85720
The minimum value of \(f(x)=x+\frac{4}{x+2}\) is
1 -1
2 -2
3 1
4 2
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{4}{\mathrm{x}+2}\) Then, \(f^{\prime}(x)=1+\frac{(x+2) \cdot 0-1 \times 4}{(x+2)^{2}}\) \(f^{\prime}(x)=1 .-\frac{4}{(x+2)^{2}}\) For minimum value \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{4}{(x+2)^{2}}=0\) \(1=\frac{4}{(\mathrm{x}+2)^{2}}\) \((\mathrm{x}+2)^{2}=4\) \(\mathrm{x}+2= \pm 2\) \(\Rightarrow \mathrm{x}+2=2 \quad \text { or } \mathrm{x}+2=-2\) \(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=-2-2\) \(\Rightarrow \mathrm{x}=0 \quad \text { or } \quad \mathrm{x}=-4\) Then, the minimum value \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) \(\mathrm{f}(0)=0+\frac{4}{0+2}\) \(\mathrm{f}(0)=2\) So, The minimum value of \(f(x)\) is 2 .
Shift-I
Application of Derivatives
85721
The condition that \(f(x)=a x^{3}+b x^{2}+\mathbf{c x}+\mathbf{d}\) has no extreme value is
1 \(b^{2}-4 a c\)
2 \(\mathrm{b}^{2}=3 \mathrm{ac}\)
3 \(\mathrm{b}^{2}\lt 3 \mathrm{ac}\)
4 \(\mathrm{b}^{2}>3 \mathrm{ac}\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\) Then, \(f^{\prime}(x)=3 a x^{2}+2 b x+c\) For no extrema, \(f\) should have zero turning points i.e, above quadratic have no real roots \(\Rightarrow(2 \mathrm{~b})^{2}-4.3 \mathrm{a} . \mathrm{c}\lt 0\) For \(4 b^{2}-12 a c\lt 0\), has no roots \(\therefore \mathrm{b}^{2}\lt 3 \mathrm{ac}\)
AP EAMCET-2021-04.07.2021
Application of Derivatives
85722
A jet of an enemy is flying along the curve \(y-2\) \(=\mathbf{x}^{2}\). A soldier is placed at the point \((3,2)\). The nearest distance between the soldier and the jet is
1 \(\sqrt{5}\)
2 5
3 6
4 \(\sqrt{6}\)
Explanation:
(A) : Given that, the curve \(y-2=x^{2}\) Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be the position of the jet and soldier at point A \((3,2)\) Then by distance formula, \(\mathrm{AP}=\sqrt{(\mathrm{x}-3)^{2}+(\mathrm{y}-2)^{2}} \tag{i}\) By curve \(y-2=x^{2}\) \(y=A P^{2}=(x-3)^{2}+x^{4}\) \(\frac{d y}{d x}=2(x-3)+4 x^{3}\) And, \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2+12 \mathrm{x}^{2}\) For nearest distance the maxima and minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2(x-3)+4 x^{3}=0\) \(\text { At, } \quad x=1\) \(\frac{d^{2} y}{d x^{2}}=2+12 \times(1)^{2}=14>0\) \(\mathrm{y}\) is minimum when \(\mathrm{x}=1\) then \(\mathrm{y}=1+2=3\) then nearest distance \(\sqrt{(1-3)^{2}+(3-2)^{2}}=\sqrt{5}\)
AMU-2004
Application of Derivatives
85723
If the distance \(s\) described in time \(t\) by a particle moving on a straight line is given by \(s\) \(=\mathbf{t}^{5}-40 t^{3}+30 t^{2}+80 t-250\), then its minimum acceleration is
1 260
2 -260
3 130
4 -130
Explanation:
