Chapter 20

Let \(f(x)=x \cdot\left[\frac{x}{2}\right]\), for \(-10

If a function \(f(x)\) defined by \(f(x)=\left\\{\begin{array}{l}a e^{x}+b e^{-x},-1 \leq x<1 \\ c x^{2}, 1 \leq x \leq 3 \\ a x^{2}+2 c x, 3

If the function \(\mathrm{f}\) defined on \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) by \(f(x)=\left\\{\begin{array}{c}\frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4}\end{array}\right.\) is continuous, then \(\mathrm{k}\) is equal to: (a) 2 (b) \(\frac{1}{2}\) (c) 1 (d) \(\frac{1}{\sqrt{2}}\)

If \(f(x)=[x]-\left[\frac{x}{4}\right], x \in \mathrm{R}\), where \([x]\) denotes the greatest integer function, then: \(\quad\) (a) \(f\) is continuous at \(x=4\). (b) \(\lim _{x \rightarrow 4+} f(x)\) exists but \(\lim _{x \rightarrow 4} f(x)\) does not exist. (c) Both \(\lim _{x \rightarrow 4-} f(x)\) and \(\lim _{x \rightarrow 4} f(x)\) exist but are not equal. (d) \(\lim _{x \rightarrow 4-} f(x)\) exists but \(\lim _{x \rightarrow 4+} f(x)\) does not exist.

If the function \(f(x)=\left\\{\begin{array}{l}a|\pi-x|+1, x \leq 5 \\ b|x-\pi|+3, x>5\end{array}\right.\) is continuous at \(x=5\), then the value of \(a-b\) is: (a) \(\frac{2}{\pi+5}\) (b) \(\frac{-2}{\pi+5}\) (c) \(\frac{2}{\pi-5}\) (d) \(\frac{2}{5-\pi}\)

Let \(f:[-1,3] \rightarrow \mathrm{R}\) be defined as \(f(x)= \begin{cases}|x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3\end{cases}\) where \([t]\) denotes the greatest integer less than or equal to \(t\). Then, \(f\) is discontinuous at: \(\quad\) (a) only one point (b) only two points (c) only three points (d) four or more points

Let \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a function defined as \(f(x)=\left\\{\begin{array}{ccc}5, & \text { if } & x \leq 1 \\\ \mathrm{a}+\mathrm{bx}, & \text { if } & 1

If the function \(f\) defined as \(f(x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1}\) \(x \neq 0\), is continuous at \(x=0\), then the ordered pair \((k, f(0))\) is equal to? \(\begin{array}{llll}\text { (a) }(3,1) & \text { (b) }(3,2) & \text { (c) }\left(\frac{1}{3}, 2\right) & \text { (d) }(2,1)\end{array}\)

Let \(f(x)=\left\\{\begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\\ k, & x=2\end{array}\right.\) The value of \(k\) for which \(f\) is continuous at \(x=2\) is |Online April \(\mathbf{1 5}, \mathbf{2 0 1 8}]\) \(\begin{array}{ll}\text { (a) } e^{-2} & \text { (b) } e\end{array}\) (c) \(e^{-1}\) (d) 1

Let \(\mathrm{a}, \mathrm{b} \in \mathrm{R},(\mathrm{a} \neq 0)\). if the function \(\mathrm{f}\) defined as \(f(x)= \begin{cases}\frac{2 x^{2}}{a} & , 0 \leq x<1 \\ a & , 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{3}} & , & \sqrt{2} \leq x<\infty\end{cases}\) is continuous in the interval \([0, \infty)\), then an ordered pair \((\mathrm{a}, \mathrm{b})\) is: (a) \((-\sqrt{2}, 1-\sqrt{3})\) (b) \((\sqrt{2},-1+\sqrt{3})\) (c) \((\sqrt{2}, 1-\sqrt{3})\) (d) \((-\sqrt{2}, 1+\sqrt{3})\)

If the function \(f(x)=\left\\{\begin{array}{l}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}, x \neq \pi \\ k \quad, x=\pi\end{array}\right.\) is continuous at \(x=\pi\), then \(\mathrm{k}\) equals: (a) 0 (b) \(\frac{1}{2}\) (c) 2 (d) \(\frac{1}{4}\)

