Question

y=x²/2x²+7x+6 and d²y/dx²=-A/(x+2)³+B/(2x+3)³ then find A and B​

Answer 1

✪ANSWER✪

• $\boxed{A = 8, B = 36}$

★EXPLAINATION★

It is given that

$y = \frac{ {x}^{2} }{2 {x}^{2} + 7x + 6}$

$y = \frac{ {x}^{2} + \frac{7}{2} x + 3 - \frac{7}{2}x - 3 }{2 {x}^{2} + 7x + 6}$

$y = \frac{ \frac{1}{2}(2 {x}^{2} + 7x + 6) - \frac{7}{2}x - 3 }{2 {x}^{2} + 7x + 6}$

$y = \frac{1}{2} - \frac{1}{2}. \frac{7x + 6}{2 {x}^{2} + 7x + 6 }$

$y = \frac{1}{2} - \frac{1}{2}. \frac{7x + 6 }{2 {x}^{2} + 4x + 3x + 6}$

$y = \frac{1}{2} - \frac{1}{2} . \frac{ 7x + 6 }{2x(x + 2) + 3(x + 2)}$

$y = \frac{1}{2} - \frac{1}{2}. \frac{ 7x + 6 }{(x + 2)(2x + 3)}$

Put

$\frac{ 7x + 6 }{(x + 2)(2x + 3)} = \frac{p}{x + 2} + \frac{q}{2x + 3}$

$7x + 6 = p(2x + 3) + q(x + 2)$

$7x + 6 = (2p + q)x + (3p + 2q)$

This equation is true for all value of x, hence co-efficients must be equal

$2p + q = 7 \:, \: 3p + 2q = 6$

$⇒p = 8 \:, \: q = - 9$

Therefore,

$y = \frac{1}{2} - \frac{1}{2} .( \frac{8}{ x+ 2} - \frac{9}{2x + 3} )$

Differenting w.r.t x we get

$\frac{dy}{dx} = 0 - \frac{1}{2} .( \frac{ - 8}{ {(x + 2)}^{2} } - \frac{ - 9 \times 2}{ {(2x + 3)}^{2} })$

$\frac{dy}{dx} = \frac{4}{ {(x + 2)}^{2} } - \frac{9}{ {(2x + 3)}^{2} }$

Differenting w.r.t x we get

$\frac{ {d}^{2} y}{ {dx}^{2} } = \frac{4.( - 2)}{ {(x + 2)}^{3} } - \frac{9.( - 2).2}{ {(2x + 3)}^{3} }$

$\frac{ {d}^{2} y}{ {dx}^{2} } = \frac{ - 8}{ {(x + 2)}^{3} } + \frac{36}{ {(2x + 3)}^{3} }$

Comparing with given expression of $\frac{ {d}^{2} y}{ {dx}^{2} }$, we get

$- A = - 8, B = 36$

$⇒ A = 8, B = 36$