85725
The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is
1 \(12 \sqrt{3} \pi\)
2 \(4 \sqrt{3} \pi\)
3 \(3 \sqrt{3} \pi\)
4 \(8 \sqrt{3} \pi\)
Explanation:
(A) : Given, sphere of diameter \(=6\) units Then, radius \((\mathrm{r})=3\) units Let \(h\) be the height and \(r\) be the radius of the cylinder inscribed in the sphere and the given sphere is of radius is \(\mathrm{R}\). Then volume of cylinder - \(\mathrm{V}=\pi \mathrm{R}^{2} \mathrm{~h}\) In right \(\triangle \mathrm{OBA}\) \(\mathrm{AB}^{2}+\mathrm{OB}^{2}=\mathrm{OA}^{2}\) \(\mathrm{R}^{2}+\frac{\mathrm{h}^{2}}{4}=\mathrm{r}^{2}\) So, \(\quad R^{2}=r^{2}-\frac{h^{2}}{4}\) Putting the value of \(\mathrm{R}^{2}\) in equation (i), we get - \(\mathrm{V}=\pi\left(\mathrm{r}^{2}-\frac{\mathrm{h}^{2}}{4}\right) \cdot \mathrm{h}\) \(\mathrm{V}=\pi\left(\mathrm{r}^{2} \mathrm{~h}-\frac{\mathrm{h}^{3}}{4}\right) \tag{i}\) \(\frac{\mathrm{dV}}{\mathrm{dh}}=\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right) \tag{iii}\) For stationary point \(\frac{\mathrm{dV}}{\mathrm{dh}}=0\) \(\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right)=0\) \(\mathrm{r}^{2}=\frac{3 \mathrm{~h}^{2}}{4} \Rightarrow \mathrm{h}^{2}=\frac{4 \mathrm{r}^{2}}{3} \Rightarrow \mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Now, \(\frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{dh}^{2}}=\pi\left(\frac{-6}{4} \mathrm{~h}\right)\) \(\therefore \quad\left[\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right]_{\mathrm{ath}=\frac{2 \mathrm{r}}{\sqrt{3}}}=\pi\left(\frac{-3}{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}\right)\lt 0\) Then, volume is maximum at \(\mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Maximum volume is, \(=\pi\left(\mathrm{r}^{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}-\frac{1}{4} \frac{8 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\pi\left(\frac{2 \mathrm{r}^{3}}{\sqrt{3}}-\frac{2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)\) \(=\pi\left(\frac{6 \mathrm{r}^{3}-2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{r}^{3}}{3 \sqrt{3}}=\frac{4 \pi(3)^{3}}{3 \sqrt{3}}\) \(=12 \sqrt{3} \pi \text { cubit units. }\)
Shift-I
Application of Derivatives
85726
The sum of the maximum and the minimum values of \(2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) is
1 \(\frac{\pi^{2}}{8}\)
2 \(\frac{3 \pi^{2}}{8}\)
3 \(\frac{3 \pi^{2}}{2}\)
4 \(4 \pi^{2}\)
Explanation:
(C) : Given, Let, \(f(x)=2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) let, \(\quad y=\cos ^{-1} x\) Then, \(f(x)=2 y^{2}-\pi y+\frac{\pi^{2}}{4}\) \(=2\left[y^{2}-\frac{\pi y}{2}+\frac{\pi^{2}}{8}\right]\) \(=2\left[y^{2}-2 \times y \times \frac{\pi}{4}+\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{-\pi^{2}+2 \pi^{2}}{16}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]=2\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) On putting the value \(y\) in above equation \(f(x)=2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(\because \quad\left(0 \leq \cos ^{-1} x \leq \pi\right)\) Then, \(\quad f(x)=2\left(0-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2 \times \frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\) \(=\frac{\pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{4} \text { (minimum value) }\) And, \(f(x)=2\left(\pi-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2\left(\frac{3 \pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(=\frac{9 \pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{10 \pi^{2}}{8}\) \(\left.