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)


ANSWER

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

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


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)


ANSWER

$$f'(1)<0$$

$$f(2)<0$$

$$f'(x)\neq0$$ 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
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