Verified Solution • 2025–26 Session moderate

QJEEM22Ma241e: JEE Main 2022 Mathematics - Class 11 JEEM22Ma241e Solution: Let f(x) and g(x) be two real polynomials of degree 2 and 1…

Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x))=8x2−2xf(g(x)) = 8{x^2} - 2x and g(f(x))=4x2+6x+1g(f(x)) = 4{x^2} + 6x + 1 , then the value of f(2)+g(2)f(2) + g(2) is _________.

Step-by-Step Verified Solution

f(g(x))=8x2−2xf(g(x))=8 x^{2}-2 x g(f(x))=4x2+6x+1g(f(x))=4 x^{2}+6 x+1

let f(x)=cx2+dx+ef(x)=c x^{2}+d x+e

g(x)=ax+bg(x)=a x+b

f(g(x))=c(ax+b)2+d(ax+b)+e≡8x2−2xf(g(x))=c(a x+b)^{2}+d(a x+b)+e \equiv 8 x^{2}-2 x

g(f(x))=a(cx2+dx+e)+b≡4x2+6x+1g(f(x))=a\left(c x^{2}+d x+e\right)+b \equiv 4 x^{2}+6 x+1

∴ac=4ad=6ae+b=1\therefore \quad a c=4 \quad a d=6 \quad a e+b=1

a2c=82abc+ad=−2cb2+bd+e=0a^{2} c=8 \quad 2 a b c+a d=-2 \quad c b^{2}+b d+e=0

By solving

a=2b=−1a=2 \quad b=-1

c=2d=3e=1c=2 \quad d=3 \quad e=1

∴f(x)=2x2+3x+1\therefore \quad f(x)=2 x^{2}+3 x+1

g(x)=2x−1g(x)=2 x-1

f(2)+g(2)=2(2)2+3(2)+1+2(2)−1f(2)+g(2)=2(2)^{2}+3(2)+1+2(2)-1

=18=18

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