=\frac{5 \pi^{2}}{4} \quad \text { (Maximum value }\right)\) So, the sum of the maximum and the minimum \(=\frac{\pi^{2}}{4}+\frac{5 \pi^{2}}{4}=\frac{6 \pi^{2}}{4}=\frac{3 \pi^{2}}{2}\)
Shift-I
Application of Derivatives
85727
If \(x \in R\) and \(1 \leq \frac{3 x^{2}-7 x+8}{x^{2}+1} \leq 2\), then the minimum and maximum values of \(x\) are respectively
85728
The minimum and maximum values of \(\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) are respectively
1 \(-(2 \sqrt{3}-1)\) and \(2 \sqrt{3}-1\)
2 \(-(1+2 \sqrt{2})\) and \(1+2 \sqrt{2}\)
3 -3 and 3
4 -2 and 2
Explanation:
(C) : Given, \(\text { Let, } f(x)=\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) We know that, For \(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}\) the maximum and minimum value are \(+\sqrt{a^{2}+b^{2}}\) and \(-\sqrt{a^{2}+b^{2}}\) Thus, the maximum value for \(\mathrm{f}(\mathrm{x})\) is - \(=\sqrt{(2 \sqrt{2})^{2}+1}\) \(=\sqrt{8+1}\) \(=\sqrt{9}\) \(=3\) And, the minimum value for \(\mathrm{f}(\mathrm{x})\) is, \(=-\sqrt{(2 \sqrt{2})^{2}+1}\) \(=-\sqrt{8+1}\) \(=-\sqrt{9}=-3\) Hence, the minimum value of \(f(x)\) is -3 and maximum value of \(f(x)\) is 3 .
85725
The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is
1 \(12 \sqrt{3} \pi\)
2 \(4 \sqrt{3} \pi\)
3 \(3 \sqrt{3} \pi\)
4 \(8 \sqrt{3} \pi\)
Explanation:
(A) : Given, sphere of diameter \(=6\) units Then, radius \((\mathrm{r})=3\) units Let \(h\) be the height and \(r\) be the radius of the cylinder inscribed in the sphere and the given sphere is of radius is \(\mathrm{R}\). Then volume of cylinder - \(\mathrm{V}=\pi \mathrm{R}^{2} \mathrm{~h}\) In right \(\triangle \mathrm{OBA}\) \(\mathrm{AB}^{2}+\mathrm{OB}^{2}=\mathrm{OA}^{2}\) \(\mathrm{R}^{2}+\frac{\mathrm{h}^{2}}{4}=\mathrm{r}^{2}\) So, \(\quad R^{2}=r^{2}-\frac{h^{2}}{4}\) Putting the value of \(\mathrm{R}^{2}\) in equation (i), we get - \(\mathrm{V}=\pi\left(\mathrm{r}^{2}-\frac{\mathrm{h}^{2}}{4}\right) \cdot \mathrm{h}\) \(\mathrm{V}=\pi\left(\mathrm{r}^{2} \mathrm{~h}-\frac{\mathrm{h}^{3}}{4}\right) \tag{i}\) \(\frac{\mathrm{dV}}{\mathrm{dh}}=\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right) \tag{iii}\) For stationary point \(\frac{\mathrm{dV}}{\mathrm{dh}}=0\) \(\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right)=0\) \(\mathrm{r}^{2}=\frac{3 \mathrm{~h}^{2}}{4} \Rightarrow \mathrm{h}^{2}=\frac{4 \mathrm{r}^{2}}{3} \Rightarrow \mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Now, \(\frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{dh}^{2}}=\pi\left(\frac{-6}{4} \mathrm{~h}\right)\) \(\therefore \quad\left[\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right]_{\mathrm{ath}=\frac{2 \mathrm{r}}{\sqrt{3}}}=\pi\left(\frac{-3}{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}\right)\lt 