StemCET Logo

Maths Question 16 – JEE-MAIN 2025

If y(x)=|sinxcosxsinx+cosx+1272827111|, xR, then d2ydx2+y is equal to

Look for column or row operations that can simplify the determinant before expansion, especially if elements are sums of other elements.

Step 1: Simplify the Determinant for y(x)✦ Active

First, simplify the given determinant for y(x) by applying column operations. Apply the operation C3C3C1C2 to the determinant:

y(x)=|sinxcosxsinx+cosx+1272827111|=|sinxcosx1272827272811111|=|sinxcosx1272828111|

Now, expand this simplified determinant (e.g., along the third row R3):

y(x)=1((cosx)(28)(1)(28))1((sinx)(28)(1)(27))+(1)((sinx)(28)(cosx)(27)) =(28cosx28)(28sinx27)(28sinx27cosx) =28cosx28+28sinx+2728sinx+27cosx =(28cosx+27cosx)+(28sinx28sinx)+(28+27) =cosx1
Step 2: Calculate the First and Second Derivatives○ Expand

Now that we have y(x)=1cosx, we can find its first and second derivatives:

dydx=ddx(1cosx)=0(sinx)=sinx d2ydx2=ddx(sinx)=cosx
Step 3: Compute the Final Expression○ Expand

Finally, substitute the expressions for y(x) and d2ydx2 into the required expression d2ydx2+y:

d2ydx2+y=cosx+(1cosx) =cosx1cosx =1
💡 Teacher's Secret Hint

Ensure all terms cancel out correctly after substitution.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.