85706
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) for every real number \(x\), then the minimum value of \(f\) is
1 does not exist
2 1
3 -1
4 2
Explanation:
(C) : Given, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}\) \(f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{x}=0\) If \(\quad \mathrm{x}\lt 0, \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) If \(\quad \mathrm{x}>0, \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\mathrm{x}\) is the point of minima Minimum value of \(\mathrm{f}(\mathrm{x})\) \(f(0)=-1\)
AMU-2007
Application of Derivatives
85708
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\mid 2 x^{2}+3 x\) \(-2 \mid+\sin x \cos x\) in the interval \([0,1]\) is :
(B) : Given function, \(f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x\) in the interval \([0,1]\) Then, \(f(x)=|(2 x-1)(x+2)|+\sin x \cos x\) \(f^{\prime}(x)=\left\{\begin{array}{l}4 x+3+\frac{\cos 2 x}{4}, \frac{1}{2}\lt x\lt 1 \\ -(4 x+3)+\frac{\cos 2 x}{4}, 0 \leq \frac{1}{2}\end{array}\right.\) For, \(0 \leq \mathrm{x}\lt \frac{1}{2} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) For, \(\frac{1}{2}\lt \mathrm{x} \leq 1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) Then \(\mathrm{f}(\mathrm{x})\) has local minima at, \(\mathrm{x}=\frac{1}{2}\) and Local maxima at, \(\mathrm{x}=1\) So, \(f\left(\frac{1}{2}\right)+f(1)=3+\frac{1}{2}(1+2 \cos 1) \sin 1\)
JEE Main-2022-24.06.2022
Application of Derivatives
85709
Through the point \((4,5)\), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the \(x\) and \(y\) axes is
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 5\)
4 \(2: 3\)
Explanation:
(C): From question, Equation of \(\mathrm{AB}\) \(\frac{\mathrm{x}}{\mathrm{m}}+\frac{\mathrm{y}}{\mathrm{n}}=1 \tag{i}\) \(\because\) Line (i) pass through \((4,5)\) \(\frac{4}{m}+\frac{5}{n}=1 \Rightarrow \frac{4}{m}=1-\frac{5}{n} \Rightarrow m=\frac{4 n}{n-5}\) \(\because\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\) \(=\frac{1}{2} \times m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n\) \(A=\frac{2 n^{2}}{n-5}\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=2\left\{\frac{(\mathrm{n}-5) \cdot 2 \mathrm{n}-\mathrm{n}^{2}(1)}{(\mathrm{n}-5)^{2}}\right\}\) \(\frac{d A}{d n}=2\left\{\frac{2 n^{2}-10 n-n^{2}}{(n-5)^{2}}\right\}=\frac{2 n^{2}-20 n}{(n-5)^{2}}\) \(\because\) Area least for some value of \(n\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=0\) \(2 \mathrm{n}^{2}-20 \mathrm{n}=0\) \(2 \mathrm{n}(\mathrm{n}-10)=0\) \(\mathrm{n}=0,10\) \(\text { At } n=10, \quad m=\frac{4 n}{n-5}=\frac{40}{5}\) \(m=8\) \(\therefore m: n=8: 10=4: 5\)
Shift-I]
Application of Derivatives
85710
If the function \(f(x)=a \sin (x)+\frac{1}{3} \sin (3 x)\) attains maximum value at \(\mathrm{x}=\frac{\pi}{3}\), then ' \(\mathrm{a}\) '
1 3
2 \(\frac{1}{3}\)
3 2
4 \(\frac{1}{2}\)
Explanation:
(C): Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\) \(\frac{d}{d x}(f(x))=a \cos x+\cos 3 x=0 \quad\left\{\begin{array}{l}\text { Slope of tangent } \\ \text { at maxima is zero }\end{array}\right.\) Then, \(\mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=\mathrm{a} \cos \frac{\pi}{3}+\cos \pi=0\) \(\frac{a}{2}-1=0\) \(a=2\)
Shift-I
Application of Derivatives
85711
If \(x, y, z\) are three positive numbers, then the minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\) is
1 1
2 2
3 3
4 6
Explanation:
(D) : We have A.M. \(\geq\) G.M between two numbers \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{b}}}=1\) Similarly, inequality between \(\frac{\mathrm{c}}{\mathrm{a}} \& \frac{\mathrm{a}}{\mathrm{c}}\) and \(\frac{\mathrm{b}}{\mathrm{c}} \& \frac{\mathrm{c}}{\mathrm{b}}\) \(\frac{1}{2}\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{c}}\right) \geq \sqrt{\frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{c}}}=1\) and \(\frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{c}} \cdot \frac{\mathrm{c}}{\mathrm{b}}}=1\) \(\frac{1}{2}\left(\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right) \geq 3\) \(\Rightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \geq 6\) Thus, the minimum value of the given expression is 6 .
