Question

The value 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. $-2$
• B. $-2+\pi$
• C. $-\pi 4$
• D. $-2-\pi 4$

Answer 1

The correct option is A $-2$

Solving the equation: $x^2+ax+{sin}^{-1}(x^2-4x+5)+{cos}^{-1}(x^2-4x+5)=0$

$x^2+ax=-\left\{{sin}^{-1}\right(x^2-4x+5)+{cos}^{-1}(x^2-4x+5\left)\right\}$

$x(x+a)=-\left\{{sin}^{-1}\right(x^2-4x+5)+{cos}^{-1}(x^2-4x+5\left)\right\}$

Since L.H.S i.e., $x(x+a)\ge 0,$

and R.H.S i.e., $-\left\{{sin}^{-1}\right(x^2-4x+5)+{cos}^{-1}(x^2-4x+5\left)\right\}<0$

$⟹x(x+a)=0$

$⟹x=0,-a$

Also,$-\left\{{sin}^{-1}\right(x^2-4x+5)+{cos}^{-1}(x^2-4x+5\left)\right\}=0$

putting $x=-a,$ then $(x^2-4x+5)$ becomes $(a^2+4a+5)$

$⟹\left\{{sin}^{-1}\right(a^2+4a+5)+{cos}^{-1}(a^2+4a+5\left)\right\}=0$

$⟹{sin}^{-1}(a^2+4a+5)=-{cos}^{-1}(a^2+4a+5)$

$⟹{tan}^{-1}(a^2+4a+5)=-1$

$⟹(a^2+4a+5)=tan(-1)$

$⟹a^2+4a+5=\frac{-\Pi }{4}$

Subtracting 5 from both the sides, we get;

$⟹a^2+4a+5-5=\frac{-\Pi }{4}-5$

$⟹a^2+4a=\frac{-\Pi }{4}-5$

$⟹a^2+4a+(\frac{\Pi }{4}+5)=0$

Now solving the quadratic equation ,

$⟹a=\frac{-4\pm \sqrt{-4-\Pi }}{2}$

$⟹a=\frac{-4\pm i\sqrt{4+\Pi }}{2}$

$⟹a=-2\pm i\frac{\sqrt{4+\Pi }}{2}$

So, the given equation has atleast one solution for $a=-2$