Q.If $f$ and $g$ are two real functions defined as $f(x) = \dfrac{x}{x+1}$ and $g(x) = \dfrac{x}{1-x}$; $x \neq 1$, then find $f \circ g(x)$ and $(f \circ g)^{-1}(x)$.

Showing the 12 most recent of 20 on this concept.

1. CBSE 2026Set CX1 mark
   Q.If the function $f:R\to R$ and $g:R\to R$ are defined by $f(x)=\cos x$ and $g(x)=3x^{2}$ respectively, find $gof$.

   ›Reveal solutionSolution

   Substitute $f$ into $g$: $(g\circ f)(x)=3\cos^{2}x$.

   Concept: In a composition $g\circ f$, apply $f$ first, then feed the result into $g$.

   Here $f(x)=\cos x$ and $g(x)=3x^{2}$, so …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $f : R \to R$ be given by $f(x) = (3 - x^3)^{1/3}$, then $f \circ f(x)$ is equal to

   (a) $x^{1/3}$

   (b) $x^3$

   (c) $x$

   (d) $(3-x^3)$

   ›Reveal solutionSolution

   Substituting $f(x)$ into itself cancels the cube and cube-root, leaving $x$.

   Given $f(x) = (3-x^3)^{1/3}$.

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.Assertion (A): Let $f(x) = x^2$, $g(x) = \cos x$, then $fog \neq gof$. Reason (R): $(fog)(x) = f(x)\,g(x)$

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   $fog(x)=\cos^2x$ and $gof(x)=\cos(x^2)$ are genuinely different (A is true), but the Reason misstates composition as ordinary multiplication (R is false).

   With $f(x)=x^2$, $g(x)=\cos x$: $(fog)(x)=f(g(x))=(\cos x)^2=\cos^2x$, while $(gof)(x)=g(f(x))=\cos(x^2)$. These are different functions (e.g. at $x=\pi/2$, $fog=0$ but $gof=\cos(\pi^2/4)\ne0$), so Assertion (A) is TRUE.

   …

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4. CBSE 2025Set ANNUAL1 mark
   Q.If $f : \{1,3,4\} \to \{1,2,5\}$ and $g : \{1,2,5\} \to \{1,3\}$ be given by $f = \{(1,2),(3,5),(4,1)\}$ and $g = \{(1,3),(2,3),(5,1)\}$, write down $gof$.

   ›Reveal solutionSolution

   Apply $f$ first, then $g$, to each element of the domain of $f$: $(gof)(x)=g(f(x))$.

   Given $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(1,3),(2,3),(5,1)\}$.

   $(gof)(x) = g(f(x))$ for each $x$ in the domain of $f$, i.e. $x\in\{1,3,4\}$.

   • $x=1$: $f(1)=2$, then $g(2)=3$. So $(gof)(1)=3$. …

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5. CBSE 2025Set ANNUAL1 mark
   Q.If f : R → R and g : R → R are two functions defined by f(x) = 2x + 1 and g(x) = x² − 2 respectively, find (g o f)(x).

   ›Reveal solutionSolution

   $(g \circ f)(x)$ means substitute $f(x)$ into $g$, i.e. compute $g(f(x))$.

   Given: $f(x) = 2x+1$, $g(x) = x^2 - 2$

   Step 1: $(g\circ f)(x) = g(f(x)) = g(2x+1)$

   Step 2 — substitute $2x+1$ in place of $x$ in $g(x) = x^2 - 2$:

   $g(2x+1) = (2x+1)^2 - 2$

   Step 3 — expand: …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $f: \mathrm{R} \to \mathrm{R}$ and $g: \mathrm{R} \to \mathrm{R}$ are the two real functions defined by $f(x) = 3x^2 + 1$ and $g(x) = 1 - x$, then $(gof)(-2)$ is ................. .

   (a) 12

   (b) 28

   (c) –12

   (d) –28

   ›Reveal solutionSolution

   Evaluate the inner function first, then apply the outer function: $(g\circ f)(x) = g(f(x))$.

