Let \(y^{2}=a\cos ^{2}x+b\sin ^{2}x\) be such that \(a+b=2024\) and \(a>0,b>0\) follows \(y^{4}+y^{3}y^{''}=k\). If \(\lambda\) is the maximum possible value of \(k\), then \(\lambda =\)
Show Hint
Show that y^4 + y^3 y'' reduces to the product ab, then maximise ab given a + b = 2024.
Step 1: Use w and Its Derivatives
With $w=y^{2}$: $y^{4}+y^{3}y''=w\left(w+\dfrac{w''}{2}\right)-\dfrac{w'^{2}}{4}$.
Step 2: Evaluate
$w+\dfrac{w''}{2}=b\cos^{2}x+a\sin^{2}x$. So $w\left(w+\dfrac{w''}{2}\right)=ab(\cos^{4}x+\sin^{4}x)+(a^{2}+b^{2})\sin^{2}x\cos^{2}x$, and $\dfrac{w'^{2}}{4}=(b-a)^{2}\sin^{2}x\cos^{2}x$.
Step 3: Subtract and Maximise
The difference is $ab(\cos^{2}x+\sin^{2}x)^{2}=ab$. Its maximum for $a+b=2024$ occurs at $a=b=1012$, giving $(1012)^{2}$.
Final Answer:
The maximum of ab is 1012 squared.
\[ \boxed{\text{(C) }(1012)^{2}} \]