80237
Let \(S=\left(t \in R: f(x)=|x-\pi| .\left(e^{|x|}-1\right) \sin |x|\right.\) is not differentiable at \(t\}\). Then, the set \(S\) is equal to
1 \(\phi\) (an empty set)
2 \(\{0\}\)
3 \(\{\pi\}\)
4 \(\{0, \pi\}\)
Explanation:
(A) : We have- \(f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x|\) We can write above function as- \(f(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \sin x, x\lt 0 \\ -(x-\pi)\left(e^{x}-1\right) \sin x, 0 \leq x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \sin x, x \geq \pi\end{array}\right.\) We check the differentiability at \(x=0\) and \(\pi\) we have- \(f^{\prime}(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \cos x+\left(e^{-x}-1\right) \sin x+(x-\pi) \sin x e^{-x}(-1), x\lt 0 \\ -\left[(x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x}\right] 0\lt x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x} x>\pi\end{array}\right.\) Clearly, \(\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)\) And \(\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) and \(\mathrm{x}=\pi\) Hence \(\mathrm{f}\) is differentiable for all \(\mathrm{x}\). So, the set of \(\mathrm{S}=\phi\)
JEE Main-2018
Limits, Continuity and Differentiability
80238
Let \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\} |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right\}\). Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then, \(S\)
1 equals \(\{-2,-1,0,1,2\}\)
2 equals \(\{-2,2\}\)
3 is an empty set
4 equals \(\{-2,-1,1,2\}\)
Explanation:
(A) : We have- \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\}, |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right.\) Let us draw the graph of \(y=f(x)\) For \(|x| \leq 2, f(x)=\max \left\{|x| x^{2}\right\}\) Let us first draw the graph of \(y=|x|\) and \(y=x^{2}\) as shown in the following figure. Clearly, \(\mathrm{y}=|\mathrm{x}|\) and \(\mathrm{y}=\mathrm{x}^{2}\) intersect at \(\mathrm{x}=-1,0,1\) Now the graph of \(y=\max \left\{|x|, x^{2}\right\}\) for \(|x| \leq 2\) is for \(|x| \in(2,4]\) \(\mathrm{f}(\mathrm{x})=8-2|\mathrm{x}|\) \(f(x)=\left\{\begin{array}{lc}8-2 x, x \in(2,4] \\ 8+2 x, x \in[-4,-2)\end{array}\right.\) Hence, the graph of \(y=f(x)\) is - Form the graph it is clear that at \(\mathrm{x}=-2,-1,0,1,2\) the curve has sharp edges and hence at these points \(f\) is not different able
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80239
Let \(f:(-1,1) \rightarrow R\) be a function defined by \(f(x)\) \(=\max \left\{-|x|,-\sqrt{1-x^{2}}\right\}\). If \(K\) is the set of all points at which \(f\) is not differentiable, then \(K\) has exactly
1 three elements
2 five elements
3 two elements
4 one elements
Explanation:
(A) : Given, f: \((-1,1) \rightarrow R\) such that- \(\mathrm{f}(\mathrm{x})=\max \left\{-|\mathrm{x}|, \sqrt{1-\mathrm{x}^{2}}\right\}\) On drawing the graph, Graph of \(y=-|x|\) is- And graph of \(y=-\sqrt{1-x^{2}}\) \(\left[\because x^{2}+y^{2}=1\right.\) represent a complete circle \(]\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}-\sqrt{1-\mathrm{x}^2} & , \quad-1\lt \mathrm{x} \leq-\frac{1}{\sqrt{2}} \\ -|\mathrm{x}| & , \quad-\frac{1}{\sqrt{2}}\lt \mathrm{x} \leq \frac{1}{\sqrt{2}} \\ -\sqrt{1-\mathrm{x}^2} & , \quad \frac{1}{\sqrt{2}}\lt \mathrm{x}\lt 1\end{array}\right.\) From the figure, it is clear that function have sharp edges, at \(x=-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\) \(\therefore\) function is not differentiable at 3 points
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80240
Let \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)\) \(=0\). If \(g(x)=|f(x)|\), then at \(x=c, g\) is
1 not differentiable
2 differentiable if \(\mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
3 not differentiable if \(f^{\prime}(\mathrm{c})=0\)
4 differentiable if \(f^{\prime}(c)=0\)
Explanation:
(D) : Given function, \(\mathrm{g}(\mathrm{x})=|\mathrm{f}(\mathrm{x})|\) Where \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)=0\), then for function \(g(x)\) at \(x=c\), \(g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} \quad[\text { where, } h>0]\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h} \quad[\text { as } f(c)=0(\text { given })]\) \(=\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \quad[\because h>0]\) \(=\left|\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\right|=\left|f^{\prime}(c)\right|\) \([\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=\mathrm{c}]\) Now if \(\mathrm{f}^{\prime}(\mathrm{c})=0\) then \(\mathrm{g}(\mathrm{x})\) is differentiable at \(\mathrm{x}=\mathrm{c}\), otherwise LHD (at \(x=c\) ) and RHD (at \(x=c\) ) is different.
