80101
Let \(f(x)=\left\{\begin{array}{ll}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{array}\right.\) then \(f\) is
1 continuous everywhere
2 discontinuous everywhere
3 continuous only at \(x=0\)
4 continuous at all rational numbers
Explanation:
(C) : Given, \(f(x)= \begin{cases}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{cases}\) LHL \(=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0\) RHL \(=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\) And, \(f(0)=0\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\)
Karnataka CET-2013
Limits, Continuity and Differentiability
80102
If\(f(x)= \begin{cases}\frac{x^2-(a+2) x+a}{x-2} & , x \neq 2 \\ 2 & , x=2\end{cases}\) continuous at \(x=2\), then the value of \(a\) is
80104
If \(f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, \text { if } x \neq 1 \\ k, \text { if } x=1\end{array}\right.\) is continuous at \(x=\) 1 , then the value of \(k\) is
1 0
2 -1
3 1
4 e
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\log x}{x-1} \text { if } x \neq 1 \\ k \text { if } x=1\end{array}\right.\) is continuous at \(\mathrm{x}=1\) Since, the function is continuous at \(x=1\) Then \(f(1)=\lim _{x \rightarrow 1} f(x)\) Apply L-Hospital rule, \(k=\lim _{x \rightarrow 1} \frac{\log x}{x-1}\) \(k=\lim _{x \rightarrow 1} \frac{1 / x}{1}\) \(k=\frac{1 / 1}{1} \Rightarrow k=1\) \(\left[\frac{0}{0} \text { from }\right]\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
80101
Let \(f(x)=\left\{\begin{array}{ll}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{array}\right.\) then \(f\) is
1 continuous everywhere
2 discontinuous everywhere
3 continuous only at \(x=0\)
4 continuous at all rational numbers
Explanation:
(C) : Given, \(f(x)= \begin{cases}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{cases}\) LHL \(=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0\) RHL \(=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\) And, \(f(0)=0\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\)
Karnataka CET-2013
Limits, Continuity and Differentiability
80102
If\(f(x)= \begin{cases}\frac{x^2-(a+2) x+a}{x-2} & , x \neq 2 \\ 2 & , x=2\end{cases}\) continuous at \(x=2\), then the value of \(a\) is
80104
If \(f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, \text { if } x \neq 1 \\ k, \text { if } x=1\end{array}\right.\) is continuous at \(x=\) 1 , then the value of \(k\) is
1 0
2 -1
3 1
4 e
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\log x}{x-1} \text { if } x \neq 1 \\ k \text { if } x=1\end{array}\right.\) is continuous at \(\mathrm{x}=1\) Since, the function is continuous at \(x=1\) Then \(f(1)=\lim _{x \rightarrow 1} f(x)\) Apply L-Hospital rule, \(k=\lim _{x \rightarrow 1} \frac{\log x}{x-1}\) \(k=\lim _{x \rightarrow 1} \frac{1 / x}{1}\) \(k=\frac{1 / 1}{1} \Rightarrow k=1\) \(\left[\frac{0}{0} \text { from }\right]\)
80101
Let \(f(x)=\left\{\begin{array}{ll}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{array}\right.\) then \(f\) is
1 continuous everywhere
2 discontinuous everywhere
3 continuous only at \(x=0\)
4 continuous at all rational numbers
Explanation:
(C) : Given, \(f(x)= \begin{cases}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{cases}\) LHL \(=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0\) RHL \(=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\) And, \(f(0)=0\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\)
Karnataka CET-2013
Limits, Continuity and Differentiability
80102
If\(f(x)= \begin{cases}\frac{x^2-(a+2) x+a}{x-2} & , x \neq 2 \\ 2 & , x=2\end{cases}\) continuous at \(x=2\), then the value of \(a\) is
80104
If \(f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, \text { if } x \neq 1 \\ k, \text { if } x=1\end{array}\right.\) is continuous at \(x=\) 1 , then the value of \(k\) is
1 0
2 -1
3 1
4 e
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\log x}{x-1} \text { if } x \neq 1 \\ k \text { if } x=1\end{array}\right.\) is continuous at \(\mathrm{x}=1\) Since, the function is continuous at \(x=1\) Then \(f(1)=\lim _{x \rightarrow 1} f(x)\) Apply L-Hospital rule, \(k=\lim _{x \rightarrow 1} \frac{\log x}{x-1}\) \(k=\lim _{x \rightarrow 1} \frac{1 / x}{1}\) \(k=\frac{1 / 1}{1} \Rightarrow k=1\) \(\left[\frac{0}{0} \text { from }\right]\)
80101
Let \(f(x)=\left\{\begin{array}{ll}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{array}\right.\) then \(f\) is
1 continuous everywhere
2 discontinuous everywhere
3 continuous only at \(x=0\)
4 continuous at all rational numbers
Explanation:
(C) : Given, \(f(x)= \begin{cases}x, \text { if } x \text { is irrational } \\ 0, \text { if } x \text { is rational }\end{cases}\) LHL \(=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0\) RHL \(=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\) And, \(f(0)=0\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\)
Karnataka CET-2013
Limits, Continuity and Differentiability
80102
If\(f(x)= \begin{cases}\frac{x^2-(a+2) x+a}{x-2} & , x \neq 2 \\ 2 & , x=2\end{cases}\) continuous at \(x=2\), then the value of \(a\) is
80104
If \(f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, \text { if } x \neq 1 \\ k, \text { if } x=1\end{array}\right.\) is continuous at \(x=\) 1 , then the value of \(k\) is
1 0
2 -1
3 1
4 e
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{ccc}\frac{\log x}{x-1} \text { if } x \neq 1 \\ k \text { if } x=1\end{array}\right.\) is continuous at \(\mathrm{x}=1\) Since, the function is continuous at \(x=1\) Then \(f(1)=\lim _{x \rightarrow 1} f(x)\) Apply L-Hospital rule, \(k=\lim _{x \rightarrow 1} \frac{\log x}{x-1}\) \(k=\lim _{x \rightarrow 1} \frac{1 / x}{1}\) \(k=\frac{1 / 1}{1} \Rightarrow k=1\) \(\left[\frac{0}{0} \text { from }\right]\)