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

For minimum,

equating

2 + x^{2} = 6 - 3x

x^{2} + 3x - 4 = 0

x = 1, -4

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

At x = 1, 6 - 3x = 3.

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

The correct option is C.

PUT THE DIFFERENT VALUE OF X

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