85706
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) for every real number \(x\), then the minimum value of \(f\) is
1 does not exist
2 1
3 -1
4 2
Explanation:
(C) : Given, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}\) \(f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{x}=0\) If \(\quad \mathrm{x}\lt 0, \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) If \(\quad \mathrm{x}>0, \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\mathrm{x}\) is the point of minima Minimum value of \(\mathrm{f}(\mathrm{x})\) \(f(0)=-1\)
AMU-2007
Application of Derivatives
85708
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\mid 2 x^{2}+3 x\) \(-2 \mid+\sin x \cos x\) in the interval \([0,1]\) is :
(B) : Given function, \(f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x\) in the interval \([0,1]\) Then, \(f(x)=|(2 x-1)(x+2)|+\sin x \cos x\) \(f^{\prime}(x)=\left\{\begin{array}{l}4 x+3+\frac{\cos 2 x}{4}, \frac{1}{2}\lt x\lt 1 \\ -(4 x+3)+\frac{\cos 2 x}{4}, 0 \leq \frac{1}{2}\end{array}\right.\) For, \(0 \leq \mathrm{x}\lt \frac{1}{2} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) For, \(\frac{1}{2}\lt \mathrm{x} \leq 1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) Then \(\mathrm{f}(\mathrm{x})\) has local minima at, \(\mathrm{x}=\frac{1}{2}\) and Local maxima at, \(\mathrm{x}=1\) So, \(f\left(\frac{1}{2}\right)+f(1)=3+\frac{1}{2}(1+2 \cos 1) \sin 1\)
JEE Main-2022-24.06.2022
Application of Derivatives
85709
Through the point \((4,5)\), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the \(x\) and \(y\) axes is
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 5\)
4 \(2: 3\)
Explanation:
(C): From question, Equation of \(\mathrm{AB}\) \(\frac{\mathrm{x}}{\mathrm{m}}+\frac{\mathrm{y}}{\mathrm{n}}=1 \tag{i}\) \(\because\) Line (i) pass through \((4,5)\) \(\frac{4}{m}+\frac{5}{n}=1 \Rightarrow \frac{4}{m}=1-\frac{5}{n} \Rightarrow m=\frac{4 n}{n-5}\) \(\because\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\) \(=\frac{1}{2} \times m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n\) \(A=\frac{2 n^{2}}{n-5}\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=2\left\{\frac{(\mathrm{n}-5) \cdot 2 \mathrm{n}-\mathrm{n}^{2}(1)}{(\mathrm{n}-5)^{2}}\right\}\) \(\frac{d A}{d n}=2\left\{\frac{2 n^{2}-10 n-n^{2}}{(n-5)^{2}}\right\}=\frac{2 n^{2}-20 n}{(n-5)^{2}}\) \(\because\) Area least for some value of \(n\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=0\) \(2 \mathrm{n}^{2}-20 \mathrm{n}=0\) \(2 \mathrm{n}(\mathrm{n}-10)=0\) \(\mathrm{n}=0,10\) \(\text { At } n=10, \quad m=\frac{4 n}{n-5}=\frac{40}{5}\) \(m=8\) \(\therefore m: n=8: 10=4: 5\)
Shift-I]
Application of Derivatives
85710
If the function \(f(x)=a \sin (x)+\frac{1}{3} \sin (3 x)\) attains maximum value at \(\mathrm{x}=\frac{\pi}{3}\), then ' \(\mathrm{a}\) '
1 3
2 \(\frac{1}{3}\)
3 2
4 \(\frac{1}{2}\)
Explanation:
(C): Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\) \(\frac{d}{d x}(f(x))=a \cos x+\cos 3 x=0 \quad\left\{\begin{array}{l}\text { Slope of tangent } \\ \text { at maxima is zero }\end{array}\right.\) Then, \(\mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=\mathrm{a} \cos \frac{\pi}{3}+\cos \pi=0\) \(\frac{a}{2}-1=0\) \(a=2\)
