Home
Class 12
MATHS
If f(x) =|{:(cosx,1,0),(1,2 cosx,1),(0,1...

If `f(x) =|{:(cosx,1,0),(1,2 cosx,1),(0,1,2 cosx):}|, "then" |overset(pi//2)underset(0)intf(x)dx|` is equal to

A

`1//4`

B

`1//3`

C

`1//2`

D

1

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to evaluate the integral of the determinant function \( f(x) \) defined as: \[ f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right| \] ### Step 1: Calculate the Determinant We will compute the determinant \( f(x) \): \[ f(x) = \cos x \cdot \left| \begin{array}{cc} 2 \cos x & 1 \\ 1 & 2 \cos x \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 1 & 1 \\ 0 & 2 \cos x \end{array} \right| + 0 \cdot \left| \begin{array}{cc} 1 & 2 \cos x \\ 0 & 1 \end{array} \right| \] Calculating the 2x2 determinants: 1. For the first determinant: \[ \left| \begin{array}{cc} 2 \cos x & 1 \\ 1 & 2 \cos x \end{array} \right| = (2 \cos x)(2 \cos x) - (1)(1) = 4 \cos^2 x - 1 \] 2. For the second determinant: \[ \left| \begin{array}{cc} 1 & 1 \\ 0 & 2 \cos x \end{array} \right| = (1)(2 \cos x) - (1)(0) = 2 \cos x \] Now substituting back into \( f(x) \): \[ f(x) = \cos x (4 \cos^2 x - 1) - 2 \cos x \] \[ = 4 \cos^3 x - \cos x - 2 \cos x \] \[ = 4 \cos^3 x - 3 \cos x \] ### Step 2: Recognize the Trigonometric Identity Notice that \( 4 \cos^3 x - 3 \cos x \) can be rewritten using the cosine triple angle identity: \[ f(x) = \cos(3x) \] ### Step 3: Set Up the Integral We need to evaluate the integral: \[ \int_0^{\frac{\pi}{2}} f(x) \, dx = \int_0^{\frac{\pi}{2}} \cos(3x) \, dx \] ### Step 4: Evaluate the Integral The integral of \( \cos(3x) \) is: \[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C \] Now, we evaluate it from 0 to \( \frac{\pi}{2} \): \[ \int_0^{\frac{\pi}{2}} \cos(3x) \, dx = \left[ \frac{1}{3} \sin(3x) \right]_0^{\frac{\pi}{2}} = \frac{1}{3} \sin\left(3 \cdot \frac{\pi}{2}\right) - \frac{1}{3} \sin(0) \] Calculating the sine values: \[ = \frac{1}{3} \sin\left(\frac{3\pi}{2}\right) - 0 = \frac{1}{3} \cdot (-1) = -\frac{1}{3} \] ### Step 5: Take the Absolute Value Since the problem asks for the absolute value: \[ \left| \int_0^{\frac{\pi}{2}} f(x) \, dx \right| = \left| -\frac{1}{3} \right| = \frac{1}{3} \] ### Final Answer Thus, the final answer is: \[ \frac{1}{3} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

Let f(x) = |{:(cos x ,1,0 ),(1,2cosx,1),(0,1,2cosx):}| then

overset(pi)underset(0)int (1)/(1+3^(cosx)) dx is equal to

int_(0)^(pi)|1+2cosx| dx is equal to :

int_(0)^(pi)(x.sin^(2)x.cosx)dx is equal to

lim_(x->0) (Cosx -1)/x is equal to

If f(x)=|(0,x^2-sinx,cosx-2),(sinx-x^2,0,1-2x),(2-cosx,2x-1,0)|, then int f(x) dx is equal to

int_0^(pi/2) cosx/(1+sinx)dx

(lim)_(x->0)(cos2x-1)/(cosx-1)

If f(x) =|{:(sinx,cosx,sinx),(cosx,-sinx,cosx),(x,1,1):}| find the value of 2[f'(0)]+[f'(1)]^(2)

If A=int_0^pi cosx/(x+2)^2 \ dx , then int_0^(pi//2) (sin 2x)/(x+1) \ dx is equal to