#### Passage

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 $\displaystyle f(x)>0 \forall x \in R$

On the basis of above information, answer the following questions :
Mathematics

# $\displaystyle \int \left ( f(x)+g(x)\right)dx$ is equal to

$\displaystyle \frac {x^{4}}{4}+\frac {x^{3}}{3}+\frac {x^{2}}{2}+x+c$

##### SOLUTION
Consider $\displaystyle 6\int f(x)g(x)dx=x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c$

$\Rightarrow \displaystyle \int f(x)g(x)dx=\dfrac { 1 }{ 6 } (x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c)$

Differentiate w.r.t $x$,

$\Rightarrow f(x)g(x)=x^{ 5 }+2x^{ 3 }+x$ ........... $(1)$

Consider, $\displaystyle 2\int \dfrac { g(x)dx }{ f(x) } =x^{ 2 }+c$

$\Rightarrow \displaystyle \int \dfrac { g(x)dx }{ f(x) } =\frac { 1 }{ 2 } (x^{ 2 }+c)$

Differentiate w.r.t $x$

$\Rightarrow \dfrac { g(x) }{ f(x) } =x$ ............ $(2)$

$(1)\times (2)\Rightarrow ({ g(x)) }^{ 2 }=x^{ 6 }+2x^{ 4 }+x^{ 2 }=({ x }^{ 3 }+x)^{ 2 }$
$\Rightarrow g(x)={ x }^{ 3 }+x$

$\dfrac { (1) }{ (2) } \Rightarrow (f(x))^{ 2 }=x^{ 4 }+2x^{ 2 }+1=({ x }^{ 2 }+1)^{ 2 }$
$\Rightarrow f(x)={ x }^{ 2 }+1$

Then, $\displaystyle \int { (g(x)+f(x))dx } =\int (x^{3}+x+{x}^{2}+1)dx$
$=\dfrac { { x }^{ 4 } }{ 4 } +\dfrac { { x }^{ 3 } }{ 3 } +\dfrac { x^{ 2 } }{ 2 } +x+c$

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Single Correct Hard Published on 17th 08, 2020
Mathematics

# $\displaystyle f'(1) + g'(2)$ is equal to

$15$

##### SOLUTION
Consider $\displaystyle 6\int f(x)g(x)dx=x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c$
$\Rightarrow \displaystyle \int f(x)g(x)dx=\dfrac { 1 }{ 6 } (x^{ 6 }+3x^{ 4 }+3x^{ 2 }+c)$

Differentiate w.r.t $x$,

$\Rightarrow f(x)g(x)=x^{ 5 }+2x^{ 3 }+x$ ........... $(1)$

Consider, $\displaystyle 2\int \dfrac { g(x)dx }{ f(x) } =x^{ 2 }+c$

$\Rightarrow \displaystyle \int \dfrac { g(x)dx }{ f(x) } =\frac { 1 }{ 2 } (x^{ 2 }+c)$

Differentiate w.r.t $x$

$\Rightarrow \dfrac { g(x) }{ f(x) } =x$ ............ $(2)$

$(1)\times (2)\Rightarrow ({ g(x)) }^{ 2 }=x^{ 6 }+2x^{ 4 }+x^{ 2 }=({ x }^{ 3 }+x)^{ 2 }$
$\Rightarrow g(x)={ x }^{ 3 }+x$

$\dfrac { (1) }{ (2) } \Rightarrow (f(x))^{ 2 }=x^{ 4 }+2x^{ 2 }+1=({ x }^{ 2 }+1)^{ 2 }$
$\Rightarrow f(x)={ x }^{ 2 }+1$

Now, $f'(x)=2x$ and $g'(x)=3x^2+1$
$\Rightarrow f'(1)+g'(2)=2+13=15$

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Single Correct Medium Published on 17th 08, 2020
Questions 244531
Subjects 8
Chapters 125
Enrolled Students 145

#### Realted Questions

Q1 One Word Medium
Evaluate the following limits:

$\displaystyle \lim_{x\rightarrow 0}\left ( \frac{\tan x}{x} \right )^{1/x^{2}}$ is $e^{1/a}$ Find a

Asked in: Mathematics - Limits and Derivatives

1 Verified Answer | Published on 17th 08, 2020

Q2 Subjective Medium
Solve :
$\displaystyle \int \dfrac {dy}{e^{(x-1)y} -1 }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle\lim _{ n\rightarrow \infty }{ \left[ \frac { 1 }{ 1+{ n }^{ 2 } } +\frac { 2 }{ { 2 }^{ 2 }+{ n }^{ 2 } } +...+\frac { n }{ { n }^{ 2 }+{ n }^{ 2 } } \right]}$ is
• A. $log 2$
• B. $2 log 2$
• C. $\displaystyle \frac{\log 2}{3}$
• D. $\displaystyle \frac{\log 2}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int{(x^{3}a+x^{2}a+xa)(2x^{2}a+3xa+6)^{1/a}dx}$=
• A. $\dfrac{1}{6(a+1)}(2x^{3a}+3x^{2a}+6x^{a})^{1+\dfrac{1}{a}}+C$
• B. $\dfrac{1}{3(a+1)}(2x^{3a}+3x^{2a}+6x^{a})^{1+\dfrac{1}{a}}+C$
• C. $none\ of\ these$
• D. $\dfrac{1}{6(a+1)}(2x^{3a}+3x^{2a}+6x^{a})^{1-\dfrac{1}{a}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int_1^{32}\dfrac{dx}{x^{1/5}\sqrt{1+x^{4/5}}}$
• A. $\dfrac{2}{5}(\sqrt{17}-\sqrt{2})$
• B. $\dfrac{5}{2}(\sqrt{17}-\sqrt{2})$
• C. $\dfrac{5}{2}(\sqrt{17}+\sqrt{2})$
• D. $\dfrac{2}{5}(\sqrt{17}+\sqrt{2})$