Let f(x) be the function given by $f(x)=3 x^5-5 x^3+21 x+3 \sin x+4 \cos x+5$. Then,

(a) f(x) is increasing on R and f(x) = 0 has exactly one negative root
(b) f(x) is decreasing on R and f(x) = 0 has exactly one positive root
(c) f(x) is decreasing and f(x) = 0 has exactly one negative root
(d) f(x) is increasing and f(x) = 0 has exactly one positive root

(a) only

We have,

$f(x)=3 x^5-5 x^3+21 x+3 \sin x+4 \cos x+5$

$\Rightarrow f'(x)=15 x^4-15 x^2+21+3 \cos x-4 \sin x$

$\Rightarrow f'(x)=\phi(x)+\psi(x)$, where

$\phi(x)=15 x^4-15 x^2+21$ and $\psi(x)=3 \cos x-4 \sin x$

Now,

$\phi(x)=15 x^4-15 x^2+21$

$\Rightarrow \phi(x)=15\left(x^4-x^2+\frac{7}{5}\right)$

$\Rightarrow \phi(x)=15\left\{\left(x^2-\frac{1}{2}\right)^2+\frac{23}{20}\right\}$

$\Rightarrow \phi(x)=15\left(x^2-\frac{1}{2}\right)^2+\frac{69}{4} \geq \frac{69}{4}$ for all $x \in R$

and, $\psi(x)=3 \cos x-4 \sin x$

$\Rightarrow -5 \leq \psi(x) \leq 5$ for all $x \in R$

∴  $f'(x) \geq \frac{69}{4}-5>0$ for all $x \in R$

⇒ f(x) is increasing on R.

We observe that f(0) = 9 > 0 and $f(-\infty)=-\infty$

Therefore, y = f(x) cuts x-axis exactly ones.

Hence, f(x) = 0 has exactly one negative root.