f(x)=x5+2x3+3x+1f′(x)=5x4+6x2+3f′(1)=5+6+3=14g(f(x))=xg′(f(x))f′(x)=1 for f(x)=7⇒x5+2x3+3x+1=7⇒x=1g′(7)f′(1)=1⇒g′(7)=f′(1)1=141 x=1,f(x)=7⇒g(7)=1g′(7)g(7)=1/141=14
Let f(x)=x^5+2 x^3+3 x+1, x ∈ R, and g(x) be a function such that g(f(x))=x for all x ∈ R. Then g(7) g^prime(7) is equal to :
JEE Main 2024 — Mathematics Algebra
Let f(x)=x5+2x3+3x+1,x∈R, and g(x) be a function such that g(f(x))=x for all x∈R. Then g′(7)g(7) is equal to :
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
14
- B
42
- C
7
- D
1
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