NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85716
The minimum value of \(9^{x}+9^{1-x}, x \in R\) is
1 2
2 3
3 6
4 9
Explanation:
(C) : Minimum value of \(9^{x}+9^{1-x}, x \in R\) We know that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{9^{x} \cdot 9^{1-x}}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{\frac{9^{x} \cdot 9}{9^{x}}}\) \(9^{x}+9^{1-x} \geq 2 \sqrt{9}\) \(9^{x}+9^{1-x} \geq 6\) Minimum value is 6
AMU-2016
Application of Derivatives
85717
The maximum value of \(\frac{\log x}{x}\), if \(x>0\) is
1 1
2 \(2 / \mathrm{e}\)
3 e
4 \(1 / \mathrm{e}\)
Explanation:
(D) : Let \(f(x)=\frac{\log x}{x}\) Differentiating w.r.t. ' \(x\) ' we get \(f^{\prime}(x)=\frac{x \times \frac{1}{x}-\log x}{x^{2}}=\frac{1-\log x}{x^{2}}\) For critical point \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\log x=0\) Now, \(f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) \times 2 x}{x^{4}}\) \(=\frac{-x-2 x+2 x \log x}{x^{4}}\) \(=\frac{-3 x+2 x \log x}{x^{4}}\) Then, \(f^{\prime \prime}(e)=\frac{-3 e+2 e}{e^{4}}\) \(\Rightarrow \quad \frac{-1}{\mathrm{e}^{3}}\lt 0\) \(\mathrm{f}(\mathrm{x})\) is maximum when \(\mathrm{x}=\mathrm{e}\) So, maximum value \(=\mathrm{f}(\mathrm{x})=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\)
Karnataka CET-2020
Application of Derivatives
85718
The difference between the greatest and the least values of the function \(f(x)=\sin 2 x-x\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is:
1 \(\pi\)
2 \(\sqrt{3}-\pi / 3\)
3 \(-\sqrt{3}+\pi / 3\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\sin 2 \mathrm{x}-\mathrm{x}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(\mathrm{y}=\sin 2 \mathrm{x}-\mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos 2 \mathrm{x}-1\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) then \(2 \cos 2 \mathrm{x}=1\) \(\cos 2 \mathrm{x}=\frac{1}{2}\) \(\cos 2 \mathrm{x}=\cos \frac{\pi}{3}\) \(2 \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}\) \(\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\) Now, \(\quad \mathrm{f}\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\) \(\mathrm{f}\left(-\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{3}\right)+\frac{\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\) \(f\left(-\frac{\pi}{2}\right)=\sin (-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=\sin \pi-\frac{\pi}{2}=-\frac{\pi}{2}\) The greatest value of \(=\frac{\pi}{2}\) And least value \(=-\frac{\pi}{2}\) Required difference \(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi\)
AMU-2013
Application of Derivatives
85719
The least value of the function \(f(x)=a x+b / x, a\) \(>0, b>0, x>0\) is
1 \(\sqrt{\mathrm{ab}}\)
2 \(2 \sqrt{\frac{a}{b}}\)
3 \(2 \sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\)
4 \(2 \sqrt{\mathrm{ab}}\)
Explanation:
(D) : Given, \(f(x)=a x+\frac{b}{x}\) Then, \(\quad f^{\prime}(x)=a-\frac{b}{x^{2}}\) For maxima and minima \(f^{\prime}(x)=0\) \(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0\) \(\mathrm{x}=\sqrt{\frac{b}{a}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}}\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{~b}}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{3 / 2}}>0\) Hence, \(f(x)\) is minimum at \(x=\sqrt{\frac{b}{a}}\) and minimum value \(a \sqrt{\frac{b}{a}}+\frac{b}{\left(\sqrt{\frac{b}{a}}\right)}\) \(=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}\)
85716
The minimum value of \(9^{x}+9^{1-x}, x \in R\) is
1 2
2 3
3 6
4 9
Explanation:
(C) : Minimum value of \(9^{x}+9^{1-x}, x \in R\) We know that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{9^{x} \cdot 9^{1-x}}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{\frac{9^{x} \cdot 9}{9^{x}}}\) \(9^{x}+9^{1-x} \geq 2 \sqrt{9}\) \(9^{x}+9^{1-x} \geq 6\) Minimum value is 6
AMU-2016
Application of Derivatives
85717
The maximum value of \(\frac{\log x}{x}\), if \(x>0\) is
1 1
2 \(2 / \mathrm{e}\)
3 e
4 \(1 / \mathrm{e}\)
Explanation:
(D) : Let \(f(x)=\frac{\log x}{x}\) Differentiating w.r.t. ' \(x\) ' we get \(f^{\prime}(x)=\frac{x \times \frac{1}{x}-\log x}{x^{2}}=\frac{1-\log x}{x^{2}}\) For critical point \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\log x=0\) Now, \(f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) \times 2 x}{x^{4}}\) \(=\frac{-x-2 x+2 x \log x}{x^{4}}\) \(=\frac{-3 x+2 x \log x}{x^{4}}\) Then, \(f^{\prime \prime}(e)=\frac{-3 e+2 e}{e^{4}}\) \(\Rightarrow \quad \frac{-1}{\mathrm{e}^{3}}\lt 0\) \(\mathrm{f}(\mathrm{x})\) is maximum when \(\mathrm{x}=\mathrm{e}\) So, maximum value \(=\mathrm{f}(\mathrm{x})=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\)
Karnataka CET-2020
Application of Derivatives
85718
The difference between the greatest and the least values of the function \(f(x)=\sin 2 x-x\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is:
1 \(\pi\)
2 \(\sqrt{3}-\pi / 3\)
3 \(-\sqrt{3}+\pi / 3\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\sin 2 \mathrm{x}-\mathrm{x}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(\mathrm{y}=\sin 2 \mathrm{x}-\mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos 2 \mathrm{x}-1\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) then \(2 \cos 2 \mathrm{x}=1\) \(\cos 2 \mathrm{x}=\frac{1}{2}\) \(\cos 2 \mathrm{x}=\cos \frac{\pi}{3}\) \(2 \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}\) \(\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\) Now, \(\quad \mathrm{f}\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\) \(\mathrm{f}\left(-\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{3}\right)+\frac{\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\) \(f\left(-\frac{\pi}{2}\right)=\sin (-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=\sin \pi-\frac{\pi}{2}=-\frac{\pi}{2}\) The greatest value of \(=\frac{\pi}{2}\) And least value \(=-\frac{\pi}{2}\) Required difference \(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi\)
AMU-2013
Application of Derivatives
85719
The least value of the function \(f(x)=a x+b / x, a\) \(>0, b>0, x>0\) is
1 \(\sqrt{\mathrm{ab}}\)
2 \(2 \sqrt{\frac{a}{b}}\)
3 \(2 \sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\)
4 \(2 \sqrt{\mathrm{ab}}\)
Explanation:
(D) : Given, \(f(x)=a x+\frac{b}{x}\) Then, \(\quad f^{\prime}(x)=a-\frac{b}{x^{2}}\) For maxima and minima \(f^{\prime}(x)=0\) \(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0\) \(\mathrm{x}=\sqrt{\frac{b}{a}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}}\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{~b}}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{3 / 2}}>0\) Hence, \(f(x)\) is minimum at \(x=\sqrt{\frac{b}{a}}\) and minimum value \(a \sqrt{\frac{b}{a}}+\frac{b}{\left(\sqrt{\frac{b}{a}}\right)}\) \(=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}\)
85716
The minimum value of \(9^{x}+9^{1-x}, x \in R\) is
1 2
2 3
3 6
4 9
Explanation:
(C) : Minimum value of \(9^{x}+9^{1-x}, x \in R\) We know that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{9^{x} \cdot 9^{1-x}}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{\frac{9^{x} \cdot 9}{9^{x}}}\) \(9^{x}+9^{1-x} \geq 2 \sqrt{9}\) \(9^{x}+9^{1-x} \geq 6\) Minimum value is 6
AMU-2016
Application of Derivatives
85717
The maximum value of \(\frac{\log x}{x}\), if \(x>0\) is
1 1
2 \(2 / \mathrm{e}\)
3 e
4 \(1 / \mathrm{e}\)
Explanation:
(D) : Let \(f(x)=\frac{\log x}{x}\) Differentiating w.r.t. ' \(x\) ' we get \(f^{\prime}(x)=\frac{x \times \frac{1}{x}-\log x}{x^{2}}=\frac{1-\log x}{x^{2}}\) For critical point \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\log x=0\) Now, \(f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) \times 2 x}{x^{4}}\) \(=\frac{-x-2 x+2 x \log x}{x^{4}}\) \(=\frac{-3 x+2 x \log x}{x^{4}}\) Then, \(f^{\prime \prime}(e)=\frac{-3 e+2 e}{e^{4}}\) \(\Rightarrow \quad \frac{-1}{\mathrm{e}^{3}}\lt 0\) \(\mathrm{f}(\mathrm{x})\) is maximum when \(\mathrm{x}=\mathrm{e}\) So, maximum value \(=\mathrm{f}(\mathrm{x})=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\)
Karnataka CET-2020
Application of Derivatives
85718
The difference between the greatest and the least values of the function \(f(x)=\sin 2 x-x\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is:
1 \(\pi\)
2 \(\sqrt{3}-\pi / 3\)
3 \(-\sqrt{3}+\pi / 3\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\sin 2 \mathrm{x}-\mathrm{x}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(\mathrm{y}=\sin 2 \mathrm{x}-\mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos 2 \mathrm{x}-1\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) then \(2 \cos 2 \mathrm{x}=1\) \(\cos 2 \mathrm{x}=\frac{1}{2}\) \(\cos 2 \mathrm{x}=\cos \frac{\pi}{3}\) \(2 \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}\) \(\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\) Now, \(\quad \mathrm{f}\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\) \(\mathrm{f}\left(-\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{3}\right)+\frac{\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\) \(f\left(-\frac{\pi}{2}\right)=\sin (-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=\sin \pi-\frac{\pi}{2}=-\frac{\pi}{2}\) The greatest value of \(=\frac{\pi}{2}\) And least value \(=-\frac{\pi}{2}\) Required difference \(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi\)