0\) Then, volume is maximum at \(\mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Maximum volume is, \(=\pi\left(\mathrm{r}^{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}-\frac{1}{4} \frac{8 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\pi\left(\frac{2 \mathrm{r}^{3}}{\sqrt{3}}-\frac{2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)\) \(=\pi\left(\frac{6 \mathrm{r}^{3}-2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{r}^{3}}{3 \sqrt{3}}=\frac{4 \pi(3)^{3}}{3 \sqrt{3}}\) \(=12 \sqrt{3} \pi \text { cubit units. }\)
Shift-I
Application of Derivatives
85726
The sum of the maximum and the minimum values of \(2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) is
1 \(\frac{\pi^{2}}{8}\)
2 \(\frac{3 \pi^{2}}{8}\)
3 \(\frac{3 \pi^{2}}{2}\)
4 \(4 \pi^{2}\)
Explanation:
(C) : Given, Let, \(f(x)=2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) let, \(\quad y=\cos ^{-1} x\) Then, \(f(x)=2 y^{2}-\pi y+\frac{\pi^{2}}{4}\) \(=2\left[y^{2}-\frac{\pi y}{2}+\frac{\pi^{2}}{8}\right]\) \(=2\left[y^{2}-2 \times y \times \frac{\pi}{4}+\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{-\pi^{2}+2 \pi^{2}}{16}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]=2\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) On putting the value \(y\) in above equation \(f(x)=2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(\because \quad\left(0 \leq \cos ^{-1} x \leq \pi\right)\) Then, \(\quad f(x)=2\left(0-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2 \times \frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\) \(=\frac{\pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{4} \text { (minimum value) }\) And, \(f(x)=2\left(\pi-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2\left(\frac{3 \pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(=\frac{9 \pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{10 \pi^{2}}{8}\) \(\left.=\frac{5 \pi^{2}}{4} \quad \text { (Maximum value }\right)\) So, the sum of the maximum and the minimum \(=\frac{\pi^{2}}{4}+\frac{5 \pi^{2}}{4}=\frac{6 \pi^{2}}{4}=\frac{3 \pi^{2}}{2}\)
Shift-I
Application of Derivatives
85727
If \(x \in R\) and \(1 \leq \frac{3 x^{2}-7 x+8}{x^{2}+1} \leq 2\), then the minimum and maximum values of \(x\) are respectively
85728
The minimum and maximum values of \(\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) are respectively
1 \(-(2 \sqrt{3}-1)\) and \(2 \sqrt{3}-1\)
2 \(-(1+2 \sqrt{2})\) and \(1+2 \sqrt{2}\)
3 -3 and 3
4 -2 and 2
Explanation:
(C) : Given, \(\text { Let, } f(x)=\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) We know that, For \(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}\) the maximum and minimum value are \(+\sqrt{a^{2}+b^{2}}\) and \(-\sqrt{a^{2}+b^{2}}\) Thus, the maximum value for \(\mathrm{f}(\mathrm{x})\) is - \(=\sqrt{(2 \sqrt{2})^{2}+1}\) \(=\sqrt{8+1}\) \(=\sqrt{9}\) \(=3\) And, the minimum value for \(\mathrm{f}(\mathrm{x})\) is, \(=-\sqrt{(2 \sqrt{2})^{2}+1}\) \(=-\sqrt{8+1}\) \(=-\sqrt{9}=-3\) Hence, the minimum value of \(f(x)\) is -3 and maximum value of \(f(x)\) is 3 .
