Question

The set of value(s) of $a$ for which $x^2+ax+{\sin }^{-1}(x^2-4x+5)+{\cos }^{-1}(x^2-4x+5)=0$ has at least one solution is

• A. $(-\infty ,\sqrt{2\pi }]\cup -[\sqrt{2\pi },\infty )$
• B. $(-\infty ,-\sqrt{2\pi })\cup (\sqrt{2\pi },\infty )$
• C. $\{2-\frac{\pi }{4}\}$
• D. $\{-2-\frac{\pi }{4}\}$

Answer 1

Answer: (D)

$x^2+ax=\{si{n}^{-1}(x^2-4x+5)+co{s}^{-1}(x^2-4x+5)\}$

$x(x+a)=\{si{n}^{-1}(x^2-4x+5)+co{s}^{-1}(x^2-4x+5)\}$

$Since,LHSi.ex(x+a)\ge 0$

$x(x+a)=0$

$x-0,-a$

$Also,-\{si{n}^{-1}(x^2-4x+5)+co{s}^{-1}(x^2-4x+5)\}=0$

$Puttingx=-a$

$Then{x}^2-4x+5becomes{a}^2+4a+5$

$si{n}^{-1}(a^2+4a+5)+co{s}^{-1}(a^2+4a+5)=0$

$ta{n}^{-1}(a^2+4a+5)=-1$

$a^2+4a+5=tan(-1)$

$a^2+4a+5=-\frac{\pi }{4}$

$Subtracting5frombothsides$

$a^2+4a+5-5=-\frac{\pi }{4}-5$

$a=\frac{-4\pm \sqrt{4-\pi }}{2}$

$a=-2\pm i\frac{\sqrt{4+\pi }}{2}$

Thus, the solution is -2