AMU-2013
Application of Derivatives
85719
The least value of the function \(f(x)=a x+b / x, a\) \(>0, b>0, x>0\) is
1 \(\sqrt{\mathrm{ab}}\)
2 \(2 \sqrt{\frac{a}{b}}\)
3 \(2 \sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\)
4 \(2 \sqrt{\mathrm{ab}}\)
Explanation:
(D) : Given, \(f(x)=a x+\frac{b}{x}\) Then, \(\quad f^{\prime}(x)=a-\frac{b}{x^{2}}\) For maxima and minima \(f^{\prime}(x)=0\) \(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0\) \(\mathrm{x}=\sqrt{\frac{b}{a}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}}\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{~b}}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{3 / 2}}>0\) Hence, \(f(x)\) is minimum at \(x=\sqrt{\frac{b}{a}}\) and minimum value \(a \sqrt{\frac{b}{a}}+\frac{b}{\left(\sqrt{\frac{b}{a}}\right)}\) \(=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85716
The minimum value of \(9^{x}+9^{1-x}, x \in R\) is
1 2
2 3
3 6
4 9
Explanation:
(C) : Minimum value of \(9^{x}+9^{1-x}, x \in R\) We know that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{9^{x} \cdot 9^{1-x}}\) \(\frac{9^{x}+9^{1-x}}{2} \geq \sqrt{\frac{9^{x} \cdot 9}{9^{x}}}\) \(9^{x}+9^{1-x} \geq 2 \sqrt{9}\) \(9^{x}+9^{1-x} \geq 6\) Minimum value is 6
AMU-2016
Application of Derivatives
85717
The maximum value of \(\frac{\log x}{x}\), if \(x>0\) is
1 1
2 \(2 / \mathrm{e}\)
3 e
4 \(1 / \mathrm{e}\)
Explanation:
(D) : Let \(f(x)=\frac{\log x}{x}\) Differentiating w.r.t. ' \(x\) ' we get \(f^{\prime}(x)=\frac{x \times \frac{1}{x}-\log x}{x^{2}}=\frac{1-\log x}{x^{2}}\) For critical point \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(1-\log x=0\) Now, \(f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) \times 2 x}{x^{4}}\) \(=\frac{-x-2 x+2 x \log x}{x^{4}}\) \(=\frac{-3 x+2 x \log x}{x^{4}}\) Then, \(f^{\prime \prime}(e)=\frac{-3 e+2 e}{e^{4}}\) \(\Rightarrow \quad \frac{-1}{\mathrm{e}^{3}}\lt 0\) \(\mathrm{f}(\mathrm{x})\) is maximum when \(\mathrm{x}=\mathrm{e}\) So, maximum value \(=\mathrm{f}(\mathrm{x})=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}\)
Karnataka CET-2020
Application of Derivatives
85718
The difference between the greatest and the least values of the function \(f(x)=\sin 2 x-x\) on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is:
1 \(\pi\)
2 \(\sqrt{3}-\pi / 3\)
3 \(-\sqrt{3}+\pi / 3\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\sin 2 \mathrm{x}-\mathrm{x}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) \(\mathrm{y}=\sin 2 \mathrm{x}-\mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos 2 \mathrm{x}-1\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) then \(2 \cos 2 \mathrm{x}=1\) \(\cos 2 \mathrm{x}=\frac{1}{2}\) \(\cos 2 \mathrm{x}=\cos \frac{\pi}{3}\) \(2 \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}\) \(\mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\) Now, \(\quad \mathrm{f}\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\) \(\mathrm{f}\left(-\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{3}\right)+\frac{\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\) \(f\left(-\frac{\pi}{2}\right)=\sin (-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=\sin \pi-\frac{\pi}{2}=-\frac{\pi}{2}\) The greatest value of \(=\frac{\pi}{2}\) And least value \(=-\frac{\pi}{2}\) Required difference \(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi\)
AMU-2013
Application of Derivatives
85719
The least value of the function \(f(x)=a x+b / x, a\) \(>0, b>0, x>0\) is
1 \(\sqrt{\mathrm{ab}}\)
2 \(2 \sqrt{\frac{a}{b}}\)
3 \(2 \sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\)
4 \(2 \sqrt{\mathrm{ab}}\)
Explanation:
(D) : Given, \(f(x)=a x+\frac{b}{x}\) Then, \(\quad f^{\prime}(x)=a-\frac{b}{x^{2}}\) For maxima and minima \(f^{\prime}(x)=0\) \(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0\) \(\mathrm{x}=\sqrt{\frac{b}{a}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}}\) \(\mathrm{f}^{\prime \prime}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{~b}}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{3 / 2}}>0\) Hence, \(f(x)\) is minimum at \(x=\sqrt{\frac{b}{a}}\) and minimum value \(a \sqrt{\frac{b}{a}}+\frac{b}{\left(\sqrt{\frac{b}{a}}\right)}\) \(=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}\)