85689
Consider the following statements: 1.The function attains local minimum value at \(\mathrm{x}=\frac{7}{5}\) 2.\(\mathbf{x}=\mathbf{2}\) is the point of inflexion. Which of the above statements is/are correct?
1 1 only
2 2 only
3 Both 1 and 2
4 Neither 1 nor 2
Explanation:
(C) : \(\text { (1) } \mathrm{f}^{\prime \prime}\left(\frac{7}{5}\right) =\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right)^{2}(5)+0+0\) \(=\left(\frac{2}{5}\right)\left(\frac{9}{25}\right)(5)>0\) So, local minimum value at \(\mathrm{x}=\frac{7}{5}\) (2) \(\mathrm{f}^{\prime \prime}(2)=0\) So, both (1) and (2) are correct.
SCRA-2015
Application of Derivatives
85690
Let \(f(x)=\left\{\begin{array}{c}x^{3}-x^{2}+10 x-7, x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\). Then the set of all values of \(b\), for which \(f(x)\) has maximum value at \(\mathrm{x}=1\), is:
1 \((-6,-2)\)
2 \((2,6)\)
3 \([-6,-2) \cup(2,6]\)
4 \([-\sqrt{6},-2) \cup(2, \sqrt{6}]\)
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{l}x^{3}-x^{2}+10 x-7, \quad x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\) Then, for \(x\lt 1, f(x)=x^{3}-x^{2}+10 x-7\) \(f^{\prime}(x)=3 x^{2}-2 x+10>0\) \(\Rightarrow \quad \mathrm{f}(\mathrm{x})\) is increasing. And, for \(x>1, f^{\prime}(x)\lt 0 \Rightarrow f(x)\) is decreasing \(\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)\) From maximum value at \(\mathrm{x}=1\) \(\Rightarrow \quad 3 \geq-2+\log _{2}\left(\mathrm{~b}^{2}-4\right)\) \(\log _{2}\left(b^{2}-4\right) \leq 5\) \(b^{2}-4>0 \text { and } b^{2}-4 \leq 32 \text { or } b^{2}-36 \leq 0\) \(b^{2}>4 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b \leq 6\) So, \(\quad b \in[-6,-2] \cup(2,6]\)
JEE Main-2022-26.07.2022
Application of Derivatives
85691
Let \(x=2\) be a local minima of the function \(f(x)\) \(=2 x^{4}-18 x^{2}+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(\mathbf{M}=\)
1 \(18 \sqrt{6}-\frac{33}{2}\)
2 \(12 \sqrt{6}-\frac{33}{2}\)
3 \(12 \sqrt{6}-\frac{31}{2}\)
4 \(18 \sqrt{6}-\frac{31}{2}\)
Explanation:
(B) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{4}-18 \mathrm{x}^{2}+8 \mathrm{x}+12, \mathrm{x} \in(-4,4)\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}^{3}-36 \mathrm{x}+8\) \(f^{\prime}(x)=4\left(2 x^{3}-9 x+2\right)\) For maxima/minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(4\left(2 \mathrm{x}^{3}-9 \mathrm{x}+2\right)=0\) \(2 \mathrm{x}^{3}-9 \mathrm{x}+2=0\) Then, Since, \(f^{\prime}(x)\) change sign from \(+v e\) to \(-v e\) passing through \(-1+\sqrt{\frac{3}{2}}\) Then, \(x=-1+\sqrt{\frac{3}{2}}\) is point of local maxima. So, \(\quad M=f\left(-1+\sqrt{\frac{3}{2}}\right)\) \(M=2\left(-1+\sqrt{\frac{3}{2}}\right)^{4}-18\left(-1+\sqrt{\frac{3}{2}}\right)^{2}+8\left(-1+\sqrt{\frac{3}{2}}\right)+12\) \(M=2\left(1-4 \sqrt{\frac{3}{2}}+6 \times \frac{3}{2}-4 \times \frac{3}{2} \sqrt{\frac{3}{2}}+\frac{3}{2} \times \frac{3}{2}\right)\) \(-18\left(1-2 \sqrt{\frac{3}{2}}+\frac{3}{2}\right)-8+8 \sqrt{\frac{3}{2}}+12\) \(M=2-4 \sqrt{6}+18-6 \sqrt{6}+\frac{9}{2}-18+18 \sqrt{6}-27-8+4 \sqrt{6}+12\) \(\mathrm{M}=12 \sqrt{6}-\frac{33}{2}\)
