Question

Let $f,g:R\to R$ be defined by $f\left(x\right)=x^2-\frac{\cos x}{2}$ and $g\left(x\right)=\frac{x\sin x}{2}$. Then

• A. $f\left(x\right)=g\left(x\right)$ has exactly one solution.
• B. $f\left(x\right)=g\left(x\right)$ has exactly two solutions.
• C. $f\left(x\right)-g\left(x\right)\to \infty$ as $x\to -\infty$
• D. The minimum value of $f\left(g\right(x\left)\right)$ is $-1.$

Answer 1

Answer: (B), (C)

$f\left(x\right)-g\left(x\right)\to \infty$ as $x\to -\infty$

Let $h\left(x\right)=f\left(x\right)-g\left(x\right)$
Then $h\left(x\right)=x^2-\frac{\cos x}{2}-\frac{x\sin x}{2}$
$h^{\prime }\left(x\right)=2x-\frac{x\cos x}{2}$
$\Rightarrow h^{\prime }\left(x\right)=\frac{x}{2}(4-\cos x)$

$h^{\prime }\left(x\right)>0\forall x>0$
and $h^{\prime }\left(x\right)<0\forall x<0$
So, $h$ is monotonically increasing for $x>0$
and $h$ is monotonically decreasing for $x<0$

Also, $h\left(x\right)\to \infty$ as $x\to \pm \infty$,
$h\left(0\right)=\frac{-1}{2}$
So, by intermediate value theorem,
$\exists$ a value of $x\in (-\infty ,0)$ such that $h\left(x\right)=0$
and $\exists$ a value of $x\in (0,\infty )$ such that $h\left(x\right)=0$

$f\left(g\right(x\left)\right)={\left(\frac{x\sin x}{2}\right)}^2-\frac{1}{2}\cos \left(\frac{x\sin x}{2}\right)$
Since, ${\left(\frac{x\sin x}{2}\right)}^2$ is a positive quantity, hence minimum value of $f\left(g\right(x\left)\right)$ is occure when $\cos \left(\frac{x\sin x}{2}\right)$ is maximum and ${\left(\frac{x\sin x}{2}\right)}^2$ is minimum.
The maximum value of $\cos \left(\frac{x\sin x}{2}\right)$ is $1$ when $x=n\pi ,n\in Z$.
And at $x=n\pi ,n\in Z,{\left(\frac{x\sin x}{2}\right)}^2=0$
$\therefore minf\left(g\right(x\left)\right)=-\frac{1}{2}$

Alternate Solution:
The graph of $y=x^2$ is a parabola passing through origin.
Since, $\cos x\in [-1,1],$ so at higher value of $x,$ $x^2-\frac{\cos x}{2}\approx x^2$
Hence, the graph of $y=x^2-\frac{\cos x}{2}$ is looks like parabola, shifted from origin.

And the graph of $y=\frac{x\sin x}{2}$ is sinusoidal with amplitude $\frac{x}{2}.$
Now, plot the graph of both the functions.
From graph, it is clear that $f\left(x\right)=g\left(x\right)$ has exactly two solutions.

$f\left(x\right)-g\left(x\right)=x^2-\frac{\cos x}{2}-\frac{x\sin x}{2}$
$=x^2(1-\frac{\cos x}{2x^2}-\frac{\sin x}{2x})$
So, as $x\to \pm \infty \Rightarrow f\left(x\right)-g\left(x\right)\to \infty$