If y(x)=| array ccc x & x & x+ x+1 27 & 28 & 27 1 & 1 & 1 array |, x ∈ ℝ , then (d^2 y)/(d x^2)+y is equal to

Correct answer: A. -1 aligned & C_3 arrow C_3-C_1 & y(x)=| array ccc x & x & 1+ x 27 & 28 & 0 1 & 1 & 0 array | & y(x)=-(1+ x) & (d y)/(d x)= x & (d^2 y)/(d x^2)= x & (d^2 y)/(d x^2)+y=-1 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Determinants

If y(x)=| array ccc x & x & x+ x+1 27 & 28 & 27 1 & 1 & 1 array |,...

\quad y(x)=\left|

\begin{array}{ccc}\sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1\end{array}

\right|, x \in \mathbb{R}

, then

\frac{d^2 y}{d x^2}+y

is equal to

Options

A $-1$
B $28$
C $27$
D $1$

Correct answer

A. $-1$

Step-by-step solution

\begin{aligned} & C_3 \rightarrow C_3-C_1 \\ & y(x)=\left|\begin{array}{ccc}\sin x & \cos x & 1+\cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0\end{array}\right| \\ & y(x)=-(1+\cos x) \\ & \frac{d y}{d x}=\sin x \\ & \frac{d^2 y}{d x^2}=\cos x \\ & \frac{d^2 y}{d x^2}+y=-1\end{aligned}

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