85725
The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is
1 \(12 \sqrt{3} \pi\)
2 \(4 \sqrt{3} \pi\)
3 \(3 \sqrt{3} \pi\)
4 \(8 \sqrt{3} \pi\)
Explanation:
(A) : Given, sphere of diameter \(=6\) units Then, radius \((\mathrm{r})=3\) units Let \(h\) be the height and \(r\) be the radius of the cylinder inscribed in the sphere and the given sphere is of radius is \(\mathrm{R}\). Then volume of cylinder - \(\mathrm{V}=\pi \mathrm{R}^{2} \mathrm{~h}\) In right \(\triangle \mathrm{OBA}\) \(\mathrm{AB}^{2}+\mathrm{OB}^{2}=\mathrm{OA}^{2}\) \(\mathrm{R}^{2}+\frac{\mathrm{h}^{2}}{4}=\mathrm{r}^{2}\) So, \(\quad R^{2}=r^{2}-\frac{h^{2}}{4}\) Putting the value of \(\mathrm{R}^{2}\) in equation (i), we get - \(\mathrm{V}=\pi\left(\mathrm{r}^{2}-\frac{\mathrm{h}^{2}}{4}\right) \cdot \mathrm{h}\) \(\mathrm{V}=\pi\left(\mathrm{r}^{2} \mathrm{~h}-\frac{\mathrm{h}^{3}}{4}\right) \tag{i}\) \(\frac{\mathrm{dV}}{\mathrm{dh}}=\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right) \tag{iii}\) For stationary point \(\frac{\mathrm{dV}}{\mathrm{dh}}=0\) \(\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right)=0\) \(\mathrm{r}^{2}=\frac{3 \mathrm{~h}^{2}}{4} \Rightarrow \mathrm{h}^{2}=\frac{4 \mathrm{r}^{2}}{3} \Rightarrow \mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Now, \(\frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{dh}^{2}}=\pi\left(\frac{-6}{4} \mathrm{~h}\right)\) \(\therefore \quad\left[\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right]_{\mathrm{ath}=\frac{2 \mathrm{r}}{\sqrt{3}}}=\pi\left(\frac{-3}{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}\right)\lt 0\) Then, volume is maximum at \(\mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Maximum volume is, \(=\pi\left(\mathrm{r}^{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}-\frac{1}{4} \frac{8 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\pi\left(\frac{2 \mathrm{r}^{3}}{\sqrt{3}}-\frac{2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)\) \(=\pi\left(\frac{6 \mathrm{r}^{3}-2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{r}^{3}}{3 \sqrt{3}}=\frac{4 \pi(3)^{3}}{3 \sqrt{3}}\) \(=12 \sqrt{3} \pi \text { cubit units. }\)
Shift-I
Application of Derivatives
85726
The sum of the maximum and the minimum values of \(2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) is
1 \(\frac{\pi^{2}}{8}\)
2 \(\frac{3 \pi^{2}}{8}\)
3 \(\frac{3 \pi^{2}}{2}\)
4 \(4 \pi^{2}\)
Explanation:
(C) : Given, Let, \(f(x)=2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) let, \(\quad y=\cos ^{-1} x\) Then, \(f(x)=2 y^{2}-\pi y+\frac{\pi^{2}}{4}\) \(=2\left[y^{2}-\frac{\pi y}{2}+\frac{\pi^{2}}{8}\right]\) \(=2\left[y^{2}-2 \times y \times \frac{\pi}{4}+\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{-\pi^{2}+2 \pi^{2}}{16}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]=2\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) On putting the value \(y\) in above equation \(f(x)=2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(\because \quad\left(0 \leq \cos ^{-1} x \leq \pi\right)\) Then, \(\quad f(x)=2\left(0-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2 \times \frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\) \(=\frac{\pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{4} \text { (minimum value) }\) And, \(f(x)=2\left(\pi-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2\left(\frac{3 \pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(=\frac{9 \pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{10 \pi^{2}}{8}\) \(\left.