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Math Notes For Class 12 Chapter 1 Ex 1.1 Solved & Quiz
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Math Notes For Class 12 Chapter 1 Ex 1.1 Solved & Quiz
In Class 12 Mathematics (FSc Pre-Engineering), Chapter 1 establishes the foundations of Calculus by introducing the concepts of Functions and Limits. A function $f$ from a set $X$ to a set $Y$ is a rule or correspondence that assigns to each element $x \in X$ a unique element $y \in Y$, denoted as $y = f(x)$.
Key mathematical definitions and rules required for Exercise 1.1 include:
| Concept / Function Type | Mathematical Form | Domain Constraint Rule | Key Identity / Property |
|---|---|---|---|
| Polynomial Function | $P(x) = a_n x^n + \dots + a_0$ | $\mathbb{R}$ (All real numbers) | Continuous over $(-\infty, \infty)$ |
| Radical Function | $f(x) = \sqrt{g(x)}$ | $g(x) \ge 0$ | Radicand must be non-negative in $\mathbb{R}$ |
| Rational Function | $f(x) = \frac{P(x)}{Q(x)}$ | $Q(x) \ne 0$ | Exclude points where denominator vanishes |
| Difference Quotient | $\frac{f(a+h) - f(a)}{h}$ | $h \ne 0$ | Used to derive derivative formulas |
| Trigonometric Difference | $\sin A - \sin B$ | $\mathbb{R}$ | $2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ |
| Cosine Difference | $\cos A - \cos B$ | $\mathbb{R}$ | $-2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ |
Question 1: Given that:
(a) $f(x) = x^2 - x$ (b) $f(x) = \sqrt{x + 6}$
Find for each case: (i) $f(-2)$, (ii) $f(0)$, (iii) $f(x - 1)$, (iv) $f(x^2 + 1)$.
Solution:
Part (a): Given $f(x) = x^2 - x$
(i) For $f(-2)$:
Substitute $x = -2$:
$f(-2) = (-2)^2 - (-2) = 4 + 2 = 6$.
Answer: $$f(-2) = 6$$
(ii) For $f(0)$:
Substitute $x = 0$:
$f(0) = (0)^2 - (0) = 0$.
Answer: $$f(0) = 0$$
(iii) For $f(x - 1)$:
Substitute $x = x - 1$:
$f(x - 1) = (x - 1)^2 - (x - 1) = (x^2 - 2x + 1) - x + 1 = x^2 - 3x + 2$.
Answer: $$f(x - 1) = x^2 - 3x + 2$$
(iv) For $f(x^2 + 1)$:
Substitute $x = x^2 + 1$:
$f(x^2 + 1) = (x^2 + 1)^2 - (x^2 + 1) = (x^4 + 2x^2 + 1) - x^2 - 1 = x^4 + x^2$.
Answer: $$f(x^2 + 1) = x^4 + x^2$$
Part (b): Given $f(x) = \sqrt{x + 6}$
(i) For $f(-2)$:
Substitute $x = -2$:
$f(-2) = \sqrt{-2 + 6} = \sqrt{4} = 2$.
Answer: $$f(-2) = 2$$
(ii) For $f(0)$:
Substitute $x = 0$:
$f(0) = \sqrt{0 + 6} = \sqrt{6}$.
Answer: $$f(0) = \sqrt{6}$$
(iii) For $f(x - 1)$:
Substitute $x = x - 1$:
$f(x - 1) = \sqrt{(x - 1) + 6} = \sqrt{x + 5}$.
Answer: $$f(x - 1) = \sqrt{x + 5}$$
(iv) For $f(x^2 + 1)$:
Substitute $x = x^2 + 1$:
$f(x^2 + 1) = \sqrt{(x^2 + 1) + 6} = \sqrt{x^2 + 7}$.
Answer: $$f(x^2 + 1) = \sqrt{x^2 + 7}$$
Question 2: Find $\frac{f(a + h) - f(a)}{h}$ and simplify for the following functions:
(i) $f(x) = 6x - 9$ (ii) $f(x) = \sin x$ (iii) $f(x) = x^3 + 2x^2 - 1$ (iv) $f(x) = \cos x$
Solution:
Part (i): $f(x) = 6x - 9$
$f(a) = 6a - 9$
$f(a + h) = 6(a + h) - 9 = 6a + 6h - 9$
Now, calculate the difference quotient:
$$\frac{f(a + h) - f(a)}{h} = \frac{(6a + 6h - 9) - (6a - 9)}{h} = \frac{6h}{h} = 6$$
Answer: $$6$$
Part (ii): $f(x) = \sin x$
$f(a) = \sin a$
$f(a + h) = \sin(a + h)$
$$\frac{f(a + h) - f(a)}{h} = \frac{\sin(a + h) - \sin a}{h}$$
Using sum-to-product identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ with $A = a+h$ and $B = a$:
$$\sin(a+h) - \sin a = 2 \cos\left(\frac{2a+h}{2}\right) \sin\left(\frac{h}{2}\right) = 2 \cos\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$
Therefore:
$$\frac{f(a + h) - f(a)}{h} = \frac{2 \cos\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} = \cos\left(a + \frac{h}{2}\right) \left[ \frac{\sin(h/2)}{h/2} \right]$$
Answer: $$\frac{2}{h} \cos\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$
Part (iii): $f(x) = x^3 + 2x^2 - 1$
$f(a) = a^3 + 2a^2 - 1$
$f(a + h) = (a + h)^3 + 2(a + h)^2 - 1 = (a^3 + 3a^2h + 3ah^2 + h^3) + 2(a^2 + 2ah + h^2) - 1$
$f(a + h) - f(a) = 3a^2h + 3ah^2 + h^3 + 4ah + 2h^2$
Divide by $h$ ($h \ne 0$):
$$\frac{f(a + h) - f(a)}{h} = \frac{h(3a^2 + 3ah + h^2 + 4a + 2h)}{h} = 3a^2 + 3ah + h^2 + 4a + 2h$$
Answer: $$3a^2 + 4a + 3ah + 2h + h^2$$
Part (iv): $f(x) = \cos x$
$f(a) = \cos a$
$f(a + h) = \cos(a + h)$
$$\frac{f(a + h) - f(a)}{h} = \frac{\cos(a + h) - \cos a}{h}$$
Using sum-to-product identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$:
$$\cos(a+h) - \cos a = -2 \sin\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$
Dividing by $h$:
$$\frac{f(a + h) - f(a)}{h} = -\frac{2}{h} \sin\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$
Answer: $$-\frac{2}{h} \sin\left(a + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$
Question 3: Express the following:
(a) The perimeter $P$ of a square as a function of its area $A$.
