85609
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}, x \in R\), then the minimum value of \(f \mathrm{i}\)
1 0
2 \(4 / 5\)
3 \(3 / 5\)
4 -1
Explanation:
(D) : We have, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) Maximum or minimum value point is given by \(f^{\prime}(x)=0\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=0\) \(\Rightarrow \frac{2 x\left(x^{2}+1-x^{2}+1\right)}{\left(x^{2}+1\right)^{2}}=0 \Rightarrow 4 x=0 \Rightarrow x=0\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)^{2}}\right)\) \(f^{\prime \prime}(x)=\frac{\left(x^{2}+1\right)^{2} 4-4 x \times 2\left(x^{2}+1\right) \times 2 x}{\left(x^{2}+1\right)^{4}}\) \(=\frac{4\left(\mathrm{x}^{2}+1\right)^{2}-16 \mathrm{x}^{2}\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}\) \(\frac{4\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+1-4 \mathrm{x}^{2}\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}=\frac{4\left(-3 \mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{3}}\) Now \(\mathrm{f}^{\prime \prime}(0)=\frac{4}{1}>0\) \(\therefore \mathrm{x}=0\) is a minimum point and minimum value of \(\mathrm{f}\) is given by \(\mathrm{f}(0)=\frac{0-1}{0+1}=-1\)
BCECE-2006
Application of Derivatives
85610
The local minimum value of \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) is
1 -1
2 -2
3 0
4 1
Explanation:
(B) : We have, \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=5\left(\mathrm{x}^{\frac{2}{3}}-1\right)\) and \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{10}{3 \mathrm{x}^{\frac{1}{3}}}\) For extreme values, put \(f^{\prime}(x)=03 x^{\frac{1}{3}}\) \(\Rightarrow \mathrm{x}^{\frac{2}{3}}=1 \Rightarrow \mathrm{x}=1\) Now, \(\mathrm{f}^{\prime \prime}(1)=\frac{10}{3}>0 \Rightarrow \mathrm{x}=1\) is a point of local minima, and minimum value \(=f(1)=3-5=-2\).
COMEDK-2018
Application of Derivatives
85611
Let \(f: R \rightarrow R\) be defined by \(f(x)=\left\{\begin{array}{l}k-2 x, x \leq-1 \\ 2 x+3, x>-1\end{array}\right.\). If \(f(x)\) has a local minimum at \(x=-1\), then a possible value of \(k i\)
1 0
2 \(-1 / 2\)
3 -1
4 1
Explanation:
(C) : We have, \(f(x)= \begin{cases}k-2 x , \quad x \leq-1 \\ 2 x+3 , \quad x>-1\end{cases}\) \(\mathrm{f}(\mathrm{x})\) will be continuous at \(\mathrm{x}=-1\) If, \(\quad \lim f(x)=\lim f(x)\) \(x \rightarrow-1 \quad x \rightarrow-1^{+}\) \(k-2(-1)=2(-1)+3\) \(k+2=-2+3\) \(k+2=1\) \(\mathrm{k}=-1\) For value of \(k, f\) is continuous at \(x=-1\) And \(\mathrm{f}^{\prime}(-1)\) does not exist \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\), at \(\mathrm{x}>-1\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=-1\) Then, possible value of \(\mathrm{k}\) is -1
SRM JEE-2019
Application of Derivatives
85612
Let \(f(x)\) be a function defined as \(f(x)= \begin{cases}\sin \left(x^{2}-3 x\right), x \leq 0 \\ 6 x+5 x^{2}, x>0\end{cases}\) Then, at \(x=0, f(x)\)
1 has a local maximum
2 has a local minimum
3 is discontinuous
4 none of these
Explanation:
(B) : \(f(x)=\left\{\begin{array}{cc}\sin \left(x^{2}-3 x\right) , \mathrm{x} \leq 0 \\ 6 x+5 x^{2} , \mathrm{x}>0\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}(2 x-3) \cos \left(x^2-3 x\right) & , x \leq 0 \\ 6+10 x & , x>0\end{array}\right.\) Since, \(\mathrm{f}^{\prime}(\mathrm{x})\) changes sign from \(-v \mathrm{e}\) to \(+v \mathrm{e}\) at \(\mathrm{x}=0\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=0\).
