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Functions question

2023 · 13 Apr · Shift 1 · Q34
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  5. /2023 · 13 Apr · Shift 1 · Q34

Functions question

2023 · 13 Apr · Shift 1 · Q34

JEE MainMathematicsFunctionsMCQ+4 / −1
For x∈Rx \in \mathbb{R}x∈R, two real valued functions f(x)f(x)f(x) and g(x)g(x)g(x) are such that, g(x)=x+1g(x)=\sqrt{x}+1g(x)=x​+1 and f∘g(x)=x+3−xf \circ g(x)=x+3-\sqrt{x}f∘g(x)=x+3−x​. Then f(0)f(0)f(0) is equal to
  1. A
    5
  2. B
    0
  3. C
    −-− 3
  4. D
    1
View written solutionFree

Correct answer: A

  1. We are given g(x)=x+1g(x)=\sqrt{x}+1g(x)=x​+1 and f∘g(x)=f(g(x))=x+3−x.f\circ g(x)=f(g(x))=x+3-\sqrt{x}.f∘g(x)=f(g(x))=x+3−x​.

  2. Substitute g(x)g(x)g(x) into the composition: f(x+1)=x+3−x.f(\sqrt{x}+1)=x+3-\sqrt{x}.f(x​+1)=x+3−x​.

  3. Let t=x+1.t=\sqrt{x}+1.t=x​+1. Then x=t−1\sqrt{x}=t-1x​=t−1 and hence x=(t−1)2.x=(t-1)^2.x=(t−1)2.

  4. Rewrite the right-hand side in terms of ttt: f(t)=x+3−x=(t−1)2+3−(t−1).f(t)=x+3-\sqrt{x}=(t-1)^2+3-(t-1).f(t)=x+3−x​=(t−1)2+3−(t−1).

  5. Simplify: f(t)=t2−2t+1+3−t+1=t2−3t+5.f(t)=t^2-2t+1+3-t+1=t^2-3t+5.f(t)=t2−2t+1+3−t+1=t2−3t+5.

  6. Therefore, f(y)=y2−3y+5f(y)=y^2-3y+5f(y)=y2−3y+5 for relevant real yyy in the range of ggg.

  7. Now compute f(0)f(0)f(0): f(0)=02−3(0)+5=5.f(0)=0^2-3(0)+5=5.f(0)=02−3(0)+5=5.

  8. Hence the correct option is A.\boxed{A}.A​.

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