85605
If \(f(x)=3 x^{3}-9 x^{2}-27 x+15\), then the maximum value of \(f(x)\) is
1 -30
2 -66
3 66
4 30
Explanation:
(D) : \(f(x)=3 x^{3}-9 x^{2}-27 x+15\) On differenting both side we get- \(f^{\prime}(x)=9 x^{2}-18 x-27\) For maximum or minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(9 \mathrm{x}^{2}-18 \mathrm{x}-27=0\) \(\mathrm{x}^{2}-2 \mathrm{x}-3=0\) \(\mathrm{x}^{2}-3 \mathrm{x}+\mathrm{x}-3=0\) \(\mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0\) \(\mathrm{x}=-1 \text { and } 3\) \(\mathrm{x}=-1 \text { maxima }\) \(\text { and } \mathrm{x}=3 \text { minima }\) For maximum value \(\mathrm{f}(-1)=3(-1)^{3}-9(-1)^{2}-27(-1)+15\) \(=-3-9+27+15\) \(=30\)
Application of Derivatives
85606
A stone is dropped into a pond. Waves in the form of circles are generated and radius of outermost ripple increases at the rate of 5 \(\mathrm{cm} / \mathrm{sec}\). Then area increased after 2 seconds is
1 \(100 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
2 \(25 \mathrm{~cm}^{2} / \mathrm{sec}\)
3 \(50 \mathrm{~cm}^{2} / \mathrm{sec}\)
4 \(40 \mathrm{~cm}^{2} / \mathrm{sec}\)
Explanation:
(A) : We have given, \(\frac{\mathrm{dr}}{\mathrm{dt}}=5 \mathrm{~cm} / \mathrm{sec}\) \(\mathrm{dr}=5 \mathrm{dt}\) By integrating, we get - Then \(\mathrm{r}=5 \mathrm{t}\) \(\text { At, } \mathrm{t}=2\) \(\mathrm{r}=10\) After 2 second \((t=2)\) Area of circle, \(\mathrm{A}=\pi \mathrm{r}^{2}\) \(\frac{\mathrm{dA}}{\mathrm{dr}}=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\) \(=2 \times \pi \times 10 \times 5\) \(=100 \pi\)
MHT CET-2019
Application of Derivatives
85607
For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
85608
If \(x+y=8\), then maximum value of \(x^{2} y\) is
1 \(\frac{2048}{3}\)
2 \(\frac{2048}{27}\)
3 \(\frac{2048}{7}\)
4 \(\frac{2048}{81}\)
Explanation:
(B) : We have Let, \(\begin{aligned} y=8-x \\ p=x^{2} y \\ p=x^{2}(8-x) \\ p=8 x^{2}-x^{3}\end{aligned}\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{\mathrm{dp}}{\mathrm{dx}}=16 x-3 \mathrm{x}^{2}\) For maxima or minima \(\frac{d p}{d x}=0\) \(16 x-3 x^{2}=0\) \(16-3 x=0\) \(x=\frac{16}{3}\) Now maximum value, \(x^{2} y=\left(\frac{16}{3}\right)^{2}\left(8-\frac{16}{3}\right)=\frac{16}{3} \times \frac{16}{3} \times \frac{8}{3}\) \(=\frac{2048}{27}\)
85605
If \(f(x)=3 x^{3}-9 x^{2}-27 x+15\), then the maximum value of \(f(x)\) is
1 -30
2 -66
3 66
4 30
Explanation:
(D) : \(f(x)=3 x^{3}-9 x^{2}-27 x+15\) On differenting both side we get- \(f^{\prime}(x)=9 x^{2}-18 x-27\) For maximum or minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(9 \mathrm{x}^{2}-18 \mathrm{x}-27=0\) \(\mathrm{x}^{2}-2 \mathrm{x}-3=0\) \(\mathrm{x}^{2}-3 \mathrm{x}+\mathrm{x}-3=0\) \(\mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0\) \(\mathrm{x}=-1 \text { and } 3\) \(\mathrm{x}=-1 \text { maxima }\) \(\text { and } \mathrm{x}=3 \text { minima }\) For maximum value \(\mathrm{f}(-1)=3(-1)^{3}-9(-1)^{2}-27(-1)+15\) \(=-3-9+27+15\) \(=30\)
Application of Derivatives
85606
