If y= sin -1(x √1-x+√x √1-x2) (d y/d x)=(1/2 √x(1-x))+p, then p=

Question

If y= sin -1(x √1-x+√x √1-x2) (d y/d x)=(1/2 √x(1-x))+p, then p= (A) 0 (B)  sin -1 x (C)  sin -1 √x (D) none of these

Tardigrade Admin · Accepted Answer

x= sin θ ; x= sin 2 φ y= sin -1( sin θ cos φ+ sin φ cos θ)= sin -1(( sin (θ+φ))=θ+φ= sin -1 x+ sin -1 √x. Dy =(1/√1- x 2)+(1/2 √ x (1- x )) assuming x 2+ x ≤ 1 i.e. 0 ≤ x <(√5-1/2) ] Note: sin -1(x √1-y2+y √1-x2) = sin -1 x+ sin -1 y text if x2+y2 ≤ 1=π-( sin -1 x+ sin -1 y) text if x2+y2 ≥ 1

Q. If $y = sin^{- 1} (x 1 - x + x 1 - x^{2}) & \frac{d y}{d x} = \frac{1}{2 x ( 1 - x )} + p$ , then $p =$

0

$sin^{- 1} x$

$sin^{- 1} x$

none of these

Q. If y=sin−1(x1−x​+x​1−x2​)&dxdy​=2x(1−x)​1​+p, then p=

0

sin−1x

sin−1x​

none of these

x=sinθ;x=sin2ϕ y=sin−1(sinθcosϕ+sinϕcosθ)=sin−1((sin(θ+ϕ))=θ+ϕ=sin−1x+sin−1x​ Dy=1−x2​1​+2x(1−x)​1​ assuming x2+x≤1 i.e. 0≤x<25​−1​ ] Note: sin−1(x1−y2​+y1−x2​) =sin−1x+sin−1y if x2+y2≤1=π−(sin−1x+sin−1y) if x2+y2≥1

Q. If $y = sin^{- 1} (x 1 - x + x 1 - x^{2}) & \frac{d y}{d x} = \frac{1}{2 x ( 1 - x )} + p$ , then $p =$

$sin^{- 1} x$

$sin^{- 1} x$