85601
If \(P\) is a point on the segment \(A B\) of length 12 \(\mathrm{cm}\), then the position of \(\mathrm{P}\) for \(\mathrm{AP}^{2}+\mathrm{BP}^{2}\) to be minimum is such that
1 \(\mathrm{P}\) divides \(\mathrm{BA}\) in the ratio 2:1 internally
2 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(4: 3\) internally
3 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(2: 3\) internally
4 \(\mathrm{P}\) is the midpoint of segment \(\mathrm{AB}\)
Explanation:
(D) : Let \(\mu=\mathrm{AP}^{2}+\mathrm{BP}^{2}\) \(\begin{aligned} & x+y=12 \\ & y=12-x\end{aligned}\) \(\mu=x^2+(12-x)^2\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{d \mu}{d x}=2 x+2(12-x)(-1)\) \(=2 x-24+2 x\) \(\frac{d \mu}{d x}=4 x-24\) For maximum or minimum \(\frac{d \mu}{d x}=0\) \(4 x-24=0\) \(x=6\) \(\frac{d^{2} \mu}{d x^{2}}=4>0\) \(\therefore \mathrm{AP}^{2}+\mathrm{BP}^{2}\) is minimum at \(\mathrm{x}=6\) Hence, \(\mathrm{P}\) is midpoint of \(\mathrm{AB}\)
MHT CET-2020
Application of Derivatives
85602
The maximum volume of a right circular cylinder if the sum of its radius and height is \(6 \mathrm{~m}\) is
85603
The minimum value of the function \(f(x)=x \log x\) is
1 \(-\mathrm{e}\)
2 e
3 \(\frac{1}{\mathrm{e}}\)
4 \(\frac{-1}{\mathrm{e}}\)
Explanation:
(D) : We have \(f(x)=x \log x\) Differentiating both side we get- \(f^{\prime}(x)=1+\log x\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1+\log \mathrm{x}=0\) \(\log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1}\) \(\mathrm{x}=\frac{1}{\mathrm{e}}\) Again differentiating w.r. \(t \mathrm{x}\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) At, \(\quad \mathrm{x}=\frac{1}{\mathrm{e}}, \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{e}>0\) Hence, at \(\mathrm{x}=\frac{1}{\mathrm{e}} \mathrm{f}(\mathrm{x})\) is minimum Minimum value of \(f(x)\) at \(x=\frac{1}{e}\) \(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \mathrm{e}^{-1}=-\frac{1}{\mathrm{e}}\)
MHT CET-2021
Application of Derivatives
85604
If \(f(x)=x+\frac{1}{x}, x \neq 0\), then local maximum and minimum value of function \(f\) are
1 1 and -1
2 2 and -2
3 -1 and 1
4 -2 and 2
Explanation:
(D) : Given, \(f(x)=x+\frac{1}{x}, x \neq 0\) On differentiating w.r.t \(\mathrm{x}\) we get- \(\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^{2}}\) Again Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(f^{\prime \prime}(x)=\frac{2}{x^{3}}\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{1}{\mathrm{x}^{2}}=0\) \(\frac{1}{\mathrm{x}^{2}}=1\) \(\mathrm{x}^{2}=1\) \(\mathrm{x}= \pm 1\) \(x=-1\) is point of local maxima and \(x=1\) is point of local minima Now, for local maxima value \(f(-1)=-1+\frac{1}{-1}=-2\) And local minima value \(f(1)=1+\frac{1}{1}=2\)
85601
If \(P\) is a point on the segment \(A B\) of length 12 \(\mathrm{cm}\), then the position of \(\mathrm{P}\) for \(\mathrm{AP}^{2}+\mathrm{BP}^{2}\) to be minimum is such that
1 \(\mathrm{P}\) divides \(\mathrm{BA}\) in the ratio 2:1 internally
2 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(4: 3\) internally
3 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(2: 3\) internally
4 \(\mathrm{P}\) is the midpoint of segment \(\mathrm{AB}\)
Explanation:
(D) : Let \(\mu=\mathrm{AP}^{2}+\mathrm{BP}^{2}\) \(\begin{aligned} & x+y=12 \\ & y=12-x\end{aligned}\) \(\mu=x^2+(12-x)^2\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{d \mu}{d x}=2 x+2(12-x)(-1)\) \(=2 x-24+2 x\) \(\frac{d \mu}{d x}=4 x-24\) For maximum or minimum \(\frac{d \mu}{d x}=0\) \(4 x-24=0\) \(x=6\) \(\frac{d^{2} \mu}{d x^{2}}=4>0\) \(\therefore \mathrm{AP}^{2}+\mathrm{BP}^{2}\) is minimum at \(\mathrm{x}=6\) Hence, \(\mathrm{P}\) is midpoint of \(\mathrm{AB}\)
