85702
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\) \((x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\). The, which of the following is NOT true?
1 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=4\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
2 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=3\), there exists \(\alpha \in(3,5)\) where \(f\) attains local minima
3 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
4 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=6\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
Explanation:
(C) : Given, \(f(x)=(x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\) \(f^{\prime}(x)=n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}}+(x-3)^{n_{1}}(x-5)^{n_{2}-1} \cdot n_{2}\) \(f^{\prime}(x)=(x-3)^{n_{1}-1}(x-5)^{n_{2}-1}\left(n_{1}+n_{2}\right)\left(x-\frac{5 n_{1}+3 n_{2}}{n_{1}+n_{2}}\right)\) For, \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\) \(f^{\prime}(x)=8(x-3)^{2}(x-5)^{4}\left(x-\frac{30}{8}\right)\) \(\operatorname{minima}\) at \(\mathrm{x}=\frac{30}{8}\) So, for \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where of attains local maxima.
JEE Main-2022-29.06.2022
Application of Derivatives
85703
The sum of the absolute minimum and the absolute maximum values of the function \(f(x)=\) \(\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) is :
1 \(\frac{\sqrt{17}+3}{2}\)
2 \(\frac{\sqrt{17}+5}{2}\)
3 5
4 \(\frac{9-\sqrt{17}}{2}\)
Explanation:
(A) : Given, function \(f(x)=\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) Then, \(f(x)=\left\{\begin{array}{l}x^2-4 x+(-2), \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^2+2 x+2, \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{array}\right.\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\) when \(\mathrm{x} \in\left(-1, \frac{3-\sqrt{17}}{2}\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4=0 \Rightarrow \mathrm{x}=2\) \(\mathrm{f}^{\prime}(\mathrm{x})=2(\mathrm{x}-2) \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\) is always decreases and \(f(2)=2, f(-1)=3\) \(\mathrm{f}\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}\) And, \(\quad f^{\prime}(x)\) when, \(x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\) \(f^{\prime}(x)=-2 x+2=-2(x-1)\). \(f^{\prime}(x)=0\) when \(x=1\) \(\mathrm{f}(1)=3\) Then, absolute minimum value \(=\frac{\sqrt{17}-3}{2}\) So, \(\text { Sum }=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}\)
JEE Main-2022-26.06.2022
Application of Derivatives
85704
If \(m\) and \(n\) respectively are the number of local maximum and local minimum points of the function \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\), then the ordered pair \((m, n)\) is equal \(\mathbf{t}\)
1 \((3,2)\)
2 \((2,3)\)
3 \((2,2)\)
4 \((3,4)\)
Explanation:
(B) : Given, And, \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\) \(\mathrm{m}=\text { local maximum }\) \(\mathrm{n}=\text { local minimum }\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\left(\mathrm{x}^{4}-5 \mathrm{x}^{2}+4\right) \cdot 2 \mathrm{x}}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}-4\right)}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(f^{\prime}(x)=\frac{2 x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^{2}}}\) So, \(\mathrm{m}=2\) and \(\mathrm{n}=3\) Hence, the ordered pair \((\mathrm{m}, \mathrm{n})=(2,3)\)
JEE Main-2022-27.06.2022
Application of Derivatives
85705
The curve \(y(x)=a x^{3}+b x^{2}+c x+5\) touches the \(x\)-axis at the point \(P(-2,0)\) and cuts the \(y\)-axis at the point \(Q\), where \(y\) is equal to 3 . Then the local maximum value of \(y(x)\) is :
1 \(\frac{27}{4}\)
2 \(\frac{29}{4}\)
3 \(\frac{37}{4}\)
4 \(\frac{9}{2}\)
Explanation:
(A) : Given, \(y(x)=a x^{3}+b x^{2}+c x+5\) is passing through \(P(-2,0)\) then \(8 a-4 b+2 c=5 \quad \ldots . .(1)\) \(y^{\prime}(x)=3 a x^{2}+2 b x+c\) touches \(x\)-axis at \((-2,0)\) \(12 a-4 b+c=0 \tag{2}\) Again for \(\mathrm{x}=0, \mathrm{y}^{\prime}(\mathrm{x})=3\) \(\mathrm{c}=3\) Solving eq. (1), (2) \& (3) \(a=-\frac{1}{2}, b=-\frac{3}{4}\) \(y^{\prime}(x)=-\frac{3}{2} x^{2}-\frac{3}{2} x+3\) \(\mathrm{y}(\mathrm{x})\) has local maxima at \(\mathrm{x}=1\) \(y(1)=\frac{27}{4}\)
85702
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\) \((x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\). The, which of the following is NOT true?
