Let |M| denote the determinant of a square matrix M. Let g: [0, π/2] → ℝ be the function defined by

Q: Wait, why is the determinant of an odd-order skew-symmetric matrix always zero?

This is a beautiful property! For any skew-symmetric matrix $A$, we have $A^T = -A$. Taking the determinant of both sides, we get $\det(A^T) = \det(-A)$. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$ for an $n \times n$ matrix, we have $\det(A) = (-1)^n \det(A)$. If $n$ is odd, then $\det(A) = -\det(A)$, which implies $2 \det(A) = 0$, so $\det(A) = 0$. It is a classic JEE trap to make you calculate the whole thing, but the symmetry kills it instantly!

Q: Why did we drop the modulus signs in $g(\theta)$?

Great catch! We never drop modulus signs blindly. We always check the domain. The problem explicitly states $\theta \in [0, \frac{\pi}{2}]$. In this first quadrant, both $\sin \theta$ and $\cos \theta$ are non-negative. Therefore, $|\sin \theta| = \sin \theta$ and $|\cos \theta| = \cos \theta$. Always verify the interval before simplifying absolute values!

Q: How did we determine the constant $k$ in $p(x) = k(x-1)(x-\sqrt{2})$?

The problem provided a specific condition: $p(2) = 2 - \sqrt{2}$. Once we identified the roots as $1$ and $\sqrt{2}$, we wrote the general form $p(x) = k(x-1)(x-\sqrt{2})$. By substituting $x=2$ into this equation, we get $k(2-1)(2-\sqrt{2}) = 2-\sqrt{2}$. This simplifies to $k(1)(2-\sqrt{2}) = 2-\sqrt{2}$, which clearly shows $k=1$. It is just a matter of using the given information to lock in the polynomial.

Question

Let $|M|$ denote the determinant of a square matrix $M$. Let $g: [0, \frac{\pi}{2}] o \mathbb{R}$ be the function defined by $g( heta) = \sqrt{f( heta) - 1} + \sqrt{f(\frac{\pi}{2} - heta) - 1}$ where $f( heta) = \frac{1}{2} \begin{vmatrix} 1 & \sin heta & 1 \ -\sin heta & 1 & \sin heta \ -1 & -\sin heta & 1 \end{vmatrix} + \begin{vmatrix} \sin \pi & \cos( heta + \frac{\pi}{4}) & an( heta - \frac{\pi}{4}) \ \sin( heta - \frac{\pi}{4}) & -\cos \frac{\pi}{2} & \log_e (\frac{4}{\pi}) \ \cot( heta + \frac{\pi}{4}) & \log_e (\frac{\pi}{4}) & an \pi \end{vmatrix}$. Let $p(x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g( heta)$, and $p(2) = 2 - \sqrt{2}$. Then, which of the following is/are TRUE ? Options: (A) $p(\frac{3+\sqrt{2}}{4}) < 0$ (B) $p(\frac{1+3\sqrt{2}}{4}) > 0$ (C) $p(\frac{5\sqrt{2}-1}{4}) > 0$ (D) $p(\frac{5-\sqrt{2}}{4}) < 0$

