The largest value of $\min (2 + x^{2} , 6 - 3x)$ when $x > 0$ is

For minimum,

equating

$2 + x2 = 6 - 3x$

$x2 + 3x - 4 = 0$

$x = 1, -4$

Since $x > 0,$ so value occurs at $x = 1.$

At $x = 1$

$2+x^{2}=3$

$6 - 3x = 3.$

it means the largest value of the function $min( 2 + x^2 , 6 − 3x)$

$min( 3, 3)$ is $3$

The correct option is C.

PUT THE DIFFERENT VALUE OF X

Aptitude Overflow