(D) : Given, \(f(x)=4 x^{3}-6 x^{2}+2 x+5, F(0)=5\) Then, for anti-derivative \(F(x)\) of \(f(x)\), Integrate both side, we get - \(\int f(x)=\int\left(4 x^{3}-6 x^{2}+2 x+5\right)\) \(F(x)=\frac{4 x^{4}}{4}-\frac{6 x^{3}}{3}+\frac{2 x^{2}}{2}+5 x+C\) \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+C \tag{i}\) Then, \(\mathrm{F}(0)=0+0+0+0+\mathrm{C}\) \(5=\mathrm{C}\) \(\mathrm{C}=5\) Put, \(\mathrm{C}=5\) in equation in equation (i), we get - So, \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+5\)
CG PET-2013
Application of Derivatives
85681
If \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), \text { if } x\lt 1 \\ 3-2 x, \text { if } x \leq 1\end{array}\right.\) then \(f(x)\) has
1 local minimum at \(x=1\)
2 local maximum at \(x=1\)
3 Both local maximum and local minimum at \(x=1\)
4 None of the above
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), & \text { if } x\lt 1 \\ 3-2 x, & \text { if } x \leq 1\end{array}\right.\) And, \(\quad f(1)=3-2=1\) \(\because \quad f(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Now, graph of function \(f(\mathrm{x})\) is From the above graph of function, we see that \(f(x)\) has local maxima at \(\mathrm{x}=1\)
CG PET-2015
Application of Derivatives
85683
The distance of the point on the curve \(x^{2}=2 y\), which is nearest to the point \((0,5)\) is
1 3
2 4
3 \(2 \sqrt{2}\)
4 None of these
Explanation:
(A) : Let, the point \((\mathrm{h}, \mathrm{k})\) lie on the curve \(\mathrm{x}^{2}=2 \mathrm{y}\) \(\therefore \quad \mathrm{h}^{2}=2 \mathrm{k}\) Distance between \((\mathrm{h}, \mathrm{k})\) and \((0,5)\) is \(\mathrm{D}=\sqrt{\mathrm{h}^{2}+(\mathrm{k}-5)^{2}}\) \(\mathrm{D}=\sqrt{2 \mathrm{k}+\mathrm{k}^{2}-10 \mathrm{k}+25} \quad\left[\therefore \mathrm{h}^{2}=2 \mathrm{k}\right]\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}^{2}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\mathrm{z}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\frac{\mathrm{dz}}{\mathrm{dk}}=2 \mathrm{k}-8\) For maxima or minima put, \(\frac{\mathrm{dz}}{\mathrm{dk}}=0\) \(2 \mathrm{k}-8=0 \Rightarrow \mathrm{k}=4\) And, \(\mathrm{h}=2 \sqrt{2}\) Now, \(\quad \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}}=2>0\) \(\therefore\) Distance is minimum at \((\mathrm{h}, \mathrm{k})=(2 \sqrt{2}, 4)\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}=\sqrt{16-32+25}=\sqrt{9}=3\)
CG PET-2017
Application of Derivatives
85684
Consider a cuboid of sides \(2 x, 4 x\) and \(5 x\) and a closed hemisphere of radius \(r\). If the sum of their surface areas is a constant \(k\), then the ratio \(x: r\), for which the sum of their volumes is maximum, is
1 \(2: 5\)
2 \(19: 45\)
3 \(3: 8\)
4 \(19: 15\)
Explanation:
(B) : Given, a cuboid of sides \(2 \mathrm{x}, 4 \mathrm{x}\) and \(5 \mathrm{x}\) Then, surface area of cuboid \(=2\left(8 x^{2}+20 x^{2}+10 x^{2}\right)\) \(=2\left(38 x^{2}\right)\) \(=76 x^{2}\) and a closed hemisphere of radius \(r\). Then, surface area of closed hemisphere \(=3 \pi \mathrm{r}^{2}\) \(\therefore\) The sum of their surface area \(=76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) And, volume \((\mathrm{V})=\) volume of cuboid + volume of closed hemisphere. \(\mathrm{V}=40 \mathrm{x}^{3}+\frac{2}{3} \pi \mathrm{r}^{3}\) From, \(76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) \(\quad r^2=\frac{k-76 x^2}{3 \pi} \Rightarrow r=\left(\frac{k-76 x^2}{3 \pi}\right)^{1 / 2}\) \(\text { Then, } V=40 x^3+\frac{2}{3} \pi\left(\frac{k-76 x^2}{3 \pi}\right)^{3 / 2}\) \(\therefore \quad \frac{d V}{d x}=120 x^2+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^2}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x)}{3 \pi}\right)\) For maxima/minima put \(\frac{\mathrm{dv}}{\mathrm{dx}}=0\) \(120 x^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^{2}}{3 \pi}\right)^{\frac{1}{2}}\left(\frac{-152 x}{3 \pi}\right)=0\) \(120 \mathrm{x}^{2}=\frac{152 \mathrm{x}}{3}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2}\) \(\frac{45}{19} \mathrm{x}^{2}=\mathrm{x}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2} ; \mathrm{n} \neq 0\) \(\frac{45}{19} x=\left(\frac{k-76 x^{2}}{3 \pi}\right)^{1 / 2}\) \(\left(\frac{45}{19}\right)^{2} x^{2}=\frac{k-76 x^{2}}{3 \pi} \Rightarrow\left(\frac{45}{19}\right)^{2} x^{2}=r^{2}\) \(\frac{x^{2}}{r^{2}}=\left(\frac{19}{45}\right)^{2}\) So, \(\frac{x}{r}=\frac{19}{45} \Rightarrow x: r=19: 45\)
(D) : Given, \(f(x)=4 x^{3}-6 x^{2}+2 x+5, F(0)=5\) Then, for anti-derivative \(F(x)\) of \(f(x)\), Integrate both side, we get - \(\int f(x)=\int\left(4 x^{3}-6 x^{2}+2 x+5\right)\) \(F(x)=\frac{4 x^{4}}{4}-\frac{6 x^{3}}{3}+\frac{2 x^{2}}{2}+5 x+C\) \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+C \tag{i}\) Then, \(\mathrm{F}(0)=0+0+0+0+\mathrm{C}\) \(5=\mathrm{C}\) \(\mathrm{C}=5\) Put, \(\mathrm{C}=5\) in equation in equation (i), we get - So, \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+5\)
CG PET-2013
Application of Derivatives
85681
If \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), \text { if } x\lt 1 \\ 3-2 x, \text { if } x \leq 1\end{array}\right.\) then \(f(x)\) has
1 local minimum at \(x=1\)
2 local maximum at \(x=1\)
3 Both local maximum and local minimum at \(x=1\)
4 None of the above
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), & \text { if } x\lt 1 \\ 3-2 x, & \text { if } x \leq 1\end{array}\right.\) And, \(\quad f(1)=3-2=1\) \(\because \quad f(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Now, graph of function \(f(\mathrm{x})\) is From the above graph of function, we see that \(f(x)\) has local maxima at \(\mathrm{x}=1\)
CG PET-2015
Application of Derivatives
85683
The distance of the point on the curve \(x^{2}=2 y\), which is nearest to the point \((0,5)\) is
1 3
2 4
3 \(2 \sqrt{2}\)
4 None of these
Explanation:
(A) : Let, the point \((\mathrm{h}, \mathrm{k})\) lie on the curve \(\mathrm{x}^{2}=2 \mathrm{y}\) \(\therefore \quad \mathrm{h}^{2}=2 \mathrm{k}\) Distance between \((\mathrm{h}, \mathrm{k})\) and \((0,5)\) is \(\mathrm{D}=\sqrt{\mathrm{h}^{2}+(\mathrm{k}-5)^{2}}\) \(\mathrm{D}=\sqrt{2 \mathrm{k}+\mathrm{k}^{2}-10 \mathrm{k}+25} \quad\left[\therefore \mathrm{h}^{2}=2 \mathrm{k}\right]\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}^{2}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\mathrm{z}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\frac{\mathrm{dz}}{\mathrm{dk}}=2 \mathrm{k}-8\) For maxima or minima put, \(\frac{\mathrm{dz}}{\mathrm{dk}}=0\) \(2 \mathrm{k}-8=0 \Rightarrow \mathrm{k}=4\) And, \(\mathrm{h}=2 \sqrt{2}\) Now, \(\quad \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}}=2>0\) \(\therefore\) Distance is minimum at \((\mathrm{h}, \mathrm{k})=(2 \sqrt{2}, 4)\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}=\sqrt{16-32+25}=\sqrt{9}=3\)
CG PET-2017
Application of Derivatives
85684
Consider a cuboid of sides \(2 x, 4 x\) and \(5 x\) and a closed hemisphere of radius \(r\). If the sum of their surface areas is a constant \(k\), then the ratio \(x: r\), for which the sum of their volumes is maximum, is
1 \(2: 5\)
2 \(19: 45\)
3 \(3: 8\)
4 \(19: 15\)
Explanation:
(B) : Given, a cuboid of sides \(2 \mathrm{x}, 4 \mathrm{x}\) and \(5 \mathrm{x}\) Then, surface area of cuboid \(=2\left(8 x^{2}+20 x^{2}+10 x^{2}\right)\) \(=2\left(38 x^{2}\right)\) \(=76 x^{2}\) and a closed hemisphere of radius \(r\). Then, surface area of closed hemisphere \(=3 \pi \mathrm{r}^{2}\) \(\therefore\) The sum of their surface area \(=76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) And, volume \((\mathrm{V})=\) volume of cuboid + volume of closed hemisphere. \(\mathrm{V}=40 \mathrm{x}^{3}+\frac{2}{3} \pi \mathrm{r}^{3}\) From, \(76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) \(\quad r^2=\frac{k-76 x^2}{3 \pi} \Rightarrow r=\left(\frac{k-76 x^2}{3 \pi}\right)^{1 / 2}\) \(\text { Then, } V=40 x^3+\frac{2}{3} \pi\left(\frac{k-76 x^2}{3 \pi}\right)^{3 / 2}\) \(\therefore \quad \frac{d V}{d x}=120 x^2+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^2}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x)}{3 \pi}\right)\) For maxima/minima put \(\frac{\mathrm{dv}}{\mathrm{dx}}=0\) \(120 x^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^{2}}{3 \pi}\right)^{\frac{1}{2}}\left(\frac{-152 x}{3 \pi}\right)=0\) \(120 \mathrm{x}^{2}=\frac{152 \mathrm{x}}{3}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2}\) \(\frac{45}{19} \mathrm{x}^{2}=\mathrm{x}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2} ; \mathrm{n} \neq 0\) \(\frac{45}{19} x=\left(\frac{k-76 x^{2}}{3 \pi}\right)^{1 / 2}\) \(\left(\frac{45}{19}\right)^{2} x^{2}=\frac{k-76 x^{2}}{3 \pi} \Rightarrow\left(\frac{45}{19}\right)^{2} x^{2}=r^{2}\) \(\frac{x^{2}}{r^{2}}=\left(\frac{19}{45}\right)^{2}\) So, \(\frac{x}{r}=\frac{19}{45} \Rightarrow x: r=19: 45\)
