85712
For the function \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5, x=2\) is a point of
1 local maxima
2 local minima
3 point of inflexion
4 none of these
Explanation:
(C) : Given, \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5\) ...(i) Differentiating with respect to \(\mathrm{x}\), we get \(f^{\prime}(x)=\frac{4}{3} 3 x^{2}-16 x+16=4 x^{2}-16 x+16\) Now for maximum/minimum we put \(f^{\prime}(\mathrm{x})=0\) \(\Rightarrow x^{2}-4 x+4=0 \Rightarrow(x-2)^{2}=0 \Rightarrow x=2\) \(f^{\prime \prime}(\mathrm{x})=8 \mathrm{x}-16,\left.f^{\prime \prime}(\mathrm{x})\right|_{\mathrm{x}=2}=0\) \(f^{\prime \prime}(\mathrm{x})=8 \geq 0\) \(\therefore \quad \mathrm{x}=2\) is the point of inflexion
AMU-2010
Application of Derivatives
85713
If the area of a circular sector of perimeter 60 \(m\) is to be maximized, then its radius must be
1 \({20}\)
2 15
3 10
4 5
Explanation:
(B): Given, the perimeter of circular sector \(=60 \mathrm{~m}\) Let radius be \(\mathrm{r} \mathrm{m}\). Length of \(\operatorname{arc} \mathrm{AB}=l\) \(\therefore\) perimeter \((\mathrm{p})=l+2 \mathrm{r}\) \(l =\mathrm{p}-2 \mathrm{r}\) \(\because \text { Area } =\frac{1}{2} l \mathrm{r}\) \(\mathrm{A}=\frac{(\mathrm{p}-2 \mathrm{r}) \mathrm{r}}{2}=\frac{\mathrm{pr}-2 \mathrm{r}^{2}}{2}\) For \(\mathrm{A}\) to be maximum, \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{d}{d r}\left(\frac{p r-2 r^{2}}{2}\right)=0 \Rightarrow \frac{p-4 r}{2}=0\) \(p=4 r \Rightarrow 60=4 r\) \(r=15 m\)
Shift-2]
Application of Derivatives
85714
For the function \(f(x)=x^{3}-6 x^{2}-12 x-3, x=2\) is
1 point of maxima
2 point of minima
3 point of inflection
4 not a critical point
Explanation:
(C): Given, \(f(x)=x^{3}-6 x^{2}-12 x-3\) Then, \(\quad f^{\prime}(x)=3 x^{2}-12 x-12\) \(f^{\prime}(x)=3\left(x^{2}-4 x-4\right)\) \(f^{\prime \prime}(\mathrm{x})=3(2 \mathrm{x}-4)\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}-4=0\) \(\Rightarrow x=\frac{4 \pm \sqrt{16+16}}{2 \times 1}=\frac{4 \pm 4 \sqrt{2}}{2}\) \(x=2 \pm 2 \sqrt{2}\) (These two points have either maxima or minima) So, \(\quad \mathrm{f}^{\prime \prime}(2)=3(2 \times 2-4)\) \(\mathrm{f}^{\prime \prime}(2)=0\) Hence, \(x=2 \text { is point of inflection. }\)
Shift-II]
Application of Derivatives
85715
Let \(f(x)=x^{2}+2 x+2, g(x)=-x^{2}+2 x-1\) and a, b be the extreme values of \(f(x), g(x)\) respectively. If \(c\) is the extreme value of \(\frac{f}{g}(x)\) (for \(x \neq 1\) ), then \(a+2 b+5 c+4=\)
85712
For the function \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5, x=2\) is a point of
1 local maxima
2 local minima
3 point of inflexion
4 none of these
Explanation:
(C) : Given, \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5\) ...(i) Differentiating with respect to \(\mathrm{x}\), we get \(f^{\prime}(x)=\frac{4}{3} 3 x^{2}-16 x+16=4 x^{2}-16 x+16\) Now for maximum/minimum we put \(f^{\prime}(\mathrm{x})=0\) \(\Rightarrow x^{2}-4 x+4=0 \Rightarrow(x-2)^{2}=0 \Rightarrow x=2\) \(f^{\prime \prime}(\mathrm{x})=8 \mathrm{x}-16,\left.f^{\prime \prime}(\mathrm{x})\right|_{\mathrm{x}=2}=0\) \(f^{\prime \prime}(\mathrm{x})=8 \geq 0\) \(\therefore \quad \mathrm{x}=2\) is the point of inflexion
AMU-2010
Application of Derivatives
85713
If the area of a circular sector of perimeter 60 \(m\) is to be maximized, then its radius must be
