrm{x}, \mathrm{y} \in \mathbb{D} .$ Then, then value of $\log _{\mathrm{e}}(f(4))$ is [JEE(Advanced)-2018]

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Sol. 2 $\mathrm{P}(\mathrm{x}, \mathrm{y}): \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{y})+\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}$ $\mathrm{P}(0,0): \mathrm{f}(0)=\mathrm{f}(0) \mathrm{f}^{\prime}(0)+\mathrm{f}^{\prime}(0) \mathrm{f}(0)$ $\Rightarrow 1=2 \mathrm{f}^{\prime}(0)$ $\Rightarrow \mathrm{f}^{\prime}(0)=\frac{1}{2}$ $\mathrm{P}(\mathrm{x}, 0): \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(0)+\mathrm{f}^{\prime}(\mathrm{x}) \cdot \mathrm{f}(0)$ $\Rightarrow \mathrm{f}(\mathrm{x})=\frac{1}{2} \mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \mathrm{f}(\mathrm{x})$ $\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\frac{1}{2} \mathrm{x}}$ $\Rightarrow \ln (\mathrm{f}(4))=2$

Q. Let $\mathrm{f}_{1}: \square \rightarrow \square, \mathrm{f}_{2}:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \square, \mathrm{f}_{3}:\left(-1, \mathrm{e}^{\frac{\pi}{2}}-2\right) \rightarrow \square$ and $\mathrm{f}_{4}: \square \rightarrow \square$ be functions defined by (i) $\mathrm{f}_{1}(\mathrm{x})=\sin (\sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}})$ [JEE(Advanced)-2018]

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Sol. (D) (i) $\mathrm{f}(\mathrm{x})=\sin \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}$ $\quad \mathrm{f}_{1}^{\prime}(\mathrm{x})=\cos \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}} \cdot \frac{1}{2 \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}}\left(0-\mathrm{e}^{-\mathrm{x}^{2}} \cdot(-2 \mathrm{x})\right)$ $\quad \quad \quad \quad$ at $\mathrm{x}=0 \quad \mathrm{f}_{1}^{\prime}(\mathrm{x})$ does not exist So. $\mathrm{P} \rightarrow 2$ (ii) $\mathrm{f}_{2}(\mathrm{x})=\left\{\begin{array}{cl}{\frac{|\sin \mathrm{x}|}{\tan ^{-1} \mathrm{x}},} & {\mathrm{x} \neq 0} \\ {0} & {\mathrm{x}=0}\end{array}\right.$ $\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x} \frac{x}{\tan ^{-1} x}=1$ $\Rightarrow \mathrm{f}_{2}(\mathrm{x})$ does not continuous at $\mathrm{x}=0$ So $\mathrm{Q} \rightarrow 1$ (iii) $\mathrm{f}_{3}(\mathrm{x})=[\sin \ell \mathrm{n}(\mathrm{x}+2)]=0$ $\quad 1<\mathrm{x}+2<\mathrm{e}^{\pi / 2}$ $\Rightarrow 0<\ell n(\mathrm{x}+2)<\frac{\pi}{2}$ $\Rightarrow \quad 0<\sin (\ln (\mathrm{x}+2)<1$ $\Rightarrow \quad \mathrm{f}_{3}(\mathrm{x})=0$ So $\mathrm{R} \rightarrow 4$ (iv) $\mathrm{f}_{4}(\mathrm{x})=\left\{\begin{array}{cl}{\mathrm{x}^{2} \sin \frac{1}{\mathrm{x}},} & {\mathrm{x} \neq 0} \\ {0} & {, \mathrm{x}=0}\end{array}\right.$ So $\mathrm{S} \rightarrow 3$

• May 26, 2021 at 8:48 pm

yes

0
• December 5, 2020 at 10:35 pm

YOU HAVE WRITTEN ONE QUESTION (your key is B,C :but i am getting ans B,D )
AND NTA KEY IS ALSO (B,D) ONLY