Question

The minimum value of $2^{\left(x^{2}-3\right)^{3}}+27$ is

• A. $1$
• B. $2$
• C. $ {{2}^{27}} $
• D. None of these

Answer 1

Answer: (A)

$2^{\left(x^{2}-3\right)^{3+27}}$ is minimum when $\left(x^{2}-3\right)^{3}+27$ is minimum Let $y=\left(x^{2}-3\right)^{3}+27$
$=x^{6}-27-9 x^{4}+27 x^{2}+27$
$ \Rightarrow y=x^{6}-9 x^{4}+27\, x^{2} $
$\Rightarrow y=x^{2}\left(x^{4}-9 \,x^{2}+27\right) $
$\Rightarrow y=x^{2}\left[x^{4}-9\, x^{2}+\frac{81}{4}-\frac{81}{4}+27\right] $
$\Rightarrow y=x^{2}\left[\left(x^{4}-9\, x^{2}+\frac{81}{4}\right)+\frac{27}{4}\right] $
$\Rightarrow y=x^{2}\left[\left(x^{2}-\frac{9}{2}\right)^{2}+\frac{27}{2}\right] \geq 0 \forall x $
$\therefore $ Minimum value of $\left(x^{2}-3\right)^{3}+27$ is 0.
Hence, minimum value of $2^{\left(x^{2}-3\right)^{3}+27}$
$=2^{0}=1 $