Question

Assertion :Consider $f\left(x\right)=si{n}^{-1}\left(sec\right(ta{n}^{-1}x)+co{s}^{-1}(cosec\left(co{t}^{-1}x\right)$
Statement-1: Domain of f(x) is a singleton. Reason: Statement-2: Range of the function $f\left(x\right)$ is a singleton.

• A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1.
• C. Statement-1 is true, Statement-2 is false.
• D. Statement-1 is false, Statement-2 is true.

Answer 1

The correct option is B Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1.

$f\left(x\right)=si{n}^{-1}\left(sec\right(ta{n}^{-1}\left(x\right)\left)\right)+co{s}^{-1}\left(cosec\right(co{t}^{-1}x\left)\right)$

$f\left(x\right)=si{n}^{-1}\left(\sqrt{ta{n}^2\left(ta{n}^{-1}\right(x\left)\right)+1}\right)+co{s}^{-1}\left(\sqrt{co{t}^2\left(co{t}^{-1}\right(x\left)\right)+1}\right)$

$f\left(x\right)=si{n}^{-1}\left(\sqrt{x^2+1}\right)+co{s}^{-1}\left(\sqrt{x^2+1}\right)$

Hence $-1\le \sqrt{x^2+1}\le 1$
$x^2+1\le 1$
$x^2\le 0$
Hence domain is singleton set {0}.
Therefore range will be singleton set $\frac{\pi }{2}$

Both the assertion and reason are correct, however, the reason is not the correct explanation for the assertion.