JEE Main-2023-25.01.2023
Application of Derivatives
85692
Let the function \(f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p\) -9) \(x-6\) have a maxima for some value of \(x\lt \) 0 and a minima for some value of \(x>0\). Then, the set of all values of \(p\) is
85693
If the function \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and \(g(x)=\frac{x^{3}}{3}+a x+b x^{2}, a \neq 2 b\) have common extreme point, then \(a+2 b+7\) is equal to :
1 \(\frac{3}{2}\)
2 3
3 6
4 4
Explanation:
(C) : Given, \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and, \(\quad g(x)=\frac{x^{3}}{3}+a x+b x^{2}\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{~b}+\mathrm{ax}\) And, \(g^{\prime}(x)=x^{2}+a+2 b x\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=0\) (for find \(\left.\mathrm{x}\right)\) \(x^{2}+2 b+a x-x^{2}-a-2 b x=0\) \((2 b-a)+a x-2 b x=0\) \((2 b-a)-x(2 b-a)=0\) \((2 b-a)(1-x)=0\) \(2 b=a\) or \(x=1\) But given \(a \neq 2 b\) then \(x=1\) Put, \(x=1\) in \(f^{\prime}(x)=0\) or \(g^{\prime}(x)=0\) \(1+2 \mathrm{~b}+\mathrm{a}=0\) \(7+2 b+a=6\) So, \(\quad a+2 b+7=6\)
85689
Consider the following statements: 1.The function attains local minimum value at \(\mathrm{x}=\frac{7}{5}\) 2.\(\mathbf{x}=\mathbf{2}\) is the point of inflexion. Which of the above statements is/are correct?
1 1 only
2 2 only
3 Both 1 and 2
4 Neither 1 nor 2
Explanation:
(C) : \(\text { (1) } \mathrm{f}^{\prime \prime}\left(\frac{7}{5}\right) =\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right)^{2}(5)+0+0\) \(=\left(\frac{2}{5}\right)\left(\frac{9}{25}\right)(5)>0\) So, local minimum value at \(\mathrm{x}=\frac{7}{5}\) (2) \(\mathrm{f}^{\prime \prime}(2)=0\) So, both (1) and (2) are correct.
SCRA-2015
Application of Derivatives
85690
Let \(f(x)=\left\{\begin{array}{c}x^{3}-x^{2}+10 x-7, x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\). Then the set of all values of \(b\), for which \(f(x)\) has maximum value at \(\mathrm{x}=1\), is:
1 \((-6,-2)\)
2 \((2,6)\)
3 \([-6,-2) \cup(2,6]\)
4 \([-\sqrt{6},-2) \cup(2, \sqrt{6}]\)
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{l}x^{3}-x^{2}+10 x-7, \quad x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\) Then, for \(x\lt 1, f(x)=x^{3}-x^{2}+10 x-7\) \(f^{\prime}(x)=3 x^{2}-2 x+10>0\) \(\Rightarrow \quad \mathrm{f}(\mathrm{x})\) is increasing. And, for \(x>1, f^{\prime}(x)\lt 0 \Rightarrow f(x)\) is decreasing \(\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)\) From maximum value at \(\mathrm{x}=1\) \(\Rightarrow \quad 3 \geq-2+\log _{2}\left(\mathrm{~b}^{2}-4\right)\) \(\log _{2}\left(b^{2}-4\right) \leq 5\) \(b^{2}-4>0 \text { and } b^{2}-4 \leq 32 \text { or } b^{2}-36 \leq 0\) \(b^{2}>4 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b \leq 6\) So, \(\quad b \in[-6,-2] \cup(2,6]\)
JEE Main-2022-26.07.2022
Application of Derivatives
85691