Shift-I
Application of Derivatives
85711
If \(x, y, z\) are three positive numbers, then the minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\) is
1 1
2 2
3 3
4 6
Explanation:
(D) : We have A.M. \(\geq\) G.M between two numbers \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{b}}}=1\) Similarly, inequality between \(\frac{\mathrm{c}}{\mathrm{a}} \& \frac{\mathrm{a}}{\mathrm{c}}\) and \(\frac{\mathrm{b}}{\mathrm{c}} \& \frac{\mathrm{c}}{\mathrm{b}}\) \(\frac{1}{2}\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{c}}\right) \geq \sqrt{\frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{c}}}=1\) and \(\frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{c}} \cdot \frac{\mathrm{c}}{\mathrm{b}}}=1\) \(\frac{1}{2}\left(\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right) \geq 3\) \(\Rightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \geq 6\) Thus, the minimum value of the given expression is 6 .
85706
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) for every real number \(x\), then the minimum value of \(f\) is
1 does not exist
2 1
3 -1
4 2
Explanation:
(C) : Given, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}\) \(f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{x}=0\) If \(\quad \mathrm{x}\lt 0, \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) If \(\quad \mathrm{x}>0, \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\mathrm{x}\) is the point of minima Minimum value of \(\mathrm{f}(\mathrm{x})\) \(f(0)=-1\)
AMU-2007
Application of Derivatives
85708
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\mid 2 x^{2}+3 x\) \(-2 \mid+\sin x \cos x\) in the interval \([0,1]\) is :
(B) : Given function, \(f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x\) in the interval \([0,1]\) Then, \(f(x)=|(2 x-1)(x+2)|+\sin x \cos x\) \(f^{\prime}(x)=\left\{\begin{array}{l}4 x+3+\frac{\cos 2 x}{4}, \frac{1}{2}\lt x\lt 1 \\ -(4 x+3)+\frac{\cos 2 x}{4}, 0 \leq \frac{1}{2}\end{array}\right.\) For, \(0 \leq \mathrm{x}\lt \frac{1}{2} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) For, \(\frac{1}{2}\lt \mathrm{x} \leq 1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) Then \(\mathrm{f}(\mathrm{x})\) has local minima at, \(\mathrm{x}=\frac{1}{2}\) and Local maxima at, \(\mathrm{x}=1\) So, \(f\left(\frac{1}{2}\right)+f(1)=3+\frac{1}{2}(1+2 \cos 1) \sin 1\)
JEE Main-2022-24.06.2022
Application of Derivatives
85709
Through the point \((4,5)\), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the \(x\) and \(y\) axes is
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 5\)
4 \(2: 3\)
Explanation:
