80096
If \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 -1
3 0
4 2
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}=\frac{2(0)+1}{0-1}=\frac{1}{-1}\) \(\lim _{x \rightarrow 0} \frac{(1+k x)-(1-k x)}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1\) \(\lim _{x \rightarrow 0} \frac{2 k x}{(\sqrt{1+k x}+\sqrt{1-k x}) x}=-1\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{k}}{(\sqrt{1+\mathrm{kx}}+\sqrt{1-\mathrm{kx}})}=-1\) So, \(\mathrm{k}=-1\)
Karnataka CET-2018
Limits, Continuity and Differentiability
80097
If \(f(x)=\left\{\begin{aligned} k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{aligned}\right.\) is continuous at \(x=2\), then the value of \(k\) is
1 \(\frac{4}{3}\)
2 \(\frac{3}{4}\)
3 3
4 4
Explanation:
(B): Given, \(f(x)=\left\{\begin{array}{rll}k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\therefore \operatorname{LHL}(\) at \(\mathrm{x}=-2)=\operatorname{RHL}(\) at \(\mathrm{x}=2)\) \(\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{kx}^{2}=\lim _{\mathrm{x} \rightarrow 2^{+}} 3\) \(\lim _{\mathrm{x} \rightarrow 0} \mathrm{k}(2-\mathrm{h})^{2}=3\) \(\mathrm{k}(2-0)^{2}=3\) \(4 \mathrm{k}=3\) So, \(\mathrm{k}=\frac{3}{4}\)
Karnataka CET-2017
Limits, Continuity and Differentiability
80098
The derivative of \(\cos ^{-1}\left(2 x^{2}-1\right)\) w.r.t. \(\cos ^{-1} x\) is
80099
\(f(x)=\left\{\begin{array}{cl}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) is continuous, find \(k\).
1 \(\frac{3}{7}\)
2 \(\frac{7}{2}\)
3 \(\frac{2}{7}\)
4 \(\frac{4}{7}\)
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) \(f(x)\) is continuous Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=5\) \(\therefore \mathrm{LHL}=\mathrm{RHL}\) \(\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{x \rightarrow 5^{+}} 2 k\) \(3 \times 5-8=2 \mathrm{k} \Rightarrow 15-8=2 \mathrm{k}\) \(7=2 \mathrm{k} \Rightarrow \mathrm{k}=\frac{7}{2}\)
Karnataka CET-2015
Limits, Continuity and Differentiability
80100
If \(f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, x \neq 0 \\ 2 k \text { is continuous at } x=0\end{array}\right.\) then the value of \(k\) is
80096
If \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 -1
3 0
4 2
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}=\frac{2(0)+1}{0-1}=\frac{1}{-1}\) \(\lim _{x \rightarrow 0} \frac{(1+k x)-(1-k x)}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1\) \(\lim _{x \rightarrow 0} \frac{2 k x}{(\sqrt{1+k x}+\sqrt{1-k x}) x}=-1\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{k}}{(\sqrt{1+\mathrm{kx}}+\sqrt{1-\mathrm{kx}})}=-1\) So, \(\mathrm{k}=-1\)
Karnataka CET-2018
Limits, Continuity and Differentiability
80097
If \(f(x)=\left\{\begin{aligned} k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{aligned}\right.\) is continuous at \(x=2\), then the value of \(k\) is
1 \(\frac{4}{3}\)
2 \(\frac{3}{4}\)
3 3
4 4
Explanation:
(B): Given, \(f(x)=\left\{\begin{array}{rll}k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\therefore \operatorname{LHL}(\) at \(\mathrm{x}=-2)=\operatorname{RHL}(\) at \(\mathrm{x}=2)\) \(\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{kx}^{2}=\lim _{\mathrm{x} \rightarrow 2^{+}} 3\) \(\lim _{\mathrm{x} \rightarrow 0} \mathrm{k}(2-\mathrm{h})^{2}=3\) \(\mathrm{k}(2-0)^{2}=3\) \(4 \mathrm{k}=3\) So, \(\mathrm{k}=\frac{3}{4}\)
Karnataka CET-2017
Limits, Continuity and Differentiability
80098
The derivative of \(\cos ^{-1}\left(2 x^{2}-1\right)\) w.r.t. \(\cos ^{-1} x\) is
80099
\(f(x)=\left\{\begin{array}{cl}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) is continuous, find \(k\).
