Q.Find the values of $x$ for which the function $f(x) = 3 + 3x - 3x^2 + x^3$ is strictly increasing.

Show $x-\log_e(1+x)>0$ on $(0,1)$ by checking it is increasing from $0$; the other options fail.

Using monotonicity to prove an inequality is a CBSE/NCERT Class 12 application of derivatives technique.

Define $g(x)=x-\log_e(1+x)$. Then

$g'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0 \quad \text{for } x\in(0,1).$

So $g$ is increasing, and $g(0)=0$, hence $g(x)>0$ for $x\in(0,1)$, i.e.

$\log_e(1+x) < x.$

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Split at $x=0$ to remove the modulus; the derivative is positive on both pieces, so $f$ is increasing on all of $\mathbb{R}$.

Monotonicity via the sign of $f'$ is a CBSE/NCERT Class 12 application of derivatives topic.

For $x\ge0$: $f(x)=\dfrac{x}{1+x}$, so

$f'(x) = \frac{(1+x)-x}{(1+x)^2} = \frac{1}{(1+x)^2} > 0.$

For $x<0$: $|x|=-x$, so $f(x)=\dfrac{x}{1-x}$, and

$f'(x) = \frac{(1-x)+x}{(1-x)^2} = \frac{1}{(1-x)^2} > 0.$

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$\dfrac{dy}{dx}=6x-1=5$ at $(1,3)$, so the normal slope is $-\tfrac{1}{5}$; the normal line is $x + 5y - 16 = 0$ — option (ii).

Differentiate the curve $y = 3x^2 - x + 1$:

$\frac{dy}{dx} = 6x - 1.$

At the point $(1,3)$ the tangent slope is

$m_t = 6(1) - 1 = 5.$

The normal is perpendicular to the tangent, so its slope is the negative reciprocal:

$m_n = -\frac{1}{m_t} = -\frac{1}{5}.$

Using the point-slope form through $(1,3)$:

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State the standard sufficient condition for strict monotonic increase.

A differentiable function $f(x)$ is strictly increasing on $R$ if its derivative is positive throughout: …

The slope of the tangent at a point on $y = f(x)$ is the value of the derivative $\dfrac{dy}{dx}$ there, also called the gradient of the curve.

For a curve $y = f(x)$, the tangent line at a point $(a, b)$ has slope

$\left.\frac{dy}{dx}\right|_{(a,\,b)} = f'(a).$

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A function is increasing at a point $c$ when its slope there is positive, i.e. $f'(c) > 0$.

The derivative $f'(c)$ measures the instantaneous rate of change (the slope of the tangent) of the function at $x = c$. Interpreting the sign:

• if $f'(c) > 0$, the tangent slopes upward, so the function values are rising as $x$ increases past $c$ — the function is increasing at $c$; …

If $f'(x) > 0$ throughout $(a, b)$, the function is increasing on that interval, not decreasing. So the statement is False.

The sign of the derivative tells us how a function behaves:

• $f'(x) > 0$ on an interval $\Rightarrow$ $f$ is increasing there,
• $f'(x) < 0$ on an interval $\Rightarrow$ $f$ is decreasing there. …