85609
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}, x \in R\), then the minimum value of \(f \mathrm{i}\)
1 0
2 \(4 / 5\)
3 \(3 / 5\)
4 -1
Explanation:
(D) : We have, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) Maximum or minimum value point is given by \(f^{\prime}(x)=0\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=0\) \(\Rightarrow \frac{2 x\left(x^{2}+1-x^{2}+1\right)}{\left(x^{2}+1\right)^{2}}=0 \Rightarrow 4 x=0 \Rightarrow x=0\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)^{2}}\right)\) \(f^{\prime \prime}(x)=\frac{\left(x^{2}+1\right)^{2} 4-4 x \times 2\left(x^{2}+1\right) \times 2 x}{\left(x^{2}+1\right)^{4}}\) \(=\frac{4\left(\mathrm{x}^{2}+1\right)^{2}-16 \mathrm{x}^{2}\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}\) \(\frac{4\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+1-4 \mathrm{x}^{2}\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}=\frac{4\left(-3 \mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{3}}\) Now \(\mathrm{f}^{\prime \prime}(0)=\frac{4}{1}>0\) \(\therefore \mathrm{x}=0\) is a minimum point and minimum value of \(\mathrm{f}\) is given by \(\mathrm{f}(0)=\frac{0-1}{0+1}=-1\)
BCECE-2006
Application of Derivatives
85610
The local minimum value of \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) is
1 -1
2 -2
3 0
4 1
Explanation:
(B) : We have, \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=5\left(\mathrm{x}^{\frac{2}{3}}-1\right)\) and \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{10}{3 \mathrm{x}^{\frac{1}{3}}}\) For extreme values, put \(f^{\prime}(x)=03 x^{\frac{1}{3}}\) \(\Rightarrow \mathrm{x}^{\frac{2}{3}}=1 \Rightarrow \mathrm{x}=1\) Now, \(\mathrm{f}^{\prime \prime}(1)=\frac{10}{3}>0 \Rightarrow \mathrm{x}=1\) is a point of local minima, and minimum value \(=f(1)=3-5=-2\).
COMEDK-2018
Application of Derivatives
85611
Let \(f: R \rightarrow R\) be defined by \(f(x)=\left\{\begin{array}{l}k-2 x, x \leq-1 \\ 2 x+3, x>-1\end{array}\right.\). If \(f(x)\) has a local minimum at \(x=-1\), then a possible value of \(k i\)
1 0
2 \(-1 / 2\)
3 -1
4 1
Explanation:
(C) : We have, \(f(x)= \begin{cases}k-2 x , \quad x \leq-1 \\ 2 x+3 , \quad x>-1\end{cases}\) \(\mathrm{f}(\mathrm{x})\) will be continuous at \(\mathrm{x}=-1\) If, \(\quad \lim f(x)=\lim f(x)\) \(x \rightarrow-1 \quad x \rightarrow-1^{+}\) \(k-2(-1)=2(-1)+3\) \(k+2=-2+3\) \(k+2=1\) \(\mathrm{k}=-1\) For value of \(k, f\) is continuous at \(x=-1\) And \(\mathrm{f}^{\prime}(-1)\) does not exist \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\), at \(\mathrm{x}>-1\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=-1\) Then, possible value of \(\mathrm{k}\) is -1
SRM JEE-2019
Application of Derivatives
85612
Let \(f(x)\) be a function defined as \(f(x)= \begin{cases}\sin \left(x^{2}-3 x\right), x \leq 0 \\ 6 x+5 x^{2}, x>0\end{cases}\) Then, at \(x=0, f(x)\)
1 has a local maximum
2 has a local minimum
3 is discontinuous
4 none of these
Explanation:
(B) : \(f(x)=\left\{\begin{array}{cc}\sin \left(x^{2}-3 x\right) , \mathrm{x} \leq 0 \\ 6 x+5 x^{2} , \mathrm{x}>0\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}(2 x-3) \cos \left(x^2-3 x\right) & , x \leq 0 \\ 6+10 x & , x>0\end{array}\right.\) Since, \(\mathrm{f}^{\prime}(\mathrm{x})\) changes sign from \(-v \mathrm{e}\) to \(+v \mathrm{e}\) at \(\mathrm{x}=0\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=0\).