MHT CET-2020
Application of Derivatives
85602
The maximum volume of a right circular cylinder if the sum of its radius and height is \(6 \mathrm{~m}\) is
85603
The minimum value of the function \(f(x)=x \log x\) is
1 \(-\mathrm{e}\)
2 e
3 \(\frac{1}{\mathrm{e}}\)
4 \(\frac{-1}{\mathrm{e}}\)
Explanation:
(D) : We have \(f(x)=x \log x\) Differentiating both side we get- \(f^{\prime}(x)=1+\log x\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1+\log \mathrm{x}=0\) \(\log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1}\) \(\mathrm{x}=\frac{1}{\mathrm{e}}\) Again differentiating w.r. \(t \mathrm{x}\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) At, \(\quad \mathrm{x}=\frac{1}{\mathrm{e}}, \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{e}>0\) Hence, at \(\mathrm{x}=\frac{1}{\mathrm{e}} \mathrm{f}(\mathrm{x})\) is minimum Minimum value of \(f(x)\) at \(x=\frac{1}{e}\) \(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \mathrm{e}^{-1}=-\frac{1}{\mathrm{e}}\)
MHT CET-2021
Application of Derivatives
85604
If \(f(x)=x+\frac{1}{x}, x \neq 0\), then local maximum and minimum value of function \(f\) are
1 1 and -1
2 2 and -2
3 -1 and 1
4 -2 and 2
Explanation:
(D) : Given, \(f(x)=x+\frac{1}{x}, x \neq 0\) On differentiating w.r.t \(\mathrm{x}\) we get- \(\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^{2}}\) Again Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(f^{\prime \prime}(x)=\frac{2}{x^{3}}\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{1}{\mathrm{x}^{2}}=0\) \(\frac{1}{\mathrm{x}^{2}}=1\) \(\mathrm{x}^{2}=1\) \(\mathrm{x}= \pm 1\) \(x=-1\) is point of local maxima and \(x=1\) is point of local minima Now, for local maxima value \(f(-1)=-1+\frac{1}{-1}=-2\) And local minima value \(f(1)=1+\frac{1}{1}=2\)
85601
If \(P\) is a point on the segment \(A B\) of length 12 \(\mathrm{cm}\), then the position of \(\mathrm{P}\) for \(\mathrm{AP}^{2}+\mathrm{BP}^{2}\) to be minimum is such that
1 \(\mathrm{P}\) divides \(\mathrm{BA}\) in the ratio 2:1 internally
2 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(4: 3\) internally
3 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(2: 3\) internally
4 \(\mathrm{P}\) is the midpoint of segment \(\mathrm{AB}\)
Explanation:
(D) : Let \(\mu=\mathrm{AP}^{2}+\mathrm{BP}^{2}\) \(\begin{aligned} & x+y=12 \\ & y=12-x\end{aligned}\) \(\mu=x^2+(12-x)^2\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{d \mu}{d x}=2 x+2(12-x)(-1)\) \(=2 x-24+2 x\) \(\frac{d \mu}{d x}=4 x-24\) For maximum or minimum \(\frac{d \mu}{d x}=0\) \(4 x-24=0\) \(x=6\) \(\frac{d^{2} \mu}{d x^{2}}=4>0\) \(\therefore \mathrm{AP}^{2}+\mathrm{BP}^{2}\) is minimum at \(\mathrm{x}=6\) Hence, \(\mathrm{P}\) is midpoint of \(\mathrm{AB}\)
MHT CET-2020
Application of Derivatives
85602
The maximum volume of a right circular cylinder if the sum of its radius and height is \(6 \mathrm{~m}\) is
85603
The minimum value of the function \(f(x)=x \log x\) is
1 \(-\mathrm{e}\)
2 e
3 \(\frac{1}{\mathrm{e}}\)
4 \(\frac{-1}{\mathrm{e}}\)
Explanation:
(D) : We have \(f(x)=x \log x\) Differentiating both side we get- \(f^{\prime}(x)=1+\log x\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1+\log \mathrm{x}=0\) \(\log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1}\) \(\mathrm{x}=\frac{1}{\mathrm{e}}\) Again differentiating w.r. \(t \mathrm{x}\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) At, \(\quad \mathrm{x}=\frac{1}{\mathrm{e}}, \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{e}>0\) Hence, at \(\mathrm{x}=\frac{1}{\mathrm{e}} \mathrm{f}(\mathrm{x})\) is minimum Minimum value of \(f(x)\) at \(x=\frac{1}{e}\) \(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \mathrm{e}^{-1}=-\frac{1}{\mathrm{e}}\)
