85628
The maximum value of \(f(x)=e^{\sin x}+e^{\cos x} ; x \in R\) is
1 \(2 \mathrm{e}\)
2 \(2 \sqrt{\mathrm{e}}\)
3 \(2 \mathrm{e}^{1 / \sqrt{2}}\)
4 \(2 \mathrm{e}^{-1 / \sqrt{2}}\)
Explanation:
(C) : Given, \(f(x)=e^{\sin x}+e^{\cos x}, x \in R\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\sin x} \cdot(\cos x)+e^{\cos x} \cdot(-\sin x)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{e}^{\sin x} \cdot(\cos x)=e^{\cos x}(\sin x)\) \(e^{(\sin x-\cos x)}=\tan x\) \(\tan x=1\) \(x=\frac{\pi}{4}\) Now maximum value \(f\left(\frac{\pi}{4}\right)=e^{\sin \frac{\pi}{4}}+e^{\cos \frac{\pi}{4}}=e^{\frac{1}{\sqrt{2}}}+e^{\frac{1}{\sqrt{2}}}\) \(f\left(\frac{\pi}{4}\right)=2 e^{\frac{1}{\sqrt{2}}}\)
WB JEE-2022
Application of Derivatives
85629
Let \(f(x)=(x-2)^{17}(x+5)^{24}\), then
1 f does not have a critical point at \(x=2\)
2 f has a minimum at \(x=2\)
3 \(f\) has neither a maximum nor a minimum at \(x\) \(=2\)
4 f has a maximum at \(\mathrm{x}=2\)
Explanation:
(C) : We have \(f(x)=(x-2)^{17}(x+5)^{24}\) On differentiating w.r.t \(x\) we get- \(f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+(x-2)^{17} \cdot(24)(x+5)^{23}\) \(=(\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]=0\) \(17 \mathrm{x}+85+24 \mathrm{x}-48=0\) \(41 \mathrm{x}=-37\) \(\mathrm{x}=-\frac{37}{41}\) \(\mathrm{f}^{\prime}\left(2^{+}\right)=\text {positive or } \mathrm{f}^{\prime}(2)>0\) \(\mathrm{f}^{\prime}\left(2^{-}\right)=\text {positive or } \mathrm{f}^{\prime}\left(2^{-}\right)>0\) \(\therefore \quad \mathrm{x}=2 \text { is neither maximum nor minimum. }\)
WB JEE-2022
Application of Derivatives
85630
Let \(\mathbf{f}: R \rightarrow R\) be given \(f(x)=\left|\mathbf{x}^{2}-1\right|, \mathbf{x} \in \mathbb{R}\). Then
1 F has a local minimum at \(\mathrm{x}= \pm 1\) but no local maximum.
2 \(\mathrm{F}\) has a local maximum at \(\mathrm{x}=0\) but on local minimum.
3 F has a local maximum at \(\mathrm{x}= \pm 1\) \& a local maxima at \(\mathrm{x}=0\)
4 F has neither a local maxima nor a local minimum at any point.
Explanation:
(C): \(\mathrm{f}\) has a local minima at \(\mathrm{x}= \pm 1\), and a local maxima at \(\mathrm{x}\) \(=0\) Hence, correct option is (c).
