JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(y(x)=\left|\begin{array}{ccc}\sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1\end{array}\right|, x \in \mathbb{R}\), then \(\frac{d^2 y}{d x^2}+y\) is equal to
- A \(-1\)
- B \(28\)
- C \(27\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & C_3 \rightarrow C_3-C_1 \\ & y(x)=\left|\begin{array}{ccc}\sin x & \cos x & 1+\cos x \\ 27 & 28 & 0 \\ 1 & 1 & 0\end{array}\right| \\ & y(x)=-(1+\cos x) \\ & \frac{d y}{d x}=\sin x \\ & \frac{d^2 y}{d x^2}=\cos x \\ & \frac{d^2 y}{d x^2}+y=-1\end{aligned}\)
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