Let $|M|$ denote the determinant of a square matrix $M$. Let $g:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be the function defined by $$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$$ where $$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right|.$$ Let $p(x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?

(A) $p\left(\frac{3+\sqrt{2}}{4}\right) 0$

(B) $p\left(\frac{1+3 \sqrt{2}}{4}\right)>0$

(D) $p\left(\frac{5-\sqrt{2}}{4}\right)<0$

Determinants — JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 Question with Solution

Let

∣ M ∣

denote the determinant of a square matrix

M

. Let

g : [0, \frac{π}{2}] \to R

be the function defined by

g (θ) = f (θ) - 1 + f (\frac{π}{2} - θ) - 1

where

f (θ) = \frac{1}{2} 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1 + sin π sin (θ - \frac{π}{4}) cot (θ + \frac{π}{4}) cos (θ + \frac{π}{4}) - cos \frac{π}{2} lo g_{e} (\frac{π}{4}) tan (θ - \frac{π}{4}) lo g_{e} (\frac{4}{π}) tan π .

Let

p (x)

be a quadratic polynomial whose roots are the maximum and minimum values of the function

g (θ)

, and

p (2) = 2 - 2

. Then, which of the following is/are TRUE ?

Answer: (A), (C)

The first determinant simplifies to

1 + sin^{2} θ

(by row operations). The second determinant is skew-symmetric (all diagonal entries are 0 and off-diagonal entries satisfy

a_{ij} = - a_{j i}

), so it equals 0.

Thus

f (θ) = 1 + sin^{2} θ

g (θ) = sin^{2} θ + cos^{2} θ = ∣ sin θ ∣ + ∣ cos θ ∣ = sin θ + cos θ

for

θ \in [0, π /2]

.

The range of

g (θ) = sin θ + cos θ = 2 sin (θ + π /4)

[0, π /2]

[1, 2]

.

So

p (x) = k (x - 1) (x - 2)

. From

p (2) = 2 - 2

k (2 - 1) (2 - 2) = 2 - 2 \Rightarrow k = 1

Thus

p (x) = (x - 1) (x - 2)

.

Checking:

Options (A) and (C) are correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options