WB JEE-2021
Application of Derivatives
85631
Let \(\mathrm{f}(\mathrm{x})=1-\sqrt{\left(\mathrm{x}^{2}\right)}\), where the square root is to be taken positive, then
1 f has no extreme at \(x=0\)
2 f has minima at \(x=0\)
3 f has maxima at \(\mathrm{x}=0\)
4 \(\mathrm{f}^{\prime}\) exists at 0
Explanation:
(C) : We have given, \(f(x)=1-\sqrt{\left(x^{2}\right)}\) \(f(x)=1-|x|\) \(f(x) \begin{cases}1-x x \geq 0 \\ 1+x x\lt 0\end{cases}\) Differenting with respect to \(x\) we get- \(f^{\prime}(x)\left\{\begin{array}{cc}-1 & x \geq 0(\text { maxima }) \\ 1 & x\lt 0(\text { minima })\end{array}\right.\) Maximum value At, \(\quad \mathrm{x}=-1\) \(\mathrm{f}(\mathrm{x})=1-\mathrm{x}\) \(\mathrm{f}(-1)=1-(-1)\) \(=2\) \(\mathrm{f}(1)=1-(1)\) \(\mathrm{f}(1)=20\) \(\mathrm{f}(\mathrm{x})\) has maxima at \(\mathrm{x}=0\)
WB JEE-2020
Application of Derivatives
85632
If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\) attains its maximum and minimum at \(p\) and \(q\) respectively, such that \(p^{2}=q\), than a equals to
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
(A) : Given, function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2}+1\) On differentiating with respect to ' \(x\) ' we get- \(f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2}=0\) Again differentiating with respect ' \(x\) ' we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-18 \mathrm{a}\) For maximum or minimum \(f^{\prime}(x)=0\) \(6 x^{2}-18 a x+12 a^{2}=0\) \(6 x^{2}-6 a x-12 a x+12 a^{2}=0\) \(6 x(x-a)-12 a(x-a)=0\) \((6 x-12 a)(x-a)=0\) \(x=2 a, a\) \(f^{\prime \prime}(a)=12 a-18 a=-6 a\lt 0 \text { (maxima) }\) \(f^{\prime \prime}(2 a)=24 a-18 a=6 a>0(\text { minima) }\) Given, condition maximum \((p)=a\) And minimum \(\mathrm{q}=2 \mathrm{a}\) \(\Rightarrow \quad \mathrm{p}^{2}=\mathrm{q}\) \(\mathrm{a}^{2}=2 \mathrm{a}\) \(\mathrm{a}^{2}-2 \mathrm{a}=0\) \(\mathrm{a}(\mathrm{a}-2)=0\) \(\mathrm{a}=0,2\) \(\mathrm{a}=2 \quad[\mathrm{a}>0]\)
85628
The maximum value of \(f(x)=e^{\sin x}+e^{\cos x} ; x \in R\) is
1 \(2 \mathrm{e}\)
2 \(2 \sqrt{\mathrm{e}}\)
3 \(2 \mathrm{e}^{1 / \sqrt{2}}\)
4 \(2 \mathrm{e}^{-1 / \sqrt{2}}\)
Explanation:
(C) : Given, \(f(x)=e^{\sin x}+e^{\cos x}, x \in R\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\sin x} \cdot(\cos x)+e^{\cos x} \cdot(-\sin x)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{e}^{\sin x} \cdot(\cos x)=e^{\cos x}(\sin x)\) \(e^{(\sin x-\cos x)}=\tan x\) \(\tan x=1\) \(x=\frac{\pi}{4}\) Now maximum value \(f\left(\frac{\pi}{4}\right)=e^{\sin \frac{\pi}{4}}+e^{\cos \frac{\pi}{4}}=e^{\frac{1}{\sqrt{2}}}+e^{\frac{1}{\sqrt{2}}}\) \(f\left(\frac{\pi}{4}\right)=2 e^{\frac{1}{\sqrt{2}}}\)
WB JEE-2022
Application of Derivatives
85629
Let \(f(x)=(x-2)^{17}(x+5)^{24}\), then
1 f does not have a critical point at \(x=2\)
2 f has a minimum at \(x=2\)
3 \(f\) has neither a maximum nor a minimum at \(x\) \(=2\)
4 f has a maximum at \(\mathrm{x}=2\)
Explanation:
(C) : We have \(f(x)=(x-2)^{17}(x+5)^{24}\) On differentiating w.r.t \(x\) we get- \(f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+(x-2)^{17} \cdot(24)(x+5)^{23}\) \(=(\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]=0\) \(17 \mathrm{x}+85+24 \mathrm{x}-48=0\) \(41 \mathrm{x}=-37\) \(\mathrm{x}=-\frac{37}{41}\) \(\mathrm{f}^{\prime}\left(2^{+}\right)=\text {positive or } \mathrm{f}^{\prime}(2)>0\) \(\mathrm{f}^{\prime}\left(2^{-}\right)=\text {positive or } \mathrm{f}^{\prime}\left(2^{-}\right)>0\) \(\therefore \quad \mathrm{x}=2 \text { is neither maximum nor minimum. }\)
WB JEE-2022
Application of Derivatives
85630
Let \(\mathbf{f}: R \rightarrow R\) be given \(f(x)=\left|\mathbf{x}^{2}-1\right|, \mathbf{x} \in \mathbb{R}\). Then
1 F has a local minimum at \(\mathrm{x}= \pm 1\) but no local maximum.
2 \(\mathrm{F}\) has a local maximum at \(\mathrm{x}=0\) but on local minimum.
