1−(y′(x))2=y(x)1−(dxdy)2=y2(dxdy)2=1−y2 1−y2dy=dx OR 1−y2dy=−dx⇒sin−1y=x+c,sin−1y=−x+cx=0,y=0⇒c=0sin−1y=x, as y≥0sinx=y⇒dxdy=cosxdx2d2y=−sinx⇒−sinx+sinx+1=1
JEE Main 2024 — Mathematics Calculus
Let ∫0x1−(y′(t))2dt=∫0xy(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 is equal to
Held on 9 Apr 2024 · Verified 6 Jul 2026.
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