Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column $I$ with the 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 $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$ $(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$ $(s)$ lies on $(4x^2-1)(y^2-1)=0$

  • A
    $A \rightarrow p; B \rightarrow q; C \rightarrow p; D \rightarrow s$
  • B
    $A \rightarrow q; B \rightarrow s; C \rightarrow s; D \rightarrow q$
  • C
    $A \rightarrow q; B \rightarrow r; C \rightarrow p; D \rightarrow r$
  • D
    $A \rightarrow r; B \rightarrow s; C \rightarrow q; D \rightarrow p$

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Which pair$(s)$ of function$(s)$ is/are equal? (where ${x}$ and $[x]$ denote the fractional part and integral part functions respectively.)

Let $S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1) \}$. Then $\sum_{x \in S} (2x - 1)^2$ is equal to . . . . . .

Match List $I$ with List $II$ and select the correct answer using the code given below the lists:
List $I$ List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is $2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is $3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is $4$. $1$

Codes: $P \quad Q \quad R \quad S$

Consider the statements:
$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.
$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.
Then the correct option among the following is:

For $k \in R$,let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k$,where $0 < |x| < \frac{1}{\sqrt{2}}$,be $\alpha$ and $\beta$,where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2}- bx -5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$,then $\frac{b}{k^{2}}$ is equal to $......$

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