3 F has a local maximum at \(\mathrm{x}= \pm 1\) \& a local maxima at \(\mathrm{x}=0\)
4 F has neither a local maxima nor a local minimum at any point.
Explanation:
(C): \(\mathrm{f}\) has a local minima at \(\mathrm{x}= \pm 1\), and a local maxima at \(\mathrm{x}\) \(=0\) Hence, correct option is (c).
WB JEE-2021
Application of Derivatives
85631
Let \(\mathrm{f}(\mathrm{x})=1-\sqrt{\left(\mathrm{x}^{2}\right)}\), where the square root is to be taken positive, then
1 f has no extreme at \(x=0\)
2 f has minima at \(x=0\)
3 f has maxima at \(\mathrm{x}=0\)
4 \(\mathrm{f}^{\prime}\) exists at 0
Explanation:
(C) : We have given, \(f(x)=1-\sqrt{\left(x^{2}\right)}\) \(f(x)=1-|x|\) \(f(x) \begin{cases}1-x x \geq 0 \\ 1+x x\lt 0\end{cases}\) Differenting with respect to \(x\) we get- \(f^{\prime}(x)\left\{\begin{array}{cc}-1 & x \geq 0(\text { maxima }) \\ 1 & x\lt 0(\text { minima })\end{array}\right.\) Maximum value At, \(\quad \mathrm{x}=-1\) \(\mathrm{f}(\mathrm{x})=1-\mathrm{x}\) \(\mathrm{f}(-1)=1-(-1)\) \(=2\) \(\mathrm{f}(1)=1-(1)\) \(\mathrm{f}(1)=20\) \(\mathrm{f}(\mathrm{x})\) has maxima at \(\mathrm{x}=0\)
WB JEE-2020
Application of Derivatives
85632
If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\) attains its maximum and minimum at \(p\) and \(q\) respectively, such that \(p^{2}=q\), than a equals to
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
(A) : Given, function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2}+1\) On differentiating with respect to ' \(x\) ' we get- \(f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2}=0\) Again differentiating with respect ' \(x\) ' we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-18 \mathrm{a}\) For maximum or minimum \(f^{\prime}(x)=0\) \(6 x^{2}-18 a x+12 a^{2}=0\) \(6 x^{2}-6 a x-12 a x+12 a^{2}=0\) \(6 x(x-a)-12 a(x-a)=0\) \((6 x-12 a)(x-a)=0\) \(x=2 a, a\) \(f^{\prime \prime}(a)=12 a-18 a=-6 a\lt 0 \text { (maxima) }\) \(f^{\prime \prime}(2 a)=24 a-18 a=6 a>0(\text { minima) }\) Given, condition maximum \((p)=a\) And minimum \(\mathrm{q}=2 \mathrm{a}\) \(\Rightarrow \quad \mathrm{p}^{2}=\mathrm{q}\) \(\mathrm{a}^{2}=2 \mathrm{a}\) \(\mathrm{a}^{2}-2 \mathrm{a}=0\) \(\mathrm{a}(\mathrm{a}-2)=0\) \(\mathrm{a}=0,2\) \(\mathrm{a}=2 \quad[\mathrm{a}>0]\)
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Application of Derivatives
85628
The maximum value of \(f(x)=e^{\sin x}+e^{\cos x} ; x \in R\) is
1 \(2 \mathrm{e}\)
2 \(2 \sqrt{\mathrm{e}}\)
3 \(2 \mathrm{e}^{1 / \sqrt{2}}\)
4 \(2 \mathrm{e}^{-1 / \sqrt{2}}\)
Explanation:
(C) : Given, \(f(x)=e^{\sin x}+e^{\cos x}, x \in R\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\sin x} \cdot(\cos x)+e^{\cos x} \cdot(-\sin x)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{e}^{\sin x} \cdot(\cos x)=e^{\cos x}(\sin x)\) \(e^{(\sin x-\cos x)}=\tan x\) \(\tan x=1\) \(x=\frac{\pi}{4}\) Now maximum value \(f\left(\frac{\pi}{4}\right)=e^{\sin \frac{\pi}{4}}+e^{\cos \frac{\pi}{4}}=e^{\frac{1}{\sqrt{2}}}+e^{\frac{1}{\sqrt{2}}}\) \(f\left(\frac{\pi}{4}\right)=2 e^{\frac{1}{\sqrt{2}}}\)