1 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=4\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
2 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=3\), there exists \(\alpha \in(3,5)\) where \(f\) attains local minima
3 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
4 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=6\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
Explanation:
(C) : Given, \(f(x)=(x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\) \(f^{\prime}(x)=n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}}+(x-3)^{n_{1}}(x-5)^{n_{2}-1} \cdot n_{2}\) \(f^{\prime}(x)=(x-3)^{n_{1}-1}(x-5)^{n_{2}-1}\left(n_{1}+n_{2}\right)\left(x-\frac{5 n_{1}+3 n_{2}}{n_{1}+n_{2}}\right)\) For, \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\) \(f^{\prime}(x)=8(x-3)^{2}(x-5)^{4}\left(x-\frac{30}{8}\right)\) \(\operatorname{minima}\) at \(\mathrm{x}=\frac{30}{8}\) So, for \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where of attains local maxima.
JEE Main-2022-29.06.2022
Application of Derivatives
85703
The sum of the absolute minimum and the absolute maximum values of the function \(f(x)=\) \(\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) is :
1 \(\frac{\sqrt{17}+3}{2}\)
2 \(\frac{\sqrt{17}+5}{2}\)
3 5
4 \(\frac{9-\sqrt{17}}{2}\)
Explanation:
(A) : Given, function \(f(x)=\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) Then, \(f(x)=\left\{\begin{array}{l}x^2-4 x+(-2), \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^2+2 x+2, \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{array}\right.\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\) when \(\mathrm{x} \in\left(-1, \frac{3-\sqrt{17}}{2}\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4=0 \Rightarrow \mathrm{x}=2\) \(\mathrm{f}^{\prime}(\mathrm{x})=2(\mathrm{x}-2) \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\) is always decreases and \(f(2)=2, f(-1)=3\) \(\mathrm{f}\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}\) And, \(\quad f^{\prime}(x)\) when, \(x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\) \(f^{\prime}(x)=-2 x+2=-2(x-1)\). \(f^{\prime}(x)=0\) when \(x=1\) \(\mathrm{f}(1)=3\) Then, absolute minimum value \(=\frac{\sqrt{17}-3}{2}\) So, \(\text { Sum }=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}\)
JEE Main-2022-26.06.2022
Application of Derivatives
85704
If \(m\) and \(n\) respectively are the number of local maximum and local minimum points of the function \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\), then the ordered pair \((m, n)\) is equal \(\mathbf{t}\)
1 \((3,2)\)
2 \((2,3)\)
3 \((2,2)\)
4 \((3,4)\)
Explanation:
(B) : Given, And, \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\) \(\mathrm{m}=\text { local maximum }\) \(\mathrm{n}=\text { local minimum }\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\left(\mathrm{x}^{4}-5 \mathrm{x}^{2}+4\right) \cdot 2 \mathrm{x}}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}-4\right)}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(f^{\prime}(x)=\frac{2 x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^{2}}}\) So, \(\mathrm{m}=2\) and \(\mathrm{n}=3\) Hence, the ordered pair \((\mathrm{m}, \mathrm{n})=(2,3)\)
JEE Main-2022-27.06.2022
Application of Derivatives
85705
The curve \(y(x)=a x^{3}+b x^{2}+c x+5\) touches the \(x\)-axis at the point \(P(-2,0)\) and cuts the \(y\)-axis at the point \(Q\), where \(y\) is equal to 3 . Then the local maximum value of \(y(x)\) is :
1 \(\frac{27}{4}\)
2 \(\frac{29}{4}\)
3 \(\frac{37}{4}\)
4 \(\frac{9}{2}\)
Explanation:
(A) : Given, \(y(x)=a x^{3}+b x^{2}+c x+5\) is passing through \(P(-2,0)\) then \(8 a-4 b+2 c=5 \quad \ldots . .(1)\) \(y^{\prime}(x)=3 a x^{2}+2 b x+c\) touches \(x\)-axis at \((-2,0)\) \(12 a-4 b+c=0 \tag{2}\) Again for \(\mathrm{x}=0, \mathrm{y}^{\prime}(\mathrm{x})=3\) \(\mathrm{c}=3\) Solving eq. (1), (2) \& (3) \(a=-\frac{1}{2}, b=-\frac{3}{4}\) \(y^{\prime}(x)=-\frac{3}{2} x^{2}-\frac{3}{2} x+3\) \(\mathrm{y}(\mathrm{x})\) has local maxima at \(\mathrm{x}=1\) \(y(1)=\frac{27}{4}\)
85702
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\) \((x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\). The, which of the following is NOT true?
