TRIGONOMETRY FUNCTIONS - 2 (Inverse Trigonometric Functions)
Hyperbolic functions
| (i). $\sin h (-x)$ | $=-\sin h x$ | odd function |
| $\cos h (-\mathrm{x})$ | $=\cos h \mathrm{x}$ | even function |
| $\tan h (-\mathrm{x})$ | $=-\tan h \mathrm{x}$ | odd function |
| (ii). Function | Domain | Range |
| $\sin h ^{-1} \mathrm{x}$ | $\mathrm{R}$ | $ \mathrm{R}$ |
| $\cos h ^{-1} \mathrm{X} $ | $(0, \infty)$ | $(1-, \infty)$ |
| $\tan h ^{-1} \mathrm{x} $ | $\mathrm{R}$ | $ (-1, 1)$ |
| $\operatorname{\cot h}^{-1} \mathrm{x}$ | $ \mathrm{R}-\{0\}$ | $ R-[1, 1]$ |
| $\operatorname{\sec h}^{-1} \mathrm{X}$ | $(0, \infty)$ | $(0, 1)$ |
| $\operatorname{\cosec h}^{-1} \mathrm{x}$ | $\mathrm{R}-\{0\} $ | $ R-\{0\}$ |
| (iii). $\sin h \left(\sin h ^{-1} \mathrm{x}\right)=\mathrm{x}$ | $\sin h ^{-1}(\sin h x)=x$ | |
| $\cos h \left(\cos h ^{-1} \mathrm{x}\right)=\mathrm{x}$ | $\cos h ^{-1}(\cos h x)=x$ | |
| $\tan h \left(\tan h ^{-1} x\right)=x$ | $\tan h ^{-1}(\tan h x)=x$ | |
| $\sin h \left(\sin ^{-1} x\right)=x^{n}$ | $\sin h \left(\sin h ^{-1} x^{n}\right)=x^{n}$ |
(iv). $\sin h ^{-1} x=\log _{e}\left(x+\sqrt{x^{2}+1}\right)$
$\cos h ^{-1} x=\log _{e}\left(x+\sqrt{x^{2}-1}\right)$
$\tan h ^{-1} x=\frac{1}{2} \log _{e}\left(\frac{x+1}{x-1}\right) x>1, x<-1$
$\operatorname{\cot h}^{-1} x=\frac{1}{2} \log _{e}\left(\frac{x-1}{x+1}\right) x>1, x<-1$
$\operatorname{\sec h}^{-1} x=\log _{e}\left(\frac{1+\sqrt{1-x^{2}}}{x}\right) 0<x \leq 1$
$\operatorname{\cosec h}^{-1} x=\left\{\begin{array}{lll}\log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right) & \text { if } & x>0 \\ \log _e\left(\frac{1-\sqrt{1+x^2}}{x}\right) & \text { if } & x<0\end{array}\right.$
(v). $\sin h ^{-1} \mathrm{x}=\operatorname{\cosec h}^{-1}\left(\frac{1}{\mathrm{x}}\right)$
$\sin h ^{-1} x=\cos h ^{-1} \sqrt{x^{2}+1}$
$\cos h ^{-1} x=\sin h ^{-1} \sqrt{x^{2}-1}$
$\sin h \left(\cos h ^{-1} x\right)=\sqrt{x^{2}-1}$
Examples
1. Total number of positive integral values of $n$ sothat the equation $\cos ^{-1} x+\left(\sin ^{-1} y\right)^{2}=\frac{n \pi^{2}}{4}$ and $\left(\sin ^{-1} \mathrm{y}\right)^{2}-\cos ^{-1} \mathrm{x}=\frac{\pi^{2}}{16}$ are consistent, is equal to
(a). 1
(b). 4
(c). 3
(d). 2
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Solution:
$ \text { we have } 2\left(\sin ^{-1} \mathrm{y}\right)^{2}=\frac{(4 \mathrm{n}+1) \pi^{2}}{16} \quad \text { On adding the two given equations } $
$0\leq \frac{(4 \mathrm{n}+1) \pi^{2}}{16} \leq \frac{2 \pi^{2}}{4} \Rightarrow-\frac{1}{4} \leq \mathrm{n} \leq \frac{7}{4}$
Also $2 \cos ^{-1} \mathrm{x}=\frac{(4 \mathrm{n}-1) \pi^{2}}{16}$ On subtracting the two given equations
$0\leq \frac{(4 n-1) \pi^{2}}{16} \leq 2 \pi \Rightarrow \frac{1}{4} \leq \mathrm{n} \leq \frac{8}{\pi}+\frac{1}{4}$
$\therefore \mathrm{n}=1$
Answer: (a)
