(C) : Given, \(\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+\mathrm{x}\) First we check the continuity at \(x=0\), R.H.L, \(\begin{aligned} \mathrm{f}(0+\mathrm{h}) =\lim _{\mathrm{h} \rightarrow 0}|0+\mathrm{h}-2|+(0+\mathrm{h}) \\ =|-2|=2\end{aligned}\) At, \(\quad \mathrm{x}=2\), R.H.L, \(f(2+h)=\lim _{h \rightarrow 0}|2+h-2|+(2+h)\) \(=0+2+0=2\) L.H.L, \(f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)\) \(=0+2-0=2\) And, \(\quad \mathrm{f}(0)=2, \quad \mathrm{f}(2)=2\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and \(\mathrm{x}=0\) Now, we check differentiability \(f(x)=\left\{\begin{array}{cc}(-x+2-x) x\lt 0 \\ (-x+2+x) 0 \leq x \leq 2 \\ (x-2+2) 2 \leq x\end{array}\right.\) \(f(x)=\left\{\begin{array}{lr}2-2 x, x\lt 0 \\ 2, \quad 0 \leq x \leq 2 \\ 2 x-2, 2 \leq x\end{array}\right.\) Hence, \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=2,0\)
Karnataka CET-2010
Limits, Continuity and Differentiability
80106
The function \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to ' \(f\) ' at \(x=0\) so that it is continuous at \(\mathbf{x}=\mathbf{0}\) is
1 \(\log \mathrm{a}+\log \mathrm{b}\)
2 0
3 \(\mathrm{a}-\mathrm{b}\)
4 \(a+b\)
Explanation:
(D) : Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}\) \(=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\mathrm{a}}{1+\mathrm{ax}}+\frac{\mathrm{b}}{1-\mathrm{bx}}}{1}=\frac{\mathrm{a}}{1+\mathrm{a} \times 0}+\frac{\mathrm{b}}{1-\mathrm{b} \times 0}=\mathrm{a}+\mathrm{b}\)
Karnataka CET-2009
Limits, Continuity and Differentiability
80107
\(f(x)=2 a-x\) in \(-a\lt x\lt a\) \(=3 x-2 a \text { in } a \leq x\) Then which of the following is true?
1 \(f(x)\) is differentiable at all \(x \geq a\)
2 \(f(x)\) is continuous at all \(x\lt a\)
3 \(f(x)\) is discontinuous at \(x=a\)
4 \(f(x)\) is not differentiable at \(x=a\)
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}2 a-x \text { in } -a\lt x\lt a \\ 3 x-2 a \text { in } a \leq x\end{array}\right.\) At, \(\mathrm{x}=\mathrm{a}\) \(\text { L.H.L }=\lim _{x \rightarrow a^{-}} f(x)\) \(=\lim _{x \rightarrow a^{-}} 2 a-x=2 a-a=a\) \(\text { R.H.L }=\lim _{x \rightarrow a^{+}} f(x)\) \(\quad=\lim _{x \rightarrow a^{+}} 3 x-2 a=3 a-2 a=a\) \(\therefore\) L.H.L \(=\) R.H.L \(=\mathrm{f}\) (a) Hence it is continuous at \(\mathrm{x}=\mathrm{a}\) Again at \(\mathrm{x}=\mathrm{a}\) L.H.D \(=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{2 a-(a-h)-a}{-h}\) \(=\frac{\mathrm{h}}{-\mathrm{h}}=-1\) And, R.H.D \(=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\) \(=\lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3\) \(\therefore\) L.H.D \(\neq\) R.H.D Hence, it is not differentiable at \(\mathrm{x}=\mathrm{a}\)
Karnataka CET-2008
Limits, Continuity and Differentiability
80108
If the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 0
3 \(\frac{1}{2}\)
4 -1
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) Since, \(x\) is continuous \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \frac{1-\cos \mathrm{x}}{\mathrm{x}^{2}}=\mathrm{k}\) \(\lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \quad\) [Using L' Hospital's rule] \(\frac{1}{2} \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=\mathrm{k}\) \(\frac{1}{2} \times 1=\mathrm{k}\) So, \(\mathrm{k}=\frac{1}{2}\)