   Given $f(x) = 3x^2+1$ and $g(x) = 1-x$.

   Step 1: Find $f(-2)$.

   $f(-2) = 3(-2)^2 + 1 = 3(4) + 1 = 13$

   …

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7. CBSE 2023Set ANNUAL1 markMCQ
   Q.If $f:R \to R, f(x)=\sin x$ and $g:R \to R, g(x)=x^2$, then $(f \circ g)(x)$ is equal to:

   (a) $\sin x^2$

   (b) $\sin x$

   (c) $\sin^2 x^2$

   (d) $\sin^2 x$

   ›Reveal solutionSolution

   Composition $(f\circ g)(x)$ means apply $g$ first, then apply $f$ to that result.

   Given $f(x)=\sin x$ and $g(x)=x^2$.

   $(f\circ g)(x)=f(g(x))=f(x^2)=\sin(x^2)$

   …

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8. CBSE 2022Set FF1 mark
   Q.If $f:R \to R$ where $f(x) = \cos x$ and $g:R \to R$ where $g(x) = x^2$, then prove that $fog \neq gof$.

   ›Reveal solutionSolution

   Composition is order-sensitive: $(f\circ g)(x)=\cos(x^2)$ but $(g\circ f)(x)=\cos^2x$, and one counterexample proves $f\circ g\ne g\circ f$.

   Concept. To disprove an equality of functions it is enough to find one input where the two sides differ.

   With $f(x)=\cos x$ and $g(x)=x^2$:

   $(f\circ g)(x)=f(g(x))=f(x^2)=\cos(x^2),$

   $(g\circ f)(x)=g(f(x))=g(\cos x)=(\cos x)^2=\cos^2 x.$

   Take $x=\sqrt{\pi}$:

   $(f\circ g)(\sqrt\pi)=\cos\pi=-1,\qquad (g\circ f)(\sqrt\pi)=\cos^2\sqrt\pi\ge 0.$ …

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9. CBSE 2022Set HE2191 markMCQ
   Q.If $f(x) = 8x^3$ and $g(x) = \dfrac{1}{x^3}$, then the value of $gof$ is:

   (a) $8x^3$

   (b) $512x^3$

   (c) $\dfrac{1}{512x^9}$

   (d) $2x$

   ›Reveal solutionSolution

   Composite function $gof(x) = g(f(x))$; substitute $f(x)=8x^3$ into $g$.

   Given $f(x) = 8x^3$ and $g(x) = \dfrac{1}{x^3}$.

   …

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10. CBSE 2022Set ANNUAL1 mark
    Q.If $f(x) = 27x^3$ and $g(x) = x^{1/3}$, then $gof(x) =$ ______.

    ›Reveal solutionSolution

    Composition means apply $f$ first, then $g$ to the result; here the cube and cube-root cancel, leaving a factor of 3.

    $f(x)=27x^3$ and $g(x)=x^{1/3}$.

    …

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11. CBSE 2022Set ANNUAL1 mark
    Q.Find $g \circ f$, if $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ are given by $f(x) = 8x^3$ and $g(x) = x^{1/3}$.

    ›Reveal solutionSolution

    Substitute $f(x)$ into $g$ and simplify the cube root.

    Given $f(x) = 8x^3$ and $g(x) = x^{1/3}$. The composition is

    $(g \circ f)(x) = g\big(f(x)\big) = g\big(8x^3\big) = \big(8x^3\big)^{1/3}.$

    …

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12. CBSE 2020Set NC1 mark
    Q.Find $g \circ f$ and $f \circ g$, if $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ are given by $f(x) = \cos x$ and $g(x) = 3x^2$.

    ›Reveal solutionSolution

    By definition, $(g\circ f)(x)=g(f(x))$ and $(f\circ g)(x)=f(g(x))$ — substitute one function's rule into the other.

    Given $f(x)=\cos x$ and $g(x)=3x^2$.

    $g\circ f$:

    $(g\circ f)(x)=g(f(x))=g(\cos x)=3(\cos x)^2=3\cos^2x$

    $f\circ g$: …

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