Question

The value ${}^{\prime }{a}^{\prime }$ for which the equation $4{cosec}^2\left(\pi (a+x)\right)+a^2-4a=0$ has a real solution is:

• A. $a=1$
• B. $a=2$
• C. $a=3$
• D. none of these

Answer 1

Answer: (B)

$a=2$

$4{\csc }^2\left(\pi (a+x)\right)+a^2-4a=0$

$[4{\csc }^2\left(\pi (a+x)\right)-4]+(a^2-4a+4)=0$

$4({\csc }^2\left(\pi (a+x)\right)-1)+{(a-2)}^2=0$

$\left(2\cot {\left(\left(\pi (a+x)\right)\right)}^2\right)+{(a-2)}^2=0$

$\Rightarrow {(a-2)}^2=0\Rightarrow a-2=0\Rightarrow a=2$