WB JEE-2022
Application of Derivatives
85629
Let \(f(x)=(x-2)^{17}(x+5)^{24}\), then
1 f does not have a critical point at \(x=2\)
2 f has a minimum at \(x=2\)
3 \(f\) has neither a maximum nor a minimum at \(x\) \(=2\)
4 f has a maximum at \(\mathrm{x}=2\)
Explanation:
(C) : We have \(f(x)=(x-2)^{17}(x+5)^{24}\) On differentiating w.r.t \(x\) we get- \(f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+(x-2)^{17} \cdot(24)(x+5)^{23}\) \(=(\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]=0\) \(17 \mathrm{x}+85+24 \mathrm{x}-48=0\) \(41 \mathrm{x}=-37\) \(\mathrm{x}=-\frac{37}{41}\) \(\mathrm{f}^{\prime}\left(2^{+}\right)=\text {positive or } \mathrm{f}^{\prime}(2)>0\) \(\mathrm{f}^{\prime}\left(2^{-}\right)=\text {positive or } \mathrm{f}^{\prime}\left(2^{-}\right)>0\) \(\therefore \quad \mathrm{x}=2 \text { is neither maximum nor minimum. }\)
WB JEE-2022
Application of Derivatives
85630
Let \(\mathbf{f}: R \rightarrow R\) be given \(f(x)=\left|\mathbf{x}^{2}-1\right|, \mathbf{x} \in \mathbb{R}\). Then
1 F has a local minimum at \(\mathrm{x}= \pm 1\) but no local maximum.
2 \(\mathrm{F}\) has a local maximum at \(\mathrm{x}=0\) but on local minimum.
3 F has a local maximum at \(\mathrm{x}= \pm 1\) \& a local maxima at \(\mathrm{x}=0\)
4 F has neither a local maxima nor a local minimum at any point.
Explanation:
(C): \(\mathrm{f}\) has a local minima at \(\mathrm{x}= \pm 1\), and a local maxima at \(\mathrm{x}\) \(=0\) Hence, correct option is (c).
WB JEE-2021
Application of Derivatives
85631
Let \(\mathrm{f}(\mathrm{x})=1-\sqrt{\left(\mathrm{x}^{2}\right)}\), where the square root is to be taken positive, then
1 f has no extreme at \(x=0\)
2 f has minima at \(x=0\)
3 f has maxima at \(\mathrm{x}=0\)
4 \(\mathrm{f}^{\prime}\) exists at 0
Explanation:
(C) : We have given, \(f(x)=1-\sqrt{\left(x^{2}\right)}\) \(f(x)=1-|x|\) \(f(x) \begin{cases}1-x x \geq 0 \\ 1+x x\lt 0\end{cases}\) Differenting with respect to \(x\) we get- \(f^{\prime}(x)\left\{\begin{array}{cc}-1 & x \geq 0(\text { maxima }) \\ 1 & x\lt 0(\text { minima })\end{array}\right.\) Maximum value At, \(\quad \mathrm{x}=-1\) \(\mathrm{f}(\mathrm{x})=1-\mathrm{x}\) \(\mathrm{f}(-1)=1-(-1)\) \(=2\) \(\mathrm{f}(1)=1-(1)\) \(\mathrm{f}(1)=20\) \(\mathrm{f}(\mathrm{x})\) has maxima at \(\mathrm{x}=0\)
WB JEE-2020
Application of Derivatives
85632
If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\) attains its maximum and minimum at \(p\) and \(q\) respectively, such that \(p^{2}=q\), than a equals to
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