Let \(x=2\) be a local minima of the function \(f(x)\) \(=2 x^{4}-18 x^{2}+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(\mathbf{M}=\)
1 \(18 \sqrt{6}-\frac{33}{2}\)
2 \(12 \sqrt{6}-\frac{33}{2}\)
3 \(12 \sqrt{6}-\frac{31}{2}\)
4 \(18 \sqrt{6}-\frac{31}{2}\)
Explanation:
(B) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{4}-18 \mathrm{x}^{2}+8 \mathrm{x}+12, \mathrm{x} \in(-4,4)\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}^{3}-36 \mathrm{x}+8\) \(f^{\prime}(x)=4\left(2 x^{3}-9 x+2\right)\) For maxima/minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(4\left(2 \mathrm{x}^{3}-9 \mathrm{x}+2\right)=0\) \(2 \mathrm{x}^{3}-9 \mathrm{x}+2=0\) Then, Since, \(f^{\prime}(x)\) change sign from \(+v e\) to \(-v e\) passing through \(-1+\sqrt{\frac{3}{2}}\) Then, \(x=-1+\sqrt{\frac{3}{2}}\) is point of local maxima. So, \(\quad M=f\left(-1+\sqrt{\frac{3}{2}}\right)\) \(M=2\left(-1+\sqrt{\frac{3}{2}}\right)^{4}-18\left(-1+\sqrt{\frac{3}{2}}\right)^{2}+8\left(-1+\sqrt{\frac{3}{2}}\right)+12\) \(M=2\left(1-4 \sqrt{\frac{3}{2}}+6 \times \frac{3}{2}-4 \times \frac{3}{2} \sqrt{\frac{3}{2}}+\frac{3}{2} \times \frac{3}{2}\right)\) \(-18\left(1-2 \sqrt{\frac{3}{2}}+\frac{3}{2}\right)-8+8 \sqrt{\frac{3}{2}}+12\) \(M=2-4 \sqrt{6}+18-6 \sqrt{6}+\frac{9}{2}-18+18 \sqrt{6}-27-8+4 \sqrt{6}+12\) \(\mathrm{M}=12 \sqrt{6}-\frac{33}{2}\)
JEE Main-2023-25.01.2023
Application of Derivatives
85692
Let the function \(f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p\) -9) \(x-6\) have a maxima for some value of \(x\lt \) 0 and a minima for some value of \(x>0\). Then, the set of all values of \(p\) is
85693
If the function \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and \(g(x)=\frac{x^{3}}{3}+a x+b x^{2}, a \neq 2 b\) have common extreme point, then \(a+2 b+7\) is equal to :
1 \(\frac{3}{2}\)
2 3
3 6
4 4
Explanation:
(C) : Given, \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and, \(\quad g(x)=\frac{x^{3}}{3}+a x+b x^{2}\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{~b}+\mathrm{ax}\) And, \(g^{\prime}(x)=x^{2}+a+2 b x\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=0\) (for find \(\left.\mathrm{x}\right)\) \(x^{2}+2 b+a x-x^{2}-a-2 b x=0\) \((2 b-a)+a x-2 b x=0\) \((2 b-a)-x(2 b-a)=0\) \((2 b-a)(1-x)=0\) \(2 b=a\) or \(x=1\) But given \(a \neq 2 b\) then \(x=1\) Put, \(x=1\) in \(f^{\prime}(x)=0\) or \(g^{\prime}(x)=0\) \(1+2 \mathrm{~b}+\mathrm{a}=0\) \(7+2 b+a=6\) So, \(\quad a+2 b+7=6\)
85689
Consider the following statements: 1.The function attains local minimum value at \(\mathrm{x}=\frac{7}{5}\) 2.\(\mathbf{x}=\mathbf{2}\) is the point of inflexion. Which of the above statements is/are correct?
1 1 only
2 2 only
3 Both 1 and 2
4 Neither 1 nor 2
Explanation:
(C) : \(\text { (1) } \mathrm{f}^{\prime \prime}\left(\frac{7}{5}\right) =\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right)^{2}(5)+0+0\) \(=\left(\frac{2}{5}\right)\left(\frac{9}{25}\right)(5)>0\) So, local minimum value at \(\mathrm{x}=\frac{7}{5}\) (2) \(\mathrm{f}^{\prime \prime}(2)=0\) So, both (1) and (2) are correct.
