#### Passage

Let $F: R\rightarrow R$ be a thrice differential function. Suppose that $F(1) = 0, F(3) = -4$ and $F'(x)<0$ for all $x\in\left(\dfrac{1}{2},3\right)$. Let $f(x) = xF(x)$ for all $x\in R$.
Mathematics

# If $\displaystyle\int_1^{3}x^2F'(x)dx = -12$ and $\displaystyle\int_1^{3}x^3F''(x)dx = 40$, then the correct expression(s) is (are)

$9f'(3)-f'(1)+32=0$

$\displaystyle\int_1^3f(x)dx = -12$

##### SOLUTION
$\int _{ 1 }^{ 3 }{ f(x)dx } =\int _{ 1 }^{ 3 }{ xF(x) } dx=(F(x).\frac { { x }^{ 2 } }{ 2 } )^{3}_{1}\quad -\int _{ 1 }^{ 3 }{ \frac { { x }^{ 2 } }{ 2 } } .F^{ ' }\left( x \right) dx\\ =\frac { 9 }{ 2 } (-4)-\frac { 1 }{ 2 } (-12)\\ =-12$
$f^{ ' }\left( 3 \right) =F(3)+3F^{ ' }(3) -1)\\ f^{ ' }\left( 1 \right) =F^{ ' }\left( 1 \right) -2)$
For option C, subsitute equation $1)$ and $2)$
$9f^{ ' }\left( 3 \right) -f^{ ' }\left( 1 \right) +32\\ \Rightarrow 9F(3)+27F^{ ' }\left( 3 \right) -F^{ ' }\left( 1 \right) +32$
$27F^{ ' }\left( 3 \right) -F^{ ' }\left( 1 \right) -4=0$
On substituting the values of F(x) we get the value as zero

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Multiple Correct Hard Published on 17th 09, 2020
Mathematics

# The correct statements is (are)

$f'(1)<0$

$f(2)<0$

$f'(x)\neq0$ for any $x\in(1,3)$

##### SOLUTION
$f(x) = xF(x)$
Differentiating both side w.r.t $x$
$f'(x) =F(x)+xF'(x)$
$\displaystyle \therefore f'(1) = F(1)+F'(1) =0+F'(1) <0 \because F'(x) < 0 \forall x \in \left(\frac{1}{2}, 3\right)$
$f(2) = 2F(2) < 2F(1) < 0$, Since $F(x)$ is decreasing function in the given interval
also $f'(x) < 0 \forall x \in (1,3)\Rightarrow f(x) \neq 0$ for any $x \epsilon (1,3)$

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Multiple Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Single Correct Hard
If $g(x)= \displaystyle \int_{0}^{x} cos4t dt,$ then g(x+) equals
• A. $\dfrac{g(x)}{g(\pi)}$
• B. $g(x)+g(\pi)$
• C. $g(x).g(\pi)$
• D. $g(x)-g(\pi)$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \frac{3x^{2}}{x^{6}+1}dx$
• A. $\displaystyle \frac{1}{2}\tan ^{-1}x^{3}.$
• B. $\displaystyle \tan ^{-1}x^{6}.$
• C. $\displaystyle \sec ^{-1}x^{3}.$
• D. $\displaystyle \tan ^{-1}x^{3}.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\int {\frac{2x+5}{x^2+5x-3}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\displaystyle \int \dfrac {4x+5}{2x^2+5x+18}dx$

The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$
The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b].