∵f′′(x)−g′′(x)=0 Integrating, f′(x)−g′(x)=c⇒f′(1)−g′(1)=c⇒4−2=c⇒c=2 ∴f′(x)−g′(x)=2; Integrating, f(x)−g(x)=2x+c1 ⇒f(2)−g(2)=4+c1⇒9−3=4+c1⇒c1=2∴f(x)−g(x)=2x+2 At x=3/2,f(x)−g(x)=3+2=5
f(x) and g(x) are two differentiable functions on [0,2] such that f^prime prime(x)-g^prime prime(x)=0 f^prime(1)=2 g^prime(1)=4 f(2)=3 g(2)=9 then…
JEE Main 2002 — Mathematics Calculus
f(x) and g(x) are two differentiable functions on [0,2] such that f′′(x)−g′′(x)=0 f′(1)=2g′(1)=4f(2)=3g(2)=9 then f(x)−g(x) at x=3/2 is
Held on 30 Apr 2002 · Verified 6 Jul 2026.
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0
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2
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10
- D
5
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