Question

# Let f:→R→R,g:R→R and h:R→R be differentiable functions such that f(x)=x3+3x+2,g(f(x))=x and h(g(g(x)))=x for all x ε R. Then

A

g(2)=115

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B

h(1)=666

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C

h(0)=16

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D

h(g(3))=36

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Solution

## The correct options are B h′(1)=666 C h(0)=16 f(x)=x3+3x+2⇒f′(x)=3x2+3Also f(0)=2,f(1)=6,f(2)=16,f(3)=38,f(6)=236And g(f(x))=x⇒g(2)=0,g(6)=1,g(16)=2,g(38)=3,g(236)=6 (a)g(f(x))=x⇒g′(f(x)).f′(x)=1For g′(2),f(x)=2⇒x=0 ∴ Putting x=0, we get g′(f(0))f′(0)=1 ⇒g′(2)=13 (b) h(g(g(x)))=x⇒h′(g(g(x))).g′(g(x)).g′(x)=1For h′(1) we need g (g(x)) = 1 ∴g(x)=6⇒x=236∴ Putting x = 236, we get h′[g(g(236))]=1g′(g(236)).g′(236)⇒h′(g(6))=1g′(6).g′(236)⇒h′(1)=1g′(f(1)).g′(f(6))=f′(1).f′(6)=6×111=666 (c) h[g(g(x))]=xFor h(0),g(g(x))=0⇒g(x)=2⇒x=16 ∴ Putting x = 16, we get h(g(g(16)))=16⇒h(0)=16 (d)h[g(g(x))]=xFor h(g(3)),we need g(x)=3⇒x=38 ∴ Putting x = 38, we get h[g(g(38))]=38⇒h(g(3))=38

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