(C) : Given, \(\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+\mathrm{x}\) First we check the continuity at \(x=0\), R.H.L, \(\begin{aligned} \mathrm{f}(0+\mathrm{h}) =\lim _{\mathrm{h} \rightarrow 0}|0+\mathrm{h}-2|+(0+\mathrm{h}) \\ =|-2|=2\end{aligned}\) At, \(\quad \mathrm{x}=2\), R.H.L, \(f(2+h)=\lim _{h \rightarrow 0}|2+h-2|+(2+h)\) \(=0+2+0=2\) L.H.L, \(f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)\) \(=0+2-0=2\) And, \(\quad \mathrm{f}(0)=2, \quad \mathrm{f}(2)=2\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and \(\mathrm{x}=0\) Now, we check differentiability \(f(x)=\left\{\begin{array}{cc}(-x+2-x) x\lt 0 \\ (-x+2+x) 0 \leq x \leq 2 \\ (x-2+2) 2 \leq x\end{array}\right.\) \(f(x)=\left\{\begin{array}{lr}2-2 x, x\lt 0 \\ 2, \quad 0 \leq x \leq 2 \\ 2 x-2, 2 \leq x\end{array}\right.\) Hence, \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=2,0\)
Karnataka CET-2010
Limits, Continuity and Differentiability
80106
The function \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to ' \(f\) ' at \(x=0\) so that it is continuous at \(\mathbf{x}=\mathbf{0}\) is
1 \(\log \mathrm{a}+\log \mathrm{b}\)
2 0
3 \(\mathrm{a}-\mathrm{b}\)
4 \(a+b\)
Explanation:
(D) : Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}\) \(=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\mathrm{a}}{1+\mathrm{ax}}+\frac{\mathrm{b}}{1-\mathrm{bx}}}{1}=\frac{\mathrm{a}}{1+\mathrm{a} \times 0}+\frac{\mathrm{b}}{1-\mathrm{b} \times 0}=\mathrm{a}+\mathrm{b}\)
Karnataka CET-2009
Limits, Continuity and Differentiability
80107
\(f(x)=2 a-x\) in \(-a\lt x\lt a\) \(=3 x-2 a \text { in } a \leq x\) Then which of the following is true?
1 \(f(x)\) is differentiable at all \(x \geq a\)
2 \(f(x)\) is continuous at all \(x\lt a\)
3 \(f(x)\) is discontinuous at \(x=a\)
4 \(f(x)\) is not differentiable at \(x=a\)
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}2 a-x \text { in } -a\lt x\lt a \\ 3 x-2 a \text { in } a \leq x\end{array}\right.\) At, \(\mathrm{x}=\mathrm{a}\) \(\text { L.H.L }=\lim _{x \rightarrow a^{-}} f(x)\) \(=\lim _{x \rightarrow a^{-}} 2 a-x=2 a-a=a\) \(\text { R.H.L }=\lim _{x \rightarrow a^{+}} f(x)\) \(\quad=\lim _{x \rightarrow a^{+}} 3 x-2 a=3 a-2 a=a\) \(\therefore\) L.H.L \(=\) R.H.L \(=\mathrm{f}\) (a) Hence it is continuous at \(\mathrm{x}=\mathrm{a}\) Again at \(\mathrm{x}=\mathrm{a}\) L.H.D \(=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{2 a-(a-h)-a}{-h}\) \(=\frac{\mathrm{h}}{-\mathrm{h}}=-1\) And, R.H.D \(=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\) \(=\lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3\) \(\therefore\) L.H.D \(\neq\) R.H.D Hence, it is not differentiable at \(\mathrm{x}=\mathrm{a}\)
Karnataka CET-2008
Limits, Continuity and Differentiability
80108
If the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 0
3 \(\frac{1}{2}\)
4 -1
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) Since, \(x\) is continuous \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \frac{1-\cos \mathrm{x}}{\mathrm{x}^{2}}=\mathrm{k}\) \(\lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \quad\) [Using L' Hospital's rule] \(\frac{1}{2} \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=\mathrm{k}\) \(\frac{1}{2} \times 1=\mathrm{k}\) So, \(\mathrm{k}=\frac{1}{2}\)