80237
Let \(S=\left(t \in R: f(x)=|x-\pi| .\left(e^{|x|}-1\right) \sin |x|\right.\) is not differentiable at \(t\}\). Then, the set \(S\) is equal to
1 \(\phi\) (an empty set)
2 \(\{0\}\)
3 \(\{\pi\}\)
4 \(\{0, \pi\}\)
Explanation:
(A) : We have- \(f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x|\) We can write above function as- \(f(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \sin x, x\lt 0 \\ -(x-\pi)\left(e^{x}-1\right) \sin x, 0 \leq x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \sin x, x \geq \pi\end{array}\right.\) We check the differentiability at \(x=0\) and \(\pi\) we have- \(f^{\prime}(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \cos x+\left(e^{-x}-1\right) \sin x+(x-\pi) \sin x e^{-x}(-1), x\lt 0 \\ -\left[(x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x}\right] 0\lt x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x} x>\pi\end{array}\right.\) Clearly, \(\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)\) And \(\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) and \(\mathrm{x}=\pi\) Hence \(\mathrm{f}\) is differentiable for all \(\mathrm{x}\). So, the set of \(\mathrm{S}=\phi\)
JEE Main-2018
Limits, Continuity and Differentiability
80238
Let \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\} |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right\}\). Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then, \(S\)
1 equals \(\{-2,-1,0,1,2\}\)
2 equals \(\{-2,2\}\)
3 is an empty set
4 equals \(\{-2,-1,1,2\}\)
Explanation:
(A) : We have- \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\}, |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right.\) Let us draw the graph of \(y=f(x)\) For \(|x| \leq 2, f(x)=\max \left\{|x| x^{2}\right\}\) Let us first draw the graph of \(y=|x|\) and \(y=x^{2}\) as shown in the following figure. Clearly, \(\mathrm{y}=|\mathrm{x}|\) and \(\mathrm{y}=\mathrm{x}^{2}\) intersect at \(\mathrm{x}=-1,0,1\) Now the graph of \(y=\max \left\{|x|, x^{2}\right\}\) for \(|x| \leq 2\) is for \(|x| \in(2,4]\) \(\mathrm{f}(\mathrm{x})=8-2|\mathrm{x}|\) \(f(x)=\left\{\begin{array}{lc}8-2 x, x \in(2,4] \\ 8+2 x, x \in[-4,-2)\end{array}\right.\) Hence, the graph of \(y=f(x)\) is - Form the graph it is clear that at \(\mathrm{x}=-2,-1,0,1,2\) the curve has sharp edges and hence at these points \(f\) is not different able
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80239
Let \(f:(-1,1) \rightarrow R\) be a function defined by \(f(x)\) \(=\max \left\{-|x|,-\sqrt{1-x^{2}}\right\}\). If \(K\) is the set of all points at which \(f\) is not differentiable, then \(K\) has exactly
1 three elements
2 five elements
3 two elements
4 one elements
Explanation:
(A) : Given, f: \((-1,1) \rightarrow R\) such that- \(\mathrm{f}(\mathrm{x})=\max \left\{-|\mathrm{x}|, \sqrt{1-\mathrm{x}^{2}}\right\}\) On drawing the graph, Graph of \(y=-|x|\) is- And graph of \(y=-\sqrt{1-x^{2}}\) \(\left[\because x^{2}+y^{2}=1\right.\) represent a complete circle \(]\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}-\sqrt{1-\mathrm{x}^2} & , \quad-1\lt \mathrm{x} \leq-\frac{1}{\sqrt{2}} \\ -|\mathrm{x}| & , \quad-\frac{1}{\sqrt{2}}\lt \mathrm{x} \leq \frac{1}{\sqrt{2}} \\ -\sqrt{1-\mathrm{x}^2} & , \quad \frac{1}{\sqrt{2}}\lt \mathrm{x}\lt 1\end{array}\right.\) From the figure, it is clear that function have sharp edges, at \(x=-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\) \(\therefore\) function is not differentiable at 3 points
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80240
