Mathematics

# $\displaystyle \frac{x^{2}-x-1}{x^{3}-8}=\displaystyle \frac{A}{x-2}+\displaystyle \frac{Bx+C}{x^{2}+2x+4}\Rightarrow A+B=$

$1$

##### SOLUTION

$LHS=\dfrac {x^{2}-x-1}{x^{3}-8}; RHS=\dfrac {A}{x-2}+\dfrac {Bx+C}{x^{2}+2x+4}$
$=\dfrac {A(x^{2}+2x+4)}{x^{3}-8}+(Bx+C)(x-2)$
$x^{2}-x-1=A(x^{2}+2x+4)+(Bx+C)(x-2)$
$LHS=RHS$
comparing the co-efficient of $x^{2}$ we get
$A+B=1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Hard
Evaluate: $\displaystyle \int \dfrac { 1 } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } } d x ; x = 0$
• A. $\dfrac { \left( x ^ { 4 } - 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$
• B. $\dfrac { \sqrt { x ^ { 4 } + 1 } } { x } + c$
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• D. $-\dfrac { \left( x ^ { 4 } + 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \int _{ 2 }^{ 8 }{ \cfrac { \sqrt { 10-x } }{ \sqrt { x } +\sqrt { 10-x } } }dx$ is
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1 Verified Answer | Published on 17th 09, 2020

Q3 Assertion & Reason Hard
##### ASSERTION

The value of $\displaystyle \int_{0}^{\pi /2}\sin ^{6}xdx=\frac{5\pi }{16}$

##### REASON

If n is even, then $\int_{0}^{\pi /2}\sin ^{n}xdx$ equals $\displaystyle \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}...\frac{1}{2}\times \frac{\pi }{2}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integral:
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Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$