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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Determinants

Among the statements : I: If | array ccc 1 & α & β α & 1 & γ β & γ...

Among the statements :
I: If \left|1cos⁡αcos⁡βcos⁡α1cos⁡γcos⁡βcos⁡γ1\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|0cos⁡αcos⁡βcos⁡α0cos⁡γcos⁡βcos⁡γ0\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|, then cos⁡2α+cos⁡2β+cos⁡2γ=32\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=\frac{3}{2}, and
II : If \left|x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q},then, then\mathrm{p}^{2}=196 \mathrm{q}^{2},

Options

  1. Aonly II is true
  2. Bboth are false
  3. Cboth are true
  4. Donly I is true

Correct answer

B. both are false

Step-by-step solution

Statement I: Let D1=∣1cos⁡αcos⁡βcos⁡α1cos⁡γcos⁡βcos⁡γ1∣=1−cos⁡2α−cos⁡2β−cos⁡2γ+2cos⁡αcos⁡βcos⁡γD_1 = \begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma.
D2=∣0cos⁡αcos⁡βcos⁡α0cos⁡γcos⁡βcos⁡γ0∣=2cos⁡αcos⁡βcos⁡γD_2 = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix} = 2\cos\alpha\cos\beta\cos\gamma.
D1=D2⇒1−cos⁡2α−cos⁡2β−cos⁡2γ=0⇒cos⁡2α+cos⁡2β+cos⁡2γ=1≠32D_1 = D_2 \Rightarrow 1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma = 0 \Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \neq \frac{3}{2}. Statement I is false.
Statement II: Evaluate by substituting x=0x = 0: determinant =−12= -12, so q=−12q = -12.
At x=1x = 1: determinant =12= 12, so p+q=12⇒p=24p + q = 12 \Rightarrow p = 24.
p2=576p^2 = 576 and 196q2=196×144=28224196q^2 = 196 \times 144 = 28224. Since p2≠196q2p^2 \neq 196q^2, Statement II is false.
Both statements are false.
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