If $f (a) = a - 2$ and $F (a, b) = b^{2} + a$ , then $F [3, f (4)]$ is :

a^{2} - 4 a + 7

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Solution

The correct option is E 7
given : - $F (q) = a - 2$
$F (a, b) = b^{2} + a$
Find : - $F (3, f (4)) = ?$
solve : -
$F (q) = a - 2$
$F (4) = 4 - 2 = 0$
$F (a, b) = b^{2} + a$
$F [3, f (4)] = b^{2} + 3$
$F (3, 2) = 2^{2} + 3$
$4 + 3$ $= 7$

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