Definite Integration
General
Grade 12
Question:
If $f(x) = \begin{vmatrix} \cos x & e^{x^2} & 2x \cos^2 x/2 \\ x^2 & \sec x & \sin x + x^3 \\ 1 & 2 & x + \tan x \end{vmatrix}$, then the value of $\int_{-\pi/2}^{\pi/2} (x^2 + 1)\{f(x) + f''(x)\}dx$ is
Step-by-Step Solution
Key Concept: General
As, $f(x) = \begin{vmatrix} \cos x & e^{x^2} & 2x \cos^2 x/2 \\ x^2 & \sec x & \sin x + x^3 \\ 1 & 2 & x + \tan x \end{vmatrix}$<br>$\Rightarrow f(-x) = -f(x) \Rightarrow f(x)$ is odd<br>$\Rightarrow f'(x)$ is even $\Rightarrow f''(x)$ is odd<br>Thus, $f(x) + f''(x)$ is odd function let,<br>$\phi(x) = (x^2 + 1)\{f(x) + f''(x)\} \Rightarrow \phi(-x) = -\phi(x)$<br>i.e. $\phi(x)$ is odd<br>So, $\int_{-\pi/2}^{\pi/2} (x^2 + 1)\{f(x) + f''(x)\}dx = 0$
Correct Answer: D