1 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=4\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
2 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=3\), there exists \(\alpha \in(3,5)\) where \(f\) attains local minima
3 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
4 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=6\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
Explanation:
(C) : Given, \(f(x)=(x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\) \(f^{\prime}(x)=n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}}+(x-3)^{n_{1}}(x-5)^{n_{2}-1} \cdot n_{2}\) \(f^{\prime}(x)=(x-3)^{n_{1}-1}(x-5)^{n_{2}-1}\left(n_{1}+n_{2}\right)\left(x-\frac{5 n_{1}+3 n_{2}}{n_{1}+n_{2}}\right)\) For, \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\) \(f^{\prime}(x)=8(x-3)^{2}(x-5)^{4}\left(x-\frac{30}{8}\right)\) \(\operatorname{minima}\) at \(\mathrm{x}=\frac{30}{8}\) So, for \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where of attains local maxima.
JEE Main-2022-29.06.2022
Application of Derivatives
85703
The sum of the absolute minimum and the absolute maximum values of the function \(f(x)=\) \(\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) is :
1 \(\frac{\sqrt{17}+3}{2}\)
2 \(\frac{\sqrt{17}+5}{2}\)
3 5
4 \(\frac{9-\sqrt{17}}{2}\)
Explanation:
(A) : Given, function \(f(x)=\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) Then, \(f(x)=\left\{\begin{array}{l}x^2-4 x+(-2), \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^2+2 x+2, \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{array}\right.\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\) when \(\mathrm{x} \in\left(-1, \frac{3-\sqrt{17}}{2}\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4=0 \Rightarrow \mathrm{x}=2\) \(\mathrm{f}^{\prime}(\mathrm{x})=2(\mathrm{x}-2) \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\) is always decreases and \(f(2)=2, f(-1)=3\) \(\mathrm{f}\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}\) And, \(\quad f^{\prime}(x)\) when, \(x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\) \(f^{\prime}(x)=-2 x+2=-2(x-1)\). \(f^{\prime}(x)=0\) when \(x=1\) \(\mathrm{f}(1)=3\) Then, absolute minimum value \(=\frac{\sqrt{17}-3}{2}\) So, \(\text { Sum }=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}\)
JEE Main-2022-26.06.2022
Application of Derivatives
85704
If \(m\) and \(n\) respectively are the number of local maximum and local minimum points of the function \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\), then the ordered pair \((m, n)\) is equal \(\mathbf{t}\)
1 \((3,2)\)
2 \((2,3)\)
3 \((2,2)\)
4 \((3,4)\)
Explanation:
(B) : Given, And, \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\) \(\mathrm{m}=\text { local maximum }\) \(\mathrm{n}=\text { local minimum }\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\left(\mathrm{x}^{4}-5 \mathrm{x}^{2}+4\right) \cdot 2 \mathrm{x}}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}-4\right)}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(f^{\prime}(x)=\frac{2 x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^{2}}}\) So, \(\mathrm{m}=2\) and \(\mathrm{n}=3\) Hence, the ordered pair \((\mathrm{m}, \mathrm{n})=(2,3)\)
JEE Main-2022-27.06.2022
Application of Derivatives
85705
The curve \(y(x)=a x^{3}+b x^{2}+c x+5\) touches the \(x\)-axis at the point \(P(-2,0)\) and cuts the \(y\)-axis at the point \(Q\), where \(y\) is equal to 3 . Then the local maximum value of \(y(x)\) is :
1 \(\frac{27}{4}\)
2 \(\frac{29}{4}\)
3 \(\frac{37}{4}\)
4 \(\frac{9}{2}\)
Explanation:
(A) : Given, \(y(x)=a x^{3}+b x^{2}+c x+5\) is passing through \(P(-2,0)\) then \(8 a-4 b+2 c=5 \quad \ldots . .(1)\) \(y^{\prime}(x)=3 a x^{2}+2 b x+c\) touches \(x\)-axis at \((-2,0)\) \(12 a-4 b+c=0 \tag{2}\) Again for \(\mathrm{x}=0, \mathrm{y}^{\prime}(\mathrm{x})=3\) \(\mathrm{c}=3\) Solving eq. (1), (2) \& (3) \(a=-\frac{1}{2}, b=-\frac{3}{4}\) \(y^{\prime}(x)=-\frac{3}{2} x^{2}-\frac{3}{2} x+3\) \(\mathrm{y}(\mathrm{x})\) has local maxima at \(\mathrm{x}=1\) \(y(1)=\frac{27}{4}\)
85702
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\) \((x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\). The, which of the following is NOT true?