SCRA-2015
Application of Derivatives
85690
Let \(f(x)=\left\{\begin{array}{c}x^{3}-x^{2}+10 x-7, x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\). Then the set of all values of \(b\), for which \(f(x)\) has maximum value at \(\mathrm{x}=1\), is:
1 \((-6,-2)\)
2 \((2,6)\)
3 \([-6,-2) \cup(2,6]\)
4 \([-\sqrt{6},-2) \cup(2, \sqrt{6}]\)
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{l}x^{3}-x^{2}+10 x-7, \quad x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\) Then, for \(x\lt 1, f(x)=x^{3}-x^{2}+10 x-7\) \(f^{\prime}(x)=3 x^{2}-2 x+10>0\) \(\Rightarrow \quad \mathrm{f}(\mathrm{x})\) is increasing. And, for \(x>1, f^{\prime}(x)\lt 0 \Rightarrow f(x)\) is decreasing \(\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)\) From maximum value at \(\mathrm{x}=1\) \(\Rightarrow \quad 3 \geq-2+\log _{2}\left(\mathrm{~b}^{2}-4\right)\) \(\log _{2}\left(b^{2}-4\right) \leq 5\) \(b^{2}-4>0 \text { and } b^{2}-4 \leq 32 \text { or } b^{2}-36 \leq 0\) \(b^{2}>4 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b \leq 6\) So, \(\quad b \in[-6,-2] \cup(2,6]\)
JEE Main-2022-26.07.2022
Application of Derivatives
85691
Let \(x=2\) be a local minima of the function \(f(x)\) \(=2 x^{4}-18 x^{2}+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(\mathbf{M}=\)
1 \(18 \sqrt{6}-\frac{33}{2}\)
2 \(12 \sqrt{6}-\frac{33}{2}\)
3 \(12 \sqrt{6}-\frac{31}{2}\)
4 \(18 \sqrt{6}-\frac{31}{2}\)
Explanation:
(B) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{4}-18 \mathrm{x}^{2}+8 \mathrm{x}+12, \mathrm{x} \in(-4,4)\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}^{3}-36 \mathrm{x}+8\) \(f^{\prime}(x)=4\left(2 x^{3}-9 x+2\right)\) For maxima/minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(4\left(2 \mathrm{x}^{3}-9 \mathrm{x}+2\right)=0\) \(2 \mathrm{x}^{3}-9 \mathrm{x}+2=0\) Then, Since, \(f^{\prime}(x)\) change sign from \(+v e\) to \(-v e\) passing through \(-1+\sqrt{\frac{3}{2}}\) Then, \(x=-1+\sqrt{\frac{3}{2}}\) is point of local maxima. So, \(\quad M=f\left(-1+\sqrt{\frac{3}{2}}\right)\) \(M=2\left(-1+\sqrt{\frac{3}{2}}\right)^{4}-18\left(-1+\sqrt{\frac{3}{2}}\right)^{2}+8\left(-1+\sqrt{\frac{3}{2}}\right)+12\) \(M=2\left(1-4 \sqrt{\frac{3}{2}}+6 \times \frac{3}{2}-4 \times \frac{3}{2} \sqrt{\frac{3}{2}}+\frac{3}{2} \times \frac{3}{2}\right)\) \(-18\left(1-2 \sqrt{\frac{3}{2}}+\frac{3}{2}\right)-8+8 \sqrt{\frac{3}{2}}+12\) \(M=2-4 \sqrt{6}+18-6 \sqrt{6}+\frac{9}{2}-18+18 \sqrt{6}-27-8+4 \sqrt{6}+12\) \(\mathrm{M}=12 \sqrt{6}-\frac{33}{2}\)
JEE Main-2023-25.01.2023
Application of Derivatives
85692
Let the function \(f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p\) -9) \(x-6\) have a maxima for some value of \(x\lt \) 0 and a minima for some value of \(x>0\). Then, the set of all values of \(p\) is
85693
If the function \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and \(g(x)=\frac{x^{3}}{3}+a x+b x^{2}, a \neq 2 b\) have common extreme point, then \(a+2 b+7\) is equal to :
1 \(\frac{3}{2}\)
2 3
3 6
4 4
Explanation:
(C) : Given, \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and, \(\quad g(x)=\frac{x^{3}}{3}+a x+b x^{2}\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{~b}+\mathrm{ax}\) And, \(g^{\prime}(x)=x^{2}+a+2 b x\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=0\) (for find \(\left.\mathrm{x}\right)\) \(x^{2}+2 b+a x-x^{2}-a-2 b x=0\) \((2 b-a)+a x-2 b x=0\) \((2 b-a)-x(2 b-a)=0\) \((2 b-a)(1-x)=0\) \(2 b=a\) or \(x=1\) But given \(a \neq 2 b\) then \(x=1\) Put, \(x=1\) in \(f^{\prime}(x)=0\) or \(g^{\prime}(x)=0\) \(1+2 \mathrm{~b}+\mathrm{a}=0\) \(7+2 b+a=6\) So, \(\quad a+2 b+7=6\)
85689
Consider the following statements: 1.The function attains local minimum value at \(\mathrm{x}=\frac{7}{5}\) 2.\(\mathbf{x}=\mathbf{2}\) is the point of inflexion. Which of the above statements is/are correct?
