NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Limits, Continuity and Differentiability
79969
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) is
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : Given, \(f(x)=\frac{x}{4+x+x^{2}}\) Differentiating both side- \(f^{\prime}(x)=\frac{\left(4+x+x^{2}\right) \cdot 1-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}\) For maximum value of \(f(x)\) \(f^{\prime}(x)=0\) \(\frac{4+x+x^{2}-x-2 x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0 \Rightarrow x^{2}=4\) \(x= \pm 2 \notin(-1,1)\) Value of \(f(x)\) at extreme points \([-1,1]\) \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=-\frac{1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value of \(f(x)\).
[JCECE-2008]
Limits, Continuity and Differentiability
79970
\(f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{\sin x}, x \neq 0 \\ k, x=0\end{array}\right.\) is continuous, if \(k\) is
79972
If \(f(9)=9, f^{\prime}(9)=4\) and \(\lim _{x \rightarrow 9} \frac{f(x)-9}{x-9}=4\), then \(\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\) is equal to:
79969
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) is
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : Given, \(f(x)=\frac{x}{4+x+x^{2}}\) Differentiating both side- \(f^{\prime}(x)=\frac{\left(4+x+x^{2}\right) \cdot 1-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}\) For maximum value of \(f(x)\) \(f^{\prime}(x)=0\) \(\frac{4+x+x^{2}-x-2 x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0 \Rightarrow x^{2}=4\) \(x= \pm 2 \notin(-1,1)\) Value of \(f(x)\) at extreme points \([-1,1]\) \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=-\frac{1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value of \(f(x)\).
[JCECE-2008]
Limits, Continuity and Differentiability
79970
\(f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{\sin x}, x \neq 0 \\ k, x=0\end{array}\right.\) is continuous, if \(k\) is
79972
If \(f(9)=9, f^{\prime}(9)=4\) and \(\lim _{x \rightarrow 9} \frac{f(x)-9}{x-9}=4\), then \(\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\) is equal to:
79969
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) is
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : Given, \(f(x)=\frac{x}{4+x+x^{2}}\) Differentiating both side- \(f^{\prime}(x)=\frac{\left(4+x+x^{2}\right) \cdot 1-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}\) For maximum value of \(f(x)\) \(f^{\prime}(x)=0\) \(\frac{4+x+x^{2}-x-2 x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0 \Rightarrow x^{2}=4\) \(x= \pm 2 \notin(-1,1)\) Value of \(f(x)\) at extreme points \([-1,1]\) \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=-\frac{1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value of \(f(x)\).
[JCECE-2008]
Limits, Continuity and Differentiability
79970
\(f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{\sin x}, x \neq 0 \\ k, x=0\end{array}\right.\) is continuous, if \(k\) is
79972
If \(f(9)=9, f^{\prime}(9)=4\) and \(\lim _{x \rightarrow 9} \frac{f(x)-9}{x-9}=4\), then \(\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\) is equal to:
79969
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) is
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : Given, \(f(x)=\frac{x}{4+x+x^{2}}\) Differentiating both side- \(f^{\prime}(x)=\frac{\left(4+x+x^{2}\right) \cdot 1-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}\) For maximum value of \(f(x)\) \(f^{\prime}(x)=0\) \(\frac{4+x+x^{2}-x-2 x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0 \Rightarrow x^{2}=4\) \(x= \pm 2 \notin(-1,1)\) Value of \(f(x)\) at extreme points \([-1,1]\) \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=-\frac{1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value of \(f(x)\).
[JCECE-2008]
Limits, Continuity and Differentiability
79970
\(f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{\sin x}, x \neq 0 \\ k, x=0\end{array}\right.\) is continuous, if \(k\) is
79972
If \(f(9)=9, f^{\prime}(9)=4\) and \(\lim _{x \rightarrow 9} \frac{f(x)-9}{x-9}=4\), then \(\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}\) is equal to:
