JEE Main 2026 28 January Morning Shift
Exam JEE Main
Year 2026
Date 28 January
Shift Morning Shift
Subject Mathematics
Chapter Functions JEE Main 2026 · 28 January · Morning Shift · Mathematics · Functions
If g(x)=3 x^ 2 +2 x-3, f(0)=-3 and 4 g(f(x))=3 x^ 2 -32 x+72 , then...
If
g ( x ) = 3 x 2 + 2 x − 3 , f ( 0 ) = − 3 g(x)=3 x^{2}+2 x-3, f(0)=-3 g ( x ) = 3 x 2 + 2 x − 3 , f ( 0 ) = − 3 and
4 g ( f ( x ) ) = 3 x 2 − 32 x + 72 4 g(f(x))=3 x^{2}-32 x+72 4 g ( f ( x )) = 3 x 2 − 32 x + 72 , then
f ( g ( 2 ) ) f(g(2)) f ( g ( 2 )) is equal to:
Options A − 7 2 -\frac{7}{2} − 2 7 B − 25 6 -\frac{25}{6} − 6 25 C 7 2 \frac{7}{2} 2 7 D 25 6 \frac{25}{6} 6 25 Correct answer C. 7 2 \frac{7}{2} 2 7
Step-by-step solution
Let
f ( x ) = a x + b f(x) = ax + b f ( x ) = a x + b . Then
4 g ( f ( x ) ) = 12 ( a x + b ) 2 + 8 ( a x + b ) − 12 4g(f(x)) = 12(ax+b)^2 + 8(ax+b) - 12 4 g ( f ( x )) = 12 ( a x + b ) 2 + 8 ( a x + b ) − 12 .
Comparing with
3 x 2 − 32 x + 72 3x^2 - 32x + 72 3 x 2 − 32 x + 72 :
x 2 x^2 x 2 :
12 a 2 = 3 ⇒ a = ± 1 2 12a^2 = 3 \Rightarrow a = \pm\frac{1}{2} 12 a 2 = 3 ⇒ a = ± 2 1 Constant:
12 b 2 + 8 b − 12 = 72 ⇒ 3 b 2 + 2 b − 21 = 0 ⇒ b = 7 3 12b^2 + 8b - 12 = 72 \Rightarrow 3b^2 + 2b - 21 = 0 \Rightarrow b = \frac{7}{3} 12 b 2 + 8 b − 12 = 72 ⇒ 3 b 2 + 2 b − 21 = 0 ⇒ b = 3 7 or
b = − 3 b = -3 b = − 3 . Since
f ( 0 ) = − 3 f(0) = -3 f ( 0 ) = − 3 ,
b = − 3 b = -3 b = − 3 .
x x x :
8 a ( 3 ( − 3 ) + 1 ) = − 32 ⇒ − 8 a = − 4 ⇒ a = 1 2 8a(3(-3)+1) = -32 \Rightarrow -8a = -4 \Rightarrow a = \frac{1}{2} 8 a ( 3 ( − 3 ) + 1 ) = − 32 ⇒ − 8 a = − 4 ⇒ a = 2 1 .
So
f ( x ) = x 2 − 3 f(x) = \frac{x}{2} - 3 f ( x ) = 2 x − 3 .
g ( 2 ) = 3 ( 4 ) + 4 − 3 = 13 g(2) = 3(4) + 4 - 3 = 13 g ( 2 ) = 3 ( 4 ) + 4 − 3 = 13 .
f ( g ( 2 ) ) = f ( 13 ) = 13 2 − 3 = 7 2 f(g(2)) = f(13) = \frac{13}{2} - 3 = \frac{7}{2} f ( g ( 2 )) = f ( 13 ) = 2 13 − 3 = 2 7 .
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