1 1 only
2 2 only
3 Both 1 and 2
4 Neither 1 nor 2
Explanation:
(C) : \(\text { (1) } \mathrm{f}^{\prime \prime}\left(\frac{7}{5}\right) =\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right)^{2}(5)+0+0\) \(=\left(\frac{2}{5}\right)\left(\frac{9}{25}\right)(5)>0\) So, local minimum value at \(\mathrm{x}=\frac{7}{5}\) (2) \(\mathrm{f}^{\prime \prime}(2)=0\) So, both (1) and (2) are correct.
SCRA-2015
Application of Derivatives
85690
Let \(f(x)=\left\{\begin{array}{c}x^{3}-x^{2}+10 x-7, x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\). Then the set of all values of \(b\), for which \(f(x)\) has maximum value at \(\mathrm{x}=1\), is:
1 \((-6,-2)\)
2 \((2,6)\)
3 \([-6,-2) \cup(2,6]\)
4 \([-\sqrt{6},-2) \cup(2, \sqrt{6}]\)
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{l}x^{3}-x^{2}+10 x-7, \quad x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\) Then, for \(x\lt 1, f(x)=x^{3}-x^{2}+10 x-7\) \(f^{\prime}(x)=3 x^{2}-2 x+10>0\) \(\Rightarrow \quad \mathrm{f}(\mathrm{x})\) is increasing. And, for \(x>1, f^{\prime}(x)\lt 0 \Rightarrow f(x)\) is decreasing \(\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)\) From maximum value at \(\mathrm{x}=1\) \(\Rightarrow \quad 3 \geq-2+\log _{2}\left(\mathrm{~b}^{2}-4\right)\) \(\log _{2}\left(b^{2}-4\right) \leq 5\) \(b^{2}-4>0 \text { and } b^{2}-4 \leq 32 \text { or } b^{2}-36 \leq 0\) \(b^{2}>4 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b \leq 6\) So, \(\quad b \in[-6,-2] \cup(2,6]\)
JEE Main-2022-26.07.2022
Application of Derivatives
85691
Let \(x=2\) be a local minima of the function \(f(x)\) \(=2 x^{4}-18 x^{2}+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(\mathbf{M}=\)
1 \(18 \sqrt{6}-\frac{33}{2}\)
2 \(12 \sqrt{6}-\frac{33}{2}\)
3 \(12 \sqrt{6}-\frac{31}{2}\)
4 \(18 \sqrt{6}-\frac{31}{2}\)
Explanation:
(B) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{4}-18 \mathrm{x}^{2}+8 \mathrm{x}+12, \mathrm{x} \in(-4,4)\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}^{3}-36 \mathrm{x}+8\) \(f^{\prime}(x)=4\left(2 x^{3}-9 x+2\right)\) For maxima/minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(4\left(2 \mathrm{x}^{3}-9 \mathrm{x}+2\right)=0\) \(2 \mathrm{x}^{3}-9 \mathrm{x}+2=0\) Then, Since, \(f^{\prime}(x)\) change sign from \(+v e\) to \(-v e\) passing through \(-1+\sqrt{\frac{3}{2}}\) Then, \(x=-1+\sqrt{\frac{3}{2}}\) is point of local maxima. So, \(\quad M=f\left(-1+\sqrt{\frac{3}{2}}\right)\) \(M=2\left(-1+\sqrt{\frac{3}{2}}\right)^{4}-18\left(-1+\sqrt{\frac{3}{2}}\right)^{2}+8\left(-1+\sqrt{\frac{3}{2}}\right)+12\) \(M=2\left(1-4 \sqrt{\frac{3}{2}}+6 \times \frac{3}{2}-4 \times \frac{3}{2} \sqrt{\frac{3}{2}}+\frac{3}{2} \times \frac{3}{2}\right)\) \(-18\left(1-2 \sqrt{\frac{3}{2}}+\frac{3}{2}\right)-8+8 \sqrt{\frac{3}{2}}+12\) \(M=2-4 \sqrt{6}+18-6 \sqrt{6}+\frac{9}{2}-18+18 \sqrt{6}-27-8+4 \sqrt{6}+12\) \(\mathrm{M}=12 \sqrt{6}-\frac{33}{2}\)
