79978
Let \(f(x)\) be a continuous function defined for 1 \(\leq \mathrm{x} \leq 3\). If \(\mathrm{f}(\mathrm{x})\) takes rational values for all \(\mathrm{x}\) and \(f(2)=10\), then what is \(f(1.5)\) equal to?
1 0
2 1
3 10
4 cannot be determined as the data is insufficient
Explanation:
(C) : Given, (i) \(\mathrm{f}(\mathrm{x})\) is continuous between \(1 \leq \mathrm{x} \leq 3\) (ii) \(\mathrm{f}(2)=10\) Hence to satisfies Rolle's Theorem ... a definite point Between the range \((1,2)\) Such that, \(\mathrm{f}(\mathrm{x})=0\) Hence \(f(1.5)=f(2)=10\)
SCRA-2012
Limits, Continuity and Differentiability
79979
\(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) is
1 continuous but not differentiable at \(x=1\)
2 differentiable at \(x=1\)
3 neither continuous nor differentiable at \(\mathrm{x}=1\)
4 continuous everywhere
Explanation:
(A) : Given, Function, \(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) At, \(\mathrm{x}=1\), LHL \(\lim _{h \rightarrow 0} \sin ^{-1}\left[\frac{1+(1-h)^{2}}{2(1-h)}\right]\) \(\sin ^{-1}\left(\frac{1+1}{2}\right) \Rightarrow \frac{\pi}{2}\) And, RHL \(\lim _{h \rightarrow 0^{+}} \sin ^{-1}\left[\frac{1+(1+h)^{2}}{2(1+h)}\right]\) \(\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) And, \(f(1)=\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) Thus, given function is continuous at \(\mathrm{x}=1\), Now, \(\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)\right\}\) \(=\frac{1}{\sqrt{1-\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)^{2}}} \cdot\left[\frac{2 \mathrm{x}(2 \mathrm{x})-\left(1+\mathrm{x}^{2}\right) \cdot 2}{(2 \mathrm{x})^{2}}\right]\) \(=\frac{2 x}{\sqrt{4 x^{2}-\left(1+x^{4}+2 x^{2}\right)}} \cdot \frac{2 x^{2}-1}{(2 x)^{2}}\) \(=\frac{2 x^{2}-1}{\sqrt{-\left(1+x^{4}-2 x^{2}\right)} \cdot 2 x}=\frac{2 x^{2}-1}{2 x \sqrt{-\left(1-x^{2}\right)^{2}}}\) Which does not exist at \(x=1\). Hence, given function is not differentiable at \(x=1\)
CG PET-2011
Limits, Continuity and Differentiability
79980
If \(f(x)=\left\{\begin{array}{cc}{[x]+[-x],} x \neq 2 \\ \lambda, x=2\end{array}\right.\), then \(f \quad\) is continuous at \(x=2\), provided \(\lambda\) is equal to
79981
The point/points of discontinuity of the function \(f(x)=\left\{\begin{array}{l}|x|+3, \text { if } \quad x \leq-3 \\ -2 x, \quad \text { if } \quad-3\lt x\lt 3 \\ 6 x+2, \text { if } \quad x \geq 3\end{array}\right.\) is/are
1 \(3,-3\)
2 3
3 -3
4 None of these
Explanation:
(B) : Given, \(f(x)= \begin{cases}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3\lt x\lt 3 \\ 6 x+2, \text { if } x \geq 3\end{cases}\) For continuity at \(x=-3\) \(\lim _{x \rightarrow-3^{+}} f(x)={ }_{x \rightarrow-3^{+}}-2 x=(-2)(-3)=6\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=-3\). For continuity at \(\mathrm{x}=3\) \(\therefore \lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}-2 x=(-2)(-3)=6\) \(\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}^{\lim }+6 x+2=[6(3)+2]=20\) \(\therefore \lim _{x \rightarrow-3^{-}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow-3^{+}} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is discontinuous of \(\mathrm{x}=3\).
