85592
For all positive value of \(x\) and \(y\), the value of \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y} \text { is: }\)
1 \(\leq 9\)
2 \(\lt 9\)
3 \(\geq 9\)
4 \(>9\)
Explanation:
(C) : Given, For all positive value of \(x\) and \(y\) We have, \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y}\) We can write as, \(\frac{1+x+x^{2}}{x y}=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\) We know that, A.M \(\geq\) G.M \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}\) \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq 3\left(\frac{1}{y^{3}}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \frac{3}{y} \tag{i}\) Similarly, \(\frac{1+y+y^{2}}{x y}=\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\) \(\frac{\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}}{3} \geq\left(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x} \geq \frac{3}{x} \tag{ii}\) On multiplying equation (i) and (ii) we get- \(\left(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\right)\left(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\right) \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)^{2}} \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)} \geq 9\)
Karnataka CET-2002
Application of Derivatives
85593
The maximum of the function \(3 \cos x-4 \sin x\) is:
1 2
2 3
3 4
4 5
Explanation:
(D) : Given, \(f(x)=3 \cos x-4 \sin x\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \cos x+b \sin x \leq \sqrt{a^{2}+b^{2}}\) \(a=3 \text { and } b=-4\) For maximum \(3 \cos x-4 \sin x \leq \sqrt{(3)^{2}+(-4)^{2}}\) \(3 \cos x-4 \sin x \leq \sqrt{25}\) \(3 \cos x-4 \sin x \leq 5\)
Karnataka CET-2004
Application of Derivatives
85594
The maximum slope of the curve \(y=-x^{3}+3 x^{2}+2 x-27 i\)
1 1
2 23
3 5
4 -23
Explanation:
(C) : Given curve, \(y=-x^{2}+3 x^{2}+2 x-27\) On differentiating both sides w.r.t \(x\) we get- \(\frac{d y}{d x}=-3 x^{2}+6 x+2\) For maximum slope \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Slope \((m)=-3 x^{2}+6 x+2=0\) \(\frac{d(m)}{d x}=-6 x+6 \Rightarrow \frac{d^{2}(m)}{d x^{2}}=-6\) For maximum, \(\frac{d(m)}{d x}=0\) \(-6 x+6=0\) \(x=1\) Slope \((m)=-3 x^{2}+6 x+2\) Value of maximum slope at \(\mathrm{x}=1\) \(\mathrm{m}=-3(1)^{2}+6(1)+2\) \(=-3+6+2\) \(\mathrm{~m}_{\max }=5\)
Application of Derivatives
85595
The minimum value of \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) if \(\mathbf{a}^{2}>\mathbf{b}^{2}\), is
1 \(b^{2}\)
2 \(a^{2}-b^{2}\)
3 \(a^{2}\)
4 \(a^{2}+b^{2}\)
Explanation:
(A) : Given, \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) On differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime}(\mathrm{x}) =\mathrm{a}^{2} 2 \cos \mathrm{x} \cdot(-\sin \mathrm{x})+\mathrm{b}^{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=-\mathrm{a}^{2} \sin 2 \mathrm{x}+\mathrm{b}^{2} \sin 2 \mathrm{x}\) \(=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) \sin 2 \mathrm{x}\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\left(b^{2}-a^{2}\right) \sin 2 x=0\) \(\sin 2 x=0\) \(2 x=0\) \(x=0\) \(f(0)=a^{2}(1)^{2}+b^{2}(0)\) \(f(0)=a^{2}(\text { maxima })\) Again on differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2 \cos 2 \mathrm{x}\) \(\mathrm{f}^{\prime \prime}(0)=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2\lt 0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{a}^{2}(0)+\mathrm{b}^{2}(1)\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}(\text { minima })\) \(\mathrm{f}^{\prime \prime}\left(\frac{\pi}{2}\right)=2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\) \(>0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}\) Hence, minimum value is \(b^{2}\)
85592
