X2-1-x2-22x2-1-k -1-x2-2-1-2x2-1- k-

The correct option is A 1/3 We have, #160; x^2+1/(x^2+2)(2x^2+1)=k #10;[ #160;1/x^2+2+ #160;1/2x^2+1 #10;] Substitute x=0 both sides ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Complex Numbers

x2+1/x2+22x2+...

Question

$\frac{x^{2} + 1}{(x^{2} + 2) (2 x^{2} + 1)} = k [\frac{1}{x^{2} + 2} + \frac{1}{2 x^{2} + 1}] \Rightarrow k =$

\frac{1}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\frac{x^{2} + 5}{(x^{2} + 2)^{2}} = \frac{1}{x^{2} + 2} + \frac{k}{(x^{2} + 2)^{2}} \Rightarrow k =

x^{2} + 2 x + 1 = (4 - x)^{2} + 3

- 2 x^{2} = (5 - x) (2 x - \frac{2}{5})

(k + 1) x^{2} + \frac{3}{2} x = 7, w h e r e k - = - 1

x^{3} - x^{2} = (x - 1)^{3}

\int \frac{(x^{2} + 1) d x}{(x^{2} + 2) (2 x^{2} + 1)} = \frac{1}{3 \sqrt{(2)}} {{tan}^{- 1} \frac{x}{\sqrt{(2)}} + {tan}^{- 1} x \sqrt{2}} .

\int \frac{x}{(x^{2} + 4) \sqrt{x^{2} + 1}} d x = \frac{1}{\sqrt{k}} {tan}^{- 1} \sqrt{\frac{x^{2} + 1}{3}} + c

k ?

\frac{(x^{2} + 1)}{(x^{2} + 2) (x^{2} + 3)} = \frac{- 1}{x^{2} + 2} + \frac{α}{x^{2} + 3}

α =

Join BYJU'S Learning Program