Let A=[ 1 tan x -tan⁡x 1 ] and B=ATA- 2 . If f(x)=det(a d j B) , then number of solution(s) of the equation 10f(x)-x=0 is (are)

Question

Let A=[ 1 tan x -tan⁡x 1 ] and B=ATA- 2 . If f(x)=det(a d j B) , then number of solution(s) of the equation 10f(x)-x=0 is (are) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(x)= operatornamedet( operatornameadj(B))=| operatornameadj(B)|=|B| =|AT | A-2|=|A| (1/ mid A) =(1/|. A |.)=(1/s e c2 x)=cos2x <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-fcfda706kfennwlq.jpg" /> 10cos2x-x=0

Q. Let $A = [1 - t an x t an x 1]$ and $B = A^{T} A^{- 2}$ . If $f (x) = d e t (a d j B)$ , then number of solution(s) of the equation $10 f (x) - x = 0$ is (are)

$f (x) = det (adj (B)) = ∣ adj (B) ∣ = ∣ B ∣$
$= ∣ ∣ A^{T} ∥ A^{- 2} ∣ ∣ = ∣ A ∣ \frac{1}{∣ A}$
$= \frac{1}{∣ A ∣} = \frac{1}{se c ^{2} x} = co s^{2} x$

$10 co s^{2} x - x = 0$

Q. Let A=[1−tan⁡x​tanx1​] and B=ATA−2 . If f(x)=det(adjB) , then number of solution(s) of the equation 10f(x)−x=0 is (are)

f(x)=det(adj(B))=∣adj(B)∣=∣B∣ =∣∣​AT∥A−2∣∣​=∣A∣∣A1​ =∣A∣1​=sec2x1​=cos2x 10cos2x−x=0

Q. Let $A = [1 - t an x t an x 1]$ and $B = A^{T} A^{- 2}$ . If $f (x) = d e t (a d j B)$ , then number of solution(s) of the equation $10 f (x) - x = 0$ is (are)

$f (x) = det (adj (B)) = ∣ adj (B) ∣ = ∣ B ∣$
$= ∣ ∣ A^{T} ∥ A^{- 2} ∣ ∣ = ∣ A ∣ \frac{1}{∣ A}$
$= \frac{1}{∣ A ∣} = \frac{1}{se c ^{2} x} = co s^{2} x$

$10 co s^{2} x - x = 0$