For all positive value of \(x\) and \(y\), the value of \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y} \text { is: }\)
1 \(\leq 9\)
2 \(\lt 9\)
3 \(\geq 9\)
4 \(>9\)
Explanation:
(C) : Given, For all positive value of \(x\) and \(y\) We have, \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y}\) We can write as, \(\frac{1+x+x^{2}}{x y}=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\) We know that, A.M \(\geq\) G.M \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}\) \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq 3\left(\frac{1}{y^{3}}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \frac{3}{y} \tag{i}\) Similarly, \(\frac{1+y+y^{2}}{x y}=\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\) \(\frac{\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}}{3} \geq\left(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x} \geq \frac{3}{x} \tag{ii}\) On multiplying equation (i) and (ii) we get- \(\left(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\right)\left(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\right) \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)^{2}} \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)} \geq 9\)
Karnataka CET-2002
Application of Derivatives
85593
The maximum of the function \(3 \cos x-4 \sin x\) is:
1 2
2 3
3 4
4 5
Explanation:
(D) : Given, \(f(x)=3 \cos x-4 \sin x\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \cos x+b \sin x \leq \sqrt{a^{2}+b^{2}}\) \(a=3 \text { and } b=-4\) For maximum \(3 \cos x-4 \sin x \leq \sqrt{(3)^{2}+(-4)^{2}}\) \(3 \cos x-4 \sin x \leq \sqrt{25}\) \(3 \cos x-4 \sin x \leq 5\)
Karnataka CET-2004
Application of Derivatives
85594
The maximum slope of the curve \(y=-x^{3}+3 x^{2}+2 x-27 i\)
1 1
2 23
3 5
4 -23
Explanation:
(C) : Given curve, \(y=-x^{2}+3 x^{2}+2 x-27\) On differentiating both sides w.r.t \(x\) we get- \(\frac{d y}{d x}=-3 x^{2}+6 x+2\) For maximum slope \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Slope \((m)=-3 x^{2}+6 x+2=0\) \(\frac{d(m)}{d x}=-6 x+6 \Rightarrow \frac{d^{2}(m)}{d x^{2}}=-6\) For maximum, \(\frac{d(m)}{d x}=0\) \(-6 x+6=0\) \(x=1\) Slope \((m)=-3 x^{2}+6 x+2\) Value of maximum slope at \(\mathrm{x}=1\) \(\mathrm{m}=-3(1)^{2}+6(1)+2\) \(=-3+6+2\) \(\mathrm{~m}_{\max }=5\)
Application of Derivatives
85595
The minimum value of \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) if \(\mathbf{a}^{2}>\mathbf{b}^{2}\), is
1 \(b^{2}\)
2 \(a^{2}-b^{2}\)
3 \(a^{2}\)
4 \(a^{2}+b^{2}\)
Explanation:
(A) : Given, \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) On differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime}(\mathrm{x}) =\mathrm{a}^{2} 2 \cos \mathrm{x} \cdot(-\sin \mathrm{x})+\mathrm{b}^{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=-\mathrm{a}^{2} \sin 2 \mathrm{x}+\mathrm{b}^{2} \sin 2 \mathrm{x}\) \(=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) \sin 2 \mathrm{x}\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\left(b^{2}-a^{2}\right) \sin 2 x=0\) \(\sin 2 x=0\) \(2 x=0\) \(x=0\) \(f(0)=a^{2}(1)^{2}+b^{2}(0)\) \(f(0)=a^{2}(\text { maxima })\) Again on differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2 \cos 2 \mathrm{x}\) \(\mathrm{f}^{\prime \prime}(0)=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2\lt 0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{a}^{2}(0)+\mathrm{b}^{2}(1)\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}(\text { minima })\) \(\mathrm{f}^{\prime \prime}\left(\frac{\pi}{2}\right)=2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\) \(>0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}\) Hence, minimum value is \(b^{2}\)
85592
For all positive value of \(x\) and \(y\), the value of \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y} \text { is: }\)