(A) : Given, function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2}+1\) On differentiating with respect to ' \(x\) ' we get- \(f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2}=0\) Again differentiating with respect ' \(x\) ' we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-18 \mathrm{a}\) For maximum or minimum \(f^{\prime}(x)=0\) \(6 x^{2}-18 a x+12 a^{2}=0\) \(6 x^{2}-6 a x-12 a x+12 a^{2}=0\) \(6 x(x-a)-12 a(x-a)=0\) \((6 x-12 a)(x-a)=0\) \(x=2 a, a\) \(f^{\prime \prime}(a)=12 a-18 a=-6 a\lt 0 \text { (maxima) }\) \(f^{\prime \prime}(2 a)=24 a-18 a=6 a>0(\text { minima) }\) Given, condition maximum \((p)=a\) And minimum \(\mathrm{q}=2 \mathrm{a}\) \(\Rightarrow \quad \mathrm{p}^{2}=\mathrm{q}\) \(\mathrm{a}^{2}=2 \mathrm{a}\) \(\mathrm{a}^{2}-2 \mathrm{a}=0\) \(\mathrm{a}(\mathrm{a}-2)=0\) \(\mathrm{a}=0,2\) \(\mathrm{a}=2 \quad[\mathrm{a}>0]\)
85628
The maximum value of \(f(x)=e^{\sin x}+e^{\cos x} ; x \in R\) is
1 \(2 \mathrm{e}\)
2 \(2 \sqrt{\mathrm{e}}\)
3 \(2 \mathrm{e}^{1 / \sqrt{2}}\)
4 \(2 \mathrm{e}^{-1 / \sqrt{2}}\)
Explanation:
(C) : Given, \(f(x)=e^{\sin x}+e^{\cos x}, x \in R\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\sin x} \cdot(\cos x)+e^{\cos x} \cdot(-\sin x)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{e}^{\sin x} \cdot(\cos x)=e^{\cos x}(\sin x)\) \(e^{(\sin x-\cos x)}=\tan x\) \(\tan x=1\) \(x=\frac{\pi}{4}\) Now maximum value \(f\left(\frac{\pi}{4}\right)=e^{\sin \frac{\pi}{4}}+e^{\cos \frac{\pi}{4}}=e^{\frac{1}{\sqrt{2}}}+e^{\frac{1}{\sqrt{2}}}\) \(f\left(\frac{\pi}{4}\right)=2 e^{\frac{1}{\sqrt{2}}}\)
WB JEE-2022
Application of Derivatives
85629
Let \(f(x)=(x-2)^{17}(x+5)^{24}\), then
1 f does not have a critical point at \(x=2\)
2 f has a minimum at \(x=2\)
3 \(f\) has neither a maximum nor a minimum at \(x\) \(=2\)
4 f has a maximum at \(\mathrm{x}=2\)
Explanation:
(C) : We have \(f(x)=(x-2)^{17}(x+5)^{24}\) On differentiating w.r.t \(x\) we get- \(f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+(x-2)^{17} \cdot(24)(x+5)^{23}\) \(=(\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]=0\) \(17 \mathrm{x}+85+24 \mathrm{x}-48=0\) \(41 \mathrm{x}=-37\) \(\mathrm{x}=-\frac{37}{41}\) \(\mathrm{f}^{\prime}\left(2^{+}\right)=\text {positive or } \mathrm{f}^{\prime}(2)>0\) \(\mathrm{f}^{\prime}\left(2^{-}\right)=\text {positive or } \mathrm{f}^{\prime}\left(2^{-}\right)>0\) \(\therefore \quad \mathrm{x}=2 \text { is neither maximum nor minimum. }\)
WB JEE-2022
Application of Derivatives
85630
Let \(\mathbf{f}: R \rightarrow R\) be given \(f(x)=\left|\mathbf{x}^{2}-1\right|, \mathbf{x} \in \mathbb{R}\). Then
1 F has a local minimum at \(\mathrm{x}= \pm 1\) but no local maximum.
2 \(\mathrm{F}\) has a local maximum at \(\mathrm{x}=0\) but on local minimum.
3 F has a local maximum at \(\mathrm{x}= \pm 1\) \& a local maxima at \(\mathrm{x}=0\)
4 F has neither a local maxima nor a local minimum at any point.
Explanation:
(C): \(\mathrm{f}\) has a local minima at \(\mathrm{x}= \pm 1\), and a local maxima at \(\mathrm{x}\) \(=0\) Hence, correct option is (c).
