Question

The complete range of values of ${}^{\prime }{a}^{\prime }$ such that ${\left(\frac{1}{2}\right)}^{|x|}=x^2-a$ is satisfied for maximum number of values of $x$ is

• A. $(-\infty ,-1)$
• B. $(-\infty ,\infty )$
• C. $(-1,1)$
• D. $(-1,\infty )$

Answer 1

Answer: (D)

$(-1,\infty )$

Clearly

$0\le {\left(\frac{1}{2}\right)}^{|x|}\le 1$

${\left(\frac{1}{2}\right)}^{|x|}$ is a decreasing function tending to 0 as x tends to $\infty$

Hence

we can check at boundary points to get values of a.

When x tends to 0 function tends to 1

So $x^2-a=1$

$\Rightarrow 0-a=1$

$\Rightarrow a=-1$

Similarly when x tends to $\infty$ function tends to 0

So

$x^2-a=0$

$\Rightarrow x^2=a$

Hence a tends to infinity.

Hence a $\in (-1,\infty )$