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∈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 Bh′(1)=666 Ch(0)=16 f(x)=x3+3x+2 f′(x)=3x2+3 Now, we have g(f(x))=x⇒g′(f(x)).f′(x)=1 We need to find g′(2)⇒f(x)=2⇒x=0 ∴g′(2).f′(0)=1⇒g′(2)=13
Now, we have h(g(g(x)))=x Replacing x by f(x) in the above expression: ⇒h(g(g(f(x))))=f(x)⇒h(g(x))=f(x)...[1] Replacing x by f(x) in the above expression: ⇒h(g(f(x)))=f(f(x))⇒h(x)=f(f(x))⇒h′(x)=f′(f(x))⋅f′(x)