1 \(\leq 9\)
2 \(\lt 9\)
3 \(\geq 9\)
4 \(>9\)
Explanation:
(C) : Given, For all positive value of \(x\) and \(y\) We have, \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y}\) We can write as, \(\frac{1+x+x^{2}}{x y}=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\) We know that, A.M \(\geq\) G.M \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}\) \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq 3\left(\frac{1}{y^{3}}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \frac{3}{y} \tag{i}\) Similarly, \(\frac{1+y+y^{2}}{x y}=\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\) \(\frac{\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}}{3} \geq\left(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x} \geq \frac{3}{x} \tag{ii}\) On multiplying equation (i) and (ii) we get- \(\left(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\right)\left(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\right) \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)^{2}} \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)} \geq 9\)
Karnataka CET-2002
Application of Derivatives
85593
The maximum of the function \(3 \cos x-4 \sin x\) is:
1 2
2 3
3 4
4 5
Explanation:
(D) : Given, \(f(x)=3 \cos x-4 \sin x\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \cos x+b \sin x \leq \sqrt{a^{2}+b^{2}}\) \(a=3 \text { and } b=-4\) For maximum \(3 \cos x-4 \sin x \leq \sqrt{(3)^{2}+(-4)^{2}}\) \(3 \cos x-4 \sin x \leq \sqrt{25}\) \(3 \cos x-4 \sin x \leq 5\)
Karnataka CET-2004
Application of Derivatives
85594
The maximum slope of the curve \(y=-x^{3}+3 x^{2}+2 x-27 i\)
1 1
2 23
3 5
4 -23
Explanation:
(C) : Given curve, \(y=-x^{2}+3 x^{2}+2 x-27\) On differentiating both sides w.r.t \(x\) we get- \(\frac{d y}{d x}=-3 x^{2}+6 x+2\) For maximum slope \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Slope \((m)=-3 x^{2}+6 x+2=0\) \(\frac{d(m)}{d x}=-6 x+6 \Rightarrow \frac{d^{2}(m)}{d x^{2}}=-6\) For maximum, \(\frac{d(m)}{d x}=0\) \(-6 x+6=0\) \(x=1\) Slope \((m)=-3 x^{2}+6 x+2\) Value of maximum slope at \(\mathrm{x}=1\) \(\mathrm{m}=-3(1)^{2}+6(1)+2\) \(=-3+6+2\) \(\mathrm{~m}_{\max }=5\)
Application of Derivatives
85595
The minimum value of \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) if \(\mathbf{a}^{2}>\mathbf{b}^{2}\), is
1 \(b^{2}\)
2 \(a^{2}-b^{2}\)
3 \(a^{2}\)
4 \(a^{2}+b^{2}\)
Explanation:
(A) : Given, \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) On differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime}(\mathrm{x}) =\mathrm{a}^{2} 2 \cos \mathrm{x} \cdot(-\sin \mathrm{x})+\mathrm{b}^{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=-\mathrm{a}^{2} \sin 2 \mathrm{x}+\mathrm{b}^{2} \sin 2 \mathrm{x}\) \(=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) \sin 2 \mathrm{x}\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\left(b^{2}-a^{2}\right) \sin 2 x=0\) \(\sin 2 x=0\) \(2 x=0\) \(x=0\) \(f(0)=a^{2}(1)^{2}+b^{2}(0)\) \(f(0)=a^{2}(\text { maxima })\) Again on differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2 \cos 2 \mathrm{x}\) \(\mathrm{f}^{\prime \prime}(0)=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2\lt 0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{a}^{2}(0)+\mathrm{b}^{2}(1)\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}(\text { minima })\) \(\mathrm{f}^{\prime \prime}\left(\frac{\pi}{2}\right)=2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\) \(>0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}\) Hence, minimum value is \(b^{2}\)
85592
For all positive value of \(x\) and \(y\), the value of \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y} \text { is: }\)
1 \(\leq 9\)