1 \({20}\)
2 15
3 10
4 5
Explanation:
(B): Given, the perimeter of circular sector \(=60 \mathrm{~m}\) Let radius be \(\mathrm{r} \mathrm{m}\). Length of \(\operatorname{arc} \mathrm{AB}=l\) \(\therefore\) perimeter \((\mathrm{p})=l+2 \mathrm{r}\) \(l =\mathrm{p}-2 \mathrm{r}\) \(\because \text { Area } =\frac{1}{2} l \mathrm{r}\) \(\mathrm{A}=\frac{(\mathrm{p}-2 \mathrm{r}) \mathrm{r}}{2}=\frac{\mathrm{pr}-2 \mathrm{r}^{2}}{2}\) For \(\mathrm{A}\) to be maximum, \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{d}{d r}\left(\frac{p r-2 r^{2}}{2}\right)=0 \Rightarrow \frac{p-4 r}{2}=0\) \(p=4 r \Rightarrow 60=4 r\) \(r=15 m\)
Shift-2]
Application of Derivatives
85714
For the function \(f(x)=x^{3}-6 x^{2}-12 x-3, x=2\) is
1 point of maxima
2 point of minima
3 point of inflection
4 not a critical point
Explanation:
(C): Given, \(f(x)=x^{3}-6 x^{2}-12 x-3\) Then, \(\quad f^{\prime}(x)=3 x^{2}-12 x-12\) \(f^{\prime}(x)=3\left(x^{2}-4 x-4\right)\) \(f^{\prime \prime}(\mathrm{x})=3(2 \mathrm{x}-4)\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}-4=0\) \(\Rightarrow x=\frac{4 \pm \sqrt{16+16}}{2 \times 1}=\frac{4 \pm 4 \sqrt{2}}{2}\) \(x=2 \pm 2 \sqrt{2}\) (These two points have either maxima or minima) So, \(\quad \mathrm{f}^{\prime \prime}(2)=3(2 \times 2-4)\) \(\mathrm{f}^{\prime \prime}(2)=0\) Hence, \(x=2 \text { is point of inflection. }\)
Shift-II]
Application of Derivatives
85715
Let \(f(x)=x^{2}+2 x+2, g(x)=-x^{2}+2 x-1\) and a, b be the extreme values of \(f(x), g(x)\) respectively. If \(c\) is the extreme value of \(\frac{f}{g}(x)\) (for \(x \neq 1\) ), then \(a+2 b+5 c+4=\)
85712
For the function \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5, x=2\) is a point of
1 local maxima
2 local minima
3 point of inflexion
4 none of these
Explanation:
(C) : Given, \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5\) ...(i) Differentiating with respect to \(\mathrm{x}\), we get \(f^{\prime}(x)=\frac{4}{3} 3 x^{2}-16 x+16=4 x^{2}-16 x+16\) Now for maximum/minimum we put \(f^{\prime}(\mathrm{x})=0\) \(\Rightarrow x^{2}-4 x+4=0 \Rightarrow(x-2)^{2}=0 \Rightarrow x=2\) \(f^{\prime \prime}(\mathrm{x})=8 \mathrm{x}-16,\left.f^{\prime \prime}(\mathrm{x})\right|_{\mathrm{x}=2}=0\) \(f^{\prime \prime}(\mathrm{x})=8 \geq 0\) \(\therefore \quad \mathrm{x}=2\) is the point of inflexion
AMU-2010
Application of Derivatives
85713
If the area of a circular sector of perimeter 60 \(m\) is to be maximized, then its radius must be
1 \({20}\)
2 15
3 10
4 5
Explanation:
(B): Given, the perimeter of circular sector \(=60 \mathrm{~m}\) Let radius be \(\mathrm{r} \mathrm{m}\). Length of \(\operatorname{arc} \mathrm{AB}=l\) \(\therefore\) perimeter \((\mathrm{p})=l+2 \mathrm{r}\) \(l =\mathrm{p}-2 \mathrm{r}\) \(\because \text { Area } =\frac{1}{2} l \mathrm{r}\) \(\mathrm{A}=\frac{(\mathrm{p}-2 \mathrm{r}) \mathrm{r}}{2}=\frac{\mathrm{pr}-2 \mathrm{r}^{2}}{2}\) For \(\mathrm{A}\) to be maximum, \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{d}{d r}\left(\frac{p r-2 r^{2}}{2}\right)=0 \Rightarrow \frac{p-4 r}{2}=0\) \(p=4 r \Rightarrow 60=4 r\) \(r=15 m\)
Shift-2]
Application of Derivatives
85714
For the function \(f(x)=x^{3}-6 x^{2}-12 x-3, x=2\) is
1 point of maxima
2 point of minima
3 point of inflection
