Q.Prove that the logarithmic function is increasing on $(0, \infty)$.

Why this approach works

We need to decide whether $f(x)$ is even, odd, increasing, or decreasing.

• Even/odd are symmetry properties: even means $f(-x)=f(x)$; odd means $f(-x)=-f(x)$.
• Increasing/decreasing are monotonicity properties: we check the sign of the derivative $f'(x)$. If $f'(x) > 0$ for all $x$, the function is strictly increasing; if $f'(x) < 0$ for all $x$, it is strictly decreasing.

The fastest path is to test symmetry first (it’s quick), then check the derivative.

Step-by-step reasoning

1. Test for evenness Compute $f(-x)$:

Compare with $f(x) = x^3 + \frac{3}{2}x^2 + 3x + 3$.

They are not equal (signs on $x^3$ and $3x$ differ), so $f$ is not even.

2. Test for oddness For oddness we need $f(-x) = -f(x)$. Compute $-f(x)$:

This is not equal to $f(-x) = -x^3 + \frac{3}{2}x^2 - 3x + 3$ (the $x^2$ and constant terms differ in sign). So $f$ is not odd.

3. Find the derivative

4. Analyze the sign of $f'(x)$ …

We are given the function

and asked for the interval where it is strictly increasing.

A function is strictly increasing where its derivative is positive (and not zero on any interval). So the natural plan is: differentiate, set the derivative > 0, and solve for $x$.

1. Differentiate using the quotient rule

Let $u = \log x$ and $v = x^3$. Then

Simplify the numerator:

So

2. Determine where the derivative is positive

Since $x^4 > 0$ for all $x > 0$ (the domain of $\log x$), the sign of $y'$ is the sign of the numerator:

Exponentiate both sides:

Also, recall the domain: $x > 0$ because $\log x$ is defined only for positive $x$.

Thus $y' > 0$ exactly when

3. Interpret the result

This means the function is strictly increasing on $(0, e^{1/3})$ and strictly decreasing for $x > e^{1/3}$.

Now check the options:

• (A) $0 < x < e^{1/3}$ — matches exactly. …

The key idea is that a function is increasing on an interval if its derivative is $\ge 0$ for every point in that interval. Here $f(x) = x^2 + a x + 1$ is a parabola opening upward; its derivative is linear, so the condition reduces to a simple inequality.

1. Find the derivative

   $f'(x) = 2x + a$.

   For $f$ to be increasing on $[1,2]$, we need $f'(x) \ge 0$ for all $x \in [1,2]$.

2. Where is the derivative smallest on $[1,2]$?

   Since $f'(x) = 2x + a$ is linear with positive slope ($2 > 0$), it is smallest at the left endpoint $x = 1$.

   So the most restrictive condition is $f'(1) \ge 0$.

3. Set up the inequality

   $f'(1) = 2(1) + a = 2 + a \ge 0 \quad \Rightarrow \quad a \ge -2$.

4. Check the other endpoint

   At $x = 2$, $f'(2) = 4 + a$. If $a \ge -2$, then $f'(2) \ge 2 > 0$, so it automatically satisfies.

   Thus the condition $a \ge -2$ is both necessary and sufficient.

5. Interpret the question …

Concept & Intuition

To decide whether a function is increasing or decreasing on an interval, we look at its derivative. If the derivative is positive everywhere (except possibly at isolated points), the function is strictly increasing; if negative, it is strictly decreasing. Here, $y = \tan x - x$ is the difference between $\tan x$ and $x$. Since $\tan x$ grows faster than $x$ for $x>0$ (its slope is $\sec^2 x > 1$), we expect the difference to increase. The derivative will confirm this.

Step‑by‑step reasoning

1. Find the derivative

Using the identity $\sec^2 x = 1 + \tan^2 x$, we can rewrite:

2. Analyze the sign of $y'$ on $(0, \pi/2)$

   • For any $x \in (0, \pi/2)$, $\tan x > 0$ (since sine and cosine are both positive, with sine > 0).
   • Therefore $\tan^2 x > 0$ for every $x$ in $(0, \pi/2)$.
   • At the endpoint $x = 0$, $\tan 0 = 0$ so $y' = 0$, but that’s a single point, not an interval.

3. Interpret the sign

   Because $y' > 0$ for all $x$ in $(0, \pi/2)$, the function is strictly increasing on the entire open interval $(0, \pi/2)$. (A derivative that is zero at an isolated endpoint does not affect monotonicity on the open interval.)

4. Match with the options …

Concept & Intuition

We want to know where $f(x)=\tan^{-1}(\sin x+\cos x)$ is increasing.

A function is increasing where its derivative is positive.

The derivative of $\tan^{-1}(u)$ is $\frac{u'}{1+u^2}$, which is always positive in denominator, so the sign of $f'(x)$ is just the sign of $u' = \frac{d}{dx}(\sin x + \cos x)$.

Thus, the problem reduces to: Where is the derivative of $\sin x + \cos x$ positive?

That derivative is $\cos x - \sin x$. So we simply need $\cos x > \sin x$.

Step-by-step reasoning

1. Differentiate

The denominator $1+(\sin x+\cos x)^2$ is always positive (since it’s $1$ plus a square). So the sign of $f'(x)$ is exactly the sign of $\cos x - \sin x$.

2. Solve $\cos x - \sin x > 0$

Divide both sides by $\cos x$ (careful with sign changes — better to use a unit circle approach).

Alternatively, rewrite as:

Because $\cos x - \sin x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x\right) = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right)$.

3. Inequality becomes

Cosine is positive when its argument is in $(-\frac{\pi}{2}, \frac{\pi}{2})$ modulo $2\pi$.

So:

Subtract $\frac{\pi}{4}$:

4. Match with given options The interval $(-\frac{3\pi}{4}, \frac{\pi}{4})$ is not directly listed, but we look for a subinterval that lies entirely inside it.
   • Option (A): $(\frac{\pi}{4}, \frac{\pi}{2})$ — outside, since $\frac{\pi}{4}$ is the right endpoint. …

Concept: a differentiable function is increasing where f'(x) >= 0.

f(x) = 2x - arctan x - log(x + sqrt(x^2 + 1)).

Derivatives:

d/dx arctan x = 1/(1 + x^2)

d/dx log(x + sqrt(x^2+1)) = 1/sqrt(x^2 + 1) (this is sinh^-1 x)

So f'(x) = 2 - 1/(1 + x^2) - 1/sqrt(1 + x^2).

Put t = 1/sqrt(1 + x^2), so 0 < t <= 1 (t = 1 only at x = 0).

Then f'(x) = 2 - t^2 - t = -(t^2 + t - 2) = -(t + 2)(t - 1) = (t + 2)(1 - t). …

Step 1 — Differentiate

Using the chain rule on each term (each is a power of $\sin x$, whose derivative is $\cos x$):

Step 2 — Factor out the common $12\cos x$

This factorisation is the key move: it separates the sign into two independent pieces.

Step 3 — Show the bracket is always positive

Put $t=\sin x$ and consider $g(t)=t^{2}-t+1$. Its discriminant is

A quadratic with negative discriminant and positive leading coefficient has no real roots and is positive for every real $t$. (Equivalently, complete the square: $t^2-t+1=\left(t-\tfrac12\right)^{2}+\tfrac34\ \ge \tfrac34>0$.)

So the bracket never changes sign.

Step 4 — The sign of $f'$ is the sign of $\cos x$

Therefore:

• $f$ is strictly increasing where $\cos x>0$;
• $f$ is strictly decreasing where $\cos x<0$. …