85614
A differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+\) \(a x+b\) has a relative minimum at \(x=0\) for
1 all a and all b
2 all \(b\) if \(a=0\)
3 all \(\mathrm{b}>0\)
4 all \(\mathrm{a}>0\)
Explanation:
(B) : \(y=f(x)+a x+b\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{a} \because\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=0}=0 \Rightarrow \mathrm{f}^{\prime}(0)+\mathrm{a}=0\) \(\Rightarrow \mathrm{a}=0\) \(\left[\because \mathrm{f}^{\prime}(0)=0\right]\) Also, \(\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=0}>0 \quad\left[\because \mathrm{f}^{\prime}(0)>0\right]\) \(\Rightarrow \mathrm{y}\) has relative minimum at \(\mathrm{x}=0\) for all \(\mathrm{b}\), If \(\mathrm{a}=0\)
SRM JEE-2009
Application of Derivatives
85615
For a curve \(y=x^{x}\), the point
1 \(x=-1\) is a point of minimum
2 \(x=0\) is a point of minimum
3 \(x=-1\) is a point of maximum
4 \(x=0\) is a point of maximum
Explanation:
(A): We have, \(\mathrm{y}=\mathrm{xe}^{\mathrm{x}} \quad\).....(i) Differentiating (i) w.r.t. \(x\), we get \(\frac{d y}{d x}=x e^{x}+e^{x}(x+1)\) Again differentiating w.r.t. \(x\), we get \(\frac{d^{2} y}{d x^{2}}=e^{x}(x+1)+e^{x}=e^{x}(x+2)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}+1=0 \quad\left[\because \mathrm{e}^{\mathrm{x}} \neq 0\right]\) \(\Rightarrow \mathrm{x}=-1\) \(\left[\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=-1}=\mathrm{e}^{-1}(-1+2)=\frac{1}{\mathrm{e}}>0\) \(\therefore \mathrm{x}=-1\) is a point of minimum.
SRM JEE-2010
Application of Derivatives
85616
If \(y=\operatorname{alog} x+b x^{2}+x\) has its extremum value at \(x=-1\) and \(x=2\) then
1 \(a=2, b=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-1 / 2, \mathrm{~b}=1 / 2\)
4 None of these
Explanation:
(B) : We have, \(y=a \log x+b x^{2}+x\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Since, function has extremum value at \(\mathrm{x}=-1\) And \(\mathrm{x}=2\). \(\therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=0\) and \(\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=2}=0\) \(\Rightarrow-\mathrm{a}-2 \mathrm{~b}+1=0\) and \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) \(\Rightarrow a+2 b=1 \ldots\)..i) and \(a+8 b=-2\) Solving (i) and (ii), we get \(\mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
BCECE-2007
Application of Derivatives
85617
If \(|z+4| \leq 3\) then the greatest and the least values of \(|\mathbf{z}+\mathbf{1}|\) are
85614
A differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+\) \(a x+b\) has a relative minimum at \(x=0\) for
1 all a and all b
2 all \(b\) if \(a=0\)
3 all \(\mathrm{b}>0\)
4 all \(\mathrm{a}>0\)
Explanation:
(B) : \(y=f(x)+a x+b\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{a} \because\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=0}=0 \Rightarrow \mathrm{f}^{\prime}(0)+\mathrm{a}=0\) \(\Rightarrow \mathrm{a}=0\) \(\left[\because \mathrm{f}^{\prime}(0)=0\right]\) Also, \(\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=0}>0 \quad\left[\because \mathrm{f}^{\prime}(0)>0\right]\) \(\Rightarrow \mathrm{y}\) has relative minimum at \(\mathrm{x}=0\) for all \(\mathrm{b}\), If \(\mathrm{a}=0\)
SRM JEE-2009
Application of Derivatives
85615
For a curve \(y=x^{x}\), the point
1 \(x=-1\) is a point of minimum
2 \(x=0\) is a point of minimum
3 \(x=-1\) is a point of maximum
4 \(x=0\) is a point of maximum
Explanation:
(A): We have, \(\mathrm{y}=\mathrm{xe}^{\mathrm{x}} \quad\).....(i) Differentiating (i) w.r.t. \(x\), we get \(\frac{d y}{d x}=x e^{x}+e^{x}(x+1)\) Again differentiating w.r.t. \(x\), we get \(\frac{d^{2} y}{d x^{2}}=e^{x}(x+1)+e^{x}=e^{x}(x+2)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}+1=0 \quad\left[\because \mathrm{e}^{\mathrm{x}} \neq 0\right]\) \(\Rightarrow \mathrm{x}=-1\) \(\left[\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=-1}=\mathrm{e}^{-1}(-1+2)=\frac{1}{\mathrm{e}}>0\) \(\therefore \mathrm{x}=-1\) is a point of minimum.