2. The minimum value of $\left(\sin ^{-1} \mathrm{x}\right)^{3}+\left(\cos ^{-1} \mathrm{x}\right)^{3}$ is equal to
(a). $\frac{\pi^{3}}{32}$
(b). $\frac{5 \pi^{3}}{32}$
(c). $\frac{9 \pi^{3}}{32}$
(d). $\frac{11 \pi^{3}}{32}$
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Solution:
$ \begin{aligned} & \text { Let } y=\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3} \\ & =\left(\sin ^{-1} x+\cos ^{-1} x\right)\left\{\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}-\sin ^{-1} x \cdot \cos ^{-1} x\right\} \\ & =\frac{\pi}{2}\left\{\left(\sin ^{-1} x+\cos ^{-1} x\right)^{2}-3 \sin ^{-1} x \cdot \cos x\right\} \\ & =\frac{\pi}{2}\left\{\frac{\pi^{2}}{4}-3 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)\right\} \\ & =\frac{\pi}{2}\left\{3\left(\sin ^{-1} x\right)^{2}-\frac{3 \pi}{2} \sin ^{-1} x+\frac{\pi^{2}}{4}\right\} \\ & =\frac{\pi}{2}\left\{3\left(\sin ^{-1} x\right)^{2}-\frac{3 \pi}{2} \sin ^{-1} x+\frac{3 \cdot \pi^{2}}{16}-\frac{3 \pi^{2}}{16}+\frac{\pi^{2}}{4}\right\} \\ & =\frac{\pi}{2}\left\{3\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right\} \end{aligned} $
Minimum value of $\mathrm{y}=\frac{\pi}{2} \cdot \frac{\pi^{2}}{16}=\frac{\pi^{3}}{32} \quad$ sin ce $\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2} \geq 0$
Answer: (a)
3. If $A=2 \tan ^{-1}(2 \sqrt{2}-1)$ and $B=3 \sin ^{-1}\left(\frac{1}{3}\right)+\sin ^{-1}\left(\frac{3}{5}\right)$. then
(a). $\mathrm{A}=\mathrm{B}$
(b). $\mathrm{A}<\mathrm{B}$
(c). $\mathrm{A}>\mathrm{B}$
(d). None of these
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Solution:
$ \mathrm{A}=2 \tan ^{-1}(2 \sqrt{2}-1)=2 \tan ^{-1}(1.828)>2 \tan ^{-1} \sqrt{3} $
$ \begin{aligned} \mathrm{A} & >\frac{2 \pi}{3} \\ 3\sin ^{-1} \frac{1}{3} & =\sin ^{-1}\left(3 \cdot \frac{1}{3}-4\left(\frac{1}{3}\right)^{3}\right)=\sin ^{-1}\left(1-\frac{4}{27}\right)=\sin ^{-1} \frac{23}{27}=\sin ^{-1}(0.852) \\ 3\sin ^{-1} \frac{1}{3} & <\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ \sin ^{-1}\left(\frac{3}{5}\right) & =\sin ^{-1}(0.6)<\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ \therefore \mathrm{B} & <\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3} \end{aligned} $
Hence $\mathrm{A}>\mathrm{B}$
Answer: (c)
4. The complete solution set of $\sin ^{-1}(\sin 5)>x^{2}-4 x$ is
(a). $|x-2|<\sqrt{9-2 \pi}$
(b). $|\mathrm{x}-2|>\sqrt{9-2 \pi}$
(c). $|x|<\sqrt{9-2 \pi}$
(d). $|\mathrm{x}|>\sqrt{9-2 \pi}$
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Solution:
$ \sin ^{-1} \sin 5>x^{2}-4 x $
$ \sin ^{-1} \sin (5-2 \pi)>x^{2}-4 x $
$ \begin{aligned} & 5-2 \pi>x^{2}-4 x \quad(5-2 \pi) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ & x^{2}-4 x+2 \pi-5<0 \\ & (x-2)^{2}<9-2 \pi \\ & |x-2|<\sqrt{9-2 \pi} \end{aligned} $
Answer: (a)
5. Let $(\mathrm{x}, \mathrm{y})$ be such that $\sin ^{-1}(\mathrm{ax})+\cos ^{-1} \mathrm{y}+\cos ^{-1} \mathrm{bxy}=\frac{\pi}{2}$.