Let \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)\) \(=0\). If \(g(x)=|f(x)|\), then at \(x=c, g\) is
1 not differentiable
2 differentiable if \(\mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
3 not differentiable if \(f^{\prime}(\mathrm{c})=0\)
4 differentiable if \(f^{\prime}(c)=0\)
Explanation:
(D) : Given function, \(\mathrm{g}(\mathrm{x})=|\mathrm{f}(\mathrm{x})|\) Where \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)=0\), then for function \(g(x)\) at \(x=c\), \(g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} \quad[\text { where, } h>0]\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h} \quad[\text { as } f(c)=0(\text { given })]\) \(=\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \quad[\because h>0]\) \(=\left|\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\right|=\left|f^{\prime}(c)\right|\) \([\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=\mathrm{c}]\) Now if \(\mathrm{f}^{\prime}(\mathrm{c})=0\) then \(\mathrm{g}(\mathrm{x})\) is differentiable at \(\mathrm{x}=\mathrm{c}\), otherwise LHD (at \(x=c\) ) and RHD (at \(x=c\) ) is different.
80237
Let \(S=\left(t \in R: f(x)=|x-\pi| .\left(e^{|x|}-1\right) \sin |x|\right.\) is not differentiable at \(t\}\). Then, the set \(S\) is equal to
1 \(\phi\) (an empty set)
2 \(\{0\}\)
3 \(\{\pi\}\)
4 \(\{0, \pi\}\)
Explanation:
(A) : We have- \(f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x|\) We can write above function as- \(f(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \sin x, x\lt 0 \\ -(x-\pi)\left(e^{x}-1\right) \sin x, 0 \leq x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \sin x, x \geq \pi\end{array}\right.\) We check the differentiability at \(x=0\) and \(\pi\) we have- \(f^{\prime}(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \cos x+\left(e^{-x}-1\right) \sin x+(x-\pi) \sin x e^{-x}(-1), x\lt 0 \\ -\left[(x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x}\right] 0\lt x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x} x>\pi\end{array}\right.\) Clearly, \(\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)\) And \(\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) and \(\mathrm{x}=\pi\) Hence \(\mathrm{f}\) is differentiable for all \(\mathrm{x}\). So, the set of \(\mathrm{S}=\phi\)
JEE Main-2018
Limits, Continuity and Differentiability
80238
Let \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\} |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right\}\). Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then, \(S\)
1 equals \(\{-2,-1,0,1,2\}\)
2 equals \(\{-2,2\}\)
3 is an empty set
4 equals \(\{-2,-1,1,2\}\)
Explanation:
(A) : We have- \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\}, |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right.\) Let us draw the graph of \(y=f(x)\) For \(|x| \leq 2, f(x)=\max \left\{|x| x^{2}\right\}\) Let us first draw the graph of \(y=|x|\) and \(y=x^{2}\) as shown in the following figure. Clearly, \(\mathrm{y}=|\mathrm{x}|\) and \(\mathrm{y}=\mathrm{x}^{2}\) intersect at \(\mathrm{x}=-1,0,1\) Now the graph of \(y=\max \left\{|x|, x^{2}\right\}\) for \(|x| \leq 2\) is for \(|x| \in(2,4]\) \(\mathrm{f}(\mathrm{x})=8-2|\mathrm{x}|\) \(f(x)=\left\{\begin{array}{lc}8-2 x, x \in(2,4] \\ 8+2 x, x \in[-4,-2)\end{array}\right.\) Hence, the graph of \(y=f(x)\) is - Form the graph it is clear that at \(\mathrm{x}=-2,-1,0,1,2\) the curve has sharp edges and hence at these points \(f\) is not different able
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80239
