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1.

Let n1  and n2  be the number of red and black balls. respectively in box I. Let  n3 and n4  be the number of red and black balls respectively in box II.

A ball is drawn at random from box  I and transferred to box II. If the probability  of drawing  a red ball from box I, after this transfer  is  $\frac{1}{3}$, then the correct option(s)  with the possible values of n1 and n2  is/are


A) $n_{1}=4$ and $n_{2}=6$

B) $n_{1}=2$ and $n_{2}=3$

C) $n_{1}=10$ and $n_{2}=20$

D) $n_{1}=3$ and $n_{2}=6$



2.

Let n1  and n2  be the number of red and black balls. respectively in box I. Let  n3 and n4  be the number of red and black balls respectively in box II.

One of the two boxes, box I and box II a was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability  that this red ball was drawn  from box II, is  $\frac{1}{3}$, then the correct option(s)  with the possible  values of n1,n2,n3  and nis/are


A) $n_{1}=3,n_{2}=3,n_{3}=5,n_{4}=15$

B) $n_{1}=3,n_{2}=6,n_{3}=10,n_{4}=50$

C) $n_{1}=8,n_{2}=6,n_{3}=5,n_{4}=20$

D) $n_{1}=6,n_{2}=12,n_{3}=5,n_{4}=20$



3.

 Let   $F:R\rightarrow R$  be a thrice  differentiable  function. Suppose that  F(1)=0,F(3)=-4  and F'(x)<0 for x ε (1,3)  , Let f(x)=xF(X) for all x ε R.

 If  $\int_{1}^{3}x^{2} F'(x)dx=-12$      and           $\int_{1}^{3} x^{3}F"(x)=dx=40$, then the correct expression(s) is/are

  


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

B) $\int_{1}^{3} f(x) dx=12$

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

D) $\int_{1}^{3} f(x) dx=-12$



4.

 Let   $F:R\rightarrow R$  be a thrice  differentiable  function. Suppose that  F(1)=0,F(3)=-4  and F'(x)<0 for x ε (1,3)  , Let f(x)=xF(X) for all x ε R.

The correct statement(s)  is/are


A) $f^{1}(1)<0$

B) $f(2)<0$

C) $ f'(x)\neq 0$ for any $x\epsilon (1,3)$

D) f'(x) =0 for same $x\epsilon (1,3)$



5.

 The options (s) with the values of a and L that  satidfy the equation  $\frac{\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{4}at)dt}{\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{4}at)dt} =L$   is\are


A) $a=2,L=\frac{e^{4\pi}-1}{e^{\pi}-1}$

B) $a=2,L=\frac{e^{4\pi}+1}{e^{\pi}+1}$

C) $a=4,L=\frac{e^{4\pi}-1}{e^{\pi}-1}$

D) $a=4,L=\frac{e^{4\pi}+1}{e^{\pi}+1}$



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