Let \(f\) be a composite function of \(x\) defined by \(f(u)=\frac{1}{u^{2}+u-2}, u(x)=\frac{1}{x-1}\). Then the number of points \(x\) where \(f\) is discontinuous is : (a) 4 (b) 3 (c) 2 (d) 1

Let \(f(x)=-1+|x-2|\), and \(g(x)=1-|x| ;\) then the set of all points where fog is discontinuous is (a) \(\\{0,2\\}\) (b) \(\\{0,1,2\\}\) (c) \(\\{0\\}\) (d) an empty set

Let \(f:[1,3] \rightarrow R\) be a function satisfying \(\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}\), for all \(x \neq 2\) and \(f(2)=1\), where \(R\) is the set of all real numbers and \([x]\) denotes the largest integer less than or equal to \(x\). Statement 1: \(f(x)\) exists. [Online May 19, 2012] \(x \rightarrow 2\) Statement 2: \(f\) is continuous at \(x=2\). (a) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 . (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1 . (d) Statement 1 is true, Statement 2 is false.

Statement 1: A function \(f: R \rightarrow R\) is continuous at \(x_{0}\) if and only if \(\lim _{x \rightarrow x_{0}} f(x)\) exists and \(\lim _{x \rightarrow x_{0}} f(x)=f\left(x_{0}\right)\) Statement 2: A function \(f: R \rightarrow R\) is discontinuous at \(x_{0}\) i and only if, \(\lim _{x \rightarrow x_{0}} f(x)\) exists and \(\lim _{x \rightarrow x_{0}} f(x) \neq f\left(x_{0}\right)\). (a) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement \(1 .\) (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1 . (d) Statement 1 is true, Statement 2 is false.

Define \(f(x)\) as the product of two real functions \([2011 \mathrm{RS}]\) \(f_{1}(x)=x, x \in R\), and \(f_{2}(x)= \begin{cases}\sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}\) as follows: \(f(x)=\left\\{\begin{array}{cl}f_{1}(x) \cdot f_{2}(x), & \text { if } x=0 \\\ 0 & \text { if } x=0\end{array}\right.\) Statement \(-1: f(x)\) is continuous on \(\mathrm{R}\). Statement \(-2: f_{1}(x)\) and \(f_{2}(x)\) are continuous on \(\mathrm{R}\). (a) Statement \(-1\) is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement- \(1 .\) (b) Statement- 1 is true, Statement- 2 is true; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement-1 is true, Statement- 2 is false (d) Statement- 1 is false, Statement- 2 is true

The values of \(p\) and \(q\) for which the function \(f(x)= \begin{cases}\frac{\sin (p+1) x+\sin x}{x}, x<0 \\ q & , x=0 \text { is continuous for all } x \text { in } R \text {, } \\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{cases}\) are (a) \(p=\frac{5}{2}, q=\frac{1}{2}\) (b) \(p=-\frac{3}{2}, q=\frac{1}{2}\) (c) \(p=\frac{1}{2}, q=\frac{3}{2}\) (d) \(p=\frac{1}{2}, q=-\frac{3}{2}\)

The function \(f: R /\\{0\\} \rightarrow R\) given by \(f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}\) can be made continuous at \(x=0\) by defining \(f(0)\) as (a) 0 (b) 1 (c) 2 (d) \(-1\)

Let \(f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]\). If \(f(x)\) is continuous in \(\left[0, \frac{\pi}{2}\right]\), then \(f\left(\frac{\pi}{4}\right)\) is (a) \(-1\) (b) \(\frac{1}{2}\) (c) \(-\frac{1}{2}\) (d) 1

\(f\) is defined in \([-5,5]\) as \(f(x)=x\) if \(x\) is rational \(=-x\) if \(x\) is irrational. Then (a) \(f(x)\) is continuous at every \(x\), except \(x=0\) (b) \(f(x)\) is discontinuous at every \(x\), except \(x=0\) (c) \(f(x)\) is continuous everywhere (d) \(f(x)\) is discontinuous everywhere

Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a function defined by \(f(x)=\max \left\\{x, x^{2}\right\\}\). Let S denote the set of all points in \(\mathrm{R}\), where \(f\) is not differentiable. Then: (a) \(\\{0,1\\}\) (b) \(\\{0\\}\) (c) \(\phi\) (an empty set) (d) \(\\{1\)

If the function \(f(x)\left\\{\begin{array}{ll}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\\ k_{2} \cos x, & x>\pi\end{array}\right.\) is twice dif- ferentiable, then the ordered pair \(\left(k_{1}, k_{2}\right)\) is equal to: (a) \(\left(\frac{1}{2}, 1\right)\) (b) \((1,0)\) (c) \(\left(\frac{1}{2},-1\right)\) (d) \((1,1)\)

Let \(f\) be a twice differentiable function on \((1,6)\). If \(f(2)=8\), \(f^{\prime}(2)=5, f^{\prime}(x) \geq 1\) and \(f^{\prime \prime}(x) \geq 4\), for all \(x \in(1,6)\), then : (a) \(f(5)+f^{\prime}(5) \leq 26\) (b) \(f(5)+f^{\prime}(5) \geq 28\) (c) \(f^{\prime}(5)+f^{\prime \prime}(5) \leq 20\) (d) \(f(5) \leq 10\)

Suppose a differentiable function \(f(x)\) satisfies the identity \(f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y\), for all real \(x\) and \(y\). If \(\lim _{x \rightarrow 0} \frac{f(x)}{x}=1\), then \(f^{\prime}(3)\) is equal to

The function \(f(x)=\left\\{\begin{array}{l}\frac{\pi}{4}+\tan ^{-1} x,|x| \leq 1 \\ \frac{1}{2}(|x|-1),|x|>1\end{array}\right.\) is: (a) continuous on \(\mathbf{R}-\\{1\\}\) and differentiable on \(\mathbf{R}-\\{-1,1\\}\) (b) both continuous and differentiable on \(\mathbf{R}-\\{1\\}\). (c) continuous on \(\mathbf{R}-\\{-1\\}\) and differentiable on \(\mathbf{R}-\\{-1,1\\}\) (d) both continuous and differentiable on \(\mathbf{R}-\\{-1\\}\).

Let \(f\) and \(g\) be differentiable functions on \(\mathbf{R}\) such that fog is the identity function. If for some \(a, b \in \mathbf{R}, g^{\prime}(a)=5\) and \(g(a)=b\), then \(f^{\prime}(b)\) is equal to: \(\quad\) [Jan. 9,2020 (II)] (a) \(\frac{1}{5}\) (b) 1 (c) 5 (d) \(\frac{2}{5}\)

Let \(S\) be the set of all functions \(f:[0,1] \rightarrow R\), which are continuous on \([0,1]\) and differentiable on \((0,1)\). Then for every \(f\) in \(S\), there exists a \(c \in(0,1)\), depending on \(f\), such that: (a) \(|f(c)-f(\mathrm{I})|<(1-c)\left|f^{\prime}(c)\right|\) (b) \(\frac{f(1)-f(c)}{1-c}=f^{\prime}(c)\) (c) \(|f(c)+f(1)|<(1+c)\left|f^{\prime}(c)\right|\) (d) \(|f(c)-f(1)|<\left|f^{\prime}(c)\right|\)

Let \(S\) be the set of points where the function, \(f(x)=|2-| x-3 \|, x \in \boldsymbol{R}\), is not differentiable. Then \(\sum_{x \in S} f(f(x))\) is equal to-

If \(\mathrm{f}(\mathrm{x})=\left\\{\begin{array}{c}\frac{\sin (\mathrm{p}+1) x+\sin x}{x}, x<0 \\ \frac{\mathrm{q}}{x}, x=0 \\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, x>0\end{array}\right.\) is continuous at \(\mathrm{x}=0\), then the ordered pair \((\mathrm{p}, \mathrm{q})\) is equal to: (a) \(\left(-\frac{3}{2},-\frac{1}{2}\right)\) (b) \(\left(-\frac{1}{2}, \frac{3}{2}\right)\) (c) \(\left(-\frac{3}{2}, \frac{1}{2}\right)\) (d) \(\left(\frac{5}{2}, \frac{1}{2}\right)\)