(D) : Given, \(f(x)=4 x^{3}-6 x^{2}+2 x+5, F(0)=5\) Then, for anti-derivative \(F(x)\) of \(f(x)\), Integrate both side, we get - \(\int f(x)=\int\left(4 x^{3}-6 x^{2}+2 x+5\right)\) \(F(x)=\frac{4 x^{4}}{4}-\frac{6 x^{3}}{3}+\frac{2 x^{2}}{2}+5 x+C\) \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+C \tag{i}\) Then, \(\mathrm{F}(0)=0+0+0+0+\mathrm{C}\) \(5=\mathrm{C}\) \(\mathrm{C}=5\) Put, \(\mathrm{C}=5\) in equation in equation (i), we get - So, \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+5\)
CG PET-2013
Application of Derivatives
85681
If \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), \text { if } x\lt 1 \\ 3-2 x, \text { if } x \leq 1\end{array}\right.\) then \(f(x)\) has
1 local minimum at \(x=1\)
2 local maximum at \(x=1\)
3 Both local maximum and local minimum at \(x=1\)
4 None of the above
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), & \text { if } x\lt 1 \\ 3-2 x, & \text { if } x \leq 1\end{array}\right.\) And, \(\quad f(1)=3-2=1\) \(\because \quad f(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Now, graph of function \(f(\mathrm{x})\) is From the above graph of function, we see that \(f(x)\) has local maxima at \(\mathrm{x}=1\)
CG PET-2015
Application of Derivatives
85683
The distance of the point on the curve \(x^{2}=2 y\), which is nearest to the point \((0,5)\) is
1 3
2 4
3 \(2 \sqrt{2}\)
4 None of these
Explanation:
(A) : Let, the point \((\mathrm{h}, \mathrm{k})\) lie on the curve \(\mathrm{x}^{2}=2 \mathrm{y}\) \(\therefore \quad \mathrm{h}^{2}=2 \mathrm{k}\) Distance between \((\mathrm{h}, \mathrm{k})\) and \((0,5)\) is \(\mathrm{D}=\sqrt{\mathrm{h}^{2}+(\mathrm{k}-5)^{2}}\) \(\mathrm{D}=\sqrt{2 \mathrm{k}+\mathrm{k}^{2}-10 \mathrm{k}+25} \quad\left[\therefore \mathrm{h}^{2}=2 \mathrm{k}\right]\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}^{2}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\mathrm{z}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\frac{\mathrm{dz}}{\mathrm{dk}}=2 \mathrm{k}-8\) For maxima or minima put, \(\frac{\mathrm{dz}}{\mathrm{dk}}=0\) \(2 \mathrm{k}-8=0 \Rightarrow \mathrm{k}=4\) And, \(\mathrm{h}=2 \sqrt{2}\) Now, \(\quad \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}}=2>0\) \(\therefore\) Distance is minimum at \((\mathrm{h}, \mathrm{k})=(2 \sqrt{2}, 4)\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}=\sqrt{16-32+25}=\sqrt{9}=3\)
CG PET-2017
Application of Derivatives
85684
Consider a cuboid of sides \(2 x, 4 x\) and \(5 x\) and a closed hemisphere of radius \(r\). If the sum of their surface areas is a constant \(k\), then the ratio \(x: r\), for which the sum of their volumes is maximum, is
1 \(2: 5\)
2 \(19: 45\)
3 \(3: 8\)
4 \(19: 15\)
Explanation:
(B) : Given, a cuboid of sides \(2 \mathrm{x}, 4 \mathrm{x}\) and \(5 \mathrm{x}\) Then, surface area of cuboid \(=2\left(8 x^{2}+20 x^{2}+10 x^{2}\right)\) \(=2\left(38 x^{2}\right)\) \(=76 x^{2}\) and a closed hemisphere of radius \(r\). Then, surface area of closed hemisphere \(=3 \pi \mathrm{r}^{2}\) \(\therefore\) The sum of their surface area \(=76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) And, volume \((\mathrm{V})=\) volume of cuboid + volume of closed hemisphere. \(\mathrm{V}=40 \mathrm{x}^{3}+\frac{2}{3} \pi \mathrm{r}^{3}\) From, \(76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) \(\quad r^2=\frac{k-76 x^2}{3 \pi} \Rightarrow r=\left(\frac{k-76 x^2}{3 \pi}\right)^{1 / 2}\) \(\text { Then, } V=40 x^3+\frac{2}{3} \pi\left(\frac{k-76 x^2}{3 \pi}\right)^{3 / 2}\) \(\therefore \quad \frac{d V}{d x}=120 x^2+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^2}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x)}{3 \pi}\right)\) For maxima/minima put \(\frac{\mathrm{dv}}{\mathrm{dx}}=0\) \(120 x^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^{2}}{3 \pi}\right)^{\frac{1}{2}}\left(\frac{-152 x}{3 \pi}\right)=0\) \(120 \mathrm{x}^{2}=\frac{152 \mathrm{x}}{3}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2}\) \(\frac{45}{19} \mathrm{x}^{2}=\mathrm{x}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2} ; \mathrm{n} \neq 0\) \(\frac{45}{19} x=\left(\frac{k-76 x^{2}}{3 \pi}\right)^{1 / 2}\) \(\left(\frac{45}{19}\right)^{2} x^{2}=\frac{k-76 x^{2}}{3 \pi} \Rightarrow\left(\frac{45}{19}\right)^{2} x^{2}=r^{2}\) \(\frac{x^{2}}{r^{2}}=\left(\frac{19}{45}\right)^{2}\) So, \(\frac{x}{r}=\frac{19}{45} \Rightarrow x: r=19: 45\)
(D) : Given, \(f(x)=4 x^{3}-6 x^{2}+2 x+5, F(0)=5\) Then, for anti-derivative \(F(x)\) of \(f(x)\), Integrate both side, we get - \(\int f(x)=\int\left(4 x^{3}-6 x^{2}+2 x+5\right)\) \(F(x)=\frac{4 x^{4}}{4}-\frac{6 x^{3}}{3}+\frac{2 x^{2}}{2}+5 x+C\) \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+C \tag{i}\) Then, \(\mathrm{F}(0)=0+0+0+0+\mathrm{C}\) \(5=\mathrm{C}\) \(\mathrm{C}=5\) Put, \(\mathrm{C}=5\) in equation in equation (i), we get - So, \(F(x)=x^{4}-2 x^{3}+x^{2}+5 x+5\)
CG PET-2013
Application of Derivatives
85681
If \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), \text { if } x\lt 1 \\ 3-2 x, \text { if } x \leq 1\end{array}\right.\) then \(f(x)\) has
1 local minimum at \(x=1\)
2 local maximum at \(x=1\)
3 Both local maximum and local minimum at \(x=1\)
4 None of the above
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cc}\sin \left(\frac{\pi x}{2}\right), & \text { if } x\lt 1 \\ 3-2 x, & \text { if } x \leq 1\end{array}\right.\) And, \(\quad f(1)=3-2=1\) \(\because \quad f(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Now, graph of function \(f(\mathrm{x})\) is From the above graph of function, we see that \(f(x)\) has local maxima at \(\mathrm{x}=1\)
CG PET-2015
Application of Derivatives
85683
The distance of the point on the curve \(x^{2}=2 y\), which is nearest to the point \((0,5)\) is
1 3
2 4
3 \(2 \sqrt{2}\)
4 None of these
Explanation:
(A) : Let, the point \((\mathrm{h}, \mathrm{k})\) lie on the curve \(\mathrm{x}^{2}=2 \mathrm{y}\) \(\therefore \quad \mathrm{h}^{2}=2 \mathrm{k}\) Distance between \((\mathrm{h}, \mathrm{k})\) and \((0,5)\) is \(\mathrm{D}=\sqrt{\mathrm{h}^{2}+(\mathrm{k}-5)^{2}}\) \(\mathrm{D}=\sqrt{2 \mathrm{k}+\mathrm{k}^{2}-10 \mathrm{k}+25} \quad\left[\therefore \mathrm{h}^{2}=2 \mathrm{k}\right]\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}^{2}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\mathrm{z}=\mathrm{k}^{2}-8 \mathrm{k}+25\) \(\frac{\mathrm{dz}}{\mathrm{dk}}=2 \mathrm{k}-8\) For maxima or minima put, \(\frac{\mathrm{dz}}{\mathrm{dk}}=0\) \(2 \mathrm{k}-8=0 \Rightarrow \mathrm{k}=4\) And, \(\mathrm{h}=2 \sqrt{2}\) Now, \(\quad \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}}=2>0\) \(\therefore\) Distance is minimum at \((\mathrm{h}, \mathrm{k})=(2 \sqrt{2}, 4)\) \(\mathrm{D}=\sqrt{\mathrm{k}^{2}-8 \mathrm{k}+25}\) \(\mathrm{D}=\sqrt{16-32+25}=\sqrt{9}=3\)
CG PET-2017
Application of Derivatives
85684
Consider a cuboid of sides \(2 x, 4 x\) and \(5 x\) and a closed hemisphere of radius \(r\). If the sum of their surface areas is a constant \(k\), then the ratio \(x: r\), for which the sum of their volumes is maximum, is
1 \(2: 5\)
2 \(19: 45\)
3 \(3: 8\)
4 \(19: 15\)
Explanation:
(B) : Given, a cuboid of sides \(2 \mathrm{x}, 4 \mathrm{x}\) and \(5 \mathrm{x}\) Then, surface area of cuboid \(=2\left(8 x^{2}+20 x^{2}+10 x^{2}\right)\) \(=2\left(38 x^{2}\right)\) \(=76 x^{2}\) and a closed hemisphere of radius \(r\). Then, surface area of closed hemisphere \(=3 \pi \mathrm{r}^{2}\) \(\therefore\) The sum of their surface area \(=76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) And, volume \((\mathrm{V})=\) volume of cuboid + volume of closed hemisphere. \(\mathrm{V}=40 \mathrm{x}^{3}+\frac{2}{3} \pi \mathrm{r}^{3}\) From, \(76 \mathrm{x}^{2}+3 \pi \mathrm{r}^{2}=\mathrm{k}\) \(\quad r^2=\frac{k-76 x^2}{3 \pi} \Rightarrow r=\left(\frac{k-76 x^2}{3 \pi}\right)^{1 / 2}\) \(\text { Then, } V=40 x^3+\frac{2}{3} \pi\left(\frac{k-76 x^2}{3 \pi}\right)^{3 / 2}\) \(\therefore \quad \frac{d V}{d x}=120 x^2+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^2}{3 \pi}\right)^{\frac{1}{2}} \cdot\left(\frac{-76(2 x)}{3 \pi}\right)\) For maxima/minima put \(\frac{\mathrm{dv}}{\mathrm{dx}}=0\) \(120 x^{2}+\frac{2}{3} \pi \cdot \frac{3}{2}\left(\frac{k-76 x^{2}}{3 \pi}\right)^{\frac{1}{2}}\left(\frac{-152 x}{3 \pi}\right)=0\) \(120 \mathrm{x}^{2}=\frac{152 \mathrm{x}}{3}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2}\) \(\frac{45}{19} \mathrm{x}^{2}=\mathrm{x}\left(\frac{\mathrm{k}-76 \mathrm{x}^{2}}{3 \pi}\right)^{1 / 2} ; \mathrm{n} \neq 0\) \(\frac{45}{19} x=\left(\frac{k-76 x^{2}}{3 \pi}\right)^{1 / 2}\) \(\left(\frac{45}{19}\right)^{2} x^{2}=\frac{k-76 x^{2}}{3 \pi} \Rightarrow\left(\frac{45}{19}\right)^{2} x^{2}=r^{2}\) \(\frac{x^{2}}{r^{2}}=\left(\frac{19}{45}\right)^{2}\) So, \(\frac{x}{r}=\frac{19}{45} \Rightarrow x: r=19: 45\)