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a terrifying beast of algebra and trigonometry. You see a massive determinant, a function defined by square roots, and a quadratic polynomial waiting at the end. It is easy to feel overwhelmed. But here is the secret of the JEE Advanced: complexity is often just a mask for elegance. Let us peel back the layers together. ### The Determinant Deception We start with the function $f( heta)$. It is composed of two determinants, $D_1$ and $D_2$. Your instinct might be to start expanding $D_1$ immediately. That is fine, but let us be strategic. $$D_1 = \begin{vmatrix} 1 & \sin heta & 1 \ -\sin heta & 1 & \sin heta \ -1 & -\sin heta & 1 \end{vmatrix}$$ Expanding along the first row, we get $1(1 + \sin^2 heta) - \sin heta(-\sin heta + \sin heta) + 1(\sin^2 heta + 1)$. Notice that the middle term vanishes completely because $-\sin heta + \sin heta = 0$. We are left with $2(1 + \sin^2 heta)$. Now, look at $D_2$. This is where the trap is set. $$D_2 = \begin{vmatrix} \sin \pi & \cos( heta + \frac{\pi}{4}) & an( heta - \frac{\pi}{4}) \ \sin( heta - \frac{\pi}{4}) & -\cos \frac{\pi}{2} & \log_e (\frac{4}{\pi}) \ \cot( heta + \frac{\pi}{4}) & \log_e (\frac{\pi}{4}) & an \pi \end{vmatrix}$$ Look at the diagonal elements: $\sin \pi = 0$, $-\cos \frac{\pi}{2} = 0$, and $ an \pi = 0$. They are all zero! Furthermore, check the off-diagonal elements. You will find that $a_{ij} = -a_{ji}$. This is the definition of a **skew-symmetric matrix**. Since it is a $3 imes 3$ matrix (odd order), its determinant is guaranteed to be zero. The entire second part of the function vanishes into thin air. We are left with $f( heta) = \frac{1}{2} [2(1 + \sin^2 heta)] + 0 = 1 + \sin^2 heta$. ### The Function's True Face Now that we have tamed the monster, let us look at $g( heta) = \sqrt{f( heta) - 1} + \sqrt{f(\frac{\pi}{2} - heta) - 1}$. Substituting our simplified $f( heta)$, we get: $$g( heta) = \sqrt{(1 + \sin^2 heta) - 1} + \sqrt{(1 + \sin^2(\frac{\pi}{2} - heta)) - 1}$$ This simplifies to $\sqrt{\sin^2 heta} + \sqrt{\cos^2 heta}$, which is $|\sin heta| + |\cos heta|$. Because our domain is $ heta \in [0, \frac{\pi}{2}]$, we are in the first quadrant where both sine and cosine are positive. Thus, $g( heta) = \sin heta + \cos heta$. ### The Range Hunt We need the maximum and minimum values of $g( heta)$. We can rewrite this as: $$g( heta) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin heta + \frac{1}{\sqrt{2}} \cos heta ight) = \sqrt{2} \sin( heta + \frac{\pi}{4})$$ Since $ heta \in [0, \frac{\pi}{2}]$, the angle $( heta + \frac{\pi}{4})$ ranges from $\frac{\pi}{4}$ to $\frac{3\pi}{4}$. In this interval, the sine function starts at $\frac{1}{\sqrt{2}}$, goes up to $1$, and comes back down to $\frac{1}{\sqrt{2}}$. Therefore, the range of $g( heta)$ is $[1, \sqrt{2}]$. The minimum value is $1$ and the maximum is $\sqrt{2}$. ### The Polynomial Construction The problem tells us that these values are the roots of our quadratic $p(x)$. So, $p(x) = k(x - 1)(x - \sqrt{2})$. We are given $p(2) = 2 - \sqrt{2}$. Plugging in $x=2$, we get $k(2 - 1)(2 - \sqrt{2}) = 2 - \sqrt{2}$, which forces $k = 1$. Our polynomial is $p(x) = (x - 1)(x - \sqrt{2})$. This is an upward-opening parabola with roots at $1$ and $\sqrt{2} \approx 1.414$. ### The Final Verdict Now, we simply test the options by checking where the input values lie relative to the roots. For Option A, the input is $\frac{3+\sqrt{2}}{4} \approx 1.10$. This is between $1$ and $1.414$, so $p(x) < 0$. This is **TRUE**. For Option B, the input is $\frac{1+3\sqrt{2}}{4} \approx 1.31$. This is also between $1$ and $1.414$, so $p(x) < 0$. The option claims it is $> 0$, so it is **FALSE**. For Option C, the input is $\frac{5\sqrt{2}-1}{4} \approx 1.51$. This is greater than $1.414$, so $p(x) > 0$. This is **TRUE**. For Option D, the input is $\frac{5-\sqrt{2}}{4} \approx 0.89$. This is less than $1$, so $p(x) > 0$. The option claims it is $< 0$, so it is **FALSE**. And there you have it! By staying calm and looking for the underlying symmetry, we turned a complex problem into a simple parabola analysis. Keep this mindset, and you will conquer any problem the exam throws at you.

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Visualized Solution

Evaluating $D_{1}$

Simplifying $D_{1}$

Analyzing $D_{2}$

Determinant of Skew-Symmetric Matrix

Simplifying $g (θ)$

Domain and Modulus

Max and Min of $g (θ)$

Forming the Polynomial $p (x)$

Visualizing $p (x)$

Checking Option A

Checking Option B

Checking Option C

Checking Option D & Conclusion

The Sigma Insight: Location of Roots

The Determinant Deception

The Function's True Face

The Range Hunt

The Polynomial Construction

The Final Verdict

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Visualized Solution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2​