MCQMediumJEE 2023Composition & Inverse Functions

JEE Mathematics 2023 Question with Solution

For xRx \in \mathbb{R}, two real valued functions f(x)f(x) and g(x)g(x) are such that,

g(x)=x+1andfg(x)=x+3x.g(x) = \sqrt{x} + 1 \quad \text{and} \quad f \circ g(x) = x + 3 - \sqrt{x}.

Then f(0)f(0) is equal to:

  • A

    55

  • B

    00

  • C

    3-3

  • D

    11

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: g(x)=x+1g(x) = \sqrt{x} + 1 and fg(x)=x+3xf \circ g(x) = x + 3 - \sqrt{x}.

Find: f(0)f(0).

Let

u=g(x)=x+1u = g(x) = \sqrt{x} + 1

Then

x=u1andx=(u1)2\sqrt{x} = u - 1 \quad \text{and} \quad x = (u-1)^2

Now

f(g(x))=x+3xf(g(x)) = x + 3 - \sqrt{x}

becomes

f(u)=(u1)2+3(u1)f(u) = (u-1)^2 + 3 - (u-1)

Simplifying,

f(u)=u22u+1+3u+1=u23u+5\begin{aligned} f(u) &= u^2 - 2u + 1 + 3 - u + 1 \\ &= u^2 - 3u + 5 \end{aligned}

Hence,

f(x)=x23x+5f(x) = x^2 - 3x + 5

Therefore,

f(0)=0230+5=5f(0) = 0^2 - 3 \cdot 0 + 5 = 5

So, the correct option is A.

Verification from the given composition

Given: g(x)=x+1g(x) = \sqrt{x} + 1.

Find: a function ff consistent with the composition.

Take

f(u)=u23u+5f(u) = u^2 - 3u + 5

Then substituting u=g(x)=x+1u = g(x) = \sqrt{x} + 1,

f(g(x))=(x+1)23(x+1)+5f(g(x)) = (\sqrt{x}+1)^2 - 3(\sqrt{x}+1) + 5

Now expand:

f(g(x))=x+2x+13x3+5=xx+3\begin{aligned} f(g(x)) &= x + 2\sqrt{x} + 1 - 3\sqrt{x} - 3 + 5 \\ &= x - \sqrt{x} + 3 \end{aligned}

This exactly matches the given expression x+3xx + 3 - \sqrt{x}. Hence,

f(x)=x23x+5f(x) = x^2 - 3x + 5

and therefore

f(0)=5f(0) = 5

Common mistakes

  • Assuming directly that f(x)=x+3xf(x) = x + 3 - \sqrt{x} is incorrect because that expression is for f(g(x))f(g(x)), not for f(x)f(x) itself. First rewrite the input of ff as a new variable.

  • Using u=xu = \sqrt{x} instead of u=x+1u = \sqrt{x} + 1 leads to the wrong substitution. The full expression g(x)g(x) must be treated as the argument of ff.

  • Forgetting to express xx in terms of uu causes an incomplete derivation. After taking u=x+1u = \sqrt{x} + 1, use x=u1\sqrt{x} = u - 1 and then x=(u1)2x = (u-1)^2.

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