Mathematics

# Resolve into partial fractions $\displaystyle \frac{23x-11x^2}{(2x-1)(9-x^2)}$

$\displaystyle \frac { 1 }{ \left( 2x-1 \right) } +\frac { -1 }{ \left( 3-x \right) } +\frac { 4 }{ \left( 3+x \right) }$

##### SOLUTION
We know $\displaystyle \frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( 9-{ x }^{ 2 } \right) } =\frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( 3-x \right) \left( 3+x \right) }$

Let $\displaystyle \frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( x-3 \right) \left( x+3 \right) } =\frac { A }{ \left( 2x-1 \right) } +\frac { B }{ \left( 3-x \right) } +\frac { C }{ \left( 3+x \right) }$ .....(1)

$23x-11{ x }^{ 2 }=A\left( 3-x \right) \left( 3+x \right) +B\left( 2x-1 \right) \left( 3+x \right) +C\left( 2x-1 \right) \left( 3-x \right) \\ 23x-11{ x }^{ 2 }=A\left( { 9-x }^{ 2 } \right) +B\left( 2{ x }^{ 2 }+5x-3 \right) +C\left( -2{ x }^{ 2 }+7x-3 \right)$
On comparing coefficients we get
$-11=-A+2B-2C,23=5B+7C,0=9A-3B-3C$
Solving them we get
$A=1,B=-1$ and $C=4$
Putting this values in equation (1), we get
$\dfrac {23x-11x^2}{(2x-1)(x-3)(x+3)}=\dfrac {1}{(2x-1)}+\dfrac {-1}{(3-x)}+\dfrac {4}{(3+x)}$
Hence, option 'A' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
$\int_{}^{} {\frac{{dx}}{{x\left( {{x^n} + 1} \right)}}}$ is equal to
• A. $\frac{1}{n}\log \left( {\frac{{{x^n}}}{{{x^n} + 1}}} \right) + c$
• B. $\log \left( {\frac{{{x^n}}}{{{x^n} + 1}}} \right) + c$
• C. None of these
• D. $-\frac{1}{n}\log \left( {\frac{{{x^n} + 1}}{{{x^n}}}} \right) + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the integrals:
$\displaystyle \int \dfrac{(9x^2-4x+5)}{(3x^3-2x^2+5x+1)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve :
$\displaystyle \int \dfrac{u}{v} dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $f(x) = \displaystyle \left\{\begin{matrix}e^{cos x}sin x, & for |x| \leq 2,\\ 2, & otherwise\end{matrix}\right.,$ then $\displaystyle \int_{-2}^3 f(x) dx=$
• A.
• B. 1
• C. 3
• D. 2

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.