Mathematics · Functions

JEE Main 2024 — 1 February, Shift 1 — Question 11

Let f:R→R\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R} and g:R→R\mathrm{g}: \mathbf{R} \rightarrow \mathbf{R} be defined

f(x)={log⁡e            ,x>0e−x           ,x≤0andg(x)={x             ,x≥0ex           ,x<0.Then, gof :R→R  is:\begin{array}{l}f(x) = \left\{ \begin{array}{l}{\log _e}\,\,\,\,\,\,\,\,\,\,\,\,,x > 0\\e - x\,\,\,\,\,\,\,\,\,\,\,,x \le 0\end{array} \right.and\\g(x) = \left\{ \begin{array}{l}x\,\,\,\,\,\,\,\,\,\,\,\,\,,x \ge 0\\{e^x}\,\,\,\,\,\,\,\,\,\,\,,x < 0\end{array} \right..Then,\,gof\,:R \to R\,\,is:\end{array}
  1. Option A:

    one-one but not onto

  2. Option B:

    neither one-one nor onto

    Correct
  3. Option C:

    onto but not one-one

  4. Option D:

    both one-one and onto

Answer: B

Step-by-step solution

g(f(x))={log⁡, f(x)≥0ef(x), f(x)<0g(x)={e−x,(−∞,0]ein x(0,1)In x,[1,∞)\begin{array}{l}g(f(x)) = \left\{ \begin{array}{l}\log ,\,f(x) \ge 0\\{e^{f(x)}},\,f(x) < 0\end{array} \right.\\g(x) = \left\{ \begin{array}{l}{e^{ - x}},\left( { - \infty ,0} \right]\\{e^{in\,x}}(0,1)\\In\,x,\left[ {1,\infty } \right)\end{array} \right.\end{array}

figure

Graph of g(f(x))

G(f(x))Many one into

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Functions
Topic
One-One, many-one, onto, into, bijective functions