(B) : Given, \(\mathrm{S}=\mathrm{t}^{5}-40 \mathrm{t}^{3}+30 \mathrm{t}^{2}+80 \mathrm{t}-250\) Then, \(\quad \mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=5 \mathrm{t}^{4}-120 \mathrm{t}^{2}+60 \mathrm{t}+80\) And, \(\quad \mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=20 \mathrm{t}^{3}-240 \mathrm{t}+60\) For minimum value of \(\frac{\mathrm{da}}{\mathrm{dt}}=0\) Then, \(\quad \frac{\mathrm{d}}{\mathrm{dt}}\left(20 \mathrm{t}^{3}-240 \mathrm{t}+60\right)=0\) \(60 \mathrm{t}^{2}-240=0\) \(60 \mathrm{t}^{2}=240\) \(\mathrm{t}^{2}=4\) \(t=2 \mathrm{sec}\) We get minimum value when, \(\mathrm{t}=2\) So, \(\quad \mathrm{a}=20(2)^{3}-240(2)+60\) \(\mathrm{a}=20 \times 8-480+60\) \(\mathrm{a}=160-480+60\) \(a=-260\) \(\therefore\) The minimum acceleration is -260
AP EAMCET-2018-24.04.2018
Application of Derivatives
85724
If \(x^{2}+y^{2}=25\), then \(\log _{5}[\max (3 x+4 y)]\) is
85720
The minimum value of \(f(x)=x+\frac{4}{x+2}\) is
1 -1
2 -2
3 1
4 2
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{4}{\mathrm{x}+2}\) Then, \(f^{\prime}(x)=1+\frac{(x+2) \cdot 0-1 \times 4}{(x+2)^{2}}\) \(f^{\prime}(x)=1 .-\frac{4}{(x+2)^{2}}\) For minimum value \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{4}{(x+2)^{2}}=0\) \(1=\frac{4}{(\mathrm{x}+2)^{2}}\) \((\mathrm{x}+2)^{2}=4\) \(\mathrm{x}+2= \pm 2\) \(\Rightarrow \mathrm{x}+2=2 \quad \text { or } \mathrm{x}+2=-2\) \(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=-2-2\) \(\Rightarrow \mathrm{x}=0 \quad \text { or } \quad \mathrm{x}=-4\) Then, the minimum value \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) \(\mathrm{f}(0)=0+\frac{4}{0+2}\) \(\mathrm{f}(0)=2\) So, The minimum value of \(f(x)\) is 2 .
Shift-I
Application of Derivatives
85721
The condition that \(f(x)=a x^{3}+b x^{2}+\mathbf{c x}+\mathbf{d}\) has no extreme value is
1 \(b^{2}-4 a c\)
2 \(\mathrm{b}^{2}=3 \mathrm{ac}\)
3 \(\mathrm{b}^{2}\lt 3 \mathrm{ac}\)
4 \(\mathrm{b}^{2}>3 \mathrm{ac}\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\) Then, \(f^{\prime}(x)=3 a x^{2}+2 b x+c\) For no extrema, \(f\) should have zero turning points i.e, above quadratic have no real roots \(\Rightarrow(2 \mathrm{~b})^{2}-4.3 \mathrm{a} . \mathrm{c}\lt 0\) For \(4 b^{2}-12 a c\lt 0\), has no roots \(\therefore \mathrm{b}^{2}\lt 3 \mathrm{ac}\)
AP EAMCET-2021-04.07.2021
Application of Derivatives
85722
A jet of an enemy is flying along the curve \(y-2\) \(=\mathbf{x}^{2}\). A soldier is placed at the point \((3,2)\). The nearest distance between the soldier and the jet is
1 \(\sqrt{5}\)
2 5
3 6
4 \(\sqrt{6}\)
Explanation:
(A) : Given that, the curve \(y-2=x^{2}\) Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be the position of the jet and soldier at point A \((3,2)\) Then by distance formula, \(\mathrm{AP}=\sqrt{(\mathrm{x}-3)^{2}+(\mathrm{y}-2)^{2}} \tag{i}\) By curve \(y-2=x^{2}\) \(y=A P^{2}=(x-3)^{2}+x^{4}\) \(\frac{d y}{d x}=2(x-3)+4 x^{3}\) And, \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2+12 \mathrm{x}^{2}\) For nearest distance the maxima and minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2(x-3)+4 x^{3}=0\) \(\text { At, } \quad x=1\) \(\frac{d^{2} y}{d x^{2}}=2+12 \times(1)^{2}=14>0\) \(\mathrm{y}\) is minimum when \(\mathrm{x}=1\) then \(\mathrm{y}=1+2=3\) then nearest distance \(\sqrt{(1-3)^{2}+(3-2)^{2}}=\sqrt{5}\)