85609
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}, x \in R\), then the minimum value of \(f \mathrm{i}\)
1 0
2 \(4 / 5\)
3 \(3 / 5\)
4 -1
Explanation:
(D) : We have, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) Maximum or minimum value point is given by \(f^{\prime}(x)=0\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=0\) \(\Rightarrow \frac{2 x\left(x^{2}+1-x^{2}+1\right)}{\left(x^{2}+1\right)^{2}}=0 \Rightarrow 4 x=0 \Rightarrow x=0\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)^{2}}\right)\) \(f^{\prime \prime}(x)=\frac{\left(x^{2}+1\right)^{2} 4-4 x \times 2\left(x^{2}+1\right) \times 2 x}{\left(x^{2}+1\right)^{4}}\) \(=\frac{4\left(\mathrm{x}^{2}+1\right)^{2}-16 \mathrm{x}^{2}\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}\) \(\frac{4\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+1-4 \mathrm{x}^{2}\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}=\frac{4\left(-3 \mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{3}}\) Now \(\mathrm{f}^{\prime \prime}(0)=\frac{4}{1}>0\) \(\therefore \mathrm{x}=0\) is a minimum point and minimum value of \(\mathrm{f}\) is given by \(\mathrm{f}(0)=\frac{0-1}{0+1}=-1\)
BCECE-2006
Application of Derivatives
85610
The local minimum value of \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) is
1 -1
2 -2
3 0
4 1
Explanation:
(B) : We have, \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=5\left(\mathrm{x}^{\frac{2}{3}}-1\right)\) and \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{10}{3 \mathrm{x}^{\frac{1}{3}}}\) For extreme values, put \(f^{\prime}(x)=03 x^{\frac{1}{3}}\) \(\Rightarrow \mathrm{x}^{\frac{2}{3}}=1 \Rightarrow \mathrm{x}=1\) Now, \(\mathrm{f}^{\prime \prime}(1)=\frac{10}{3}>0 \Rightarrow \mathrm{x}=1\) is a point of local minima, and minimum value \(=f(1)=3-5=-2\).
COMEDK-2018
Application of Derivatives
85611
Let \(f: R \rightarrow R\) be defined by \(f(x)=\left\{\begin{array}{l}k-2 x, x \leq-1 \\ 2 x+3, x>-1\end{array}\right.\). If \(f(x)\) has a local minimum at \(x=-1\), then a possible value of \(k i\)
1 0
2 \(-1 / 2\)
3 -1
4 1
Explanation:
(C) : We have, \(f(x)= \begin{cases}k-2 x , \quad x \leq-1 \\ 2 x+3 , \quad x>-1\end{cases}\) \(\mathrm{f}(\mathrm{x})\) will be continuous at \(\mathrm{x}=-1\) If, \(\quad \lim f(x)=\lim f(x)\) \(x \rightarrow-1 \quad x \rightarrow-1^{+}\) \(k-2(-1)=2(-1)+3\) \(k+2=-2+3\) \(k+2=1\) \(\mathrm{k}=-1\) For value of \(k, f\) is continuous at \(x=-1\) And \(\mathrm{f}^{\prime}(-1)\) does not exist \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\), at \(\mathrm{x}>-1\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=-1\) Then, possible value of \(\mathrm{k}\) is -1
SRM JEE-2019
Application of Derivatives
85612
Let \(f(x)\) be a function defined as \(f(x)= \begin{cases}\sin \left(x^{2}-3 x\right), x \leq 0 \\ 6 x+5 x^{2}, x>0\end{cases}\) Then, at \(x=0, f(x)\)
1 has a local maximum
2 has a local minimum
3 is discontinuous
4 none of these
Explanation:
(B) : \(f(x)=\left\{\begin{array}{cc}\sin \left(x^{2}-3 x\right) , \mathrm{x} \leq 0 \\ 6 x+5 x^{2} , \mathrm{x}>0\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}(2 x-3) \cos \left(x^2-3 x\right) & , x \leq 0 \\ 6+10 x & , x>0\end{array}\right.\) Since, \(\mathrm{f}^{\prime}(\mathrm{x})\) changes sign from \(-v \mathrm{e}\) to \(+v \mathrm{e}\) at \(\mathrm{x}=0\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=0\).
