Q15 of 45 Page 1

If y = Pe^ax + Qe^bx, show that

To prove that

y = Pe^ax + Qe^bx

Differentiating with respect to x

Differentiate again

Put values of y, and in LHS

⇒ LHS= Pa²e^ax + Qb²e^bx – (a + b)(Pae^ax + Qbe^bx) + ab(Pe^ax + Qe^bx)

⇒ LHS= Pa²e^ax + Qb²e^bx – Pa²e^ax – Qabe^bx – Pabe^ax – Qb²e^bx + Pabe^ax + Qabe^bx

⇒ LHS = 0

⇒ LHS = RHS

Hence proved.

Using properties of determinants, prove that

Find the value of at if and

Find the value(s) of x for which y = [x(x - 2)]² is an increasing function.

Find the equations of the tangent and normal to the curve at the point

Evaluate:

If y = Peax + Qebx, show that