AMU-2004
Application of Derivatives
85723
If the distance \(s\) described in time \(t\) by a particle moving on a straight line is given by \(s\) \(=\mathbf{t}^{5}-40 t^{3}+30 t^{2}+80 t-250\), then its minimum acceleration is
1 260
2 -260
3 130
4 -130
Explanation:
(B) : Given, \(\mathrm{S}=\mathrm{t}^{5}-40 \mathrm{t}^{3}+30 \mathrm{t}^{2}+80 \mathrm{t}-250\) Then, \(\quad \mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=5 \mathrm{t}^{4}-120 \mathrm{t}^{2}+60 \mathrm{t}+80\) And, \(\quad \mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=20 \mathrm{t}^{3}-240 \mathrm{t}+60\) For minimum value of \(\frac{\mathrm{da}}{\mathrm{dt}}=0\) Then, \(\quad \frac{\mathrm{d}}{\mathrm{dt}}\left(20 \mathrm{t}^{3}-240 \mathrm{t}+60\right)=0\) \(60 \mathrm{t}^{2}-240=0\) \(60 \mathrm{t}^{2}=240\) \(\mathrm{t}^{2}=4\) \(t=2 \mathrm{sec}\) We get minimum value when, \(\mathrm{t}=2\) So, \(\quad \mathrm{a}=20(2)^{3}-240(2)+60\) \(\mathrm{a}=20 \times 8-480+60\) \(\mathrm{a}=160-480+60\) \(a=-260\) \(\therefore\) The minimum acceleration is -260
AP EAMCET-2018-24.04.2018
Application of Derivatives
85724
If \(x^{2}+y^{2}=25\), then \(\log _{5}[\max (3 x+4 y)]\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85720
The minimum value of \(f(x)=x+\frac{4}{x+2}\) is
1 -1
2 -2
3 1
4 2
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{4}{\mathrm{x}+2}\) Then, \(f^{\prime}(x)=1+\frac{(x+2) \cdot 0-1 \times 4}{(x+2)^{2}}\) \(f^{\prime}(x)=1 .-\frac{4}{(x+2)^{2}}\) For minimum value \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{4}{(x+2)^{2}}=0\) \(1=\frac{4}{(\mathrm{x}+2)^{2}}\) \((\mathrm{x}+2)^{2}=4\) \(\mathrm{x}+2= \pm 2\) \(\Rightarrow \mathrm{x}+2=2 \quad \text { or } \mathrm{x}+2=-2\) \(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=-2-2\) \(\Rightarrow \mathrm{x}=0 \quad \text { or } \quad \mathrm{x}=-4\) Then, the minimum value \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) \(\mathrm{f}(0)=0+\frac{4}{0+2}\) \(\mathrm{f}(0)=2\) So, The minimum value of \(f(x)\) is 2 .
Shift-I
Application of Derivatives
85721
The condition that \(f(x)=a x^{3}+b x^{2}+\mathbf{c x}+\mathbf{d}\) has no extreme value is
1 \(b^{2}-4 a c\)
2 \(\mathrm{b}^{2}=3 \mathrm{ac}\)
3 \(\mathrm{b}^{2}\lt 3 \mathrm{ac}\)
4 \(\mathrm{b}^{2}>3 \mathrm{ac}\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\) Then, \(f^{\prime}(x)=3 a x^{2}+2 b x+c\) For no extrema, \(f\) should have zero turning points i.e, above quadratic have no real roots \(\Rightarrow(2 \mathrm{~b})^{2}-4.3 \mathrm{a} . \mathrm{c}\lt 0\) For \(4 b^{2}-12 a c\lt 0\), has no roots \(\therefore \mathrm{b}^{2}\lt 3 \mathrm{ac}\)
AP EAMCET-2021-04.07.2021
Application of Derivatives
85722