A stone is dropped into a pond. Waves in the form of circles are generated and radius of outermost ripple increases at the rate of 5 \(\mathrm{cm} / \mathrm{sec}\). Then area increased after 2 seconds is
1 \(100 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
2 \(25 \mathrm{~cm}^{2} / \mathrm{sec}\)
3 \(50 \mathrm{~cm}^{2} / \mathrm{sec}\)
4 \(40 \mathrm{~cm}^{2} / \mathrm{sec}\)
Explanation:
(A) : We have given, \(\frac{\mathrm{dr}}{\mathrm{dt}}=5 \mathrm{~cm} / \mathrm{sec}\) \(\mathrm{dr}=5 \mathrm{dt}\) By integrating, we get - Then \(\mathrm{r}=5 \mathrm{t}\) \(\text { At, } \mathrm{t}=2\) \(\mathrm{r}=10\) After 2 second \((t=2)\) Area of circle, \(\mathrm{A}=\pi \mathrm{r}^{2}\) \(\frac{\mathrm{dA}}{\mathrm{dr}}=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\) \(=2 \times \pi \times 10 \times 5\) \(=100 \pi\)
MHT CET-2019
Application of Derivatives
85607
For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
85608
If \(x+y=8\), then maximum value of \(x^{2} y\) is
1 \(\frac{2048}{3}\)
2 \(\frac{2048}{27}\)
3 \(\frac{2048}{7}\)
4 \(\frac{2048}{81}\)
Explanation:
(B) : We have Let, \(\begin{aligned} y=8-x \\ p=x^{2} y \\ p=x^{2}(8-x) \\ p=8 x^{2}-x^{3}\end{aligned}\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{\mathrm{dp}}{\mathrm{dx}}=16 x-3 \mathrm{x}^{2}\) For maxima or minima \(\frac{d p}{d x}=0\) \(16 x-3 x^{2}=0\) \(16-3 x=0\) \(x=\frac{16}{3}\) Now maximum value, \(x^{2} y=\left(\frac{16}{3}\right)^{2}\left(8-\frac{16}{3}\right)=\frac{16}{3} \times \frac{16}{3} \times \frac{8}{3}\) \(=\frac{2048}{27}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85605
If \(f(x)=3 x^{3}-9 x^{2}-27 x+15\), then the maximum value of \(f(x)\) is
1 -30
2 -66
3 66
4 30
Explanation:
(D) : \(f(x)=3 x^{3}-9 x^{2}-27 x+15\) On differenting both side we get- \(f^{\prime}(x)=9 x^{2}-18 x-27\) For maximum or minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(9 \mathrm{x}^{2}-18 \mathrm{x}-27=0\) \(\mathrm{x}^{2}-2 \mathrm{x}-3=0\) \(\mathrm{x}^{2}-3 \mathrm{x}+\mathrm{x}-3=0\) \(\mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0\) \(\mathrm{x}=-1 \text { and } 3\) \(\mathrm{x}=-1 \text { maxima }\) \(\text { and } \mathrm{x}=3 \text { minima }\) For maximum value \(\mathrm{f}(-1)=3(-1)^{3}-9(-1)^{2}-27(-1)+15\) \(=-3-9+27+15\) \(=30\)
Application of Derivatives
85606
A stone is dropped into a pond. Waves in the form of circles are generated and radius of outermost ripple increases at the rate of 5 \(\mathrm{cm} / \mathrm{sec}\). Then area increased after 2 seconds is
1 \(100 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
2 \(25 \mathrm{~cm}^{2} / \mathrm{sec}\)
3 \(50 \mathrm{~cm}^{2} / \mathrm{sec}\)
4 \(40 \mathrm{~cm}^{2} / \mathrm{sec}\)
Explanation:
(A) : We have given, \(\frac{\mathrm{dr}}{\mathrm{dt}}=5 \mathrm{~cm} / \mathrm{sec}\) \(\mathrm{dr}=5 \mathrm{dt}\) By integrating, we get - Then \(\mathrm{r}=5 \mathrm{t}\) \(\text { At, } \mathrm{t}=2\) \(\mathrm{r}=10\) After 2 second \((t=2)\) Area of circle, \(\mathrm{A}=\pi \mathrm{r}^{2}\) \(\frac{\mathrm{dA}}{\mathrm{dr}}=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\) \(=2 \times \pi \times 10 \times 5\) \(=100 \pi\)
