Calculus — NCERT Solutions

(i) Left-hand limit (as $x \to 3^-$):
$$\lim_{x \to 3^-} f(x) = -1$$
The function approaches $y = -1$ from the left. (ii) Right-hand limit (as $x \to 3^+$):
$$\lim_{x \to 3^+} f(x) = 2$$
The function approaches $y = 2$ from the right. (iii) Actual value at $x = 3$:
The actual value of the function at $x = 3$ is $f(3) = 2$. Conclusion: Since the left-hand limit $(-1) \neq$ right-hand limit $(2)$, the limit of the function at $x = 3$ does not exist. Therefore, the function is not continuous at $x = 3$.

Discussion/Reflection:

(i) $f(x) = c$ (constant function):
For any point $a$, $\lim_{x \to a} c = c = f(a)$. So LHL = RHL = $f(a)$. Hence continuous everywhere.

(ii) $f(x) = x^n$, $n \in \mathbb{N}$:
For any point $a$, $\lim_{x \to a} x^n = a^n = f(a)$ by direct substitution. Hence continuous on $\mathbb{R}$.

(iii) $\sin x$ and $\cos x$:
Both are defined for all real $x$ and their limits equal their values at every point. Hence continuous on $\mathbb{R}$.

(iv) $f(x) = |x|$:
At $x = 0$: LHL $= \lim_{x \to 0^-}(-x) = 0$, RHL $= \lim_{x \to 0^+}(x) = 0$, and $f(0) = 0$. So continuous at $x = 0$ and clearly continuous elsewhere. Hence continuous on $\mathbb{R}$.

(v) Polynomial functions:
Every polynomial $p(x) = a_0x^n + \cdots + a_n$ satisfies $\lim_{x \to a} p(x) = p(a)$ for all $a \in \mathbb{R}$ (by direct substitution). Hence always continuous.

Part 1: Derivative of a constant $f(x) = c$

Using the first principle:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0$$

$$\boxed{\frac{d}{dx}(c) = 0}$$

Part 2: Derivative of $f(x) = ax$

Using the first principle:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{a(x+h) - ax}{h}$$
$$= \lim_{h \to 0} \frac{ax + ah - ax}{h} = \lim_{h \to 0} \frac{ah}{h} = \lim_{h \to 0} a = a$$

$$\boxed{\frac{d}{dx}(ax) = a}$$

Given: Area of square $= y = x^2$, where $x$ is the side length. At a given moment, $x = a$.

Concept: The instantaneous rate of change of area with respect to side length is the derivative $\dfrac{dy}{dx}$ evaluated at $x = a$.

Finding the derivative using first principle:
$$\frac{dy}{dx} = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$$
$$= \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x$$

At $x = a$:
$$\left.\frac{dy}{dx}\right|_{x=a} = 2a$$

Conclusion: The tendency (instantaneous rate of change) of the area at the moment when the side is $a$ is $\mathbf{2a}$ square units per unit length.

Given: Area of a circle $A = \pi r^2$. Find $\dfrac{dA}{dr}$ at $r = 5$ cm.

Concept: Rate of change of area with respect to radius = $\dfrac{dA}{dr}$.

Working:
$$A = \pi r^2$$
$$\frac{dA}{dr} = 2\pi r$$

At $r = 5$ cm:
$$\left.\frac{dA}{dr}\right|_{r=5} = 2\pi (5) = 10\pi \text{ cm}^2/\text{cm}$$

Answer: The rate of change of area with respect to radius when $r = 5$ cm is $\mathbf{10\pi}$ cm²/cm $\approx 31.4$ cm²/cm.

Given:

• Volume of cube: $V = x^3$, where $x$ = edge length.
• $\dfrac{dV}{dt} = 9$ cm³/s.
• Find $\dfrac{dS}{dt}$ when $x = 10$ cm.

Step 1: Relate $V$ and $x$.
$$V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$
$$9 = 3(10)^2 \frac{dx}{dt} = 300\frac{dx}{dt}$$
$$\frac{dx}{dt} = \frac{9}{300} = \frac{3}{100} \text{ cm/s}$$

Step 2: Surface area of cube.
$$S = 6x^2 \implies \frac{dS}{dt} = 12x \frac{dx}{dt}$$

Step 3: Substitute $x = 10$ and $\dfrac{dx}{dt} = \dfrac{3}{100}$:
$$\frac{dS}{dt} = 12(10) \times \frac{3}{100} = 120 \times \frac{3}{100} = \frac{360}{100} = 3.6 \text{ cm}^2/\text{s}$$

Answer: The surface area is increasing at the rate of $\mathbf{3.6}$ cm²/s.

Given: $f(x) = x^3 + 3x^2 + 1$.

Step 1: Find the derivative.
$$f'(x) = 3x^2 + 6x$$

At $x = -2$:

• Slope: $f'(-2) = 3(4) + 6(-2) = 12 - 12 = 0$
• $y$-coordinate: $f(-2) = (-8) + 3(4) + 1 = -8 + 12 + 1 = 5$
• Equation of tangent: $y = 0(x - (-2)) + 5 \implies y = 5$

At $x = 1$:

• Slope: $f'(1) = 3(1) + 6(1) = 9$
• $y$-coordinate: $f(1) = 1 + 3 + 1 = 5$
• Equation of tangent: $y = 9(x - 1) + 5 = 9x - 9 + 5 \implies y = 9x - 4$

(Use GeoGebra to visually verify these tangent lines on the graph of $f(x)$.)

Given: $y = f(x) = x^3 - 2x^2 + x - 3$. Find the tangent at $x = 1$.

Step 1: Find the derivative $f'(x)$.
$$f'(x) = 3x^2 - 4x + 1$$

Step 2: Find the slope at $x = 1$.
$$m = f'(1) = 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0$$

Step 3: Find the $y$-coordinate at $x = 1$.
$$b = f(1) = (1)^3 - 2(1)^2 + (1) - 3 = 1 - 2 + 1 - 3 = -3$$

Step 4: Write the equation of the tangent line.
$$y = m(x - a) + b = 0(x - 1) + (-3)$$
$$\boxed{y = -3}$$

Answer: The equation of the tangent to the curve at $x = 1$ is $y = -3$ (a horizontal line).