1 \(\frac{3}{7}\)
2 \(\frac{7}{2}\)
3 \(\frac{2}{7}\)
4 \(\frac{4}{7}\)
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) \(f(x)\) is continuous Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=5\) \(\therefore \mathrm{LHL}=\mathrm{RHL}\) \(\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{x \rightarrow 5^{+}} 2 k\) \(3 \times 5-8=2 \mathrm{k} \Rightarrow 15-8=2 \mathrm{k}\) \(7=2 \mathrm{k} \Rightarrow \mathrm{k}=\frac{7}{2}\)
Karnataka CET-2015
Limits, Continuity and Differentiability
80100
If \(f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, x \neq 0 \\ 2 k \text { is continuous at } x=0\end{array}\right.\) then the value of \(k\) is
80096
If \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 -1
3 0
4 2
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}=\frac{2(0)+1}{0-1}=\frac{1}{-1}\) \(\lim _{x \rightarrow 0} \frac{(1+k x)-(1-k x)}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1\) \(\lim _{x \rightarrow 0} \frac{2 k x}{(\sqrt{1+k x}+\sqrt{1-k x}) x}=-1\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{k}}{(\sqrt{1+\mathrm{kx}}+\sqrt{1-\mathrm{kx}})}=-1\) So, \(\mathrm{k}=-1\)
Karnataka CET-2018
Limits, Continuity and Differentiability
80097
If \(f(x)=\left\{\begin{aligned} k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{aligned}\right.\) is continuous at \(x=2\), then the value of \(k\) is
1 \(\frac{4}{3}\)
2 \(\frac{3}{4}\)
3 3
4 4
Explanation:
(B): Given, \(f(x)=\left\{\begin{array}{rll}k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\therefore \operatorname{LHL}(\) at \(\mathrm{x}=-2)=\operatorname{RHL}(\) at \(\mathrm{x}=2)\) \(\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{kx}^{2}=\lim _{\mathrm{x} \rightarrow 2^{+}} 3\) \(\lim _{\mathrm{x} \rightarrow 0} \mathrm{k}(2-\mathrm{h})^{2}=3\) \(\mathrm{k}(2-0)^{2}=3\) \(4 \mathrm{k}=3\) So, \(\mathrm{k}=\frac{3}{4}\)
Karnataka CET-2017
Limits, Continuity and Differentiability
80098
The derivative of \(\cos ^{-1}\left(2 x^{2}-1\right)\) w.r.t. \(\cos ^{-1} x\) is
80099
\(f(x)=\left\{\begin{array}{cl}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) is continuous, find \(k\).
1 \(\frac{3}{7}\)
2 \(\frac{7}{2}\)
3 \(\frac{2}{7}\)
4 \(\frac{4}{7}\)
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) \(f(x)\) is continuous Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=5\) \(\therefore \mathrm{LHL}=\mathrm{RHL}\) \(\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{x \rightarrow 5^{+}} 2 k\) \(3 \times 5-8=2 \mathrm{k} \Rightarrow 15-8=2 \mathrm{k}\) \(7=2 \mathrm{k} \Rightarrow \mathrm{k}=\frac{7}{2}\)
Karnataka CET-2015
Limits, Continuity and Differentiability
80100
If \(f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, x \neq 0 \\ 2 k \text { is continuous at } x=0\end{array}\right.\) then the value of \(k\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Limits, Continuity and Differentiability
80096
If \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 -1
3 0
4 2
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}=\frac{2(0)+1}{0-1}=\frac{1}{-1}\) \(\lim _{x \rightarrow 0} \frac{(1+k x)-(1-k x)}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1\) \(\lim _{x \rightarrow 0} \frac{2 k x}{(\sqrt{1+k x}+\sqrt{1-k x}) x}=-1\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{k}}{(\sqrt{1+\mathrm{kx}}+\sqrt{1-\mathrm{kx}})}=-1\) So, \(\mathrm{k}=-1\)
Karnataka CET-2018
Limits, Continuity and Differentiability
80097
If \(f(x)=\left\{\begin{aligned} k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{aligned}\right.\) is continuous at \(x=2\), then the value of \(k\) is
1 \(\frac{4}{3}\)
2 \(\frac{3}{4}\)
3 3
4 4
Explanation:
(B): Given, \(f(x)=\left\{\begin{array}{rll}k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\therefore \operatorname{LHL}(\) at \(\mathrm{x}=-2)=\operatorname{RHL}(\) at \(\mathrm{x}=2)\) \(\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{kx}^{2}=\lim _{\mathrm{x} \rightarrow 2^{+}} 3\) \(\lim _{\mathrm{x} \rightarrow 0} \mathrm{k}(2-\mathrm{h})^{2}=3\) \(\mathrm{k}(2-0)^{2}=3\) \(4 \mathrm{k}=3\) So, \(\mathrm{k}=\frac{3}{4}\)
Karnataka CET-2017
Limits, Continuity and Differentiability
80098
The derivative of \(\cos ^{-1}\left(2 x^{2}-1\right)\) w.r.t. \(\cos ^{-1} x\) is
80099
\(f(x)=\left\{\begin{array}{cl}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) is continuous, find \(k\).
