${1 \over {x({x^2} + 1)}} = {A \over x} + {{Bx + C} \over {({x^2} + 1)}}$, તો $(A,\,B,\,C) = $
IIT 1995, Easy
Download our app for free and get started
a
(a) \(A({x^2} + 1) + (Bx + C)x = 1\)
(a) \(A({x^2} + 1) + (Bx + C)x = 1\)
For \(x = 0,\,A = 1\) and for \(x = i\), \( - B + Ci = 1\)
\( \Rightarrow B = - 1,\,C = 0\) \( \Rightarrow \) \((A,B,C) = (1,\, - 1,\,0)\).
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1${1 \over {\sqrt {(11 - 2\sqrt {30} )} }} - {3 \over {\sqrt {(7 - 2\sqrt {10} )} }} - {4 \over {\sqrt {(8 + 4\sqrt 3 )} }} = $View Solution
- 2$log_{(4-x)}(x^2 -14x + 45)$ ના વ્યાખિયાતિત થવા માટેની બધી પ્રાકૃતિક સંખ્યાઓનો સરવાળો મેળવો.View Solution
- 3${{{x^2} - 5} \over {{x^2} - 3x + 2}}$ નું આંશિક અપૂર્ણાક મેળવો.View Solution
- 4જો ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0$ તો $x = . . . .$View Solution
- 5${\log _2}.{\log _3}....{\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{{.2}^1}}}}}}}}}}}= . . . $.View Solution
- 6$(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . to \infty\right)}$ ની કિમત શોધોView Solution
- 7સમીકરણ ${(x)^{x\sqrt x }} = {(x\sqrt x )^x}$ ના ઉકેલની સંખ્યા મેળવો.View Solution
- 8સરવાળો $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}= ..............$View Solution
- 9${{6{x^4} + 5{x^3} + {x^2} + 5x + 2} \over {1 + 5x + 6{x^2}}} $ નું આંશિક અપૂર્ણાક મેળવો.View Solution
- 10${\log _2}7$ એ . . . . થાય.View Solution