If f(x) = 2x + 1 and g(x) = x² - 2, find (f ° g)(x – Class 12 Mathematics | ConceptFirst

Exercise: Composite Functions · Q20

f(x) = 2x + 1

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✓ Free question

$(f\circ g)(x) = f(g(x)) = f(x^2-2) = 2(x^2-2)+1 = 2x^2-4+1 = 2x^2-3$ .

$(g\circ f)(x) = g(f(x)) = g(2x+1) = (2x+1)^2-2 = (4x^2+4x+1)-2 = 4x^2+4x-1$ .

Comparing, $2x^2-3 \neq 4x^2+4x-1$ (their coefficients of $x^2$ and $x$ differ), confirming $(f\circ g)(x)\neq(g\circ f)(x)$ .

✓Final answer

(f\circ g)(x)=2x^2-3

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$\mathbb{R}$
(a) $9$
(b) $-9$
(c) $6$
(d) $-6$
›Reveal solutionSolution

Composite $(g\circ f)(-3)=g(f(-3))$ : first $f(-3)=-6$ , then $g(-6)=-9$ .

Evaluating a composition of functions is a CBSE/NCERT Class 12 relations and functions basic.

First apply $f$ :

$f(-3) = |-3| - (-3)^2 = 3 - 9 = -6.$

Then apply $g$ to this result:

$g(-6) = 2(-6) + 3 = -12 + 3 = -9.$

So $(g\circ f)(-3) = -9$ .

✓Final answer
$(g\circ f)(-3) = -9$
CBSE 2022Set ANNUAL1 markMCQ
$Q. If the functions f : R \to R and g : R \to R defined as f(x) = 3x + 2 and g(x) = 2x - 3, [R = set of Real numbers] then (g\circf)(x) is: OR The value of tan(π/2 - tan⁻¹ 1/3) is equal to:$
(a) 6x - 7
(b) 6x + 1
(c) 3x + 5
(d) 6x + 4
›Reveal solutionSolution

$(g\circ f)(x)$ means apply $f$ first, then feed the result into $g$ — substitute $f(x)$ into $g$ .

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

Compute $(g\circ f)(x) = g(f(x))$ :

$g(f(x)) = 2\cdot f(x) - 3 = 2(3x+2) - 3$

$= 6x+4-3$

$= 6x+1$

This matches option (b).

✓Final answer
$(g\circ f)(x) = 6x+1$

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