A jet of an enemy is flying along the curve \(y-2\) \(=\mathbf{x}^{2}\). A soldier is placed at the point \((3,2)\). The nearest distance between the soldier and the jet is
1 \(\sqrt{5}\)
2 5
3 6
4 \(\sqrt{6}\)
Explanation:
(A) : Given that, the curve \(y-2=x^{2}\) Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be the position of the jet and soldier at point A \((3,2)\) Then by distance formula, \(\mathrm{AP}=\sqrt{(\mathrm{x}-3)^{2}+(\mathrm{y}-2)^{2}} \tag{i}\) By curve \(y-2=x^{2}\) \(y=A P^{2}=(x-3)^{2}+x^{4}\) \(\frac{d y}{d x}=2(x-3)+4 x^{3}\) And, \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2+12 \mathrm{x}^{2}\) For nearest distance the maxima and minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2(x-3)+4 x^{3}=0\) \(\text { At, } \quad x=1\) \(\frac{d^{2} y}{d x^{2}}=2+12 \times(1)^{2}=14>0\) \(\mathrm{y}\) is minimum when \(\mathrm{x}=1\) then \(\mathrm{y}=1+2=3\) then nearest distance \(\sqrt{(1-3)^{2}+(3-2)^{2}}=\sqrt{5}\)
AMU-2004
Application of Derivatives
85723
If the distance \(s\) described in time \(t\) by a particle moving on a straight line is given by \(s\) \(=\mathbf{t}^{5}-40 t^{3}+30 t^{2}+80 t-250\), then its minimum acceleration is
1 260
2 -260
3 130
4 -130
Explanation:
(B) : Given, \(\mathrm{S}=\mathrm{t}^{5}-40 \mathrm{t}^{3}+30 \mathrm{t}^{2}+80 \mathrm{t}-250\) Then, \(\quad \mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=5 \mathrm{t}^{4}-120 \mathrm{t}^{2}+60 \mathrm{t}+80\) And, \(\quad \mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=20 \mathrm{t}^{3}-240 \mathrm{t}+60\) For minimum value of \(\frac{\mathrm{da}}{\mathrm{dt}}=0\) Then, \(\quad \frac{\mathrm{d}}{\mathrm{dt}}\left(20 \mathrm{t}^{3}-240 \mathrm{t}+60\right)=0\) \(60 \mathrm{t}^{2}-240=0\) \(60 \mathrm{t}^{2}=240\) \(\mathrm{t}^{2}=4\) \(t=2 \mathrm{sec}\) We get minimum value when, \(\mathrm{t}=2\) So, \(\quad \mathrm{a}=20(2)^{3}-240(2)+60\) \(\mathrm{a}=20 \times 8-480+60\) \(\mathrm{a}=160-480+60\) \(a=-260\) \(\therefore\) The minimum acceleration is -260
AP EAMCET-2018-24.04.2018
Application of Derivatives
85724
If \(x^{2}+y^{2}=25\), then \(\log _{5}[\max (3 x+4 y)]\) is
85720
The minimum value of \(f(x)=x+\frac{4}{x+2}\) is
1 -1
2 -2
3 1
4 2
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{4}{\mathrm{x}+2}\) Then, \(f^{\prime}(x)=1+\frac{(x+2) \cdot 0-1 \times 4}{(x+2)^{2}}\) \(f^{\prime}(x)=1 .-\frac{4}{(x+2)^{2}}\) For minimum value \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{4}{(x+2)^{2}}=0\) \(1=\frac{4}{(\mathrm{x}+2)^{2}}\) \((\mathrm{x}+2)^{2}=4\) \(\mathrm{x}+2= \pm 2\) \(\Rightarrow \mathrm{x}+2=2 \quad \text { or } \mathrm{x}+2=-2\) \(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=-2-2\) \(\Rightarrow \mathrm{x}=0 \quad \text { or } \quad \mathrm{x}=-4\) Then, the minimum value \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) \(\mathrm{f}(0)=0+\frac{4}{0+2}\) \(\mathrm{f}(0)=2\) So, The minimum value of \(f(x)\) is 2 .