85609
If \(f(x)=\frac{x^{2}-1}{x^{2}+1}, x \in R\), then the minimum value of \(f \mathrm{i}\)
1 0
2 \(4 / 5\)
3 \(3 / 5\)
4 -1
Explanation:
(D) : We have, \(f(x)=\frac{x^{2}-1}{x^{2}+1}\) Maximum or minimum value point is given by \(f^{\prime}(x)=0\) \(f^{\prime}(x)=\frac{\left(x^{2}+1\right) 2 x-\left(x^{2}-1\right) 2 x}{\left(x^{2}+1\right)^{2}}=0\) \(\Rightarrow \frac{2 x\left(x^{2}+1-x^{2}+1\right)}{\left(x^{2}+1\right)^{2}}=0 \Rightarrow 4 x=0 \Rightarrow x=0\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)^{2}}\right)\) \(f^{\prime \prime}(x)=\frac{\left(x^{2}+1\right)^{2} 4-4 x \times 2\left(x^{2}+1\right) \times 2 x}{\left(x^{2}+1\right)^{4}}\) \(=\frac{4\left(\mathrm{x}^{2}+1\right)^{2}-16 \mathrm{x}^{2}\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}\) \(\frac{4\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+1-4 \mathrm{x}^{2}\right)}{\left(\mathrm{x}^{2}+1\right)^{4}}=\frac{4\left(-3 \mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)^{3}}\) Now \(\mathrm{f}^{\prime \prime}(0)=\frac{4}{1}>0\) \(\therefore \mathrm{x}=0\) is a minimum point and minimum value of \(\mathrm{f}\) is given by \(\mathrm{f}(0)=\frac{0-1}{0+1}=-1\)
BCECE-2006
Application of Derivatives
85610
The local minimum value of \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) is
1 -1
2 -2
3 0
4 1
Explanation:
(B) : We have, \(f(x)=3 x^{\frac{5}{3}}-5 x, x>0\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=5\left(\mathrm{x}^{\frac{2}{3}}-1\right)\) and \(\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{10}{3 \mathrm{x}^{\frac{1}{3}}}\) For extreme values, put \(f^{\prime}(x)=03 x^{\frac{1}{3}}\) \(\Rightarrow \mathrm{x}^{\frac{2}{3}}=1 \Rightarrow \mathrm{x}=1\) Now, \(\mathrm{f}^{\prime \prime}(1)=\frac{10}{3}>0 \Rightarrow \mathrm{x}=1\) is a point of local minima, and minimum value \(=f(1)=3-5=-2\).
COMEDK-2018
Application of Derivatives
85611
Let \(f: R \rightarrow R\) be defined by \(f(x)=\left\{\begin{array}{l}k-2 x, x \leq-1 \\ 2 x+3, x>-1\end{array}\right.\). If \(f(x)\) has a local minimum at \(x=-1\), then a possible value of \(k i\)
1 0
2 \(-1 / 2\)
3 -1
4 1
Explanation:
(C) : We have, \(f(x)= \begin{cases}k-2 x , \quad x \leq-1 \\ 2 x+3 , \quad x>-1\end{cases}\) \(\mathrm{f}(\mathrm{x})\) will be continuous at \(\mathrm{x}=-1\) If, \(\quad \lim f(x)=\lim f(x)\) \(x \rightarrow-1 \quad x \rightarrow-1^{+}\) \(k-2(-1)=2(-1)+3\) \(k+2=-2+3\) \(k+2=1\) \(\mathrm{k}=-1\) For value of \(k, f\) is continuous at \(x=-1\) And \(\mathrm{f}^{\prime}(-1)\) does not exist \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\), at \(\mathrm{x}>-1\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=-1\) Then, possible value of \(\mathrm{k}\) is -1
SRM JEE-2019
Application of Derivatives
85612
Let \(f(x)\) be a function defined as \(f(x)= \begin{cases}\sin \left(x^{2}-3 x\right), x \leq 0 \\ 6 x+5 x^{2}, x>0\end{cases}\) Then, at \(x=0, f(x)\)
1 has a local maximum
2 has a local minimum
3 is discontinuous
4 none of these
Explanation:
(B) : \(f(x)=\left\{\begin{array}{cc}\sin \left(x^{2}-3 x\right) , \mathrm{x} \leq 0 \\ 6 x+5 x^{2} , \mathrm{x}>0\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}(2 x-3) \cos \left(x^2-3 x\right) & , x \leq 0 \\ 6+10 x & , x>0\end{array}\right.\) Since, \(\mathrm{f}^{\prime}(\mathrm{x})\) changes sign from \(-v \mathrm{e}\) to \(+v \mathrm{e}\) at \(\mathrm{x}=0\) Hence, \(\mathrm{f}(\mathrm{x})\) has a local minimum at \(\mathrm{x}=0\).
