MCQMediumJEE 2026Determinants Basics

JEE Mathematics 2026 Question with Solution

Among the statements : I: If the given determinants are equal, then cos2α+cos2β+cos2γ=32\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2}, and II: If the polynomial determinant equals px+qpx + q, then p2=196q2p^2 = 196q^2, identify the truth value.

  • A

    both are true

  • B

    only I is true

  • C

    both are false

  • D

    only II is true

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Two statements are to be checked for truth.

Find: Which of the statements is true.

For Statement I, the solution expands both sides of the determinant equality.

LHS becomes

1(cos2α+cos2β+cos2γ)+2cosαcosβcosγ1 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) + 2\cos\alpha\cos\beta\cos\gamma

RHS becomes

2cosαcosβcosγ2\cos\alpha\cos\beta\cos\gamma

Equating them gives

cos2α+cos2β+cos2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

So the claim cos2α+cos2β+cos2γ=32\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2} is false. Hence, Statement I is false.

For Statement II, the solution uses row operations

R2R22R1,R3R3R1R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - R_1

After these transformations, the x2x^2 terms vanish and the determinant reduces to a linear polynomial of the form

px+qpx + q

The computation then confirms that

p2=196q2p^2 = 196q^2

Therefore, Statement II is true.

Hence, only Statement II is true. The correct option is D.

Explanation from the extracted solution

Given: The truth values of Statement I and Statement II must be identified.

Find: The correct option.

The extracted solution states that this problem requires expanding determinants. Statement I uses a trigonometric identity obtained after determinant expansion, while Statement II uses coefficient comparison for a determinant that simplifies to a linear polynomial.

For Statement I, the extracted working is:

LHS=1(cos2α+cos2β+cos2γ)+2cosαcosβcosγ\text{LHS} = 1 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) + 2\cos\alpha\cos\beta\cos\gamma

and

RHS=2cosαcosβcosγ\text{RHS} = 2\cos\alpha\cos\beta\cos\gamma

Therefore,

cos2α+cos2β+cos2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

So Statement I is false.

For Statement II, the extracted solution gives the row transformations

R2R22R1R_2 \to R_2 - 2R_1

and

R3R3R1R_3 \to R_3 - R_1

which remove the quadratic terms. The determinant becomes px+qpx + q, and the stated computation confirms

p2=196q2p^2 = 196q^2

So Statement II is true.

Therefore, only Statement II is true, so the correct option is D.

Common mistakes

  • Assuming Statement I is true because the expression looks like a standard trigonometric identity. This is wrong because the determinant expansion given in the solution leads to cos2α+cos2β+cos2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1, not 32\frac{3}{2}. Always equate the expanded LHS and RHS carefully.

  • Failing to use row transformations in Statement II. This is wrong because without simplifying the determinant first, the polynomial form is harder to identify. Use the row operations R2R22R1R_2 \to R_2 - 2R_1 and R3R3R1R_3 \to R_3 - R_1 to eliminate the x2x^2 terms.

  • Using the answer key key without checking the solution logic. This is risky because determinant questions often depend on correct expansion and simplification. Verify the truth of each statement from the working before selecting the option.

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