Let `f(x)=|(cos x,e^(x^2) , 2x cos^2x/2),(x^2, sec x, sinx+x^3),(1,2,x+tan x)|` then the value of `int_(-pi/2)^(pi/2)(x^2+1)(f(x)+f''(x))dx `

Options

A. (a)1
B. (b)`-1`
C. (c)2
D. (d)None of these

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^2 + 1)(f(x) + f''(x)) \, dx, \]

where

\[ f(x) = \begin{pmatrix} \cos x & e^{x^2} & \frac{2x \cos^2 \frac{x}{2}}{x^2} \\ x^2 & \sec x & \sin x + x^3 \\ 1 & 2 & x + \tan x \end{pmatrix}, \]

we will analyze the properties of the function \( f(x) \) and its derivatives.

### Step 1: Determine \( f(-x) \)

First, we compute \( f(-x) \):

\[ f(-x) = \begin{pmatrix} \cos(-x) & e^{(-x)^2} & \frac{2(-x) \cos^2 \frac{-x}{2}}{(-x)^2} \\ (-x)^2 & \sec(-x) & \sin(-x) + (-x)^3 \\ 1 & 2 & -x + \tan(-x) \end{pmatrix}. \]

Using the properties of trigonometric and exponential functions, we have:

- \( \cos(-x) = \cos x \)
- \( e^{(-x)^2} = e^{x^2} \)
- \( \sec(-x) = \sec x \)
- \( \sin(-x) = -\sin x \)
- \( \tan(-x) = -\tan x \)

Thus, we can simplify:

\[ f(-x) = \begin{pmatrix} \cos x & e^{x^2} & -\frac{2x \cos^2 \frac{x}{2}}{x^2} \\ x^2 & \sec x & -\sin x - x^3 \\ 1 & 2 & -x - \tan x \end{pmatrix}. \]

### Step 2: Determine \( f''(x) \)

Next, we find \( f''(x) \). The second derivative \( f''(x) \) will have similar properties, and we can analyze it in a similar manner as \( f(x) \).

### Step 3: Analyze \( f(x) + f''(x) \)

Now, we want to analyze the expression \( f(x) + f''(x) \) under the integral.

### Step 4: Check symmetry

We check if \( f(-x) + f''(-x) \) has any symmetry:

\[ g(x) = (x^2 + 1)(f(x) + f''(x)). \]

Now, we compute \( g(-x) \):

\[ g(-x) = ((-x)^2 + 1)(f(-x) + f''(-x)) = (x^2 + 1)(f(-x) + f''(-x)). \]

Using the properties of \( f(-x) \) and \( f''(-x) \), we find that:

\[ g(-x) = (x^2 + 1)(-f(x) - f''(x)). \]

### Step 5: Integral evaluation

Now, we can conclude that:

\[ g(-x) = -g(x). \]

This means that \( g(x) \) is an odd function. The integral of an odd function over a symmetric interval around zero is zero:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} g(x) \, dx = 0. \]

Thus, the value of the integral is:

\[ \boxed{0}. \]