1 \(\frac{3}{7}\)
2 \(\frac{7}{2}\)
3 \(\frac{2}{7}\)
4 \(\frac{4}{7}\)
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) \(f(x)\) is continuous Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=5\) \(\therefore \mathrm{LHL}=\mathrm{RHL}\) \(\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{x \rightarrow 5^{+}} 2 k\) \(3 \times 5-8=2 \mathrm{k} \Rightarrow 15-8=2 \mathrm{k}\) \(7=2 \mathrm{k} \Rightarrow \mathrm{k}=\frac{7}{2}\)
Karnataka CET-2015
Limits, Continuity and Differentiability
80100
If \(f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, x \neq 0 \\ 2 k \text { is continuous at } x=0\end{array}\right.\) then the value of \(k\) is
80096
If \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 -1
3 0
4 2
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \text { if } -1 \leq x\lt 0 \\ \frac{2 x+1}{x-1} \text { if } 0 \leq x \leq 1\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}=\frac{2(0)+1}{0-1}=\frac{1}{-1}\) \(\lim _{x \rightarrow 0} \frac{(1+k x)-(1-k x)}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1\) \(\lim _{x \rightarrow 0} \frac{2 k x}{(\sqrt{1+k x}+\sqrt{1-k x}) x}=-1\) \(\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{k}}{(\sqrt{1+\mathrm{kx}}+\sqrt{1-\mathrm{kx}})}=-1\) So, \(\mathrm{k}=-1\)
Karnataka CET-2018
Limits, Continuity and Differentiability
80097
If \(f(x)=\left\{\begin{aligned} k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{aligned}\right.\) is continuous at \(x=2\), then the value of \(k\) is
1 \(\frac{4}{3}\)
2 \(\frac{3}{4}\)
3 3
4 4
Explanation:
(B): Given, \(f(x)=\left\{\begin{array}{rll}k x^{2} \text { if } x \leq 2 \\ 3 \text { if } x>2\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\therefore \operatorname{LHL}(\) at \(\mathrm{x}=-2)=\operatorname{RHL}(\) at \(\mathrm{x}=2)\) \(\lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{kx}^{2}=\lim _{\mathrm{x} \rightarrow 2^{+}} 3\) \(\lim _{\mathrm{x} \rightarrow 0} \mathrm{k}(2-\mathrm{h})^{2}=3\) \(\mathrm{k}(2-0)^{2}=3\) \(4 \mathrm{k}=3\) So, \(\mathrm{k}=\frac{3}{4}\)
Karnataka CET-2017
Limits, Continuity and Differentiability
80098
The derivative of \(\cos ^{-1}\left(2 x^{2}-1\right)\) w.r.t. \(\cos ^{-1} x\) is
80099
\(f(x)=\left\{\begin{array}{cl}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) is continuous, find \(k\).
1 \(\frac{3}{7}\)
2 \(\frac{7}{2}\)
3 \(\frac{2}{7}\)
4 \(\frac{4}{7}\)
Explanation:
(B) : Given, \(f(x)=\left\{\begin{array}{cll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5\end{array}\right.\) \(f(x)\) is continuous Since, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=5\) \(\therefore \mathrm{LHL}=\mathrm{RHL}\) \(\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{x \rightarrow 5^{+}} 2 k\) \(3 \times 5-8=2 \mathrm{k} \Rightarrow 15-8=2 \mathrm{k}\) \(7=2 \mathrm{k} \Rightarrow \mathrm{k}=\frac{7}{2}\)
Karnataka CET-2015
Limits, Continuity and Differentiability
80100
If \(f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, x \neq 0 \\ 2 k \text { is continuous at } x=0\end{array}\right.\) then the value of \(k\) is