(C): From question, Equation of \(\mathrm{AB}\) \(\frac{\mathrm{x}}{\mathrm{m}}+\frac{\mathrm{y}}{\mathrm{n}}=1 \tag{i}\) \(\because\) Line (i) pass through \((4,5)\) \(\frac{4}{m}+\frac{5}{n}=1 \Rightarrow \frac{4}{m}=1-\frac{5}{n} \Rightarrow m=\frac{4 n}{n-5}\) \(\because\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\) \(=\frac{1}{2} \times m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n\) \(A=\frac{2 n^{2}}{n-5}\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=2\left\{\frac{(\mathrm{n}-5) \cdot 2 \mathrm{n}-\mathrm{n}^{2}(1)}{(\mathrm{n}-5)^{2}}\right\}\) \(\frac{d A}{d n}=2\left\{\frac{2 n^{2}-10 n-n^{2}}{(n-5)^{2}}\right\}=\frac{2 n^{2}-20 n}{(n-5)^{2}}\) \(\because\) Area least for some value of \(n\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=0\) \(2 \mathrm{n}^{2}-20 \mathrm{n}=0\) \(2 \mathrm{n}(\mathrm{n}-10)=0\) \(\mathrm{n}=0,10\) \(\text { At } n=10, \quad m=\frac{4 n}{n-5}=\frac{40}{5}\) \(m=8\) \(\therefore m: n=8: 10=4: 5\)
Shift-I]
Application of Derivatives
85710
If the function \(f(x)=a \sin (x)+\frac{1}{3} \sin (3 x)\) attains maximum value at \(\mathrm{x}=\frac{\pi}{3}\), then ' \(\mathrm{a}\) '
1 3
2 \(\frac{1}{3}\)
3 2
4 \(\frac{1}{2}\)
Explanation:
(C): Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\) \(\frac{d}{d x}(f(x))=a \cos x+\cos 3 x=0 \quad\left\{\begin{array}{l}\text { Slope of tangent } \\ \text { at maxima is zero }\end{array}\right.\) Then, \(\mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=\mathrm{a} \cos \frac{\pi}{3}+\cos \pi=0\) \(\frac{a}{2}-1=0\) \(a=2\)
Shift-I
Application of Derivatives
85711
If \(x, y, z\) are three positive numbers, then the minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\) is
1 1
2 2
3 3
4 6
Explanation:
(D) : We have A.M. \(\geq\) G.M between two numbers \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{b}}}=1\) Similarly, inequality between \(\frac{\mathrm{c}}{\mathrm{a}} \& \frac{\mathrm{a}}{\mathrm{c}}\) and \(\frac{\mathrm{b}}{\mathrm{c}} \& \frac{\mathrm{c}}{\mathrm{b}}\) \(\frac{1}{2}\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{c}}\right) \geq \sqrt{\frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{c}}}=1\) and \(\frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{c}} \cdot \frac{\mathrm{c}}{\mathrm{b}}}=1\) \(\frac{1}{2}\left(\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right) \geq 3\) \(\Rightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \geq 6\) Thus, the minimum value of the given expression is 6 .
85706
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) for every real number \(x\), then the minimum value of \(f\) is
1 does not exist
2 1
3 -1
4 2
Explanation:
(C) : Given, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}\) \(f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{x}=0\) If \(\quad \mathrm{x}\lt 0, \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) If \(\quad \mathrm{x}>0, \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\mathrm{x}\) is the point of minima Minimum value of \(\mathrm{f}(\mathrm{x})\) \(f(0)=-1\)
AMU-2007
Application of Derivatives
85708
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\mid 2 x^{2}+3 x\) \(-2 \mid+\sin x \cos x\) in the interval \([0,1]\) is :
(B) : Given function, \(f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x\) in the interval \([0,1]\) Then, \(f(x)=|(2 x-1)(x+2)|+\sin x \cos x\) \(f^{\prime}(x)=\left\{\begin{array}{l}4 x+3+\frac{\cos 2 x}{4}, \frac{1}{2}\lt x\lt 1 \\ -(4 x+3)+\frac{\cos 2 x}{4}, 0 \leq \frac{1}{2}\end{array}\right.\) For, \(0 \leq \mathrm{x}\lt \frac{1}{2} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) For, \(\frac{1}{2}\lt \mathrm{x} \leq 1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) Then \(\mathrm{f}(\mathrm{x})\) has local minima at, \(\mathrm{x}=\frac{1}{2}\) and Local maxima at, \(\mathrm{x}=1\) So, \(f\left(\frac{1}{2}\right)+f(1)=3+\frac{1}{2}(1+2 \cos 1) \sin 1\)
JEE Main-2022-24.06.2022
Application of Derivatives
85709
Through the point \((4,5)\), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the \(x\) and \(y\) axes is
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 5\)
4 \(2: 3\)
Explanation:
(C): From question, Equation of \(\mathrm{AB}\) \(\frac{\mathrm{x}}{\mathrm{m}}+\frac{\mathrm{y}}{\mathrm{n}}=1 \tag{i}\) \(\because\) Line (i) pass through \((4,5)\) \(\frac{4}{m}+\frac{5}{n}=1 \Rightarrow \frac{4}{m}=1-\frac{5}{n} \Rightarrow m=\frac{4 n}{n-5}\) \(\because\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\) \(=\frac{1}{2} \times m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n\) \(A=\frac{2 n^{2}}{n-5}\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=2\left\{\frac{(\mathrm{n}-5) \cdot 2 \mathrm{n}-\mathrm{n}^{2}(1)}{(\mathrm{n}-5)^{2}}\right\}\) \(\frac{d A}{d n}=2\left\{\frac{2 n^{2}-10 n-n^{2}}{(n-5)^{2}}\right\}=\frac{2 n^{2}-20 n}{(n-5)^{2}}\) \(\because\) Area least for some value of \(n\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=0\) \(2 \mathrm{n}^{2}-20 \mathrm{n}=0\) \(2 \mathrm{n}(\mathrm{n}-10)=0\) \(\mathrm{n}=0,10\) \(\text { At } n=10, \quad m=\frac{4 n}{n-5}=\frac{40}{5}\) \(m=8\) \(\therefore m: n=8: 10=4: 5\)
Shift-I]
Application of Derivatives
85710
If the function \(f(x)=a \sin (x)+\frac{1}{3} \sin (3 x)\) attains maximum value at \(\mathrm{x}=\frac{\pi}{3}\), then ' \(\mathrm{a}\) '
1 3
2 \(\frac{1}{3}\)
3 2
4 \(\frac{1}{2}\)
Explanation:
(C): Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\) \(\frac{d}{d x}(f(x))=a \cos x+\cos 3 x=0 \quad\left\{\begin{array}{l}\text { Slope of tangent } \\ \text { at maxima is zero }\end{array}\right.\) Then, \(\mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=\mathrm{a} \cos \frac{\pi}{3}+\cos \pi=0\) \(\frac{a}{2}-1=0\) \(a=2\)
Shift-I
Application of Derivatives
85711
If \(x, y, z\) are three positive numbers, then the minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\) is
1 1
2 2
3 3
4 6
Explanation:
(D) : We have A.M. \(\geq\) G.M between two numbers \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{b}}}=1\) Similarly, inequality between \(\frac{\mathrm{c}}{\mathrm{a}} \& \frac{\mathrm{a}}{\mathrm{c}}\) and \(\frac{\mathrm{b}}{\mathrm{c}} \& \frac{\mathrm{c}}{\mathrm{b}}\) \(\frac{1}{2}\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{c}}\right) \geq \sqrt{\frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{c}}}=1\) and \(\frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{c}} \cdot \frac{\mathrm{c}}{\mathrm{b}}}=1\) \(\frac{1}{2}\left(\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right) \geq 3\) \(\Rightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \geq 6\) Thus, the minimum value of the given expression is 6 .
