Let f:R→ R be a differentiable function and f(1)=4. Then the value of undersetx→ 1 mathop lim ∫4f(x)(2t/x-1)dt, if f'(1)=2 is:

Question

Let f:R→ R be a differentiable function and f(1)=4. Then the value of undersetx→ 1 mathop lim ∫4f(x)(2t/x-1)dt, if f'(1)=2 is: (A) 16 (B) 8 (C) 4 (D) 2

Tardigrade Admin · Accepted Answer

undersetx→ 1 mathop lim (∫4f(x)2t dt/x-1) = undersetx→ 1 mathop lim (2f(x).f(x)/1) =2f(1).f(1)=2.4.2=16

Q. Let $f : R \to R$ be a differentiable function and $f (1) = 4.$ Then the value of $x \to 1 lim \int_{4}^{f (x)} \frac{2 t}{x - 1} d t,$ if $f^{'} (1) = 2$ is:

16

8

4

2

1

$x \to 1 lim \frac{\int _{4}^{f (x)} 2 t d t}{x - 1}$
$= x \to 1 lim \frac{2 f ( x ) . f ( x )}{1}$
$= 2 f (1) . f (1) = 2.4.2 = 16$

Q. Let f:R→R be a differentiable function and f(1)=4. Then the value of x→1lim​∫4f(x)​x−12t​dt, if f′(1)=2 is:

16

8

4

2

1

x→1lim​x−1∫4f(x)​2tdt​ =x→1lim​12f(x).f(x)​ =2f(1).f(1)=2.4.2=16

Q. Let $f : R \to R$ be a differentiable function and $f (1) = 4.$ Then the value of $x \to 1 lim \int_{4}^{f (x)} \frac{2 t}{x - 1} d t,$ if $f^{'} (1) = 2$ is:

$x \to 1 lim \frac{\int _{4}^{f (x)} 2 t d t}{x - 1}$
$= x \to 1 lim \frac{2 f ( x ) . f ( x )}{1}$
$= 2 f (1) . f (1) = 2.4.2 = 16$