Match the statements in column I with statements in column II
| Column I | Column II |
|---|---|
| (a). If $a=1$ and $b=0$, then $(x, y)$ | (p). Lies on the circle $x^{2}+y^{2}=1$ |
| (b). If $a=1$ and $b=1$, then ( $x, y)$ | (q). Lies on the $\left(x^{2}-1\right)\left(y^{2}-1\right)=0$ |
| (c). If $\mathrm{a}=1$ and $\mathrm{b}=2$, then $(\mathrm{x}, \mathrm{y})$ | (r). Lies on $y=x$ |
| (d). If $a=2$ and $b=2$, then ( $x, y)$ | (s). Lies on the $(4x^2-1)(y^2-1)=0$ |
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Solution:
$ \begin{aligned} & \cos ^{-1} y+\cos ^{-1} b x y=\frac{\pi}{2}-\sin ^{-1} a x \\ & \cos ^{-1} y+\cos ^{-1} b x y=\cos ^{-1} a x \end{aligned} $
$\begin{aligned} & \text { Let } \cos ^{-1} y=A, \cos ^{-1} b x y=B \text { and } \cos ^{-1} a x=C \\ & A+B=C \\ & B=A-C \\ & \cos (A-C)=\cos B \\ & \cos A \cos C+\sin A \sin C=\cos B \\ & y a x+\sin A \sin C=b x y \\ & \sin A \sin C=b x y-a x y \\ & \sin ^2 A \sin ^2 C=(b-a)^2 x^2 y^2 \\ & \left(1-a^2 x^2\right)\left(1-y^2\right)=x^2 y^2(b-a)^2\end{aligned}$
(a). Put $\mathrm{a}=1$ and $\mathrm{b}=0$
$ \begin{aligned} & \left(1-x^{2}\right)\left(1-y^{2}\right)=x^{2} y^{2} \\ & x^{2}+y^{2}=1 \end{aligned} $
(b). Put $\mathrm{a}=1$ and $\mathrm{b}=1$
$ \begin{aligned} & \left(1-x^{2}\right)\left(1-y^{2}\right)=0 \\ & \left(x^{2}-1\right)\left(y^{2}-1\right)=0 \end{aligned} $
(c). Put $\mathrm{a}=1$ and $\mathrm{b}=2$
$ \begin{gathered} \left(1-x^{2}\right)\left(1-y^{2}\right)=x^{2} y^{2} \\ x^{2}+y^{2}=1 \end{gathered} $
(d). Put $\mathrm{a}=2$ and $\mathrm{b}=2$
$ \left(1-4 x^{2}\right)\left(1-y^{2}\right)=0 $
$ \left(4 x^{2}-1\right)\left(y^{2}-1\right)=0 $
(a) $\rightarrow$ (p), (b) $\rightarrow$ (q), (c) $\rightarrow$ (p), (d) $\rightarrow$ (s)
6. If $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4} \ldots \ldots \ldots.\right)+\cos ^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4} \ldots \ldots \ldots.\right)=\frac{\pi}{2}$ for $0<|x|<\sqrt{2}$, then $x$ equals
(a). $\frac{1}{2}$
(b). $1$
(c). $-\frac{1}{2}$
(d). $-1$
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Solution:
$ \begin{aligned} & \sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4} \ldots \ldots \ldots\right)=\frac{\pi}{2}-\cos ^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4} \ldots \ldots \ldots\right) \\ & \sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4} \ldots \ldots \ldots\right)=\sin ^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4} \ldots \ldots \ldots . .\right) \end{aligned} $
or $\sin^{-1} \theta +\cos^{-1} \theta =1 / 2$
so,
$x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4} \ldots \ldots \ldots=x-\frac{x^{2}}{2}+\frac{x^{3}}{4} \ldots \ldots \ldots$
$ \begin{aligned} & \frac{\mathrm{x}^{2}}{1-\left(-\frac{\mathrm{x}^{2}}{2}\right)}=\frac{\mathrm{x}}{1-\left(-\frac{\mathrm{x}}{2}\right)} \quad \because \mathrm{S} _{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}} \quad|\mathrm{r}|<1\\ & \frac{2 \mathrm{x}^{2}}{2+\mathrm{x}^{2}}=\frac{2 \mathrm{x}}{2+\mathrm{x}} \\ & 2 \mathrm{x}^{2}+\mathrm{x}^{3}=2 \mathrm{x}+\mathrm{x}^{3} \\ & 2 \mathrm{x}(\mathrm{x}-1)=0 \\ & \mathrm{x}=0,1 \\ & \mathrm{x}=1 \quad (0<|x|\sqrt{2})\\ \end{aligned} $
Answer: (b)
7. Let $a, b, c$ be positive real numbers. Let
$ \theta=\tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{c a}}+\tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}}, \text { then } \tan \theta= $
(a). $\frac{\pi}{4}$
(b). $\frac{\pi}{2}$
(c). $\pi$
(d). None of these
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Solution:
Let $\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{u}$
$ \begin{aligned} & \theta=\tan ^{-1} \sqrt{\frac{\mathrm{au}}{\mathrm{bc}}}+\tan ^{-1} \sqrt{\frac{\mathrm{bu}}{\mathrm{ca}}}+\tan ^{-1} \sqrt{\frac{\mathrm{cu}}{\mathrm{ab}}} \quad\left[\begin{array}{l} \text { Also you can use } \tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}+\tan ^{-1} \mathrm{z} \\ =\tan ^{-1}\left[\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{xyz}}{1-\mathrm{xy-yz-zx}}\right] \end{array}\right] \\ & \theta=\pi+\tan ^{-1}\left(\frac{\sqrt{\frac{\mathrm{au}}{\mathrm{bc}}}+\sqrt{\frac{\mathrm{bu}}{\mathrm{ca}}}}{1-\sqrt{\frac{\mathrm{au}}{\mathrm{bc}}} \sqrt{\frac{\mathrm{bc}}{\mathrm{ac}}}}\right)+\tan ^{-1} \sqrt{\frac{\mathrm{cu}}{\mathrm{ab}}}, \sqrt{\frac{\mathrm{au}}{\mathrm{bc}}} \sqrt{\frac{\mathrm{bu}}{\mathrm{ca}}}=\frac{\mathrm{u}}{\mathrm{c}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}+1>1 \\ & \theta=\pi+\tan ^{-1}\left(\frac{\frac{(\mathrm{a}+\mathrm{b}) \sqrt{\mathrm{u}}}{\sqrt{\mathrm{abc}}}}{1-\frac{\mathrm{u}}{\mathrm{c}}}\right)+\tan ^{-1} \sqrt{\frac{\mathrm{cu}}{\mathrm{ab}}} \\ & =\pi+\tan ^{-1}\left(\frac{(\mathrm{u}-\mathrm{c}) \sqrt{\mathrm{u}}}{\sqrt{\mathrm{abc}}} \cdot \frac{\mathrm{c}}{(\mathrm{c}-\mathrm{u})}\right)+\tan ^{-1} \sqrt{\frac{\mathrm{cu}}{\mathrm{ab}}} \\ & =\pi-\tan ^{-1} \sqrt{\frac{\mathrm{uc}}{\mathrm{ab}}}+\tan ^{-1} \sqrt{\frac{\mathrm{cu}}{\mathrm{ab}}} \\ & =\pi \end{aligned} $
Answer: (c)
Practice questions
1. If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$ or $\frac{\pi}{2}$ then
(a). $x+y+z=3 x y z$
(b). $x+y+z=2 x y z$
(c). $x y+y z+z x=1$
(d). None of these
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Answer: (c)2. If $\left[\cos ^{-1} x\right]+\left[\cot ^{-1} x\right]=0$, where $x$ is a nonnegative real number and [.] denotes the greatest integer function then complete set of values of $x$ is
(a). $(\cos 1,1)$
(b). $(\cot 1,1)$
(c). $(\cos 1, \cot 1)$
(d). None of there
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Answer: (c)3. Range of the function $\mathrm{f}(\mathrm{x})=\cos ^{-1}(-\{\mathrm{x}\})$, where $\{.\}$ is fractional part function is
(a). $\left(\frac{\pi}{2}, \pi\right)$
(b). $\left(\frac{\pi}{2}, \pi\right]$
(c). $\left[\frac{\pi}{2}, \pi\right)$
(d). $\left(0, \frac{\pi}{2}\right]$
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Answer: (c)4. The sum of solutions of the equation $2 \sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}+\cos ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}}=\frac{3 \pi}{2}$ is
(a). $0$
(b). $-1$
(c). $1$
(d). $2$
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Answer: (b)5. Which of the following is a rational number
(a). $\sin \left(\tan ^{-1} 3+\tan ^{-1} \frac{1}{3}\right)$
(b). $\cos \left(\frac{\pi}{2}-\sin ^{-1} \frac{3}{4}\right)$
(c). $\log _{2}\left(\sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)\right)$
(d). $\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)$
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Answer: (a, b, c)Assertion and Reasoning
6. Statement 1: $\sin ^{-1}\left(\frac{1}{\sqrt{\mathrm{e}}}\right)>\tan ^{-1}\left(\frac{1}{\sqrt{\pi}}\right)$
Statement 2: $\sin ^{-1} \mathrm{x}>\tan ^{-1} \mathrm{y}$ for $\mathrm{x}>\mathrm{y}, \forall \mathrm{x}, \mathrm{y} \in(0,1)$
(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 correct explanation for Statement - 1
(c). Statement-1 is True, Statement - 2 is False
(d). Statement 1 is False, Statement -2 is True.
