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Question

Assertion(A): Let f(x) be twice differentiable function such that f′′(x)=f(x) and f(x)=g(x). lf h(x)=[f(x)]2+[g(x)]2 and h(1)=8, then h(2)=8


Reason (R): Derivative of a constant function is zero.

A
Both A and R are true R is correct reason of A
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B
Both A and R are true R is not correct reason of A
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C
A is true but R is false
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D
A is false but R is true
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Solution

The correct option is A Both A and R are true R is correct reason of A
Let f(x)=c
f(x)=0 So Reason is True

Now, h(x)=[f(x)]2+[g(x)]2
h(x)=2f(x)f(x)+2g(x)g(x)
=2f(x)g(x)+2g(x)f′′(x) ...since f(x)=g(x)f′′(x)=g(x)
=2g(x)[f(x)+f′′(x)]
=2g(x)(f(x)f(x))
h(x)=0
means h(x)=C (a constant function)
Given h(1)=8
h(x)=8
Hence, f(2)=8
So, both A and R are correct and R is the correct explanation of A.

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