Q10. Let $f(x)$ be the function given by $f(x) = 3x^5 - 5x^3 + 21x + 3\sin x + 4\cos x + 5$. Then: (a) $f(x)$ is increasing on $\mathbb{R}$ and $f(x) = 0$ has exactly one negative root (b) $f(x)$ is decreasing on $\mathbb{R}$ 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.

1. Key idea – monotonicity via the derivative

A function $f(x)$ is increasing on (\mathbb R) if its derivative $f'(x) > 0$ for every real $x$.

2. Differentiate the given $f(x)$

3. Bounding the trigonometric part

The combination $3\cos x - 4\sin x$ looks scary, but its size is limited:

so

4. Make a lower bound for the whole derivative

Add this worst-case scenario to the algebraic part:

Set $t = x^2 \;(t \ge 0)$ to view it as a quadratic:

Its discriminant is

so the quadratic is always positive. Consequently $f'(x) > 0$ for every $x$.
That proves $f(x)$ is strictly increasing on $\mathbb R$.

5. How many zeros and where?

A continuous strictly increasing function can cut the $x$-axis at most once.
• Evaluate near $x=0$:

• As $x \to -\infty$ the dominant term $3x^5 \to -\infty$, hence $f(x) \to -\infty$.

Because the function climbs from negative values (far left) to a positive value at $x=0$, the single crossing must be somewhere left of 0, i.e.

6. Match to the options

Only option (a) says "increasing" and "exactly one negative root" – so (a) is correct.

Think of the graph as a car on a road

1. Steering wheel = derivative. If the steering wheel always points uphill, the car forever climbs – that means the function is always getting bigger (increasing).
2. We look at the steering wheel by finding $f'(x)$.
3. We notice the worst the steering wheel can tilt down because of the wiggly $\sin$ and $\cos$ terms is limited – they can only change the slope a little bit (at most $\pm5$).
4. The big $3x^5$ term is like a huge engine: for most $x$ it easily overpowers the tiny wiggles, so the slope stays positive.
5. If the slope never turns negative, the road only climbs. A road that only climbs can cross sea-level (the $x$-axis) once at most.
6. Checking heights at $x=0$ (positive) and far on the left side (negative) shows the crossing is on the left, i.e., at a negative $x$.
   So the right choice is the one saying "always increasing" and "exactly one negative root".

Step 1 – Differentiate

Step 2 – Lower bound for $f'(x)$

Because $|3\cos x - 4\sin x| \le 5$,

Let $t = x^2 \;(t \ge 0)$. Then

Discriminant:

With $a=15>0$ and $\Delta < 0$, the quadratic is positive for every $t$, so

Hence $f(x)$ is strictly increasing on $\mathbb R$.

Step 3 – Locate the zero(s)

• At $x=0$:

• As $x \to -\infty$:

By the Intermediate Value Theorem, a root exists in $(-\infty,0)$.
Because $f$ is strictly increasing, this root is unique and it lies on the negative side.

Therefore the correct option is (a):
$\boxed{\text{(a)}}$.