=\frac{5 \pi^{2}}{4} \quad \text { (Maximum value }\right)\) So, the sum of the maximum and the minimum \(=\frac{\pi^{2}}{4}+\frac{5 \pi^{2}}{4}=\frac{6 \pi^{2}}{4}=\frac{3 \pi^{2}}{2}\)
Shift-I
Application of Derivatives
85727
If \(x \in R\) and \(1 \leq \frac{3 x^{2}-7 x+8}{x^{2}+1} \leq 2\), then the minimum and maximum values of \(x\) are respectively
85728
The minimum and maximum values of \(\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) are respectively
1 \(-(2 \sqrt{3}-1)\) and \(2 \sqrt{3}-1\)
2 \(-(1+2 \sqrt{2})\) and \(1+2 \sqrt{2}\)
3 -3 and 3
4 -2 and 2
Explanation:
(C) : Given, \(\text { Let, } f(x)=\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) We know that, For \(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}\) the maximum and minimum value are \(+\sqrt{a^{2}+b^{2}}\) and \(-\sqrt{a^{2}+b^{2}}\) Thus, the maximum value for \(\mathrm{f}(\mathrm{x})\) is - \(=\sqrt{(2 \sqrt{2})^{2}+1}\) \(=\sqrt{8+1}\) \(=\sqrt{9}\) \(=3\) And, the minimum value for \(\mathrm{f}(\mathrm{x})\) is, \(=-\sqrt{(2 \sqrt{2})^{2}+1}\) \(=-\sqrt{8+1}\) \(=-\sqrt{9}=-3\) Hence, the minimum value of \(f(x)\) is -3 and maximum value of \(f(x)\) is 3 .
85725
The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is
1 \(12 \sqrt{3} \pi\)
2 \(4 \sqrt{3} \pi\)
3 \(3 \sqrt{3} \pi\)
4 \(8 \sqrt{3} \pi\)
Explanation:
(A) : Given, sphere of diameter \(=6\) units Then, radius \((\mathrm{r})=3\) units Let \(h\) be the height and \(r\) be the radius of the cylinder inscribed in the sphere and the given sphere is of radius is \(\mathrm{R}\). Then volume of cylinder - \(\mathrm{V}=\pi \mathrm{R}^{2} \mathrm{~h}\) In right \(\triangle \mathrm{OBA}\) \(\mathrm{AB}^{2}+\mathrm{OB}^{2}=\mathrm{OA}^{2}\) \(\mathrm{R}^{2}+\frac{\mathrm{h}^{2}}{4}=\mathrm{r}^{2}\) So, \(\quad R^{2}=r^{2}-\frac{h^{2}}{4}\) Putting the value of \(\mathrm{R}^{2}\) in equation (i), we get - \(\mathrm{V}=\pi\left(\mathrm{r}^{2}-\frac{\mathrm{h}^{2}}{4}\right) \cdot \mathrm{h}\) \(\mathrm{V}=\pi\left(\mathrm{r}^{2} \mathrm{~h}-\frac{\mathrm{h}^{3}}{4}\right) \tag{i}\) \(\frac{\mathrm{dV}}{\mathrm{dh}}=\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right) \tag{iii}\) For stationary point \(\frac{\mathrm{dV}}{\mathrm{dh}}=0\) \(\pi\left(\mathrm{r}^{2}-\frac{3 \mathrm{~h}^{2}}{4}\right)=0\) \(\mathrm{r}^{2}=\frac{3 \mathrm{~h}^{2}}{4} \Rightarrow \mathrm{h}^{2}=\frac{4 \mathrm{r}^{2}}{3} \Rightarrow \mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Now, \(\frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{dh}^{2}}=\pi\left(\frac{-6}{4} \mathrm{~h}\right)\) \(\therefore \quad\left[\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dh}^{2}}\right]_{\mathrm{ath}=\frac{2 \mathrm{r}}{\sqrt{3}}}=\pi\left(\frac{-3}{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}\right)\lt 0\) Then, volume is maximum at \(\mathrm{h}=\frac{2 \mathrm{r}}{\sqrt{3}}\) Maximum volume is, \(=\pi\left(\mathrm{r}^{2} \cdot \frac{2 \mathrm{r}}{\sqrt{3}}-\frac{1}{4} \frac{8 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\pi\left(\frac{2 \mathrm{r}^{3}}{\sqrt{3}}-\frac{2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)\) \(=\pi\left(\frac{6 \mathrm{r}^{3}-2 \mathrm{r}^{3}}{3 \sqrt{3}}\right)=\frac{4 \pi \mathrm{r}^{3}}{3 \sqrt{3}}=\frac{4 \pi(3)^{3}}{3 \sqrt{3}}\) \(=12 \sqrt{3} \pi \text { cubit units. }\)
Shift-I
Application of Derivatives
85726
The sum of the maximum and the minimum values of \(2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) is
1 \(\frac{\pi^{2}}{8}\)
2 \(\frac{3 \pi^{2}}{8}\)
3 \(\frac{3 \pi^{2}}{2}\)
4 \(4 \pi^{2}\)
Explanation:
(C) : Given, Let, \(f(x)=2\left(\cos ^{-1} x\right)^{2}-\pi \cos ^{-1} x+\frac{\pi^{2}}{4}\) let, \(\quad y=\cos ^{-1} x\) Then, \(f(x)=2 y^{2}-\pi y+\frac{\pi^{2}}{4}\) \(=2\left[y^{2}-\frac{\pi y}{2}+\frac{\pi^{2}}{8}\right]\) \(=2\left[y^{2}-2 \times y \times \frac{\pi}{4}+\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{-\pi^{2}+2 \pi^{2}}{16}\right]\) \(=2\left[\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]=2\left(y-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) On putting the value \(y\) in above equation \(f(x)=2\left(\cos ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(\because \quad\left(0 \leq \cos ^{-1} x \leq \pi\right)\) Then, \(\quad f(x)=2\left(0-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2 \times \frac{\pi^{2}}{16}+\frac{\pi^{2}}{8}\) \(=\frac{\pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{4} \text { (minimum value) }\) And, \(f(x)=2\left(\pi-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=2\left(\frac{3 \pi}{4}\right)^{2}+\frac{\pi^{2}}{8}\) \(=\frac{9 \pi^{2}}{8}+\frac{\pi^{2}}{8}=\frac{10 \pi^{2}}{8}\) \(\left.=\frac{5 \pi^{2}}{4} \quad \text { (Maximum value }\right)\) So, the sum of the maximum and the minimum \(=\frac{\pi^{2}}{4}+\frac{5 \pi^{2}}{4}=\frac{6 \pi^{2}}{4}=\frac{3 \pi^{2}}{2}\)
Shift-I
Application of Derivatives
85727
If \(x \in R\) and \(1 \leq \frac{3 x^{2}-7 x+8}{x^{2}+1} \leq 2\), then the minimum and maximum values of \(x\) are respectively
85728
The minimum and maximum values of \(\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) are respectively
1 \(-(2 \sqrt{3}-1)\) and \(2 \sqrt{3}-1\)
2 \(-(1+2 \sqrt{2})\) and \(1+2 \sqrt{2}\)
3 -3 and 3
4 -2 and 2
Explanation:
(C) : Given, \(\text { Let, } f(x)=\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)\) We know that, For \(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}\) the maximum and minimum value are \(+\sqrt{a^{2}+b^{2}}\) and \(-\sqrt{a^{2}+b^{2}}\) Thus, the maximum value for \(\mathrm{f}(\mathrm{x})\) is - \(=\sqrt{(2 \sqrt{2})^{2}+1}\) \(=\sqrt{8+1}\) \(=\sqrt{9}\) \(=3\) And, the minimum value for \(\mathrm{f}(\mathrm{x})\) is, \(=-\sqrt{(2 \sqrt{2})^{2}+1}\) \(=-\sqrt{8+1}\) \(=-\sqrt{9}=-3\) Hence, the minimum value of \(f(x)\) is -3 and maximum value of \(f(x)\) is 3 .
