Question

If y = asinx+bcosx,y²+(y₁)²=........(a²+b²≠0),Select Proper option from the given options.
(a) acosx-bsinx
(b) (asinx-bcosx)²
(c) a²+b²
(d) 0

Answer 1

Given, y = asinx + bcosx

differentiate both sides with respect to x,

dy/dx = d(asinx + bcosx)/dx

= a.d(sinx)/dx + b.d(cosx)/dx

= a.cosx + b(-sinx)

= acosx - bsinx

hence, y₁ = (acosx - bsinx)

taking square both sides,

(y₁)² = (acosx - bsinx)²


now, y² + (y₁)² = (asinx + bcosx)² + (acosx - bsinx)²

= a²sin²x + b²cos²x + 2absinx.cosx + a²cos²x + b²sin²x - 2absinx.cosx

= (a² + b²)sin²x + (a² + b²)cos²x

= (a² + b²){sin²x + cos²x}

but sin²x + cos²x = 1 [ from identities]

= (a² + b²) × 1 = (a² + b²)

hence, option (c) is correct

Answer 2

In the attachment I have answered this problem.

The solution is simple and easy to understand.

Correct option is C

See the attachment for detailed solution.