Maths · Derivatives of order upto two
Assertion(A): Let \) be twice differentiable function such that =-\boldsymbol{f}
Assertion(A): Let \( \boldsymbol{f}(\boldsymbol{x}) \) be twice differentiable function such that \( \boldsymbol{f}^{\prime \prime}(\boldsymbol{x})=-\boldsymbol{f}(\boldsymbol{x}) \) and \( \boldsymbol{f}^{\prime}(\boldsymbol{x})=\boldsymbol{g}(\boldsymbol{x}) . \) If \( \boldsymbol{h}(\boldsymbol{x})=[\boldsymbol{f}(\boldsymbol{x})]^{2}+[\boldsymbol{g}(\boldsymbol{x})]^{2} \) and \( \boldsymbol{h}(\mathbf{1})=\mathbf{8} \) \( \operatorname{then} h(2)=8 \) Reason (R): Derivative of a constant function is zero.
- A. Both A and R are true R is correct reason of A
- B. Both A and R are true R is not correct reason of A
- C. A is true but R is false
- D. A is false but R is true
Step-by-step solution
We are given f''(x) = -f(x) and g(x) = f'(x). Define h(x) = [f(x)]^2 + [g(x)]^2. Then h'(x) = 2f(x)f'(x) + 2g(x)g'(x) = 2f(x)g(x) + 2g(x)f''(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 0. Thus h(x) is constant. Since h(1) = 8, we have h(2) = 8. The assertion is true. The reason states that the derivative of a constant function is zero, which is the fundamental fact used to conclude that h is constant (since h'(x)=0). Hence reason is true and correctly explains the assertion.