MHT CET-2019
Application of Derivatives
85607
For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
85608
If \(x+y=8\), then maximum value of \(x^{2} y\) is
1 \(\frac{2048}{3}\)
2 \(\frac{2048}{27}\)
3 \(\frac{2048}{7}\)
4 \(\frac{2048}{81}\)
Explanation:
(B) : We have Let, \(\begin{aligned} y=8-x \\ p=x^{2} y \\ p=x^{2}(8-x) \\ p=8 x^{2}-x^{3}\end{aligned}\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{\mathrm{dp}}{\mathrm{dx}}=16 x-3 \mathrm{x}^{2}\) For maxima or minima \(\frac{d p}{d x}=0\) \(16 x-3 x^{2}=0\) \(16-3 x=0\) \(x=\frac{16}{3}\) Now maximum value, \(x^{2} y=\left(\frac{16}{3}\right)^{2}\left(8-\frac{16}{3}\right)=\frac{16}{3} \times \frac{16}{3} \times \frac{8}{3}\) \(=\frac{2048}{27}\)
85605
If \(f(x)=3 x^{3}-9 x^{2}-27 x+15\), then the maximum value of \(f(x)\) is
1 -30
2 -66
3 66
4 30
Explanation:
(D) : \(f(x)=3 x^{3}-9 x^{2}-27 x+15\) On differenting both side we get- \(f^{\prime}(x)=9 x^{2}-18 x-27\) For maximum or minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(9 \mathrm{x}^{2}-18 \mathrm{x}-27=0\) \(\mathrm{x}^{2}-2 \mathrm{x}-3=0\) \(\mathrm{x}^{2}-3 \mathrm{x}+\mathrm{x}-3=0\) \(\mathrm{x}(\mathrm{x}-3)+1(\mathrm{x}-3)=0\) \(\mathrm{x}=-1 \text { and } 3\) \(\mathrm{x}=-1 \text { maxima }\) \(\text { and } \mathrm{x}=3 \text { minima }\) For maximum value \(\mathrm{f}(-1)=3(-1)^{3}-9(-1)^{2}-27(-1)+15\) \(=-3-9+27+15\) \(=30\)
Application of Derivatives
85606
A stone is dropped into a pond. Waves in the form of circles are generated and radius of outermost ripple increases at the rate of 5 \(\mathrm{cm} / \mathrm{sec}\). Then area increased after 2 seconds is
1 \(100 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
2 \(25 \mathrm{~cm}^{2} / \mathrm{sec}\)
3 \(50 \mathrm{~cm}^{2} / \mathrm{sec}\)
4 \(40 \mathrm{~cm}^{2} / \mathrm{sec}\)
Explanation:
(A) : We have given, \(\frac{\mathrm{dr}}{\mathrm{dt}}=5 \mathrm{~cm} / \mathrm{sec}\) \(\mathrm{dr}=5 \mathrm{dt}\) By integrating, we get - Then \(\mathrm{r}=5 \mathrm{t}\) \(\text { At, } \mathrm{t}=2\) \(\mathrm{r}=10\) After 2 second \((t=2)\) Area of circle, \(\mathrm{A}=\pi \mathrm{r}^{2}\) \(\frac{\mathrm{dA}}{\mathrm{dr}}=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\) \(=2 \times \pi \times 10 \times 5\) \(=100 \pi\)
MHT CET-2019
Application of Derivatives
85607
For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
85608
If \(x+y=8\), then maximum value of \(x^{2} y\) is
1 \(\frac{2048}{3}\)
2 \(\frac{2048}{27}\)
3 \(\frac{2048}{7}\)
4 \(\frac{2048}{81}\)
Explanation:
(B) : We have Let, \(\begin{aligned} y=8-x \\ p=x^{2} y \\ p=x^{2}(8-x) \\ p=8 x^{2}-x^{3}\end{aligned}\) Differentiating both sides w.r.t. \(\mathrm{x}\), we get - \(\frac{\mathrm{dp}}{\mathrm{dx}}=16 x-3 \mathrm{x}^{2}\) For maxima or minima \(\frac{d p}{d x}=0\) \(16 x-3 x^{2}=0\) \(16-3 x=0\) \(x=\frac{16}{3}\) Now maximum value, \(x^{2} y=\left(\frac{16}{3}\right)^{2}\left(8-\frac{16}{3}\right)=\frac{16}{3} \times \frac{16}{3} \times \frac{8}{3}\) \(=\frac{2048}{27}\)