(C) : Given, \(\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+\mathrm{x}\) First we check the continuity at \(x=0\), R.H.L, \(\begin{aligned} \mathrm{f}(0+\mathrm{h}) =\lim _{\mathrm{h} \rightarrow 0}|0+\mathrm{h}-2|+(0+\mathrm{h}) \\ =|-2|=2\end{aligned}\) At, \(\quad \mathrm{x}=2\), R.H.L, \(f(2+h)=\lim _{h \rightarrow 0}|2+h-2|+(2+h)\) \(=0+2+0=2\) L.H.L, \(f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)\) \(=0+2-0=2\) And, \(\quad \mathrm{f}(0)=2, \quad \mathrm{f}(2)=2\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and \(\mathrm{x}=0\) Now, we check differentiability \(f(x)=\left\{\begin{array}{cc}(-x+2-x) x\lt 0 \\ (-x+2+x) 0 \leq x \leq 2 \\ (x-2+2) 2 \leq x\end{array}\right.\) \(f(x)=\left\{\begin{array}{lr}2-2 x, x\lt 0 \\ 2, \quad 0 \leq x \leq 2 \\ 2 x-2, 2 \leq x\end{array}\right.\) Hence, \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=2,0\)
Karnataka CET-2010
Limits, Continuity and Differentiability
80106
The function \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to ' \(f\) ' at \(x=0\) so that it is continuous at \(\mathbf{x}=\mathbf{0}\) is
1 \(\log \mathrm{a}+\log \mathrm{b}\)
2 0
3 \(\mathrm{a}-\mathrm{b}\)
4 \(a+b\)
Explanation:
(D) : Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}\) \(=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\mathrm{a}}{1+\mathrm{ax}}+\frac{\mathrm{b}}{1-\mathrm{bx}}}{1}=\frac{\mathrm{a}}{1+\mathrm{a} \times 0}+\frac{\mathrm{b}}{1-\mathrm{b} \times 0}=\mathrm{a}+\mathrm{b}\)
Karnataka CET-2009
Limits, Continuity and Differentiability
80107
\(f(x)=2 a-x\) in \(-a\lt x\lt a\) \(=3 x-2 a \text { in } a \leq x\) Then which of the following is true?
1 \(f(x)\) is differentiable at all \(x \geq a\)
2 \(f(x)\) is continuous at all \(x\lt a\)
3 \(f(x)\) is discontinuous at \(x=a\)
4 \(f(x)\) is not differentiable at \(x=a\)
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}2 a-x \text { in } -a\lt x\lt a \\ 3 x-2 a \text { in } a \leq x\end{array}\right.\) At, \(\mathrm{x}=\mathrm{a}\) \(\text { L.H.L }=\lim _{x \rightarrow a^{-}} f(x)\) \(=\lim _{x \rightarrow a^{-}} 2 a-x=2 a-a=a\) \(\text { R.H.L }=\lim _{x \rightarrow a^{+}} f(x)\) \(\quad=\lim _{x \rightarrow a^{+}} 3 x-2 a=3 a-2 a=a\) \(\therefore\) L.H.L \(=\) R.H.L \(=\mathrm{f}\) (a) Hence it is continuous at \(\mathrm{x}=\mathrm{a}\) Again at \(\mathrm{x}=\mathrm{a}\) L.H.D \(=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{2 a-(a-h)-a}{-h}\) \(=\frac{\mathrm{h}}{-\mathrm{h}}=-1\) And, R.H.D \(=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\) \(=\lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3\) \(\therefore\) L.H.D \(\neq\) R.H.D Hence, it is not differentiable at \(\mathrm{x}=\mathrm{a}\)
Karnataka CET-2008
Limits, Continuity and Differentiability
80108
If the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 0
3 \(\frac{1}{2}\)
4 -1
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) Since, \(x\) is continuous \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \frac{1-\cos \mathrm{x}}{\mathrm{x}^{2}}=\mathrm{k}\) \(\lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \quad\) [Using L' Hospital's rule] \(\frac{1}{2} \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=\mathrm{k}\) \(\frac{1}{2} \times 1=\mathrm{k}\) So, \(\mathrm{k}=\frac{1}{2}\)