WB JEE-2021
Application of Derivatives
85631
Let \(\mathrm{f}(\mathrm{x})=1-\sqrt{\left(\mathrm{x}^{2}\right)}\), where the square root is to be taken positive, then
1 f has no extreme at \(x=0\)
2 f has minima at \(x=0\)
3 f has maxima at \(\mathrm{x}=0\)
4 \(\mathrm{f}^{\prime}\) exists at 0
Explanation:
(C) : We have given, \(f(x)=1-\sqrt{\left(x^{2}\right)}\) \(f(x)=1-|x|\) \(f(x) \begin{cases}1-x x \geq 0 \\ 1+x x\lt 0\end{cases}\) Differenting with respect to \(x\) we get- \(f^{\prime}(x)\left\{\begin{array}{cc}-1 & x \geq 0(\text { maxima }) \\ 1 & x\lt 0(\text { minima })\end{array}\right.\) Maximum value At, \(\quad \mathrm{x}=-1\) \(\mathrm{f}(\mathrm{x})=1-\mathrm{x}\) \(\mathrm{f}(-1)=1-(-1)\) \(=2\) \(\mathrm{f}(1)=1-(1)\) \(\mathrm{f}(1)=20\) \(\mathrm{f}(\mathrm{x})\) has maxima at \(\mathrm{x}=0\)
WB JEE-2020
Application of Derivatives
85632
If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\) attains its maximum and minimum at \(p\) and \(q\) respectively, such that \(p^{2}=q\), than a equals to
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
(A) : Given, function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2}+1\) On differentiating with respect to ' \(x\) ' we get- \(f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2}=0\) Again differentiating with respect ' \(x\) ' we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-18 \mathrm{a}\) For maximum or minimum \(f^{\prime}(x)=0\) \(6 x^{2}-18 a x+12 a^{2}=0\) \(6 x^{2}-6 a x-12 a x+12 a^{2}=0\) \(6 x(x-a)-12 a(x-a)=0\) \((6 x-12 a)(x-a)=0\) \(x=2 a, a\) \(f^{\prime \prime}(a)=12 a-18 a=-6 a\lt 0 \text { (maxima) }\) \(f^{\prime \prime}(2 a)=24 a-18 a=6 a>0(\text { minima) }\) Given, condition maximum \((p)=a\) And minimum \(\mathrm{q}=2 \mathrm{a}\) \(\Rightarrow \quad \mathrm{p}^{2}=\mathrm{q}\) \(\mathrm{a}^{2}=2 \mathrm{a}\) \(\mathrm{a}^{2}-2 \mathrm{a}=0\) \(\mathrm{a}(\mathrm{a}-2)=0\) \(\mathrm{a}=0,2\) \(\mathrm{a}=2 \quad[\mathrm{a}>0]\)
85628
The maximum value of \(f(x)=e^{\sin x}+e^{\cos x} ; x \in R\) is
1 \(2 \mathrm{e}\)
2 \(2 \sqrt{\mathrm{e}}\)
3 \(2 \mathrm{e}^{1 / \sqrt{2}}\)
4 \(2 \mathrm{e}^{-1 / \sqrt{2}}\)
Explanation:
(C) : Given, \(f(x)=e^{\sin x}+e^{\cos x}, x \in R\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\sin x} \cdot(\cos x)+e^{\cos x} \cdot(-\sin x)\) For maxima and minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{e}^{\sin x} \cdot(\cos x)=e^{\cos x}(\sin x)\) \(e^{(\sin x-\cos x)}=\tan x\) \(\tan x=1\) \(x=\frac{\pi}{4}\) Now maximum value \(f\left(\frac{\pi}{4}\right)=e^{\sin \frac{\pi}{4}}+e^{\cos \frac{\pi}{4}}=e^{\frac{1}{\sqrt{2}}}+e^{\frac{1}{\sqrt{2}}}\) \(f\left(\frac{\pi}{4}\right)=2 e^{\frac{1}{\sqrt{2}}}\)
WB JEE-2022
Application of Derivatives
85629
Let \(f(x)=(x-2)^{17}(x+5)^{24}\), then
1 f does not have a critical point at \(x=2\)
2 f has a minimum at \(x=2\)
3 \(f\) has neither a maximum nor a minimum at \(x\) \(=2\)
4 f has a maximum at \(\mathrm{x}=2\)
Explanation:
(C) : We have \(f(x)=(x-2)^{17}(x+5)^{24}\) On differentiating w.r.t \(x\) we get- \(f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+(x-2)^{17} \cdot(24)(x+5)^{23}\) \(=(\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-2)^{16}(\mathrm{x}+5)^{23}[17(\mathrm{x}+5)+24(\mathrm{x}-2)]=0\) \(17 \mathrm{x}+85+24 \mathrm{x}-48=0\) \(41 \mathrm{x}=-37\) \(\mathrm{x}=-\frac{37}{41}\) \(\mathrm{f}^{\prime}\left(2^{+}\right)=\text {positive or } \mathrm{f}^{\prime}(2)>0\) \(\mathrm{f}^{\prime}\left(2^{-}\right)=\text {positive or } \mathrm{f}^{\prime}\left(2^{-}\right)>0\) \(\therefore \quad \mathrm{x}=2 \text { is neither maximum nor minimum. }\)
WB JEE-2022
Application of Derivatives
85630
Let \(\mathbf{f}: R \rightarrow R\) be given \(f(x)=\left|\mathbf{x}^{2}-1\right|, \mathbf{x} \in \mathbb{R}\). Then