SRM JEE-2010
Application of Derivatives
85616
If \(y=\operatorname{alog} x+b x^{2}+x\) has its extremum value at \(x=-1\) and \(x=2\) then
1 \(a=2, b=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-1 / 2, \mathrm{~b}=1 / 2\)
4 None of these
Explanation:
(B) : We have, \(y=a \log x+b x^{2}+x\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Since, function has extremum value at \(\mathrm{x}=-1\) And \(\mathrm{x}=2\). \(\therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=0\) and \(\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=2}=0\) \(\Rightarrow-\mathrm{a}-2 \mathrm{~b}+1=0\) and \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) \(\Rightarrow a+2 b=1 \ldots\)..i) and \(a+8 b=-2\) Solving (i) and (ii), we get \(\mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
BCECE-2007
Application of Derivatives
85617
If \(|z+4| \leq 3\) then the greatest and the least values of \(|\mathbf{z}+\mathbf{1}|\) are
85614
A differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+\) \(a x+b\) has a relative minimum at \(x=0\) for
1 all a and all b
2 all \(b\) if \(a=0\)
3 all \(\mathrm{b}>0\)
4 all \(\mathrm{a}>0\)
Explanation:
(B) : \(y=f(x)+a x+b\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{a} \because\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=0}=0 \Rightarrow \mathrm{f}^{\prime}(0)+\mathrm{a}=0\) \(\Rightarrow \mathrm{a}=0\) \(\left[\because \mathrm{f}^{\prime}(0)=0\right]\) Also, \(\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=0}>0 \quad\left[\because \mathrm{f}^{\prime}(0)>0\right]\) \(\Rightarrow \mathrm{y}\) has relative minimum at \(\mathrm{x}=0\) for all \(\mathrm{b}\), If \(\mathrm{a}=0\)
SRM JEE-2009
Application of Derivatives
85615
For a curve \(y=x^{x}\), the point
1 \(x=-1\) is a point of minimum
2 \(x=0\) is a point of minimum
3 \(x=-1\) is a point of maximum
4 \(x=0\) is a point of maximum
Explanation:
(A): We have, \(\mathrm{y}=\mathrm{xe}^{\mathrm{x}} \quad\).....(i) Differentiating (i) w.r.t. \(x\), we get \(\frac{d y}{d x}=x e^{x}+e^{x}(x+1)\) Again differentiating w.r.t. \(x\), we get \(\frac{d^{2} y}{d x^{2}}=e^{x}(x+1)+e^{x}=e^{x}(x+2)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}+1=0 \quad\left[\because \mathrm{e}^{\mathrm{x}} \neq 0\right]\) \(\Rightarrow \mathrm{x}=-1\) \(\left[\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=-1}=\mathrm{e}^{-1}(-1+2)=\frac{1}{\mathrm{e}}>0\) \(\therefore \mathrm{x}=-1\) is a point of minimum.
SRM JEE-2010
Application of Derivatives
85616
If \(y=\operatorname{alog} x+b x^{2}+x\) has its extremum value at \(x=-1\) and \(x=2\) then
1 \(a=2, b=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-1 / 2, \mathrm{~b}=1 / 2\)
4 None of these
Explanation:
(B) : We have, \(y=a \log x+b x^{2}+x\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Since, function has extremum value at \(\mathrm{x}=-1\) And \(\mathrm{x}=2\). \(\therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=0\) and \(\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=2}=0\) \(\Rightarrow-\mathrm{a}-2 \mathrm{~b}+1=0\) and \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) \(\Rightarrow a+2 b=1 \ldots\)..i) and \(a+8 b=-2\) Solving (i) and (ii), we get \(\mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
BCECE-2007
Application of Derivatives
85617
If \(|z+4| \leq 3\) then the greatest and the least values of \(|\mathbf{z}+\mathbf{1}|\) are
85614
A differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+\) \(a x+b\) has a relative minimum at \(x=0\) for
1 all a and all b
2 all \(b\) if \(a=0\)
3 all \(\mathrm{b}>0\)
4 all \(\mathrm{a}>0\)
Explanation:
(B) : \(y=f(x)+a x+b\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{a} \because\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=0}=0 \Rightarrow \mathrm{f}^{\prime}(0)+\mathrm{a}=0\) \(\Rightarrow \mathrm{a}=0\) \(\left[\because \mathrm{f}^{\prime}(0)=0\right]\) Also, \(\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=0}>0 \quad\left[\because \mathrm{f}^{\prime}(0)>0\right]\) \(\Rightarrow \mathrm{y}\) has relative minimum at \(\mathrm{x}=0\) for all \(\mathrm{b}\), If \(\mathrm{a}=0\)
SRM JEE-2009
Application of Derivatives
85615
For a curve \(y=x^{x}\), the point
1 \(x=-1\) is a point of minimum
2 \(x=0\) is a point of minimum
3 \(x=-1\) is a point of maximum
4 \(x=0\) is a point of maximum
Explanation:
(A): We have, \(\mathrm{y}=\mathrm{xe}^{\mathrm{x}} \quad\).....(i) Differentiating (i) w.r.t. \(x\), we get \(\frac{d y}{d x}=x e^{x}+e^{x}(x+1)\) Again differentiating w.r.t. \(x\), we get \(\frac{d^{2} y}{d x^{2}}=e^{x}(x+1)+e^{x}=e^{x}(x+2)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}+1=0 \quad\left[\because \mathrm{e}^{\mathrm{x}} \neq 0\right]\) \(\Rightarrow \mathrm{x}=-1\) \(\left[\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=-1}=\mathrm{e}^{-1}(-1+2)=\frac{1}{\mathrm{e}}>0\) \(\therefore \mathrm{x}=-1\) is a point of minimum.