(C) : Given, \(\mathrm{f}(\mathrm{x})=|\mathrm{x}-2|+\mathrm{x}\) First we check the continuity at \(x=0\), R.H.L, \(\begin{aligned} \mathrm{f}(0+\mathrm{h}) =\lim _{\mathrm{h} \rightarrow 0}|0+\mathrm{h}-2|+(0+\mathrm{h}) \\ =|-2|=2\end{aligned}\) At, \(\quad \mathrm{x}=2\), R.H.L, \(f(2+h)=\lim _{h \rightarrow 0}|2+h-2|+(2+h)\) \(=0+2+0=2\) L.H.L, \(f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)\) \(=0+2-0=2\) And, \(\quad \mathrm{f}(0)=2, \quad \mathrm{f}(2)=2\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) and \(\mathrm{x}=0\) Now, we check differentiability \(f(x)=\left\{\begin{array}{cc}(-x+2-x) x\lt 0 \\ (-x+2+x) 0 \leq x \leq 2 \\ (x-2+2) 2 \leq x\end{array}\right.\) \(f(x)=\left\{\begin{array}{lr}2-2 x, x\lt 0 \\ 2, \quad 0 \leq x \leq 2 \\ 2 x-2, 2 \leq x\end{array}\right.\) Hence, \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=2,0\)
Karnataka CET-2010
Limits, Continuity and Differentiability
80106
The function \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to ' \(f\) ' at \(x=0\) so that it is continuous at \(\mathbf{x}=\mathbf{0}\) is
1 \(\log \mathrm{a}+\log \mathrm{b}\)
2 0
3 \(\mathrm{a}-\mathrm{b}\)
4 \(a+b\)
Explanation:
(D) : Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\) is not defined at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}\) \(=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\mathrm{a}}{1+\mathrm{ax}}+\frac{\mathrm{b}}{1-\mathrm{bx}}}{1}=\frac{\mathrm{a}}{1+\mathrm{a} \times 0}+\frac{\mathrm{b}}{1-\mathrm{b} \times 0}=\mathrm{a}+\mathrm{b}\)
Karnataka CET-2009
Limits, Continuity and Differentiability
80107
\(f(x)=2 a-x\) in \(-a\lt x\lt a\) \(=3 x-2 a \text { in } a \leq x\) Then which of the following is true?
1 \(f(x)\) is differentiable at all \(x \geq a\)
2 \(f(x)\) is continuous at all \(x\lt a\)
3 \(f(x)\) is discontinuous at \(x=a\)
4 \(f(x)\) is not differentiable at \(x=a\)
Explanation:
(D) : Given, \(f(x)=\left\{\begin{array}{ccc}2 a-x \text { in } -a\lt x\lt a \\ 3 x-2 a \text { in } a \leq x\end{array}\right.\) At, \(\mathrm{x}=\mathrm{a}\) \(\text { L.H.L }=\lim _{x \rightarrow a^{-}} f(x)\) \(=\lim _{x \rightarrow a^{-}} 2 a-x=2 a-a=a\) \(\text { R.H.L }=\lim _{x \rightarrow a^{+}} f(x)\) \(\quad=\lim _{x \rightarrow a^{+}} 3 x-2 a=3 a-2 a=a\) \(\therefore\) L.H.L \(=\) R.H.L \(=\mathrm{f}\) (a) Hence it is continuous at \(\mathrm{x}=\mathrm{a}\) Again at \(\mathrm{x}=\mathrm{a}\) L.H.D \(=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{2 a-(a-h)-a}{-h}\) \(=\frac{\mathrm{h}}{-\mathrm{h}}=-1\) And, R.H.D \(=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\) \(=\lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3\) \(\therefore\) L.H.D \(\neq\) R.H.D Hence, it is not differentiable at \(\mathrm{x}=\mathrm{a}\)
Karnataka CET-2008
Limits, Continuity and Differentiability
80108
If the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) is continuous at \(x=0\), then the value of \(k\) is
1 1
2 0
3 \(\frac{1}{2}\)
4 -1
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{1-\cos x}{x^{2}} \text { for } x \neq 0 \\ k \text { for } x=0\end{array}\right.\) \(f(x)\) is continuous at \(x=0\) Since, \(x\) is continuous \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \frac{1-\cos \mathrm{x}}{\mathrm{x}^{2}}=\mathrm{k}\) \(\lim _{x \rightarrow 0} \frac{-(-\sin x)}{2 x}=k \quad\) [Using L' Hospital's rule] \(\frac{1}{2} \lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=\mathrm{k}\) \(\frac{1}{2} \times 1=\mathrm{k}\) So, \(\mathrm{k}=\frac{1}{2}\)
