85650
The point of inflection of the function \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is
1 \(\left(\frac{3}{2}, \frac{3}{4}\right)\)
2 \(\left(-\frac{3}{2},-\frac{3}{4}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Explanation:
(A) : \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) Differentiating both side w.r.t \(x\), \(\frac{d y}{d x}=x^{2}-3 x+2 \Rightarrow \frac{d^{2} y}{d x^{2}}=2 x-3\) For point of inflection, \(\frac{d^{2} y}{d x^{2}}=0 \Rightarrow 2 x-3=0\) \(x=\frac{3}{2}\) Now, \(y=\int_{0}^{3 / 2}\left(t^{2}-3 t+2\right) d t=\left[\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+2 t\right]_{0}^{3 / 2}\) \(y=\frac{3}{4}\) \(\therefore\left(\frac{3}{2}, \frac{3}{4}\right)\) is point of inflection.
UPSEE-2017
Application of Derivatives
85651
If \(f(\theta)=2\left(\sec ^{2} \theta+\cos ^{2} \theta\right)\), then its value always
85652
If \(f(x)=\frac{80}{3 \mathrm{x}^{4}+8 \mathrm{x}^{3}-18 \mathrm{x}^{2}+60}\), then the points of local maxima for the function \(f(x)\) are
1 1,3
2 \(-3,1\)
3 \(-1,3\)
4 \(-1,-3\)
Explanation:
(B) : \(f(x)=\frac{80}{3 x^{4}+8 x^{3}-18 x^{2}+60}\) \(f^{\prime}(x)= \frac{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right) \frac{d}{d x}(80)-80 \cdot \frac{d}{d x}\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)^{2}}\) \(f^{\prime}(x)= -80\left(12 x^{3}+24 x^{2}-36 x\right)\) \(=-80 \times 12 x\left(x^{2}+2 x-3\right)\) \(=\frac{-80(12 x)(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{3}+60\right)^{2}}\) For maxima or minima, \(\frac{(-80) \cdot 12 x(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}=0\) \(x=0, \quad x=1 \text { and } x=-3\) The sigh scheme of \(\mathrm{f}^{\prime}(\mathrm{x})\) is We can say that \(\mathrm{x}=-3\) and \(\mathrm{x}=1\) are the points of local maxima.
UPSEE-2015
Application of Derivatives
85653
The adjacent sides of a rectangle with given parameter as \(200 \mathrm{~cm}\) and enclosing minimum area are
1 \(20 \mathrm{~cm}\) and \(80 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\)
3 \(50 \mathrm{~cm}\) and \(50 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\) and \(70 \mathrm{~cm}\)
Explanation:
(C) : Let a and \(b\) be the adjacent sides of rectangle perimeter \(=200\) \(2(a+b)=200\) \(a+b=100\) \(b=100-a\) Area of rectangle, \(\mathrm{A}=\mathrm{ab}\) \(A=a(100-a)\) \(A=100 a-a^{2}\) \(\frac{\mathrm{dA}}{\mathrm{da}}=100-2 \mathrm{a}\) For minima, \(\frac{d A}{d a}=0\) \(100-2 a=0\) \(a=50 \mathrm{~cm} \text { and } \quad b=50 \mathrm{~cm}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85650
The point of inflection of the function \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is
1 \(\left(\frac{3}{2}, \frac{3}{4}\right)\)
2 \(\left(-\frac{3}{2},-\frac{3}{4}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Explanation:
(A) : \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) Differentiating both side w.r.t \(x\), \(\frac{d y}{d x}=x^{2}-3 x+2 \Rightarrow \frac{d^{2} y}{d x^{2}}=2 x-3\) For point of inflection, \(\frac{d^{2} y}{d x^{2}}=0 \Rightarrow 2 x-3=0\) \(x=\frac{3}{2}\) Now, \(y=\int_{0}^{3 / 2}\left(t^{2}-3 t+2\right) d t=\left[\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+2 t\right]_{0}^{3 / 2}\) \(y=\frac{3}{4}\) \(\therefore\left(\frac{3}{2}, \frac{3}{4}\right)\) is point of inflection.