1 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=4\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
2 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=3\), there exists \(\alpha \in(3,5)\) where \(f\) attains local minima
3 For \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
4 For \(\mathrm{n}_{1}=4, \mathrm{n}_{2}=6\), there exists \(\alpha \in(3,5)\) where \(f\) attains local maxima
Explanation:
(C) : Given, \(f(x)=(x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in N\) \(f^{\prime}(x)=n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}}+(x-3)^{n_{1}}(x-5)^{n_{2}-1} \cdot n_{2}\) \(f^{\prime}(x)=(x-3)^{n_{1}-1}(x-5)^{n_{2}-1}\left(n_{1}+n_{2}\right)\left(x-\frac{5 n_{1}+3 n_{2}}{n_{1}+n_{2}}\right)\) For, \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\) \(f^{\prime}(x)=8(x-3)^{2}(x-5)^{4}\left(x-\frac{30}{8}\right)\) \(\operatorname{minima}\) at \(\mathrm{x}=\frac{30}{8}\) So, for \(\mathrm{n}_{1}=3, \mathrm{n}_{2}=5\), there exists \(\alpha \in(3,5)\) where of attains local maxima.
JEE Main-2022-29.06.2022
Application of Derivatives
85703
The sum of the absolute minimum and the absolute maximum values of the function \(f(x)=\) \(\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) is :
1 \(\frac{\sqrt{17}+3}{2}\)
2 \(\frac{\sqrt{17}+5}{2}\)
3 5
4 \(\frac{9-\sqrt{17}}{2}\)
Explanation:
(A) : Given, function \(f(x)=\left|3 x-x^{2}+2\right|-x\) in the interval \([-1,2]\) Then, \(f(x)=\left\{\begin{array}{l}x^2-4 x+(-2), \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^2+2 x+2, \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{array}\right.\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\) when \(\mathrm{x} \in\left(-1, \frac{3-\sqrt{17}}{2}\right)\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-4=0 \Rightarrow \mathrm{x}=2\) \(\mathrm{f}^{\prime}(\mathrm{x})=2(\mathrm{x}-2) \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\) is always decreases and \(f(2)=2, f(-1)=3\) \(\mathrm{f}\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}\) And, \(\quad f^{\prime}(x)\) when, \(x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\) \(f^{\prime}(x)=-2 x+2=-2(x-1)\). \(f^{\prime}(x)=0\) when \(x=1\) \(\mathrm{f}(1)=3\) Then, absolute minimum value \(=\frac{\sqrt{17}-3}{2}\) So, \(\text { Sum }=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}\)
JEE Main-2022-26.06.2022
Application of Derivatives
85704
If \(m\) and \(n\) respectively are the number of local maximum and local minimum points of the function \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\), then the ordered pair \((m, n)\) is equal \(\mathbf{t}\)
1 \((3,2)\)
2 \((2,3)\)
3 \((2,2)\)
4 \((3,4)\)
Explanation:
(B) : Given, And, \(f(x)=\int_{0}^{x^{2}} \frac{t^{2}-5 t+4}{2+e^{t}} d t\) \(\mathrm{m}=\text { local maximum }\) \(\mathrm{n}=\text { local minimum }\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\left(\mathrm{x}^{4}-5 \mathrm{x}^{2}+4\right) \cdot 2 \mathrm{x}}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2 \mathrm{x}\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}-4\right)}{2+\mathrm{e}^{\mathrm{x}^{2}}}\) \(f^{\prime}(x)=\frac{2 x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^{2}}}\) So, \(\mathrm{m}=2\) and \(\mathrm{n}=3\) Hence, the ordered pair \((\mathrm{m}, \mathrm{n})=(2,3)\)
JEE Main-2022-27.06.2022
Application of Derivatives
85705
The curve \(y(x)=a x^{3}+b x^{2}+c x+5\) touches the \(x\)-axis at the point \(P(-2,0)\) and cuts the \(y\)-axis at the point \(Q\), where \(y\) is equal to 3 . Then the local maximum value of \(y(x)\) is :
1 \(\frac{27}{4}\)
2 \(\frac{29}{4}\)
3 \(\frac{37}{4}\)
4 \(\frac{9}{2}\)
Explanation:
(A) : Given, \(y(x)=a x^{3}+b x^{2}+c x+5\) is passing through \(P(-2,0)\) then \(8 a-4 b+2 c=5 \quad \ldots . .(1)\) \(y^{\prime}(x)=3 a x^{2}+2 b x+c\) touches \(x\)-axis at \((-2,0)\) \(12 a-4 b+c=0 \tag{2}\) Again for \(\mathrm{x}=0, \mathrm{y}^{\prime}(\mathrm{x})=3\) \(\mathrm{c}=3\) Solving eq. (1), (2) \& (3) \(a=-\frac{1}{2}, b=-\frac{3}{4}\) \(y^{\prime}(x)=-\frac{3}{2} x^{2}-\frac{3}{2} x+3\) \(\mathrm{y}(\mathrm{x})\) has local maxima at \(\mathrm{x}=1\) \(y(1)=\frac{27}{4}\)
