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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Sets Relations and Functions

For

x \in R

, two real valued functions

f (x)

and

g (x)

are such that

g (x) = x + 1

and

f \circ g (x) = x + 3 - x

. Then

f (0)

is equal to

(A)1
(B)-3
(C)5✓ Correct
(D)0

Answer: (C)

Solution

Put

t = g (x) = x + 1

, so that

x = t - 1

and

x = (t - 1)^{2}

. Substituting into

f (g (x)) = x + 3 - x

gives

f (t) = (t - 1)^{2} + 3 - (t - 1) = t^{2} - 3 t + 5

. Setting

t = 0

,

f (0) = 0 - 0 + 5 = 5

.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option

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