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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Functions

BITSAT 2024 · Paper 2 · Mathematics · Functions

If f: R arrow R, g: R arrow R are defined by f(x)=5 x-3, g(x)=x^2+3...

If f:R→R,g:R→Rf: R \rightarrow R, g: R \rightarrow \mathrm{R} are defined by f(x)=5x−3,g(x)=x2+3f(x)=5 x-3, g(x)=x^2+3, then gof−1(3)g o f^{-1}(3) is equal to

Options

  1. A253\frac{25}{3}
  2. B11125\frac{111}{25}
  3. C925\frac{9}{25}
  4. D25111\frac{25}{111}

Correct answer

B. 11125\frac{111}{25}

Step-by-step solution

Given, f(x)=5x−3\quad f(x)=5 x-3 and g(x)=x2+3g(x)=x^2+3 Let, y=f(x),∴y=5x−3y+3=5x⇒x=y+35∴f−1(y)=y+35⇒f−1(x)=x+35y=f(x), \therefore y=5 x-3y+3=5 x \Rightarrow x=\frac{y+3}{5}\therefore \quad f^{-1}(y)=\frac{y+3}{5} \Rightarrow f^{-1}(x)=\frac{x+3}{5} Now, g(x)=x2+3\quad g(x)=x^2+3 So, g∘f−1(3)=g[f−1(3)]=g(3+35)=g(65)=(6)2(5)2+3=3625+3=11125g {}^{\circ} f^{-1}(3)=g\left[f^{-1}(3)\right]=g\left(\frac{3+3}{5}\right)=g\left(\frac{6}{5}\right)=\frac{(6)^2}{(5)^2}+3=\frac{36}{25}+3=\frac{111}{25}
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