4 not a critical point
Explanation:
(C): Given, \(f(x)=x^{3}-6 x^{2}-12 x-3\) Then, \(\quad f^{\prime}(x)=3 x^{2}-12 x-12\) \(f^{\prime}(x)=3\left(x^{2}-4 x-4\right)\) \(f^{\prime \prime}(\mathrm{x})=3(2 \mathrm{x}-4)\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}-4=0\) \(\Rightarrow x=\frac{4 \pm \sqrt{16+16}}{2 \times 1}=\frac{4 \pm 4 \sqrt{2}}{2}\) \(x=2 \pm 2 \sqrt{2}\) (These two points have either maxima or minima) So, \(\quad \mathrm{f}^{\prime \prime}(2)=3(2 \times 2-4)\) \(\mathrm{f}^{\prime \prime}(2)=0\) Hence, \(x=2 \text { is point of inflection. }\)
Shift-II]
Application of Derivatives
85715
Let \(f(x)=x^{2}+2 x+2, g(x)=-x^{2}+2 x-1\) and a, b be the extreme values of \(f(x), g(x)\) respectively. If \(c\) is the extreme value of \(\frac{f}{g}(x)\) (for \(x \neq 1\) ), then \(a+2 b+5 c+4=\)
85712
For the function \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5, x=2\) is a point of
1 local maxima
2 local minima
3 point of inflexion
4 none of these
Explanation:
(C) : Given, \(f(x)=\frac{4}{3} x^{3}-8 x^{2}+16 x+5\) ...(i) Differentiating with respect to \(\mathrm{x}\), we get \(f^{\prime}(x)=\frac{4}{3} 3 x^{2}-16 x+16=4 x^{2}-16 x+16\) Now for maximum/minimum we put \(f^{\prime}(\mathrm{x})=0\) \(\Rightarrow x^{2}-4 x+4=0 \Rightarrow(x-2)^{2}=0 \Rightarrow x=2\) \(f^{\prime \prime}(\mathrm{x})=8 \mathrm{x}-16,\left.f^{\prime \prime}(\mathrm{x})\right|_{\mathrm{x}=2}=0\) \(f^{\prime \prime}(\mathrm{x})=8 \geq 0\) \(\therefore \quad \mathrm{x}=2\) is the point of inflexion
AMU-2010
Application of Derivatives
85713
If the area of a circular sector of perimeter 60 \(m\) is to be maximized, then its radius must be
1 \({20}\)
2 15
3 10
4 5
Explanation:
(B): Given, the perimeter of circular sector \(=60 \mathrm{~m}\) Let radius be \(\mathrm{r} \mathrm{m}\). Length of \(\operatorname{arc} \mathrm{AB}=l\) \(\therefore\) perimeter \((\mathrm{p})=l+2 \mathrm{r}\) \(l =\mathrm{p}-2 \mathrm{r}\) \(\because \text { Area } =\frac{1}{2} l \mathrm{r}\) \(\mathrm{A}=\frac{(\mathrm{p}-2 \mathrm{r}) \mathrm{r}}{2}=\frac{\mathrm{pr}-2 \mathrm{r}^{2}}{2}\) For \(\mathrm{A}\) to be maximum, \(\frac{\mathrm{dA}}{\mathrm{dr}}=0\) \(\frac{d}{d r}\left(\frac{p r-2 r^{2}}{2}\right)=0 \Rightarrow \frac{p-4 r}{2}=0\) \(p=4 r \Rightarrow 60=4 r\) \(r=15 m\)
Shift-2]
Application of Derivatives
85714
For the function \(f(x)=x^{3}-6 x^{2}-12 x-3, x=2\) is
1 point of maxima
2 point of minima
3 point of inflection
4 not a critical point
Explanation:
(C): Given, \(f(x)=x^{3}-6 x^{2}-12 x-3\) Then, \(\quad f^{\prime}(x)=3 x^{2}-12 x-12\) \(f^{\prime}(x)=3\left(x^{2}-4 x-4\right)\) \(f^{\prime \prime}(\mathrm{x})=3(2 \mathrm{x}-4)\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \mathrm{x}^{2}-4 \mathrm{x}-4=0\) \(\Rightarrow x=\frac{4 \pm \sqrt{16+16}}{2 \times 1}=\frac{4 \pm 4 \sqrt{2}}{2}\) \(x=2 \pm 2 \sqrt{2}\) (These two points have either maxima or minima) So, \(\quad \mathrm{f}^{\prime \prime}(2)=3(2 \times 2-4)\) \(\mathrm{f}^{\prime \prime}(2)=0\) Hence, \(x=2 \text { is point of inflection. }\)
Shift-II]
Application of Derivatives
85715
Let \(f(x)=x^{2}+2 x+2, g(x)=-x^{2}+2 x-1\) and a, b be the extreme values of \(f(x), g(x)\) respectively. If \(c\) is the extreme value of \(\frac{f}{g}(x)\) (for \(x \neq 1\) ), then \(a+2 b+5 c+4=\)