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Answer: (a)Comprehension (Que. no. 7 to 9)
It is given that $\mathrm{A}=\left(\tan ^{-1} \mathrm{x}\right)^{3}+\left(\cot ^{-1} \mathrm{x}\right)^{3}$ where $\mathrm{x}>0$ and $\mathrm{B}=\left(\cos ^{-1} \mathrm{t}\right)^{2}+\left(\sin ^{-1} \mathrm{t}\right)^{2}$ where $\mathrm{t} \in\left[0, \frac{1}{\sqrt{2}}\right]$ and $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ for $-1 \leq x \leq 1$ and $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ for all $x \in R$.
7. The interval in which $A$ lies is
(a). $\left[\frac{\pi^{3}}{7}, \frac{\pi^{3}}{2}\right]$
(b). $\left(\frac{\pi^{3}}{40}, \frac{\pi^{3}}{10}\right)$
(c). $\left[\frac{\pi^{3}}{32}, \frac{\pi^{3}}{8}\right]$
(d). None of these
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Answer: (c)8. The maximum value of $B$ is
(a). $\frac{\pi^{2}}{8}$
(b). $\frac{\pi^{2}}{16}$
(c). $\frac{\pi^{2}}{4}$
(d). None of these
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Answer: (c)9. If least value of $A$ is $m$ and maximum value of $B$ is $M$ then $\cot ^{-1} \cot \left(\frac{m-\pi M}{M}\right)=$
(a). $-\frac{7 \pi}{8}$
(b). $\frac{7 \pi}{8}$
(c). $-\frac{\pi}{8}$
(d). $\frac{\pi}{8}$
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Answer: (d)Single Integer Answer type Question
10. The number of all positive integral solutions of the equation $\tan ^{-1} \mathrm{x}+\cos ^{-1} \frac{\mathrm{y}}{\sqrt{1+\mathrm{y}^{2}}}=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$, are…………..
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Answer: $2$11. If $\cos ^{-1}\left(4 x^{3}-3 x\right)=a+b$
$\cos ^{-1} \mathrm{x}$, for $-1<\mathrm{x}<-\frac{1}{2}$ then $[\mathrm{a}+\mathrm{b}+2]$ is…………
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Answer: $-2$12. Match the statement of column I with values of column II
| Column I | Column II |
|---|---|
| (a). Absolute difference of greatest and least valu$\text { of } \sqrt{2}(\sin 2 x-\cos 2 x)$ | (p). $\frac{\pi}{4}$ |
| (b). Absolute difference of greatest and least valu$\text { of } x^{2}-4 x+3, x \in[1,3] \text { is }$ | (q). $\frac{\pi}{6}$ |
| (c). Greatest value of $\tan ^{-1} \frac{1-\mathrm{x}}{1+\mathrm{x}}, \mathrm{x} \in[0,1]$, is | (r). $4$ |
| (d). Absolute difference of greatest and leas$\text { value of } \cos ^{-1} x^{2}, x \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \text {, is }$ | (s). $1$ |
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Answer: a $\rarr$ r, b $\rarr$ s, c $\rarr$ 0, d $\rarr$ q13. If $a \leq \tan ^{-1} x+\cot ^{-1} x+\sin ^{-1} x \leq b$, then
(a). $\mathrm{a}=\frac{\pi}{4}$
(b). $\mathrm{a}=0$
(c). $\mathrm{b}=\frac{\pi}{2}$
(d). $\mathrm{b}=\pi$
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Answer: (b, d)14. $\cot ^{-1}\left(2.1^{2}\right)+\cot ^{-1}\left(2.2^{2}\right)+\cot ^{-1}\left(2.3^{2}\right)+………$ upto $\infty$ is equal to
(a). $\frac{\pi}{4}$
(b). $\frac{\pi}{3}$
(c). $\frac{\pi}{2}$
(d). $\pi$
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Answer: (a)15. Number of solutions of the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)^{(c)}+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^{2}}\right)$ is
(a). $1$
(b). $2$
(c). $3$
(d). $4$