2 \(\lt 9\)
3 \(\geq 9\)
4 \(>9\)
Explanation:
(C) : Given, For all positive value of \(x\) and \(y\) We have, \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{x y}\) We can write as, \(\frac{1+x+x^{2}}{x y}=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\) We know that, A.M \(\geq\) G.M \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \sqrt[3]{\frac{1}{x y} \times \frac{1}{y} \times \frac{x}{y}}\) \(=\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq 3\left(\frac{1}{y^{3}}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y} \geq \frac{3}{y} \tag{i}\) Similarly, \(\frac{1+y+y^{2}}{x y}=\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\) \(\frac{\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}}{3} \geq\left(\frac{1}{x y} \times \frac{1}{x} \times \frac{y}{x}\right)^{\frac{1}{3}}\) \(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x} \geq \frac{3}{x} \tag{ii}\) On multiplying equation (i) and (ii) we get- \(\left(\frac{1}{x y}+\frac{1}{y}+\frac{x}{y}\right)\left(\frac{1}{x y}+\frac{1}{x}+\frac{y}{x}\right) \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)^{2}} \geq \frac{9}{x y}\) \(\frac{\left(1+x+x^{2}\right)\left(1+y+y^{2}\right)}{(x y)} \geq 9\)
Karnataka CET-2002
Application of Derivatives
85593
The maximum of the function \(3 \cos x-4 \sin x\) is:
1 2
2 3
3 4
4 5
Explanation:
(D) : Given, \(f(x)=3 \cos x-4 \sin x\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \cos x+b \sin x \leq \sqrt{a^{2}+b^{2}}\) \(a=3 \text { and } b=-4\) For maximum \(3 \cos x-4 \sin x \leq \sqrt{(3)^{2}+(-4)^{2}}\) \(3 \cos x-4 \sin x \leq \sqrt{25}\) \(3 \cos x-4 \sin x \leq 5\)
Karnataka CET-2004
Application of Derivatives
85594
The maximum slope of the curve \(y=-x^{3}+3 x^{2}+2 x-27 i\)
1 1
2 23
3 5
4 -23
Explanation:
(C) : Given curve, \(y=-x^{2}+3 x^{2}+2 x-27\) On differentiating both sides w.r.t \(x\) we get- \(\frac{d y}{d x}=-3 x^{2}+6 x+2\) For maximum slope \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Slope \((m)=-3 x^{2}+6 x+2=0\) \(\frac{d(m)}{d x}=-6 x+6 \Rightarrow \frac{d^{2}(m)}{d x^{2}}=-6\) For maximum, \(\frac{d(m)}{d x}=0\) \(-6 x+6=0\) \(x=1\) Slope \((m)=-3 x^{2}+6 x+2\) Value of maximum slope at \(\mathrm{x}=1\) \(\mathrm{m}=-3(1)^{2}+6(1)+2\) \(=-3+6+2\) \(\mathrm{~m}_{\max }=5\)
Application of Derivatives
85595
The minimum value of \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) if \(\mathbf{a}^{2}>\mathbf{b}^{2}\), is
1 \(b^{2}\)
2 \(a^{2}-b^{2}\)
3 \(a^{2}\)
4 \(a^{2}+b^{2}\)
Explanation:
(A) : Given, \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) On differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime}(\mathrm{x}) =\mathrm{a}^{2} 2 \cos \mathrm{x} \cdot(-\sin \mathrm{x})+\mathrm{b}^{2} 2 \sin \mathrm{x} \cos \mathrm{x}\) \(=-\mathrm{a}^{2} \sin 2 \mathrm{x}+\mathrm{b}^{2} \sin 2 \mathrm{x}\) \(=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) \sin 2 \mathrm{x}\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\left(b^{2}-a^{2}\right) \sin 2 x=0\) \(\sin 2 x=0\) \(2 x=0\) \(x=0\) \(f(0)=a^{2}(1)^{2}+b^{2}(0)\) \(f(0)=a^{2}(\text { maxima })\) Again on differentiating both sides w.r.t. \(\mathrm{x}\), we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2 \cos 2 \mathrm{x}\) \(\mathrm{f}^{\prime \prime}(0)=\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right) 2\lt 0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{a}^{2}(0)+\mathrm{b}^{2}(1)\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}(\text { minima })\) \(\mathrm{f}^{\prime \prime}\left(\frac{\pi}{2}\right)=2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)\) \(>0\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{b}^{2}\) Hence, minimum value is \(b^{2}\)