1 F has a local minimum at \(\mathrm{x}= \pm 1\) but no local maximum.
2 \(\mathrm{F}\) has a local maximum at \(\mathrm{x}=0\) but on local minimum.
3 F has a local maximum at \(\mathrm{x}= \pm 1\) \& a local maxima at \(\mathrm{x}=0\)
4 F has neither a local maxima nor a local minimum at any point.
Explanation:
(C): \(\mathrm{f}\) has a local minima at \(\mathrm{x}= \pm 1\), and a local maxima at \(\mathrm{x}\) \(=0\) Hence, correct option is (c).
WB JEE-2021
Application of Derivatives
85631
Let \(\mathrm{f}(\mathrm{x})=1-\sqrt{\left(\mathrm{x}^{2}\right)}\), where the square root is to be taken positive, then
1 f has no extreme at \(x=0\)
2 f has minima at \(x=0\)
3 f has maxima at \(\mathrm{x}=0\)
4 \(\mathrm{f}^{\prime}\) exists at 0
Explanation:
(C) : We have given, \(f(x)=1-\sqrt{\left(x^{2}\right)}\) \(f(x)=1-|x|\) \(f(x) \begin{cases}1-x x \geq 0 \\ 1+x x\lt 0\end{cases}\) Differenting with respect to \(x\) we get- \(f^{\prime}(x)\left\{\begin{array}{cc}-1 & x \geq 0(\text { maxima }) \\ 1 & x\lt 0(\text { minima })\end{array}\right.\) Maximum value At, \(\quad \mathrm{x}=-1\) \(\mathrm{f}(\mathrm{x})=1-\mathrm{x}\) \(\mathrm{f}(-1)=1-(-1)\) \(=2\) \(\mathrm{f}(1)=1-(1)\) \(\mathrm{f}(1)=20\) \(\mathrm{f}(\mathrm{x})\) has maxima at \(\mathrm{x}=0\)
WB JEE-2020
Application of Derivatives
85632
If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\) attains its maximum and minimum at \(p\) and \(q\) respectively, such that \(p^{2}=q\), than a equals to
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
(A) : Given, function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2}+1\) On differentiating with respect to ' \(x\) ' we get- \(f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2}=0\) Again differentiating with respect ' \(x\) ' we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-18 \mathrm{a}\) For maximum or minimum \(f^{\prime}(x)=0\) \(6 x^{2}-18 a x+12 a^{2}=0\) \(6 x^{2}-6 a x-12 a x+12 a^{2}=0\) \(6 x(x-a)-12 a(x-a)=0\) \((6 x-12 a)(x-a)=0\) \(x=2 a, a\) \(f^{\prime \prime}(a)=12 a-18 a=-6 a\lt 0 \text { (maxima) }\) \(f^{\prime \prime}(2 a)=24 a-18 a=6 a>0(\text { minima) }\) Given, condition maximum \((p)=a\) And minimum \(\mathrm{q}=2 \mathrm{a}\) \(\Rightarrow \quad \mathrm{p}^{2}=\mathrm{q}\) \(\mathrm{a}^{2}=2 \mathrm{a}\) \(\mathrm{a}^{2}-2 \mathrm{a}=0\) \(\mathrm{a}(\mathrm{a}-2)=0\) \(\mathrm{a}=0,2\) \(\mathrm{a}=2 \quad[\mathrm{a}>0]\)
