Q.Show that the function $f(x) = x^3 - 3x^2 + 4x$ is increasing on $\mathbf{R}$.

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.

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

Checking the others: $e^x>1+x$ (not $<$), $\sin x < x$ (not $>$), and $\log_e x < 0 < x$ on $(0,1)$ (not $>$). Only (b) holds.

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

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

The function is continuous at $x=0$, and $f'>0$ on both sides, so $f$ is monotonically increasing on the whole real line $\mathbb{R}$.

$\dfrac{dy}{dx}=6x-1=5$ 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 $(1,3)$:

$y - 3 = -\frac{1}{5}(x - 1) \;\Rightarrow\; 5(y - 3) = -(x - 1) \;\Rightarrow\; 5y - 15 = -x + 1,$

$x + 5y - 16 = 0.$

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:

$f'(x)>0 \quad \text{for all } x\in R$

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

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

This slope — the rate of change of $y$ with respect to $x$ at that point — is commonly referred to as the gradient of the curve at $(a, b)$.

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 $f'(c) < 0$, the function is decreasing at $c$;
• if $f'(c) = 0$, $c$ is a stationary (critical) point.

Therefore the condition for $f$ to be increasing at $c$ is $f'(c) > 0$.

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.

The statement claims that $f'(x) > 0$ makes $f$ decreasing, which is the opposite of the correct rule. A positive slope means the graph rises as $x$ increases, i.e. $f$ increases.