MHT CET-2021
Application of Derivatives
85604
If \(f(x)=x+\frac{1}{x}, x \neq 0\), then local maximum and minimum value of function \(f\) are
1 1 and -1
2 2 and -2
3 -1 and 1
4 -2 and 2
Explanation:
(D) : Given, \(f(x)=x+\frac{1}{x}, x \neq 0\) On differentiating w.r.t \(\mathrm{x}\) we get- \(\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^{2}}\) Again Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(f^{\prime \prime}(x)=\frac{2}{x^{3}}\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{1}{\mathrm{x}^{2}}=0\) \(\frac{1}{\mathrm{x}^{2}}=1\) \(\mathrm{x}^{2}=1\) \(\mathrm{x}= \pm 1\) \(x=-1\) is point of local maxima and \(x=1\) is point of local minima Now, for local maxima value \(f(-1)=-1+\frac{1}{-1}=-2\) And local minima value \(f(1)=1+\frac{1}{1}=2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85601
If \(P\) is a point on the segment \(A B\) of length 12 \(\mathrm{cm}\), then the position of \(\mathrm{P}\) for \(\mathrm{AP}^{2}+\mathrm{BP}^{2}\) to be minimum is such that
1 \(\mathrm{P}\) divides \(\mathrm{BA}\) in the ratio 2:1 internally
2 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(4: 3\) internally
3 \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(2: 3\) internally
4 \(\mathrm{P}\) is the midpoint of segment \(\mathrm{AB}\)
Explanation:
(D) : Let \(\mu=\mathrm{AP}^{2}+\mathrm{BP}^{2}\) \(\begin{aligned} & x+y=12 \\ & y=12-x\end{aligned}\) \(\mu=x^2+(12-x)^2\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{d \mu}{d x}=2 x+2(12-x)(-1)\) \(=2 x-24+2 x\) \(\frac{d \mu}{d x}=4 x-24\) For maximum or minimum \(\frac{d \mu}{d x}=0\) \(4 x-24=0\) \(x=6\) \(\frac{d^{2} \mu}{d x^{2}}=4>0\) \(\therefore \mathrm{AP}^{2}+\mathrm{BP}^{2}\) is minimum at \(\mathrm{x}=6\) Hence, \(\mathrm{P}\) is midpoint of \(\mathrm{AB}\)
MHT CET-2020
Application of Derivatives
85602
The maximum volume of a right circular cylinder if the sum of its radius and height is \(6 \mathrm{~m}\) is
85603
The minimum value of the function \(f(x)=x \log x\) is
1 \(-\mathrm{e}\)
2 e
3 \(\frac{1}{\mathrm{e}}\)
4 \(\frac{-1}{\mathrm{e}}\)
Explanation:
(D) : We have \(f(x)=x \log x\) Differentiating both side we get- \(f^{\prime}(x)=1+\log x\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1+\log \mathrm{x}=0\) \(\log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1}\) \(\mathrm{x}=\frac{1}{\mathrm{e}}\) Again differentiating w.r. \(t \mathrm{x}\) we get- \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) At, \(\quad \mathrm{x}=\frac{1}{\mathrm{e}}, \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{e}>0\) Hence, at \(\mathrm{x}=\frac{1}{\mathrm{e}} \mathrm{f}(\mathrm{x})\) is minimum Minimum value of \(f(x)\) at \(x=\frac{1}{e}\) \(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \left(\frac{1}{\mathrm{e}}\right)=\frac{1}{\mathrm{e}} \log \mathrm{e}^{-1}=-\frac{1}{\mathrm{e}}\)
MHT CET-2021
Application of Derivatives
85604
If \(f(x)=x+\frac{1}{x}, x \neq 0\), then local maximum and minimum value of function \(f\) are
1 1 and -1
2 2 and -2
3 -1 and 1
4 -2 and 2
Explanation:
(D) : Given, \(f(x)=x+\frac{1}{x}, x \neq 0\) On differentiating w.r.t \(\mathrm{x}\) we get- \(\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^{2}}\) Again Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(f^{\prime \prime}(x)=\frac{2}{x^{3}}\) For maxima or minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{1}{\mathrm{x}^{2}}=0\) \(\frac{1}{\mathrm{x}^{2}}=1\) \(\mathrm{x}^{2}=1\) \(\mathrm{x}= \pm 1\) \(x=-1\) is point of local maxima and \(x=1\) is point of local minima Now, for local maxima value \(f(-1)=-1+\frac{1}{-1}=-2\) And local minima value \(f(1)=1+\frac{1}{1}=2\)
