(B) : Given function, \(y=a \log _{e} x+b x^{2}+x\) Differentiating with respect to \(x\) we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times \frac{1}{\mathrm{x}}+2 \mathrm{bx}+1\) For maxima or minima \(\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1=0\) For given condition, the function has extremum at point \(\mathrm{x}_{1}=1\) and \(\mathrm{x}_{2}=2\) Now, at \(\mathrm{x}_{1}=1\) then \(a+2 b+1=0 \tag{i}\) And, at \(\mathrm{x}_{2}=2\) \(\frac{a}{2}+2 b \times 2+1=0\) \(a+8 b+2=0 \tag{ii}\) On solving (i) and (ii) we get- \(\mathrm{a}=-\frac{2}{3}\) \(3 a+2=0\) Putting the value of a in equation (i) \(-\frac{2}{3}+2 b+1=0\) \(2 b=-\frac{1}{3} \Rightarrow b=-\frac{1}{6}\) Hence, \(a=-\frac{2}{3}\) and \(b=-\frac{1}{6}\)
UPSEE-2017
Application of Derivatives
85649
The maximum value of the function \(y=2 \tan x-\tan ^{2} x \operatorname{over}\left[0, \frac{\pi}{2}\right]\) is
1 \(\infty\)
2 1
3 3
4 2
Explanation:
(B) : \(y=2 \tan x-\tan ^{2} x, \quad x \in\left[0, \frac{\pi}{2}\right]\) \(\frac{d y}{d x}=2 \sec ^{2} x-2 \tan x \cdot \sec ^{2} x\) \(=2 \sec ^{2} x(1-\tan x)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2 \sec ^{2} x(1-\tan x)=0\) \(\sec x=0 \text { or, } 1-\tan x=0\) \(\sec x \neq 0, \text { or } x=\frac{\pi}{4}\) Now, \(\frac{d^{2} y}{d x^{2}}=4 \sec ^{2} x \cdot \tan x(1-\tan x)-2 \sec ^{4} x\) At \(x=\frac{\pi}{4}, \frac{d^{2} y}{d^{2}}=4 \sec ^{2} \frac{\pi}{4} \cdot \tan \frac{\pi}{4}\left(1-\tan \frac{\pi}{4}\right)-2 \sec ^{4} \frac{\pi}{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4(\sqrt{2})^{2} \cdot 1(1-1)-2(\sqrt{2})^{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-8\lt 0\) \(\because \mathrm{x}=\frac{\pi}{4}\) is point of local maxima. \(\therefore\) maximum value of function at point \(\mathrm{x}=\frac{\pi}{4}\) \(y=2 \tan \frac{\pi}{4}-\tan ^{2} \frac{\pi}{4}\) \(=2(1)-(1)^{2}=1\)
(B) : Given function, \(y=a \log _{e} x+b x^{2}+x\) Differentiating with respect to \(x\) we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times \frac{1}{\mathrm{x}}+2 \mathrm{bx}+1\) For maxima or minima \(\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1=0\) For given condition, the function has extremum at point \(\mathrm{x}_{1}=1\) and \(\mathrm{x}_{2}=2\) Now, at \(\mathrm{x}_{1}=1\) then \(a+2 b+1=0 \tag{i}\) And, at \(\mathrm{x}_{2}=2\) \(\frac{a}{2}+2 b \times 2+1=0\) \(a+8 b+2=0 \tag{ii}\) On solving (i) and (ii) we get- \(\mathrm{a}=-\frac{2}{3}\) \(3 a+2=0\) Putting the value of a in equation (i) \(-\frac{2}{3}+2 b+1=0\) \(2 b=-\frac{1}{3} \Rightarrow b=-\frac{1}{6}\) Hence, \(a=-\frac{2}{3}\) and \(b=-\frac{1}{6}\)
UPSEE-2017
Application of Derivatives
85649
The maximum value of the function \(y=2 \tan x-\tan ^{2} x \operatorname{over}\left[0, \frac{\pi}{2}\right]\) is
1 \(\infty\)
2 1
3 3
4 2
Explanation:
(B) : \(y=2 \tan x-\tan ^{2} x, \quad x \in\left[0, \frac{\pi}{2}\right]\) \(\frac{d y}{d x}=2 \sec ^{2} x-2 \tan x \cdot \sec ^{2} x\) \(=2 \sec ^{2} x(1-\tan x)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2 \sec ^{2} x(1-\tan x)=0\) \(\sec x=0 \text { or, } 1-\tan x=0\) \(\sec x \neq 0, \text { or } x=\frac{\pi}{4}\) Now, \(\frac{d^{2} y}{d x^{2}}=4 \sec ^{2} x \cdot \tan x(1-\tan x)-2 \sec ^{4} x\) At \(x=\frac{\pi}{4}, \frac{d^{2} y}{d^{2}}=4 \sec ^{2} \frac{\pi}{4} \cdot \tan \frac{\pi}{4}\left(1-\tan \frac{\pi}{4}\right)-2 \sec ^{4} \frac{\pi}{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4(\sqrt{2})^{2} \cdot 1(1-1)-2(\sqrt{2})^{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-8\lt 0\) \(\because \mathrm{x}=\frac{\pi}{4}\) is point of local maxima. \(\therefore\) maximum value of function at point \(\mathrm{x}=\frac{\pi}{4}\) \(y=2 \tan \frac{\pi}{4}-\tan ^{2} \frac{\pi}{4}\) \(=2(1)-(1)^{2}=1\)
(B) : Given function, \(y=a \log _{e} x+b x^{2}+x\) Differentiating with respect to \(x\) we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times \frac{1}{\mathrm{x}}+2 \mathrm{bx}+1\) For maxima or minima \(\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1=0\) For given condition, the function has extremum at point \(\mathrm{x}_{1}=1\) and \(\mathrm{x}_{2}=2\) Now, at \(\mathrm{x}_{1}=1\) then \(a+2 b+1=0 \tag{i}\) And, at \(\mathrm{x}_{2}=2\) \(\frac{a}{2}+2 b \times 2+1=0\) \(a+8 b+2=0 \tag{ii}\) On solving (i) and (ii) we get- \(\mathrm{a}=-\frac{2}{3}\) \(3 a+2=0\) Putting the value of a in equation (i) \(-\frac{2}{3}+2 b+1=0\) \(2 b=-\frac{1}{3} \Rightarrow b=-\frac{1}{6}\) Hence, \(a=-\frac{2}{3}\) and \(b=-\frac{1}{6}\)
UPSEE-2017
Application of Derivatives
85649
The maximum value of the function \(y=2 \tan x-\tan ^{2} x \operatorname{over}\left[0, \frac{\pi}{2}\right]\) is
1 \(\infty\)
2 1
3 3
4 2
Explanation:
(B) : \(y=2 \tan x-\tan ^{2} x, \quad x \in\left[0, \frac{\pi}{2}\right]\) \(\frac{d y}{d x}=2 \sec ^{2} x-2 \tan x \cdot \sec ^{2} x\) \(=2 \sec ^{2} x(1-\tan x)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2 \sec ^{2} x(1-\tan x)=0\) \(\sec x=0 \text { or, } 1-\tan x=0\) \(\sec x \neq 0, \text { or } x=\frac{\pi}{4}\) Now, \(\frac{d^{2} y}{d x^{2}}=4 \sec ^{2} x \cdot \tan x(1-\tan x)-2 \sec ^{4} x\) At \(x=\frac{\pi}{4}, \frac{d^{2} y}{d^{2}}=4 \sec ^{2} \frac{\pi}{4} \cdot \tan \frac{\pi}{4}\left(1-\tan \frac{\pi}{4}\right)-2 \sec ^{4} \frac{\pi}{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4(\sqrt{2})^{2} \cdot 1(1-1)-2(\sqrt{2})^{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-8\lt 0\) \(\because \mathrm{x}=\frac{\pi}{4}\) is point of local maxima. \(\therefore\) maximum value of function at point \(\mathrm{x}=\frac{\pi}{4}\) \(y=2 \tan \frac{\pi}{4}-\tan ^{2} \frac{\pi}{4}\) \(=2(1)-(1)^{2}=1\)
(B) : Given function, \(y=a \log _{e} x+b x^{2}+x\) Differentiating with respect to \(x\) we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times \frac{1}{\mathrm{x}}+2 \mathrm{bx}+1\) For maxima or minima \(\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1=0\) For given condition, the function has extremum at point \(\mathrm{x}_{1}=1\) and \(\mathrm{x}_{2}=2\) Now, at \(\mathrm{x}_{1}=1\) then \(a+2 b+1=0 \tag{i}\) And, at \(\mathrm{x}_{2}=2\) \(\frac{a}{2}+2 b \times 2+1=0\) \(a+8 b+2=0 \tag{ii}\) On solving (i) and (ii) we get- \(\mathrm{a}=-\frac{2}{3}\) \(3 a+2=0\) Putting the value of a in equation (i) \(-\frac{2}{3}+2 b+1=0\) \(2 b=-\frac{1}{3} \Rightarrow b=-\frac{1}{6}\) Hence, \(a=-\frac{2}{3}\) and \(b=-\frac{1}{6}\)
