Question

Let x>0, and $\left|{\int }_x^{x+1}\sin t^{2}dt\right|\le \frac{1}{g\left(x\right)}$, then g(x) equals

• A. 2x
• B. 1/x
• C. x
• D. 1/2x

Answer 1

The correct option is B 1/x

Solution:

We have:

$x>0$

$\left|{\int }_x^{x+1}\sin t^{2}dt\right|\le \frac{1}{g\left(x\right)}-------f\left(1\right)$

We know that,

$\sin t^2⩽t^{}(t>0)$

Integrating both side applying limits concerned

${\int }_x^{x+1}\sin t^{2}dt\le {\int }_x^{x+1}{t}^{}dt$

From eq (i) we can deduce that:

${\int }_x^{x+1}{t}^{}dt\le \frac{1}{g\left(x\right)}$

${\left[\frac{x^2}{2}\right)}_x^{x+1}\le \frac{1}{g\left(x\right)}$

$\frac{(x+1{)}^2-x^2}{2}⩽\frac{1}{g\left(x\right)}$

$\frac{2x+1}{2}⩽\frac{1}{g\left(x\right)}$

Now, considering boundary situation

$\frac{2x+1}{2}=\frac{1}{g\left(x\right)}$

Now, differentiate both sides w.r.t x,

1 =$\frac{g^{\prime }\left(x\right)}{g(x{)}^2}$

Integrating both sides;

$\int 1.dx=\int \frac{g^{\prime }\left(x\right)}{g(x{)}^2}dx$

$\left[\begin{array}{c}g\left(x\right)=t\\ g\left(x\right)dx=dx\end{array}\right]$

$x=-\frac{t^{-2+1}}{-2+1}$

$x=+\frac{1}{t}$

$\Rightarrow t=1/x$

Or

Answer: $g\left(x\right)=1/x$