Shift-I
Application of Derivatives
85721
The condition that \(f(x)=a x^{3}+b x^{2}+\mathbf{c x}+\mathbf{d}\) has no extreme value is
1 \(b^{2}-4 a c\)
2 \(\mathrm{b}^{2}=3 \mathrm{ac}\)
3 \(\mathrm{b}^{2}\lt 3 \mathrm{ac}\)
4 \(\mathrm{b}^{2}>3 \mathrm{ac}\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\) Then, \(f^{\prime}(x)=3 a x^{2}+2 b x+c\) For no extrema, \(f\) should have zero turning points i.e, above quadratic have no real roots \(\Rightarrow(2 \mathrm{~b})^{2}-4.3 \mathrm{a} . \mathrm{c}\lt 0\) For \(4 b^{2}-12 a c\lt 0\), has no roots \(\therefore \mathrm{b}^{2}\lt 3 \mathrm{ac}\)
AP EAMCET-2021-04.07.2021
Application of Derivatives
85722
A jet of an enemy is flying along the curve \(y-2\) \(=\mathbf{x}^{2}\). A soldier is placed at the point \((3,2)\). The nearest distance between the soldier and the jet is
1 \(\sqrt{5}\)
2 5
3 6
4 \(\sqrt{6}\)
Explanation:
(A) : Given that, the curve \(y-2=x^{2}\) Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be the position of the jet and soldier at point A \((3,2)\) Then by distance formula, \(\mathrm{AP}=\sqrt{(\mathrm{x}-3)^{2}+(\mathrm{y}-2)^{2}} \tag{i}\) By curve \(y-2=x^{2}\) \(y=A P^{2}=(x-3)^{2}+x^{4}\) \(\frac{d y}{d x}=2(x-3)+4 x^{3}\) And, \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2+12 \mathrm{x}^{2}\) For nearest distance the maxima and minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2(x-3)+4 x^{3}=0\) \(\text { At, } \quad x=1\) \(\frac{d^{2} y}{d x^{2}}=2+12 \times(1)^{2}=14>0\) \(\mathrm{y}\) is minimum when \(\mathrm{x}=1\) then \(\mathrm{y}=1+2=3\) then nearest distance \(\sqrt{(1-3)^{2}+(3-2)^{2}}=\sqrt{5}\)
AMU-2004
Application of Derivatives
85723
If the distance \(s\) described in time \(t\) by a particle moving on a straight line is given by \(s\) \(=\mathbf{t}^{5}-40 t^{3}+30 t^{2}+80 t-250\), then its minimum acceleration is
1 260
2 -260
3 130
4 -130
Explanation:
(B) : Given, \(\mathrm{S}=\mathrm{t}^{5}-40 \mathrm{t}^{3}+30 \mathrm{t}^{2}+80 \mathrm{t}-250\) Then, \(\quad \mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=5 \mathrm{t}^{4}-120 \mathrm{t}^{2}+60 \mathrm{t}+80\) And, \(\quad \mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=20 \mathrm{t}^{3}-240 \mathrm{t}+60\) For minimum value of \(\frac{\mathrm{da}}{\mathrm{dt}}=0\) Then, \(\quad \frac{\mathrm{d}}{\mathrm{dt}}\left(20 \mathrm{t}^{3}-240 \mathrm{t}+60\right)=0\) \(60 \mathrm{t}^{2}-240=0\) \(60 \mathrm{t}^{2}=240\) \(\mathrm{t}^{2}=4\) \(t=2 \mathrm{sec}\) We get minimum value when, \(\mathrm{t}=2\) So, \(\quad \mathrm{a}=20(2)^{3}-240(2)+60\) \(\mathrm{a}=20 \times 8-480+60\) \(\mathrm{a}=160-480+60\) \(a=-260\) \(\therefore\) The minimum acceleration is -260
AP EAMCET-2018-24.04.2018
Application of Derivatives
85724
If \(x^{2}+y^{2}=25\), then \(\log _{5}[\max (3 x+4 y)]\) is
85720
The minimum value of \(f(x)=x+\frac{4}{x+2}\) is
1 -1
2 -2
3 1
4 2
Explanation:
(D) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{4}{\mathrm{x}+2}\) Then, \(f^{\prime}(x)=1+\frac{(x+2) \cdot 0-1 \times 4}{(x+2)^{2}}\) \(f^{\prime}(x)=1 .-\frac{4}{(x+2)^{2}}\) For minimum value \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\frac{4}{(x+2)^{2}}=0\) \(1=\frac{4}{(\mathrm{x}+2)^{2}}\) \((\mathrm{x}+2)^{2}=4\) \(\mathrm{x}+2= \pm 2\) \(\Rightarrow \mathrm{x}+2=2 \quad \text { or } \mathrm{x}+2=-2\) \(\Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=-2-2\) \(\Rightarrow \mathrm{x}=0 \quad \text { or } \quad \mathrm{x}=-4\) Then, the minimum value \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) \(\mathrm{f}(0)=0+\frac{4}{0+2}\) \(\mathrm{f}(0)=2\) So, The minimum value of \(f(x)\) is 2 .