UPSEE-2017
Application of Derivatives
85649
The maximum value of the function \(y=2 \tan x-\tan ^{2} x \operatorname{over}\left[0, \frac{\pi}{2}\right]\) is
1 \(\infty\)
2 1
3 3
4 2
Explanation:
(B) : \(y=2 \tan x-\tan ^{2} x, \quad x \in\left[0, \frac{\pi}{2}\right]\) \(\frac{d y}{d x}=2 \sec ^{2} x-2 \tan x \cdot \sec ^{2} x\) \(=2 \sec ^{2} x(1-\tan x)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2 \sec ^{2} x(1-\tan x)=0\) \(\sec x=0 \text { or, } 1-\tan x=0\) \(\sec x \neq 0, \text { or } x=\frac{\pi}{4}\) Now, \(\frac{d^{2} y}{d x^{2}}=4 \sec ^{2} x \cdot \tan x(1-\tan x)-2 \sec ^{4} x\) At \(x=\frac{\pi}{4}, \frac{d^{2} y}{d^{2}}=4 \sec ^{2} \frac{\pi}{4} \cdot \tan \frac{\pi}{4}\left(1-\tan \frac{\pi}{4}\right)-2 \sec ^{4} \frac{\pi}{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4(\sqrt{2})^{2} \cdot 1(1-1)-2(\sqrt{2})^{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-8\lt 0\) \(\because \mathrm{x}=\frac{\pi}{4}\) is point of local maxima. \(\therefore\) maximum value of function at point \(\mathrm{x}=\frac{\pi}{4}\) \(y=2 \tan \frac{\pi}{4}-\tan ^{2} \frac{\pi}{4}\) \(=2(1)-(1)^{2}=1\)
(B) : Given function, \(y=a \log _{e} x+b x^{2}+x\) Differentiating with respect to \(x\) we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times \frac{1}{\mathrm{x}}+2 \mathrm{bx}+1\) For maxima or minima \(\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1=0\) For given condition, the function has extremum at point \(\mathrm{x}_{1}=1\) and \(\mathrm{x}_{2}=2\) Now, at \(\mathrm{x}_{1}=1\) then \(a+2 b+1=0 \tag{i}\) And, at \(\mathrm{x}_{2}=2\) \(\frac{a}{2}+2 b \times 2+1=0\) \(a+8 b+2=0 \tag{ii}\) On solving (i) and (ii) we get- \(\mathrm{a}=-\frac{2}{3}\) \(3 a+2=0\) Putting the value of a in equation (i) \(-\frac{2}{3}+2 b+1=0\) \(2 b=-\frac{1}{3} \Rightarrow b=-\frac{1}{6}\) Hence, \(a=-\frac{2}{3}\) and \(b=-\frac{1}{6}\)
UPSEE-2017
Application of Derivatives
85649
The maximum value of the function \(y=2 \tan x-\tan ^{2} x \operatorname{over}\left[0, \frac{\pi}{2}\right]\) is
1 \(\infty\)
2 1
3 3
4 2
Explanation:
(B) : \(y=2 \tan x-\tan ^{2} x, \quad x \in\left[0, \frac{\pi}{2}\right]\) \(\frac{d y}{d x}=2 \sec ^{2} x-2 \tan x \cdot \sec ^{2} x\) \(=2 \sec ^{2} x(1-\tan x)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2 \sec ^{2} x(1-\tan x)=0\) \(\sec x=0 \text { or, } 1-\tan x=0\) \(\sec x \neq 0, \text { or } x=\frac{\pi}{4}\) Now, \(\frac{d^{2} y}{d x^{2}}=4 \sec ^{2} x \cdot \tan x(1-\tan x)-2 \sec ^{4} x\) At \(x=\frac{\pi}{4}, \frac{d^{2} y}{d^{2}}=4 \sec ^{2} \frac{\pi}{4} \cdot \tan \frac{\pi}{4}\left(1-\tan \frac{\pi}{4}\right)-2 \sec ^{4} \frac{\pi}{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4(\sqrt{2})^{2} \cdot 1(1-1)-2(\sqrt{2})^{4}\) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-8\lt 0\) \(\because \mathrm{x}=\frac{\pi}{4}\) is point of local maxima. \(\therefore\) maximum value of function at point \(\mathrm{x}=\frac{\pi}{4}\) \(y=2 \tan \frac{\pi}{4}-\tan ^{2} \frac{\pi}{4}\) \(=2(1)-(1)^{2}=1\)
