JEE Main 2025 — Differentiation Question with Solution

To solve for $y(x)$, we start with the determinant of the given matrix:

$\left|\begin{array}{ccc}\sin x& \cos x& \sin x+\cos x+1\\ 27& 28& 27\\ 1& 1& 1\end{array}\right|$

Using the formula for a $3\times 3$ determinant, we expand along the first row:

$f(x)=\sin x\cdot \left|\begin{array}{cc}28& 27\\ 1& 1\end{array}\right|-\cos x\cdot \left|\begin{array}{cc}27& 27\\ 1& 1\end{array}\right|+(\sin x+\cos x+1)\cdot \left|\begin{array}{cc}27& 28\\ 1& 1\end{array}\right|$

Calculating the smaller $2\times 2$ determinants:

$\left|\begin{array}{cc}28& 27\\ 1& 1\end{array}\right|=28\cdot 1-27\cdot 1=1$

$\left|\begin{array}{cc}27& 27\\ 1& 1\end{array}\right|=27\cdot 1-27\cdot 1=0$

$\left|\begin{array}{cc}27& 28\\ 1& 1\end{array}\right|=27\cdot 1-28\cdot 1=-1$

Substituting these into the determinant calculation gives:

$f(x)=\sin x\cdot 1-\cos x\cdot 0+(\sin x+\cos x+1)\cdot (-1)$

$f(x)=\sin x-(\sin x+\cos x+1)$

$f(x)=\sin x-\sin x-\cos x-1$

$f(x)=-\cos x-1$

Now, calculate the derivatives:

First derivative $f^{\prime }(x)$:

$f^{\prime }(x)=\frac{d}{dx}(-\cos x-1)=\sin x$

Second derivative $\frac{d^{2}f}{d{x}^2}$:

$\frac{d^{2}f}{d{x}^2}=\frac{d}{dx}(\sin x)=\cos x$

Finally, compute $\frac{d^{2}y}{d{x}^2}+y$:

$\frac{d^{2}f}{d{x}^2}+f(x)=\cos x-\cos x-1=-1$

Thus, $\frac{d^{2}y}{d{x}^2}+y$ is equal to $-1$.