Question

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 \(\lambda =\)

• A. \((506)^{2}\)
• B. \(506\)
• C. \((1012)^{2}\)
• D. \(1012\)

Hint:

Show that y^4 + y^3 y'' reduces to the product ab, then maximise ab given a + b = 2024.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let \(w=y^{2}=a\cos^{2}x+b\sin^{2}x\). Then \(2yy'=w'=(b-a)\sin2x\), and differentiating again gives \(2y'^{2}+2yy''=2(b-a)\cos2x\).

Step 2: Form the Expression
Multiply \(yy''+y'^{2}=(b-a)\cos2x\) by \(y^{2}\). Then \(y^{3}y''+y^{2}y'^{2}=(b-a)y^{2}\cos2x\), where \(y^{2}y'^{2}=\dfrac{(b-a)^{2}\sin^{2}2x}{4}\).

Step 3: Simplify
Using \(y^{2}=\dfrac{(a+b)-(b-a)\cos2x}{2}\), the sum \(y^{4}+y^{3}y''\) simplifies to \(\dfrac{(a+b)^{2}-(b-a)^{2}}{4}=ab\). So \(k=ab\).

Step 4: Maximise
With \(a+b=2024\) and \(a,b\gt0\), AM-GM gives \(ab\leq\left(\dfrac{a+b}{2}\right)^{2}=(1012)^{2}\).

Final Answer:
The maximum value of k is 1012 squared. \[ \boxed{\text{(C) }(1012)^{2}} \]