UPSEE-2017
Application of Derivatives
85651
If \(f(\theta)=2\left(\sec ^{2} \theta+\cos ^{2} \theta\right)\), then its value always
85652
If \(f(x)=\frac{80}{3 \mathrm{x}^{4}+8 \mathrm{x}^{3}-18 \mathrm{x}^{2}+60}\), then the points of local maxima for the function \(f(x)\) are
1 1,3
2 \(-3,1\)
3 \(-1,3\)
4 \(-1,-3\)
Explanation:
(B) : \(f(x)=\frac{80}{3 x^{4}+8 x^{3}-18 x^{2}+60}\) \(f^{\prime}(x)= \frac{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right) \frac{d}{d x}(80)-80 \cdot \frac{d}{d x}\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)^{2}}\) \(f^{\prime}(x)= -80\left(12 x^{3}+24 x^{2}-36 x\right)\) \(=-80 \times 12 x\left(x^{2}+2 x-3\right)\) \(=\frac{-80(12 x)(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{3}+60\right)^{2}}\) For maxima or minima, \(\frac{(-80) \cdot 12 x(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}=0\) \(x=0, \quad x=1 \text { and } x=-3\) The sigh scheme of \(\mathrm{f}^{\prime}(\mathrm{x})\) is We can say that \(\mathrm{x}=-3\) and \(\mathrm{x}=1\) are the points of local maxima.
UPSEE-2015
Application of Derivatives
85653
The adjacent sides of a rectangle with given parameter as \(200 \mathrm{~cm}\) and enclosing minimum area are
1 \(20 \mathrm{~cm}\) and \(80 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\)
3 \(50 \mathrm{~cm}\) and \(50 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\) and \(70 \mathrm{~cm}\)
Explanation:
(C) : Let a and \(b\) be the adjacent sides of rectangle perimeter \(=200\) \(2(a+b)=200\) \(a+b=100\) \(b=100-a\) Area of rectangle, \(\mathrm{A}=\mathrm{ab}\) \(A=a(100-a)\) \(A=100 a-a^{2}\) \(\frac{\mathrm{dA}}{\mathrm{da}}=100-2 \mathrm{a}\) For minima, \(\frac{d A}{d a}=0\) \(100-2 a=0\) \(a=50 \mathrm{~cm} \text { and } \quad b=50 \mathrm{~cm}\)
85650
The point of inflection of the function \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is
1 \(\left(\frac{3}{2}, \frac{3}{4}\right)\)
2 \(\left(-\frac{3}{2},-\frac{3}{4}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Explanation:
(A) : \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) Differentiating both side w.r.t \(x\), \(\frac{d y}{d x}=x^{2}-3 x+2 \Rightarrow \frac{d^{2} y}{d x^{2}}=2 x-3\) For point of inflection, \(\frac{d^{2} y}{d x^{2}}=0 \Rightarrow 2 x-3=0\) \(x=\frac{3}{2}\) Now, \(y=\int_{0}^{3 / 2}\left(t^{2}-3 t+2\right) d t=\left[\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+2 t\right]_{0}^{3 / 2}\) \(y=\frac{3}{4}\) \(\therefore\left(\frac{3}{2}, \frac{3}{4}\right)\) is point of inflection.
UPSEE-2017
Application of Derivatives
85651
If \(f(\theta)=2\left(\sec ^{2} \theta+\cos ^{2} \theta\right)\), then its value always
85652
If \(f(x)=\frac{80}{3 \mathrm{x}^{4}+8 \mathrm{x}^{3}-18 \mathrm{x}^{2}+60}\), then the points of local maxima for the function \(f(x)\) are
1 1,3
2 \(-3,1\)
3 \(-1,3\)
4 \(-1,-3\)
Explanation:
(B) : \(f(x)=\frac{80}{3 x^{4}+8 x^{3}-18 x^{2}+60}\) \(f^{\prime}(x)= \frac{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right) \frac{d}{d x}(80)-80 \cdot \frac{d}{d x}\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)^{2}}\) \(f^{\prime}(x)= -80\left(12 x^{3}+24 x^{2}-36 x\right)\) \(=-80 \times 12 x\left(x^{2}+2 x-3\right)\) \(=\frac{-80(12 x)(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{3}+60\right)^{2}}\) For maxima or minima, \(\frac{(-80) \cdot 12 x(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}=0\) \(x=0, \quad x=1 \text { and } x=-3\) The sigh scheme of \(\mathrm{f}^{\prime}(\mathrm{x})\) is We can say that \(\mathrm{x}=-3\) and \(\mathrm{x}=1\) are the points of local maxima.