79978
Let \(f(x)\) be a continuous function defined for 1 \(\leq \mathrm{x} \leq 3\). If \(\mathrm{f}(\mathrm{x})\) takes rational values for all \(\mathrm{x}\) and \(f(2)=10\), then what is \(f(1.5)\) equal to?
1 0
2 1
3 10
4 cannot be determined as the data is insufficient
Explanation:
(C) : Given, (i) \(\mathrm{f}(\mathrm{x})\) is continuous between \(1 \leq \mathrm{x} \leq 3\) (ii) \(\mathrm{f}(2)=10\) Hence to satisfies Rolle's Theorem ... a definite point Between the range \((1,2)\) Such that, \(\mathrm{f}(\mathrm{x})=0\) Hence \(f(1.5)=f(2)=10\)
SCRA-2012
Limits, Continuity and Differentiability
79979
\(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) is
1 continuous but not differentiable at \(x=1\)
2 differentiable at \(x=1\)
3 neither continuous nor differentiable at \(\mathrm{x}=1\)
4 continuous everywhere
Explanation:
(A) : Given, Function, \(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) At, \(\mathrm{x}=1\), LHL \(\lim _{h \rightarrow 0} \sin ^{-1}\left[\frac{1+(1-h)^{2}}{2(1-h)}\right]\) \(\sin ^{-1}\left(\frac{1+1}{2}\right) \Rightarrow \frac{\pi}{2}\) And, RHL \(\lim _{h \rightarrow 0^{+}} \sin ^{-1}\left[\frac{1+(1+h)^{2}}{2(1+h)}\right]\) \(\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) And, \(f(1)=\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) Thus, given function is continuous at \(\mathrm{x}=1\), Now, \(\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)\right\}\) \(=\frac{1}{\sqrt{1-\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)^{2}}} \cdot\left[\frac{2 \mathrm{x}(2 \mathrm{x})-\left(1+\mathrm{x}^{2}\right) \cdot 2}{(2 \mathrm{x})^{2}}\right]\) \(=\frac{2 x}{\sqrt{4 x^{2}-\left(1+x^{4}+2 x^{2}\right)}} \cdot \frac{2 x^{2}-1}{(2 x)^{2}}\) \(=\frac{2 x^{2}-1}{\sqrt{-\left(1+x^{4}-2 x^{2}\right)} \cdot 2 x}=\frac{2 x^{2}-1}{2 x \sqrt{-\left(1-x^{2}\right)^{2}}}\) Which does not exist at \(x=1\). Hence, given function is not differentiable at \(x=1\)
CG PET-2011
Limits, Continuity and Differentiability
79980
If \(f(x)=\left\{\begin{array}{cc}{[x]+[-x],} x \neq 2 \\ \lambda, x=2\end{array}\right.\), then \(f \quad\) is continuous at \(x=2\), provided \(\lambda\) is equal to
79981
The point/points of discontinuity of the function \(f(x)=\left\{\begin{array}{l}|x|+3, \text { if } \quad x \leq-3 \\ -2 x, \quad \text { if } \quad-3\lt x\lt 3 \\ 6 x+2, \text { if } \quad x \geq 3\end{array}\right.\) is/are
1 \(3,-3\)
2 3
3 -3
4 None of these
Explanation:
(B) : Given, \(f(x)= \begin{cases}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3\lt x\lt 3 \\ 6 x+2, \text { if } x \geq 3\end{cases}\) For continuity at \(x=-3\) \(\lim _{x \rightarrow-3^{+}} f(x)={ }_{x \rightarrow-3^{+}}-2 x=(-2)(-3)=6\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=-3\). For continuity at \(\mathrm{x}=3\) \(\therefore \lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}-2 x=(-2)(-3)=6\) \(\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}^{\lim }+6 x+2=[6(3)+2]=20\) \(\therefore \lim _{x \rightarrow-3^{-}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow-3^{+}} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is discontinuous of \(\mathrm{x}=3\).
