The area bounded by the x-axis, the curve y =f(x) and the lines x = 1, x = b is equal to √b2+1-√2 for all b 1, then f(x) is

Question

The area bounded by the x-axis, the curve y =f(x) and the lines x = 1, x = b is equal to √b2+1-√2 for all b> 1, then f(x) is (A) √x-1 (B) √x+1 (C) √x2+1 (D) x/ √x2+1

Tardigrade Admin · Accepted Answer

We have ∫ limits1b f(x)dx =√b2+1-√2 On differentiating w.r.t. to b, we get f(b)=(2b/2√b2+1) ⇒ f(x)=(x/√x2+1)

Q. The area bounded by the $x$ -axis, the curve $y = f (x)$ and the lines $x = 1$ , $x = b$ is equal to $b^{2} + 1 - 2$ for all $b > 1$ , then $f (x)$ is

$x - 1$

$x + 1$

$x^{2} + 1$

$x / x^{2} + 1$

We have $1 \int b f (x) d x = b^{2} + 1 - 2$
On differentiating w.r.t. to $b$ , we get
$f (b) = \frac{2 b}{2 b ^{2} + 1}$
$\Rightarrow f (x) = \frac{x}{x ^{2} + 1}$

Q. The area bounded by the x-axis, the curve y=f(x) and the lines x=1, x=b is equal to b2+1​−2​ for all b>1, then f(x) is

x−1​

x+1​

x2+1​

x/x2+1​