Question

The value of $a$ for which the equation $a^2+2a+{\csc }^2\pi (a+x)=0$ has a solution, is/are

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

$S=a^2+2a+cose{c}^2\pi (x+a)=0$

$cose{c}^2\pi (x+a)⩾1$

$cose{c}^2\pi (x+a)=1+K⩾1,K⩾0$

$S=a^2+2a+(1+K),K⩾0$

$S=(a+1{)}^2+K=0$

$K=0\Rightarrow a=-1$ ___ (1)

& $K=0\Rightarrow cose{c}^2\pi (x+a)=1+K=0$

$\Rightarrow cose{c}^2\pi (x+a)=1$

$\Rightarrow x+a=(2n+1)\pi /2$

$a=(2n+1)\pi /2-x$

$a=(2n+1)\pi /2$ ___ (2)

If $x=0$

& from $\left(1\right),a=-1$

so, no soln. exists for $K=0$

$K>0$

for $K>0$

$S=(a+1{)}^2+K=0$

$\Rightarrow (a+1{)}^2=-K$

$\Rightarrow a+1=\pm \sqrt{-K}$

$=\pm i\sqrt{K}$

$\Rightarrow$ Imaginary soln.

$\therefore$ No real soln. exists

if & only if $x=0$

if $x\ne 0$, a has 1 soln. i.e. $a=-1$