JEE Main-2023-25.01.2023
Application of Derivatives
85692
Let the function \(f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p\) -9) \(x-6\) have a maxima for some value of \(x\lt \) 0 and a minima for some value of \(x>0\). Then, the set of all values of \(p\) is
85693
If the function \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and \(g(x)=\frac{x^{3}}{3}+a x+b x^{2}, a \neq 2 b\) have common extreme point, then \(a+2 b+7\) is equal to :
1 \(\frac{3}{2}\)
2 3
3 6
4 4
Explanation:
(C) : Given, \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and, \(\quad g(x)=\frac{x^{3}}{3}+a x+b x^{2}\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{~b}+\mathrm{ax}\) And, \(g^{\prime}(x)=x^{2}+a+2 b x\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=0\) (for find \(\left.\mathrm{x}\right)\) \(x^{2}+2 b+a x-x^{2}-a-2 b x=0\) \((2 b-a)+a x-2 b x=0\) \((2 b-a)-x(2 b-a)=0\) \((2 b-a)(1-x)=0\) \(2 b=a\) or \(x=1\) But given \(a \neq 2 b\) then \(x=1\) Put, \(x=1\) in \(f^{\prime}(x)=0\) or \(g^{\prime}(x)=0\) \(1+2 \mathrm{~b}+\mathrm{a}=0\) \(7+2 b+a=6\) So, \(\quad a+2 b+7=6\)
85689
Consider the following statements: 1.The function attains local minimum value at \(\mathrm{x}=\frac{7}{5}\) 2.\(\mathbf{x}=\mathbf{2}\) is the point of inflexion. Which of the above statements is/are correct?
1 1 only
2 2 only
3 Both 1 and 2
4 Neither 1 nor 2
Explanation:
(C) : \(\text { (1) } \mathrm{f}^{\prime \prime}\left(\frac{7}{5}\right) =\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right)^{2}(5)+0+0\) \(=\left(\frac{2}{5}\right)\left(\frac{9}{25}\right)(5)>0\) So, local minimum value at \(\mathrm{x}=\frac{7}{5}\) (2) \(\mathrm{f}^{\prime \prime}(2)=0\) So, both (1) and (2) are correct.
SCRA-2015
Application of Derivatives
85690
Let \(f(x)=\left\{\begin{array}{c}x^{3}-x^{2}+10 x-7, x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\). Then the set of all values of \(b\), for which \(f(x)\) has maximum value at \(\mathrm{x}=1\), is:
1 \((-6,-2)\)
2 \((2,6)\)
3 \([-6,-2) \cup(2,6]\)
4 \([-\sqrt{6},-2) \cup(2, \sqrt{6}]\)
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{l}x^{3}-x^{2}+10 x-7, \quad x \leq 1 \\ -2 x+\log _{2}\left(b^{2}-4\right), x>1\end{array}\right.\) Then, for \(x\lt 1, f(x)=x^{3}-x^{2}+10 x-7\) \(f^{\prime}(x)=3 x^{2}-2 x+10>0\) \(\Rightarrow \quad \mathrm{f}(\mathrm{x})\) is increasing. And, for \(x>1, f^{\prime}(x)\lt 0 \Rightarrow f(x)\) is decreasing \(\lim _{x \rightarrow 1^{+}} f(x)=-2+\log _{2}\left(b^{2}-4\right)\) From maximum value at \(\mathrm{x}=1\) \(\Rightarrow \quad 3 \geq-2+\log _{2}\left(\mathrm{~b}^{2}-4\right)\) \(\log _{2}\left(b^{2}-4\right) \leq 5\) \(b^{2}-4>0 \text { and } b^{2}-4 \leq 32 \text { or } b^{2}-36 \leq 0\) \(b^{2}>4 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b^{2} \leq 36\) \(b> \pm 2 \text { and } b \leq 6\) So, \(\quad b \in[-6,-2] \cup(2,6]\)