Shift-I
Application of Derivatives
85721
The condition that \(f(x)=a x^{3}+b x^{2}+\mathbf{c x}+\mathbf{d}\) has no extreme value is
1 \(b^{2}-4 a c\)
2 \(\mathrm{b}^{2}=3 \mathrm{ac}\)
3 \(\mathrm{b}^{2}\lt 3 \mathrm{ac}\)
4 \(\mathrm{b}^{2}>3 \mathrm{ac}\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}\) Then, \(f^{\prime}(x)=3 a x^{2}+2 b x+c\) For no extrema, \(f\) should have zero turning points i.e, above quadratic have no real roots \(\Rightarrow(2 \mathrm{~b})^{2}-4.3 \mathrm{a} . \mathrm{c}\lt 0\) For \(4 b^{2}-12 a c\lt 0\), has no roots \(\therefore \mathrm{b}^{2}\lt 3 \mathrm{ac}\)
AP EAMCET-2021-04.07.2021
Application of Derivatives
85722
A jet of an enemy is flying along the curve \(y-2\) \(=\mathbf{x}^{2}\). A soldier is placed at the point \((3,2)\). The nearest distance between the soldier and the jet is
1 \(\sqrt{5}\)
2 5
3 6
4 \(\sqrt{6}\)
Explanation:
(A) : Given that, the curve \(y-2=x^{2}\) Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be the position of the jet and soldier at point A \((3,2)\) Then by distance formula, \(\mathrm{AP}=\sqrt{(\mathrm{x}-3)^{2}+(\mathrm{y}-2)^{2}} \tag{i}\) By curve \(y-2=x^{2}\) \(y=A P^{2}=(x-3)^{2}+x^{4}\) \(\frac{d y}{d x}=2(x-3)+4 x^{3}\) And, \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2+12 \mathrm{x}^{2}\) For nearest distance the maxima and minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(2(x-3)+4 x^{3}=0\) \(\text { At, } \quad x=1\) \(\frac{d^{2} y}{d x^{2}}=2+12 \times(1)^{2}=14>0\) \(\mathrm{y}\) is minimum when \(\mathrm{x}=1\) then \(\mathrm{y}=1+2=3\) then nearest distance \(\sqrt{(1-3)^{2}+(3-2)^{2}}=\sqrt{5}\)
AMU-2004
Application of Derivatives
85723
If the distance \(s\) described in time \(t\) by a particle moving on a straight line is given by \(s\) \(=\mathbf{t}^{5}-40 t^{3}+30 t^{2}+80 t-250\), then its minimum acceleration is
1 260
2 -260
3 130
4 -130
Explanation:
(B) : Given, \(\mathrm{S}=\mathrm{t}^{5}-40 \mathrm{t}^{3}+30 \mathrm{t}^{2}+80 \mathrm{t}-250\) Then, \(\quad \mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=5 \mathrm{t}^{4}-120 \mathrm{t}^{2}+60 \mathrm{t}+80\) And, \(\quad \mathrm{a}=\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}=20 \mathrm{t}^{3}-240 \mathrm{t}+60\) For minimum value of \(\frac{\mathrm{da}}{\mathrm{dt}}=0\) Then, \(\quad \frac{\mathrm{d}}{\mathrm{dt}}\left(20 \mathrm{t}^{3}-240 \mathrm{t}+60\right)=0\) \(60 \mathrm{t}^{2}-240=0\) \(60 \mathrm{t}^{2}=240\) \(\mathrm{t}^{2}=4\) \(t=2 \mathrm{sec}\) We get minimum value when, \(\mathrm{t}=2\) So, \(\quad \mathrm{a}=20(2)^{3}-240(2)+60\) \(\mathrm{a}=20 \times 8-480+60\) \(\mathrm{a}=160-480+60\) \(a=-260\) \(\therefore\) The minimum acceleration is -260
AP EAMCET-2018-24.04.2018
Application of Derivatives
85724
If \(x^{2}+y^{2}=25\), then \(\log _{5}[\max (3 x+4 y)]\) is