Let \(f:(-1,1) \rightarrow R\) be a function defined by \(f(x)\) \(=\max \left\{-|x|,-\sqrt{1-x^{2}}\right\}\). If \(K\) is the set of all points at which \(f\) is not differentiable, then \(K\) has exactly
1 three elements
2 five elements
3 two elements
4 one elements
Explanation:
(A) : Given, f: \((-1,1) \rightarrow R\) such that- \(\mathrm{f}(\mathrm{x})=\max \left\{-|\mathrm{x}|, \sqrt{1-\mathrm{x}^{2}}\right\}\) On drawing the graph, Graph of \(y=-|x|\) is- And graph of \(y=-\sqrt{1-x^{2}}\) \(\left[\because x^{2}+y^{2}=1\right.\) represent a complete circle \(]\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}-\sqrt{1-\mathrm{x}^2} & , \quad-1\lt \mathrm{x} \leq-\frac{1}{\sqrt{2}} \\ -|\mathrm{x}| & , \quad-\frac{1}{\sqrt{2}}\lt \mathrm{x} \leq \frac{1}{\sqrt{2}} \\ -\sqrt{1-\mathrm{x}^2} & , \quad \frac{1}{\sqrt{2}}\lt \mathrm{x}\lt 1\end{array}\right.\) From the figure, it is clear that function have sharp edges, at \(x=-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\) \(\therefore\) function is not differentiable at 3 points
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80240
Let \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)\) \(=0\). If \(g(x)=|f(x)|\), then at \(x=c, g\) is
1 not differentiable
2 differentiable if \(\mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
3 not differentiable if \(f^{\prime}(\mathrm{c})=0\)
4 differentiable if \(f^{\prime}(c)=0\)
Explanation:
(D) : Given function, \(\mathrm{g}(\mathrm{x})=|\mathrm{f}(\mathrm{x})|\) Where \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)=0\), then for function \(g(x)\) at \(x=c\), \(g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} \quad[\text { where, } h>0]\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h} \quad[\text { as } f(c)=0(\text { given })]\) \(=\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \quad[\because h>0]\) \(=\left|\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\right|=\left|f^{\prime}(c)\right|\) \([\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=\mathrm{c}]\) Now if \(\mathrm{f}^{\prime}(\mathrm{c})=0\) then \(\mathrm{g}(\mathrm{x})\) is differentiable at \(\mathrm{x}=\mathrm{c}\), otherwise LHD (at \(x=c\) ) and RHD (at \(x=c\) ) is different.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Limits, Continuity and Differentiability
80237
Let \(S=\left(t \in R: f(x)=|x-\pi| .\left(e^{|x|}-1\right) \sin |x|\right.\) is not differentiable at \(t\}\). Then, the set \(S\) is equal to
1 \(\phi\) (an empty set)
2 \(\{0\}\)
3 \(\{\pi\}\)
4 \(\{0, \pi\}\)
Explanation:
(A) : We have- \(f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x|\) We can write above function as- \(f(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \sin x, x\lt 0 \\ -(x-\pi)\left(e^{x}-1\right) \sin x, 0 \leq x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \sin x, x \geq \pi\end{array}\right.\) We check the differentiability at \(x=0\) and \(\pi\) we have- \(f^{\prime}(x)=\left\{\begin{array}{cc}(x-\pi)\left(e^{-x}-1\right) \cos x+\left(e^{-x}-1\right) \sin x+(x-\pi) \sin x e^{-x}(-1), x\lt 0 \\ -\left[(x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x}\right] 0\lt x\lt \pi \\ (x-\pi)\left(e^{x}-1\right) \cos x+\left(e^{x}-1\right) \sin x+(x-\pi) \sin x e^{x} x>\pi\end{array}\right.\) Clearly, \(\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)\) And \(\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=0=\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) and \(\mathrm{x}=\pi\) Hence \(\mathrm{f}\) is differentiable for all \(\mathrm{x}\). So, the set of \(\mathrm{S}=\phi\)
JEE Main-2018