JEE Main-2022-26.07.2022
Application of Derivatives
85691
Let \(x=2\) be a local minima of the function \(f(x)\) \(=2 x^{4}-18 x^{2}+8 x+12, x \in(-4,4)\). If \(M\) is local maximum value of the function \(f\) in \((-4,4)\), then \(\mathbf{M}=\)
1 \(18 \sqrt{6}-\frac{33}{2}\)
2 \(12 \sqrt{6}-\frac{33}{2}\)
3 \(12 \sqrt{6}-\frac{31}{2}\)
4 \(18 \sqrt{6}-\frac{31}{2}\)
Explanation:
(B) : \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{4}-18 \mathrm{x}^{2}+8 \mathrm{x}+12, \mathrm{x} \in(-4,4)\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=8 \mathrm{x}^{3}-36 \mathrm{x}+8\) \(f^{\prime}(x)=4\left(2 x^{3}-9 x+2\right)\) For maxima/minima \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(4\left(2 \mathrm{x}^{3}-9 \mathrm{x}+2\right)=0\) \(2 \mathrm{x}^{3}-9 \mathrm{x}+2=0\) Then, Since, \(f^{\prime}(x)\) change sign from \(+v e\) to \(-v e\) passing through \(-1+\sqrt{\frac{3}{2}}\) Then, \(x=-1+\sqrt{\frac{3}{2}}\) is point of local maxima. So, \(\quad M=f\left(-1+\sqrt{\frac{3}{2}}\right)\) \(M=2\left(-1+\sqrt{\frac{3}{2}}\right)^{4}-18\left(-1+\sqrt{\frac{3}{2}}\right)^{2}+8\left(-1+\sqrt{\frac{3}{2}}\right)+12\) \(M=2\left(1-4 \sqrt{\frac{3}{2}}+6 \times \frac{3}{2}-4 \times \frac{3}{2} \sqrt{\frac{3}{2}}+\frac{3}{2} \times \frac{3}{2}\right)\) \(-18\left(1-2 \sqrt{\frac{3}{2}}+\frac{3}{2}\right)-8+8 \sqrt{\frac{3}{2}}+12\) \(M=2-4 \sqrt{6}+18-6 \sqrt{6}+\frac{9}{2}-18+18 \sqrt{6}-27-8+4 \sqrt{6}+12\) \(\mathrm{M}=12 \sqrt{6}-\frac{33}{2}\)
JEE Main-2023-25.01.2023
Application of Derivatives
85692
Let the function \(f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p\) -9) \(x-6\) have a maxima for some value of \(x\lt \) 0 and a minima for some value of \(x>0\). Then, the set of all values of \(p\) is
85693
If the function \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and \(g(x)=\frac{x^{3}}{3}+a x+b x^{2}, a \neq 2 b\) have common extreme point, then \(a+2 b+7\) is equal to :
1 \(\frac{3}{2}\)
2 3
3 6
4 4
Explanation:
(C) : Given, \(f(x)=\frac{x^{3}}{3}+2 b x+\frac{a x^{2}}{2}\) and, \(\quad g(x)=\frac{x^{3}}{3}+a x+b x^{2}\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2}+2 \mathrm{~b}+\mathrm{ax}\) And, \(g^{\prime}(x)=x^{2}+a+2 b x\) Then, \(\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=0\) (for find \(\left.\mathrm{x}\right)\) \(x^{2}+2 b+a x-x^{2}-a-2 b x=0\) \((2 b-a)+a x-2 b x=0\) \((2 b-a)-x(2 b-a)=0\) \((2 b-a)(1-x)=0\) \(2 b=a\) or \(x=1\) But given \(a \neq 2 b\) then \(x=1\) Put, \(x=1\) in \(f^{\prime}(x)=0\) or \(g^{\prime}(x)=0\) \(1+2 \mathrm{~b}+\mathrm{a}=0\) \(7+2 b+a=6\) So, \(\quad a+2 b+7=6\)