85706
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) for every real number \(x\), then the minimum value of \(f\) is
1 does not exist
2 1
3 -1
4 2
Explanation:
(C) : Given, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}\) \(f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\mathrm{x}=0\) If \(\quad \mathrm{x}\lt 0, \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) If \(\quad \mathrm{x}>0, \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\mathrm{x}\) is the point of minima Minimum value of \(\mathrm{f}(\mathrm{x})\) \(f(0)=-1\)
AMU-2007
Application of Derivatives
85708
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\mid 2 x^{2}+3 x\) \(-2 \mid+\sin x \cos x\) in the interval \([0,1]\) is :
(B) : Given function, \(f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x\) in the interval \([0,1]\) Then, \(f(x)=|(2 x-1)(x+2)|+\sin x \cos x\) \(f^{\prime}(x)=\left\{\begin{array}{l}4 x+3+\frac{\cos 2 x}{4}, \frac{1}{2}\lt x\lt 1 \\ -(4 x+3)+\frac{\cos 2 x}{4}, 0 \leq \frac{1}{2}\end{array}\right.\) For, \(0 \leq \mathrm{x}\lt \frac{1}{2} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) For, \(\frac{1}{2}\lt \mathrm{x} \leq 1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) Then \(\mathrm{f}(\mathrm{x})\) has local minima at, \(\mathrm{x}=\frac{1}{2}\) and Local maxima at, \(\mathrm{x}=1\) So, \(f\left(\frac{1}{2}\right)+f(1)=3+\frac{1}{2}(1+2 \cos 1) \sin 1\)
JEE Main-2022-24.06.2022
Application of Derivatives
85709
Through the point \((4,5)\), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least, when the ratio of the intercepts on the \(x\) and \(y\) axes is
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 5\)
4 \(2: 3\)
Explanation:
(C): From question, Equation of \(\mathrm{AB}\) \(\frac{\mathrm{x}}{\mathrm{m}}+\frac{\mathrm{y}}{\mathrm{n}}=1 \tag{i}\) \(\because\) Line (i) pass through \((4,5)\) \(\frac{4}{m}+\frac{5}{n}=1 \Rightarrow \frac{4}{m}=1-\frac{5}{n} \Rightarrow m=\frac{4 n}{n-5}\) \(\because\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\) \(=\frac{1}{2} \times m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n\) \(A=\frac{2 n^{2}}{n-5}\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=2\left\{\frac{(\mathrm{n}-5) \cdot 2 \mathrm{n}-\mathrm{n}^{2}(1)}{(\mathrm{n}-5)^{2}}\right\}\) \(\frac{d A}{d n}=2\left\{\frac{2 n^{2}-10 n-n^{2}}{(n-5)^{2}}\right\}=\frac{2 n^{2}-20 n}{(n-5)^{2}}\) \(\because\) Area least for some value of \(n\) \(\therefore \frac{\mathrm{dA}}{\mathrm{dn}}=0\) \(2 \mathrm{n}^{2}-20 \mathrm{n}=0\) \(2 \mathrm{n}(\mathrm{n}-10)=0\) \(\mathrm{n}=0,10\) \(\text { At } n=10, \quad m=\frac{4 n}{n-5}=\frac{40}{5}\) \(m=8\) \(\therefore m: n=8: 10=4: 5\)
Shift-I]
Application of Derivatives
85710
If the function \(f(x)=a \sin (x)+\frac{1}{3} \sin (3 x)\) attains maximum value at \(\mathrm{x}=\frac{\pi}{3}\), then ' \(\mathrm{a}\) '
1 3
2 \(\frac{1}{3}\)
3 2
4 \(\frac{1}{2}\)
Explanation:
(C): Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\) \(\frac{d}{d x}(f(x))=a \cos x+\cos 3 x=0 \quad\left\{\begin{array}{l}\text { Slope of tangent } \\ \text { at maxima is zero }\end{array}\right.\) Then, \(\mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=\mathrm{a} \cos \frac{\pi}{3}+\cos \pi=0\) \(\frac{a}{2}-1=0\) \(a=2\)
Shift-I
Application of Derivatives
85711
If \(x, y, z\) are three positive numbers, then the minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\) is
1 1
2 2
3 3
4 6
Explanation:
(D) : We have A.M. \(\geq\) G.M between two numbers \(\frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}} \Rightarrow \frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{b}}}=1\) Similarly, inequality between \(\frac{\mathrm{c}}{\mathrm{a}} \& \frac{\mathrm{a}}{\mathrm{c}}\) and \(\frac{\mathrm{b}}{\mathrm{c}} \& \frac{\mathrm{c}}{\mathrm{b}}\) \(\frac{1}{2}\left(\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{c}}\right) \geq \sqrt{\frac{\mathrm{c}}{\mathrm{a}} \cdot \frac{\mathrm{a}}{\mathrm{c}}}=1\) and \(\frac{1}{2}\left(\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{b}}\right) \geq \sqrt{\frac{\mathrm{b}}{\mathrm{c}} \cdot \frac{\mathrm{c}}{\mathrm{b}}}=1\) \(\frac{1}{2}\left(\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right) \geq 3\) \(\Rightarrow \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \geq 6\) Thus, the minimum value of the given expression is 6 .