UPSEE-2015
Application of Derivatives
85653
The adjacent sides of a rectangle with given parameter as \(200 \mathrm{~cm}\) and enclosing minimum area are
1 \(20 \mathrm{~cm}\) and \(80 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\)
3 \(50 \mathrm{~cm}\) and \(50 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\) and \(70 \mathrm{~cm}\)
Explanation:
(C) : Let a and \(b\) be the adjacent sides of rectangle perimeter \(=200\) \(2(a+b)=200\) \(a+b=100\) \(b=100-a\) Area of rectangle, \(\mathrm{A}=\mathrm{ab}\) \(A=a(100-a)\) \(A=100 a-a^{2}\) \(\frac{\mathrm{dA}}{\mathrm{da}}=100-2 \mathrm{a}\) For minima, \(\frac{d A}{d a}=0\) \(100-2 a=0\) \(a=50 \mathrm{~cm} \text { and } \quad b=50 \mathrm{~cm}\)
85650
The point of inflection of the function \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is
1 \(\left(\frac{3}{2}, \frac{3}{4}\right)\)
2 \(\left(-\frac{3}{2},-\frac{3}{4}\right)\)
3 \(\left(-\frac{1}{2},-\frac{3}{2}\right)\)
4 \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Explanation:
(A) : \(y=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) Differentiating both side w.r.t \(x\), \(\frac{d y}{d x}=x^{2}-3 x+2 \Rightarrow \frac{d^{2} y}{d x^{2}}=2 x-3\) For point of inflection, \(\frac{d^{2} y}{d x^{2}}=0 \Rightarrow 2 x-3=0\) \(x=\frac{3}{2}\) Now, \(y=\int_{0}^{3 / 2}\left(t^{2}-3 t+2\right) d t=\left[\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+2 t\right]_{0}^{3 / 2}\) \(y=\frac{3}{4}\) \(\therefore\left(\frac{3}{2}, \frac{3}{4}\right)\) is point of inflection.
UPSEE-2017
Application of Derivatives
85651
If \(f(\theta)=2\left(\sec ^{2} \theta+\cos ^{2} \theta\right)\), then its value always
85652
If \(f(x)=\frac{80}{3 \mathrm{x}^{4}+8 \mathrm{x}^{3}-18 \mathrm{x}^{2}+60}\), then the points of local maxima for the function \(f(x)\) are
1 1,3
2 \(-3,1\)
3 \(-1,3\)
4 \(-1,-3\)
Explanation:
(B) : \(f(x)=\frac{80}{3 x^{4}+8 x^{3}-18 x^{2}+60}\) \(f^{\prime}(x)= \frac{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right) \frac{d}{d x}(80)-80 \cdot \frac{d}{d x}\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)^{2}}\) \(f^{\prime}(x)= -80\left(12 x^{3}+24 x^{2}-36 x\right)\) \(=-80 \times 12 x\left(x^{2}+2 x-3\right)\) \(=\frac{-80(12 x)(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{3}+60\right)^{2}}\) For maxima or minima, \(\frac{(-80) \cdot 12 x(x-1)(x+3)}{\left(3 x^{4}+8 x^{3}-18 x^{2}+60\right)}=0\) \(x=0, \quad x=1 \text { and } x=-3\) The sigh scheme of \(\mathrm{f}^{\prime}(\mathrm{x})\) is We can say that \(\mathrm{x}=-3\) and \(\mathrm{x}=1\) are the points of local maxima.
UPSEE-2015
Application of Derivatives
85653
The adjacent sides of a rectangle with given parameter as \(200 \mathrm{~cm}\) and enclosing minimum area are
1 \(20 \mathrm{~cm}\) and \(80 \mathrm{~cm}\)
2 \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\)
3 \(50 \mathrm{~cm}\) and \(50 \mathrm{~cm}\)
4 \(30 \mathrm{~cm}\) and \(70 \mathrm{~cm}\)
Explanation:
(C) : Let a and \(b\) be the adjacent sides of rectangle perimeter \(=200\) \(2(a+b)=200\) \(a+b=100\) \(b=100-a\) Area of rectangle, \(\mathrm{A}=\mathrm{ab}\) \(A=a(100-a)\) \(A=100 a-a^{2}\) \(\frac{\mathrm{dA}}{\mathrm{da}}=100-2 \mathrm{a}\) For minima, \(\frac{d A}{d a}=0\) \(100-2 a=0\) \(a=50 \mathrm{~cm} \text { and } \quad b=50 \mathrm{~cm}\)