79978
Let \(f(x)\) be a continuous function defined for 1 \(\leq \mathrm{x} \leq 3\). If \(\mathrm{f}(\mathrm{x})\) takes rational values for all \(\mathrm{x}\) and \(f(2)=10\), then what is \(f(1.5)\) equal to?
1 0
2 1
3 10
4 cannot be determined as the data is insufficient
Explanation:
(C) : Given, (i) \(\mathrm{f}(\mathrm{x})\) is continuous between \(1 \leq \mathrm{x} \leq 3\) (ii) \(\mathrm{f}(2)=10\) Hence to satisfies Rolle's Theorem ... a definite point Between the range \((1,2)\) Such that, \(\mathrm{f}(\mathrm{x})=0\) Hence \(f(1.5)=f(2)=10\)
SCRA-2012
Limits, Continuity and Differentiability
79979
\(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) is
1 continuous but not differentiable at \(x=1\)
2 differentiable at \(x=1\)
3 neither continuous nor differentiable at \(\mathrm{x}=1\)
4 continuous everywhere
Explanation:
(A) : Given, Function, \(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) At, \(\mathrm{x}=1\), LHL \(\lim _{h \rightarrow 0} \sin ^{-1}\left[\frac{1+(1-h)^{2}}{2(1-h)}\right]\) \(\sin ^{-1}\left(\frac{1+1}{2}\right) \Rightarrow \frac{\pi}{2}\) And, RHL \(\lim _{h \rightarrow 0^{+}} \sin ^{-1}\left[\frac{1+(1+h)^{2}}{2(1+h)}\right]\) \(\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) And, \(f(1)=\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) Thus, given function is continuous at \(\mathrm{x}=1\), Now, \(\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)\right\}\) \(=\frac{1}{\sqrt{1-\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)^{2}}} \cdot\left[\frac{2 \mathrm{x}(2 \mathrm{x})-\left(1+\mathrm{x}^{2}\right) \cdot 2}{(2 \mathrm{x})^{2}}\right]\) \(=\frac{2 x}{\sqrt{4 x^{2}-\left(1+x^{4}+2 x^{2}\right)}} \cdot \frac{2 x^{2}-1}{(2 x)^{2}}\) \(=\frac{2 x^{2}-1}{\sqrt{-\left(1+x^{4}-2 x^{2}\right)} \cdot 2 x}=\frac{2 x^{2}-1}{2 x \sqrt{-\left(1-x^{2}\right)^{2}}}\) Which does not exist at \(x=1\). Hence, given function is not differentiable at \(x=1\)
CG PET-2011
Limits, Continuity and Differentiability
79980
If \(f(x)=\left\{\begin{array}{cc}{[x]+[-x],} x \neq 2 \\ \lambda, x=2\end{array}\right.\), then \(f \quad\) is continuous at \(x=2\), provided \(\lambda\) is equal to
79981
The point/points of discontinuity of the function \(f(x)=\left\{\begin{array}{l}|x|+3, \text { if } \quad x \leq-3 \\ -2 x, \quad \text { if } \quad-3\lt x\lt 3 \\ 6 x+2, \text { if } \quad x \geq 3\end{array}\right.\) is/are
1 \(3,-3\)
2 3
3 -3
4 None of these
Explanation:
(B) : Given, \(f(x)= \begin{cases}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3\lt x\lt 3 \\ 6 x+2, \text { if } x \geq 3\end{cases}\) For continuity at \(x=-3\) \(\lim _{x \rightarrow-3^{+}} f(x)={ }_{x \rightarrow-3^{+}}-2 x=(-2)(-3)=6\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=-3\). For continuity at \(\mathrm{x}=3\) \(\therefore \lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}-2 x=(-2)(-3)=6\) \(\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}^{\lim }+6 x+2=[6(3)+2]=20\) \(\therefore \lim _{x \rightarrow-3^{-}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow-3^{+}} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is discontinuous of \(\mathrm{x}=3\).
79978
Let \(f(x)\) be a continuous function defined for 1 \(\leq \mathrm{x} \leq 3\). If \(\mathrm{f}(\mathrm{x})\) takes rational values for all \(\mathrm{x}\) and \(f(2)=10\), then what is \(f(1.5)\) equal to?
