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MCQEasyJEE 2025Determinants Basics

JEE Mathematics 2025 Question with Solution

If y(x)=∣sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111∣y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1\\ 27 & 28 & 27\\ 1 & 1 & 1 \end{vmatrix}, x∈Rx \in \mathbb{R}, then d2ydx2+y\frac{d^2y}{dx^2} + y is equal to

  • A

    −1-1

  • B

    2828

  • C

    2727

  • D

    11

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

y(x)=∣sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111∣y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1\\ 27 & 28 & 27\\ 1 & 1 & 1 \end{vmatrix}

Find: d2ydx2+y\frac{d^2y}{dx^2} + y

Expand the determinant along the first row:

y(x)=sin⁡x∣282711∣−cos⁡x∣272711∣+(sin⁡x+cos⁡x+1)∣272811∣y(x) = \sin x \begin{vmatrix} 28 & 27\\ 1 & 1 \end{vmatrix} - \cos x \begin{vmatrix} 27 & 27\\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \begin{vmatrix} 27 & 28\\ 1 & 1 \end{vmatrix}

Now evaluate the minors:

∣282711∣=28−27=1\begin{vmatrix} 28 & 27\\ 1 & 1 \end{vmatrix} = 28 - 27 = 1 ∣272711∣=27−27=0\begin{vmatrix} 27 & 27\\ 1 & 1 \end{vmatrix} = 27 - 27 = 0 ∣272811∣=27−28=−1\begin{vmatrix} 27 & 28\\ 1 & 1 \end{vmatrix} = 27 - 28 = -1

Substitute these values:

y(x)=sin⁡x(1)−cos⁡x(0)+(sin⁡x+cos⁡x+1)(−1)y(x) = \sin x (1) - \cos x (0) + (\sin x + \cos x + 1)(-1) y(x)=sin⁡x−(sin⁡x+cos⁡x+1)y(x) = \sin x - (\sin x + \cos x + 1) y(x)=−cos⁡x−1y(x) = -\cos x - 1

Differentiate twice:

dydx=sin⁡x\frac{dy}{dx} = \sin x d2ydx2=cos⁡x\frac{d^2y}{dx^2} = \cos x

Therefore,

d2ydx2+y=cos⁡x+(−cos⁡x−1)=−1\frac{d^2y}{dx^2} + y = \cos x + (-\cos x - 1) = -1

So, the correct option is A.

The solution also confirms: The Correct Option is A.

Use row-based minors directly

Given: the determinant has second row [27,28,27][27, 28, 27] and third row [1,1,1][1, 1, 1].

Find: d2ydx2+y\frac{d^2y}{dx^2} + y

Observe the cofactors of the first row entries:

C11=∣282711∣=1C12=−∣272711∣=0C13=∣272811∣=−1\begin{aligned} C_{11} &= \begin{vmatrix} 28 & 27\\ 1 & 1 \end{vmatrix} = 1 \\ C_{12} &= -\begin{vmatrix} 27 & 27\\ 1 & 1 \end{vmatrix} = 0 \\ C_{13} &= \begin{vmatrix} 27 & 28\\ 1 & 1 \end{vmatrix} = -1 \end{aligned}

Hence the determinant immediately becomes

y(x)=sin⁡x(1)+cos⁡x(0)+(sin⁡x+cos⁡x+1)(−1)y(x) = \sin x (1) + \cos x (0) + (\sin x + \cos x + 1)(-1) y(x)=−cos⁡x−1y(x) = -\cos x - 1

Then

y′′=cos⁡xy'' = \cos x

So,

y′′+y=cos⁡x−cos⁡x−1=−1y'' + y = \cos x - \cos x - 1 = -1

Therefore, the correct option is A.

Common mistakes

  • A common mistake is forgetting the sign pattern while expanding the determinant along the first row. The middle term carries a negative sign, so use +,−,++,-,+. Otherwise the simplified expression for y(x)y(x) becomes incorrect.

  • Some students differentiate −cos⁡x−1-\cos x - 1 incorrectly and write dydx=−sin⁡x\frac{dy}{dx} = -\sin x. This is wrong because ddx(−cos⁡x)=sin⁡x\frac{d}{dx}(-\cos x) = \sin x. Differentiate carefully before taking the second derivative.

  • Another mistake is not simplifying the determinant fully before differentiation. First reduce the determinant to y(x)=−cos⁡x−1y(x) = -\cos x - 1, then compute derivatives. Direct differentiation of the determinant without simplification is unnecessarily error-prone.

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