SRM JEE-2010
Application of Derivatives
85616
If \(y=\operatorname{alog} x+b x^{2}+x\) has its extremum value at \(x=-1\) and \(x=2\) then
1 \(a=2, b=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-1 / 2, \mathrm{~b}=1 / 2\)
4 None of these
Explanation:
(B) : We have, \(y=a \log x+b x^{2}+x\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Since, function has extremum value at \(\mathrm{x}=-1\) And \(\mathrm{x}=2\). \(\therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=0\) and \(\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=2}=0\) \(\Rightarrow-\mathrm{a}-2 \mathrm{~b}+1=0\) and \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) \(\Rightarrow a+2 b=1 \ldots\)..i) and \(a+8 b=-2\) Solving (i) and (ii), we get \(\mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
BCECE-2007
Application of Derivatives
85617
If \(|z+4| \leq 3\) then the greatest and the least values of \(|\mathbf{z}+\mathbf{1}|\) are
85614
A differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+\) \(a x+b\) has a relative minimum at \(x=0\) for
1 all a and all b
2 all \(b\) if \(a=0\)
3 all \(\mathrm{b}>0\)
4 all \(\mathrm{a}>0\)
Explanation:
(B) : \(y=f(x)+a x+b\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{a} \because\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=0}=0 \Rightarrow \mathrm{f}^{\prime}(0)+\mathrm{a}=0\) \(\Rightarrow \mathrm{a}=0\) \(\left[\because \mathrm{f}^{\prime}(0)=0\right]\) Also, \(\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=0}>0 \quad\left[\because \mathrm{f}^{\prime}(0)>0\right]\) \(\Rightarrow \mathrm{y}\) has relative minimum at \(\mathrm{x}=0\) for all \(\mathrm{b}\), If \(\mathrm{a}=0\)
SRM JEE-2009
Application of Derivatives
85615
For a curve \(y=x^{x}\), the point
1 \(x=-1\) is a point of minimum
2 \(x=0\) is a point of minimum
3 \(x=-1\) is a point of maximum
4 \(x=0\) is a point of maximum
Explanation:
(A): We have, \(\mathrm{y}=\mathrm{xe}^{\mathrm{x}} \quad\).....(i) Differentiating (i) w.r.t. \(x\), we get \(\frac{d y}{d x}=x e^{x}+e^{x}(x+1)\) Again differentiating w.r.t. \(x\), we get \(\frac{d^{2} y}{d x^{2}}=e^{x}(x+1)+e^{x}=e^{x}(x+2)\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\Rightarrow \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)=0 \Rightarrow \mathrm{x}+1=0 \quad\left[\because \mathrm{e}^{\mathrm{x}} \neq 0\right]\) \(\Rightarrow \mathrm{x}=-1\) \(\left[\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right]_{\mathrm{x}=-1}=\mathrm{e}^{-1}(-1+2)=\frac{1}{\mathrm{e}}>0\) \(\therefore \mathrm{x}=-1\) is a point of minimum.
SRM JEE-2010
Application of Derivatives
85616
If \(y=\operatorname{alog} x+b x^{2}+x\) has its extremum value at \(x=-1\) and \(x=2\) then
1 \(a=2, b=-1\)
2 \(\mathrm{a}=2, \mathrm{~b}=-1 / 2\)
3 \(\mathrm{a}=-1 / 2, \mathrm{~b}=1 / 2\)
4 None of these
Explanation:
(B) : We have, \(y=a \log x+b x^{2}+x\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=\frac{a}{x}+2 b x+1\) Since, function has extremum value at \(\mathrm{x}=-1\) And \(\mathrm{x}=2\). \(\therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=0\) and \(\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=2}=0\) \(\Rightarrow-\mathrm{a}-2 \mathrm{~b}+1=0\) and \(\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0\) \(\Rightarrow a+2 b=1 \ldots\)..i) and \(a+8 b=-2\) Solving (i) and (ii), we get \(\mathrm{a}=2, \mathrm{~b}=\frac{-1}{2}\)
BCECE-2007
Application of Derivatives
85617
If \(|z+4| \leq 3\) then the greatest and the least values of \(|\mathbf{z}+\mathbf{1}|\) are