Limits, Continuity and Differentiability
80238
Let \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\} |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right\}\). Let \(S\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then, \(S\)
1 equals \(\{-2,-1,0,1,2\}\)
2 equals \(\{-2,2\}\)
3 is an empty set
4 equals \(\{-2,-1,1,2\}\)
Explanation:
(A) : We have- \(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\}, |x| \leq 2 \\ 8-2|x|, 2\lt |x| \leq 4\end{array}\right.\) Let us draw the graph of \(y=f(x)\) For \(|x| \leq 2, f(x)=\max \left\{|x| x^{2}\right\}\) Let us first draw the graph of \(y=|x|\) and \(y=x^{2}\) as shown in the following figure. Clearly, \(\mathrm{y}=|\mathrm{x}|\) and \(\mathrm{y}=\mathrm{x}^{2}\) intersect at \(\mathrm{x}=-1,0,1\) Now the graph of \(y=\max \left\{|x|, x^{2}\right\}\) for \(|x| \leq 2\) is for \(|x| \in(2,4]\) \(\mathrm{f}(\mathrm{x})=8-2|\mathrm{x}|\) \(f(x)=\left\{\begin{array}{lc}8-2 x, x \in(2,4] \\ 8+2 x, x \in[-4,-2)\end{array}\right.\) Hence, the graph of \(y=f(x)\) is - Form the graph it is clear that at \(\mathrm{x}=-2,-1,0,1,2\) the curve has sharp edges and hence at these points \(f\) is not different able
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80239
Let \(f:(-1,1) \rightarrow R\) be a function defined by \(f(x)\) \(=\max \left\{-|x|,-\sqrt{1-x^{2}}\right\}\). If \(K\) is the set of all points at which \(f\) is not differentiable, then \(K\) has exactly
1 three elements
2 five elements
3 two elements
4 one elements
Explanation:
(A) : Given, f: \((-1,1) \rightarrow R\) such that- \(\mathrm{f}(\mathrm{x})=\max \left\{-|\mathrm{x}|, \sqrt{1-\mathrm{x}^{2}}\right\}\) On drawing the graph, Graph of \(y=-|x|\) is- And graph of \(y=-\sqrt{1-x^{2}}\) \(\left[\because x^{2}+y^{2}=1\right.\) represent a complete circle \(]\) \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}-\sqrt{1-\mathrm{x}^2} & , \quad-1\lt \mathrm{x} \leq-\frac{1}{\sqrt{2}} \\ -|\mathrm{x}| & , \quad-\frac{1}{\sqrt{2}}\lt \mathrm{x} \leq \frac{1}{\sqrt{2}} \\ -\sqrt{1-\mathrm{x}^2} & , \quad \frac{1}{\sqrt{2}}\lt \mathrm{x}\lt 1\end{array}\right.\) From the figure, it is clear that function have sharp edges, at \(x=-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\) \(\therefore\) function is not differentiable at 3 points
JEE Main-2019-10.01.2019
Limits, Continuity and Differentiability
80240
Let \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)\) \(=0\). If \(g(x)=|f(x)|\), then at \(x=c, g\) is
1 not differentiable
2 differentiable if \(\mathrm{f}^{\prime}(\mathrm{c}) \neq 0\)
3 not differentiable if \(f^{\prime}(\mathrm{c})=0\)
4 differentiable if \(f^{\prime}(c)=0\)
Explanation:
(D) : Given function, \(\mathrm{g}(\mathrm{x})=|\mathrm{f}(\mathrm{x})|\) Where \(f: R \rightarrow R\) be differentiable at \(c \in R\) and \(f(c)=0\), then for function \(g(x)\) at \(x=c\), \(g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} \quad[\text { where, } h>0]\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}\) \(=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h} \quad[\text { as } f(c)=0(\text { given })]\) \(=\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \quad[\because h>0]\) \(=\left|\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\right|=\left|f^{\prime}(c)\right|\) \([\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=\mathrm{c}]\) Now if \(\mathrm{f}^{\prime}(\mathrm{c})=0\) then \(\mathrm{g}(\mathrm{x})\) is differentiable at \(\mathrm{x}=\mathrm{c}\), otherwise LHD (at \(x=c\) ) and RHD (at \(x=c\) ) is different.
