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1)

 Let f:R→ R be a continuous  odd function, which vanishes exactly  at one point   f(1)=12 . Suppose that F(x)=x1f(t)dt    for all x ε [-1,2] and    G(x)=x1t|f{f(t)}|dt for all  x ε [-1,2] If   limx1F(x)G(x)=114 , then the value of    f(12)  is


A) 7

B) 4

C) 8

D) 6

Answer:

Option A

Explanation:

Here,   limx1F(x)G(x)=114

           limx1F(x)G(x)=114

                                                   [ Using  L' Hospital's rule ]...........(i)

As                  F(x)=x1f(t)dt

    F'(x)= f(x)     .........(ii)

and      G(x)=x1t|f{f(t)}|dt

            G(x)=x|f{f(x)}|            ...........(iii)

          \lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\lim_{x \rightarrow 1}\frac{F'(x)}{G'(x)}=\lim_{x \rightarrow 1}\frac{f(x)}{x|f\left\{f(x)\right\}|}

                              =   \frac{f(1)}{1|f\left\{f(1)\right\}|}=\frac{1/2}{|f(1/2)|}  ........(iv)

 Given,     \lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}

  \therefore    \frac{\frac{1}{2}}{|f\left(\frac{1}{2}\right)|}=\frac{1}{14}\Rightarrow |f\left(\frac{1}{2}\right)|=7