\(\operatorname{Let} f(x)=\log _{e}(\sin x),(0

Let \(\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}\) be differentiable at \(\mathrm{c} \in \mathbf{R}\) and \(\mathrm{f}(\mathrm{c})=0\). If \(g(x)=|f(x)|\), then at \(x=c\), gis: \(\quad[\) (a) not differentiable if \(f^{\prime}(c)=0\) (b) differentiable if \(\mathrm{f}^{\prime \prime}(\mathrm{c}) \neq 0\) (c) differentiable if \(\mathrm{f}^{\prime}(\mathrm{c})=0\) (d) not differentiable

Let \(f(x)=15-|x-10| ; x \in R\). Then the set of all values of \(x\), at which the function, \(g(x)=f(f(x))\) is not differentiable, is: \(\quad\) (a) \(\\{5,10,15\\}\) (b) \(\\{10,15\\}\) (c) \(\\{5,10,15,20\\}\) (d) \(\\{10\\}\)

Let \(f\) be a differentiable function such that \(f(1)=2\) and \(f^{\prime}(x)=f(x)\) for all \(x \in R\). If \(h(x)=f(f(x))\), then \(h^{\prime}(1)\) is equal to (a) \(2 \mathrm{e}^{2}\) (b) \(4 \mathrm{e}\) (c) \(2 \mathrm{e}\) (d) \(4 \mathrm{e}^{2}\)

Let \(f(x)=\left\\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.\) \(g(x)=|f(x)|+f(x))\). Then, in the interval \((-2,2), g\) is: (a) differentiable at all points (b) not continuous (c) not differentiable at two points (d) not differentiable at one point

If \(x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0)\), then \(\frac{d y}{d x}\) at \(x=e\) is equal to: \(\quad\) (a) \(\frac{(1+2 e)}{2 \sqrt{4+e^{2}}}\) (b) \(\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}\) (c) \(\frac{(1+2 e)}{\sqrt{4+e^{2}}}\) (d) \(\frac{e}{\sqrt{4+e^{2}}}\)

Let \(\mathrm{K}\) be the set of all real values of \(x\) where the function \(f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|\) is not differentiable. Then the set \(K\) is equal to: \(\quad\) (a) \(\phi\) (an empty set) (b) \(\\{\pi\\}\) (c) \(\\{0\\}\) (d) \(\\{0, \pi\\}\)

Let \(f(x)=\left\\{\begin{array}{cc}\max \left\\{|x|, x^{2}\right\\} & |x| \leq 2 \\ 8-2|x|, & 2<|x| \leq 4\end{array}\right.\) Let \(\mathrm{S}\) be the set of points in the interval \((-4,4)\) at which \(f\) is not differentiable. Then S: \(\quad\) (a) is an empty set (b) equals \(\\{-2,-1,0,1,2\\}\) (c) equals \(\\{-2,-1,1,2\\}\) (d) equals \(\\{-2,2\\}\)

Let \(f:(-1,1) \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\max\) \(\left\\{-|x|,-\sqrt{1-x^{2}}\right\\}\). If \(\mathrm{K}\) be the set of all points at which \(f\) is not differentiable, then \(K\) has exactly: (a) five elements (b) one element (c) three elements (d) two elements

Let \(\mathrm{S}=\left\\{\mathrm{t} \in \mathrm{R}: \mathrm{f}(\mathrm{x})=|\mathrm{x}-\pi|\left(\mathrm{e}^{|\mathrm{x}|}-1\right) \sin |\mathrm{x}|\right.\) is not differentiable at \(\mathrm{t}\\}\). Then the set \(\mathrm{S}\) is equal to: (a) \(\\{0\\}\) (b) \(\\{\pi\\}\) (c) \(\\{0, \pi\\}\) (d) \(\phi\) (an empty set)

Let \(S=\left\\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| \mathrm{e}^{|l|}-\mu\right) \cdot \sin (2 \mid t), t \in R\right.\), is a differentiable function\\}. Then \(S\) is a subest of? (a) \(R \times[0, \infty)\) (b) \((-\infty, 0) \times R\) (c) \([0, \infty) \times R\) (d) \(R \times(-\infty, 0)\)