1 0
2 1
3 10
4 cannot be determined as the data is insufficient
Explanation:
(C) : Given, (i) \(\mathrm{f}(\mathrm{x})\) is continuous between \(1 \leq \mathrm{x} \leq 3\) (ii) \(\mathrm{f}(2)=10\) Hence to satisfies Rolle's Theorem ... a definite point Between the range \((1,2)\) Such that, \(\mathrm{f}(\mathrm{x})=0\) Hence \(f(1.5)=f(2)=10\)
SCRA-2012
Limits, Continuity and Differentiability
79979
\(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) is
1 continuous but not differentiable at \(x=1\)
2 differentiable at \(x=1\)
3 neither continuous nor differentiable at \(\mathrm{x}=1\)
4 continuous everywhere
Explanation:
(A) : Given, Function, \(\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) At, \(\mathrm{x}=1\), LHL \(\lim _{h \rightarrow 0} \sin ^{-1}\left[\frac{1+(1-h)^{2}}{2(1-h)}\right]\) \(\sin ^{-1}\left(\frac{1+1}{2}\right) \Rightarrow \frac{\pi}{2}\) And, RHL \(\lim _{h \rightarrow 0^{+}} \sin ^{-1}\left[\frac{1+(1+h)^{2}}{2(1+h)}\right]\) \(\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) And, \(f(1)=\sin ^{-1}\left(\frac{2}{2}\right)=\frac{\pi}{2}\) Thus, given function is continuous at \(\mathrm{x}=1\), Now, \(\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin ^{-1}\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)\right\}\) \(=\frac{1}{\sqrt{1-\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)^{2}}} \cdot\left[\frac{2 \mathrm{x}(2 \mathrm{x})-\left(1+\mathrm{x}^{2}\right) \cdot 2}{(2 \mathrm{x})^{2}}\right]\) \(=\frac{2 x}{\sqrt{4 x^{2}-\left(1+x^{4}+2 x^{2}\right)}} \cdot \frac{2 x^{2}-1}{(2 x)^{2}}\) \(=\frac{2 x^{2}-1}{\sqrt{-\left(1+x^{4}-2 x^{2}\right)} \cdot 2 x}=\frac{2 x^{2}-1}{2 x \sqrt{-\left(1-x^{2}\right)^{2}}}\) Which does not exist at \(x=1\). Hence, given function is not differentiable at \(x=1\)
CG PET-2011
Limits, Continuity and Differentiability
79980
If \(f(x)=\left\{\begin{array}{cc}{[x]+[-x],} x \neq 2 \\ \lambda, x=2\end{array}\right.\), then \(f \quad\) is continuous at \(x=2\), provided \(\lambda\) is equal to
79981
The point/points of discontinuity of the function \(f(x)=\left\{\begin{array}{l}|x|+3, \text { if } \quad x \leq-3 \\ -2 x, \quad \text { if } \quad-3\lt x\lt 3 \\ 6 x+2, \text { if } \quad x \geq 3\end{array}\right.\) is/are
1 \(3,-3\)
2 3
3 -3
4 None of these
Explanation:
(B) : Given, \(f(x)= \begin{cases}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3\lt x\lt 3 \\ 6 x+2, \text { if } x \geq 3\end{cases}\) For continuity at \(x=-3\) \(\lim _{x \rightarrow-3^{+}} f(x)={ }_{x \rightarrow-3^{+}}-2 x=(-2)(-3)=6\) \(\therefore \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=-3\). For continuity at \(\mathrm{x}=3\) \(\therefore \lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}-2 x=(-2)(-3)=6\) \(\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}^{\lim }+6 x+2=[6(3)+2]=20\) \(\therefore \lim _{x \rightarrow-3^{-}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow-3^{+}} \mathrm{f}(\mathrm{x})\) Hence, \(\mathrm{f}(\mathrm{x})\) is discontinuous of \(\mathrm{x}=3\).