(b) The area $A$ of a circle as a function of its circumference $C$.
(a) The volume $V$ of a cube as a function of the area $A$ of its base.
Solution:
Part (a): Let $x$ be the side length of the square.
Perimeter $P = 4x \implies x = \frac{P}{4}$.
Area $A = x^2 \implies A = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16}$.
Solving for $P$ in terms of $A$:
$P^2 = 16A \implies P = 4\sqrt{A}$.
Answer: $$P(A) = 4\sqrt{A}$$
Part (b): Let $r$ be the radius of the circle.
Circumference $C = 2\pi r \implies r = \frac{C}{2\pi}$.
Area $A = \pi r^2$. Substitute $r = \frac{C}{2\pi}$:
$$A = \pi \left(\frac{C}{2\pi}\right)^2 = \pi \cdot \frac{C^2}{4\pi^2} = \frac{C^2}{4\pi}$$
Answer: $$A(C) = \frac{C^2}{4\pi}$$
Part (c): Let $x$ be the length of an edge of the cube.
Area of base $A = x^2 \implies x = \sqrt{A} = A^{1/2}$.
Volume $V = x^3$. Substitute $x = A^{1/2}$:
$$V = (A^{1/2})^3 = A^{3/2}$$
Answer: $$V(A) = A^{3/2}$$
Question 4: Find the domain and range of the following functions $g(x)$:
(i) $g(x) = 2x - 5$
(ii) $g(x) = \sqrt{x^2 - 4}$
(iii) $g(x) = \sqrt{x + 1}$
(iv) $g(x) = |x - 3|$
(v) $g(x) = \begin{cases} 6x + 7, & x \le -2 \\ 4 - 3x, & x > -2 \end{cases}$
(vi) $g(x) = \frac{x^2 - 16}{x - 4}, \quad x \ne 4$
(vii) $g(x) = \frac{x^2 + 3x + 2}{x + 1}, \quad x \ne -1$
(viii) $g(x) = \sqrt{x^2 - 16}$
Solution:
Part (i): $g(x) = 2x - 5$
This is a linear polynomial, defined for all real numbers $x$.
Domain: $(-\infty, \infty)$ or $\mathbb{R}$
Range: $(-\infty, \infty)$ or $\mathbb{R}$
Answer: $$\text{Domain} = \mathbb{R}, \quad \text{Range} = \mathbb{R}$$
Part (ii): $g(x) = \sqrt{x^2 - 4}$
For real output, radicand must be non-negative: $x^2 - 4 \ge 0 \implies (x - 2)(x + 2) \ge 0$.
This holds when $x \le -2$ or $x \ge 2$.
The principal square root yields non-negative real values starting at $0$.
Domain: $(-\infty, -2] \cup [2, \infty)$
Range: $[0, \infty)$
Answer: $$\text{Domain} = (-\infty, -2] \cup [2, \infty), \quad \text{Range} = [0, \infty)$$
Part (iii): $g(x) = \sqrt{x + 1}$
Radicand constraint: $x + 1 \ge 0 \implies x \ge -1$.
The range of $\sqrt{x+1}$ starts at $\sqrt{-1+1} = 0$ and grows infinitely.
Domain: $[-1, \infty)$
Range: $[0, \infty)$
Answer: $$\text{Domain} = [-1, \infty), \quad \text{Range} = [0, \infty)$$
Part (iv): $g(x) = |x - 3|$
Absolute value is defined for every real number $x$.
Since absolute value cannot be negative, $|x - 3| \ge 0$.
Domain: $(-\infty, \infty)$ or $\mathbb{R}$
Range: $[0, \infty)$
Answer: $$\text{Domain} = \mathbb{R}, \quad \text{Range} = [0, \infty)$$
Part (v): Piecewise function $g(x) = \begin{cases} 6x + 7, & x \le -2 \\ 4 - 3x, & x > -2 \end{cases}$
The domain covers $(-\infty, -2] \cup (-2, \infty) = (-\infty, \infty)$.
For $x \le -2$: $g(x) \le 6(-2) + 7 = -5$, so output is $(-\infty, -5]$.
For $x > -2$: as $x \to -2^+$, $4 - 3x \to 4 - 3(-2) = 10$, and as $x \to \infty$, $4 - 3x \to -\infty$, giving output $(-\infty, 10)$.
Combining outputs: $(-\infty, -5] \cup (-\infty, 10) = (-\infty, 10)$.
Domain: $(-\infty, \infty)$
Range: $(-\infty, 10)$
Answer: $$\text{Domain} = \mathbb{R}, \quad \text{Range} = (-\infty, 10)$$
Part (vi): $g(x) = \frac{x^2 - 16}{x - 4}, \quad x \