If the function \(f(x)=\left\\{\begin{array}{ll}-x, & x<1 \\ a+\cos ^{-1}(x+b), & 1 \leq x \leq 2\end{array}\right.\) is differentiable at \(\mathrm{x}=1\), then \(\frac{\mathrm{a}}{\mathrm{b}}\) is equal to: (a) \(\frac{\pi+2}{2}\) (b) \(\frac{\pi-2}{2}\) (c) \(\frac{-\pi-2}{2}\) (d) \(-1-\cos ^{-1}(2)\)

If the function. \(g(x)=\left\\{\begin{array}{l}k \sqrt{x+1}, 0 \leq x \leq 3 \\ m x+2,3

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a function such that \(|f(x)| \leq x^{2}\), for all \(x \in R .\) Then, at \(x=0\), fis: \(\quad\) (a) continuous but not differentiable. (b) continuous as well as differentiable. (c) neither continuous nor differentiable. (d) differentiable but not continuous.

Let \(\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) be two functions defined by \(f(x)=\left\\{\begin{array}{l}x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, \quad, x=0\end{array}\right.\), and \(g(x)=x f(x)\) Statement I: \(f\) is a continuous function at \(\mathrm{x}=0\). Statement II: \(g\) is a differentiable function at \(x=0\). (a) Both statement I and II are false. (b) Both statement \(I\) and II are true. (c) Statement I is true, statement II is false. (d) Statement I is false, statement II is true.

Consider the function, \(f(x)=|x-2|+|x-5|, x \in R\) Statement-1: \(f^{\prime}(4)=0\) Statement- \(2: f\) is continuous in \([2,5]\), differentiable in \((2,5)\) and \(f(2)=f(5)\) (a) Statement- 1 is false, Statement- 2 is true. (b) Statement- 1 is true, statement- 2 is true; statement- 2 is a correct explanation for Statement-1. (c) Statement- 1 is true, statement- 2 is true, statement- 2 is not a correct explanation for Statement-1. (d) Statement- 1 is true, statement- 2 is false.

If \(f(x)=a|\sin x|+b e^{|x|}+c|x|^{3}\), where \(a, b, c \in R\), is differentiable at \(x=0\), then \(\quad\) (a) \(a=0, b\) and \(c\) are any real numbers (b) \(c=0, a=0, b\) is anyreal number (c) \(b=0, c=0, a\) is any real number (d) \(a=0, b=0, c\) is any real number

If \(x+|y|=2 y\), then \(y\) as a function of \(x\), at \(x=0\) is (a) differentiable but not continuous (b) continuous but not differentiable (c) continuous as well as differentiable (d) neither continuous nor differentiable

If function \(f(x)\) is differentiable at \(x=a\), then \(\lim _{x \rightarrow a} \frac{x^{2} f(a)-a^{2} f(x)}{x-a}\) is: \(\quad\) (a) \(-a^{2} f^{\prime}(a)\) (b) \(a f(a)-a^{2} f^{\prime}(a)\) (c) \(2 a f(a)-a^{2} f^{\prime}(a)\) (d) \(2 a f(a)+a^{2} f^{\prime}(a)\)

Let \(f(x)=\left\\{\begin{array}{c}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \quad \text { if } x=1\end{array} \quad[\mathbf{2 0 0 8 ]}\right.\) Then which one of the following is true? (a) \(f\) is neither differentiable at \(x=0\) nor at \(x=1\) (b) \(f\) is differentiable at \(x=0\) and at \(x=1\) (c) \(f\) is differentiable at \(x=0\) but not at \(x=1\) (d) \(f\) is differentiable at \(x=1\) but not at \(x=0\)

Let \(f: R \rightarrow R\) bea function defined by \(f(x)=\min \\{x+1,|x|+1\\}\), Then which of the following is true? (a) \(f(x)\) is differentiable everywhere (b) \(f(x)\) is not differentiable at \(x=0\) (c) \(f(x) \geq 1\) for all \(x \in R\) (d) \(f(x)\) is not differentiable at \(x=1\)