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Punjab Board Class 11 Mathematics

Exercise 2.1 Solutions — Functions and Graphs

Complete worked solutions for Exercise 2.1 from Chapter 2 (Functions and Graphs), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 2.1 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 2: Functions and Graphs.

Question 1: Given (a) \(f(x) = x^2 - 1\) and (b) \(f(x) = \sqrt{2x + 3}\), find (i) \(f(-3)\) (ii) \(f(0)\) (iii) \(f(x-2)\) (iv) \(f(x^2+3)\).

Is sawal mein function mein \(x\) ki jagah di gayi values substitute karni hain. Part (b) mein square root hai, is liye check karenge ke andar wali value negative to nahi — negative radicand ka matlab value real numbers mein exist nahi karti.

Part (a): \(f(x) = x^2 - 1\)

(i) \(f(-3) = (-3)^2 - 1 = 9 - 1 = 8\)

(ii) \(f(0) = 0^2 - 1 = -1\)

(iii) \(f(x-2) = (x-2)^2 - 1 = x^2 - 4x + 4 - 1 = x^2 - 4x + 3\)

(iv) \(f(x^2+3) = (x^2+3)^2 - 1 = x^4 + 6x^2 + 9 - 1 = x^4 + 6x^2 + 8\)

Part (b): \(f(x) = \sqrt{2x + 3}\)

(i) \(f(-3) = \sqrt{2(-3) + 3} = \sqrt{-3}\). Since the radicand is negative, \(f(-3)\) is not a real number (\(-3\) lies outside the domain \(x \ge -\frac{3}{2}\)).

(ii) \(f(0) = \sqrt{2(0) + 3} = \sqrt{3}\)

(iii) \(f(x-2) = \sqrt{2(x-2) + 3} = \sqrt{2x - 1}\)

(iv) \(f(x^2+3) = \sqrt{2(x^2+3) + 3} = \sqrt{2x^2 + 9}\)

Answer: (a) (i) \(8\) (ii) \(-1\) (iii) \(x^2 - 4x + 3\) (iv) \(x^4 + 6x^2 + 8\); (b) (i) not real (ii) \(\sqrt{3}\) (iii) \(\sqrt{2x-1}\) (iv) \(\sqrt{2x^2+9}\)

Common mistake: Writing \(f(-3) = \sqrt{-3}\) as a final answer without noticing it is not real. Always check the domain first for root functions.

Question 2: Find \(\frac{f(a+h) - f(a)}{h}\) and simplify, where (i) \(f(x) = 4x + 7\) (ii) \(f(x) = \sin x\) (iii) \(f(x) = x^3 + x^2 - 1\) (iv) \(f(x) = \tan x\).

Yeh difference quotient hai. Pehle \(f(a+h)\) expand karenge, phir \(f(a)\) subtract karke \(h\) se divide karenge. Trig parts mein sum-to-product identities kaam aayengi: \(\sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\).

(i) \(f(x) = 4x + 7\)

\[ \frac{f(a+h) - f(a)}{h} = \frac{[4(a+h) + 7] - [4a + 7]}{h} = \frac{4h}{h} = 4 \]

(ii) \(f(x) = \sin x\)

\[ \frac{\sin(a+h) - \sin a}{h} \]

Apply the identity \(\sin C - \sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\) with \(C = a+h\), \(D = a\):

\[ = \frac{2\cos\left(a + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \]

(iii) \(f(x) = x^3 + x^2 - 1\)

Expand \(f(a+h) = (a+h)^3 + (a+h)^2 - 1 = a^3 + 3a^2h + 3ah^2 + h^3 + a^2 + 2ah + h^2 - 1\). Subtracting \(f(a) = a^3 + a^2 - 1\):

\[ \frac{f(a+h) - f(a)}{h} = \frac{3a^2h + 3ah^2 + h^3 + 2ah + h^2}{h} = 3a^2 + 3ah + h^2 + 2a + h \]

(iv) \(f(x) = \tan x\)

Write \(\tan = \frac{\sin}{\cos}\) and combine over a common denominator:

\[ \frac{\tan(a+h) - \tan a}{h} = \frac{\frac{\sin(a+h)}{\cos(a+h)} - \frac{\sin a}{\cos a}}{h} = \frac{\sin(a+h)\cos a - \cos(a+h)\sin a}{h\cos(a+h)\cos a} \]

Apply the identity \(\sin A\cos B - \cos A\sin B = \sin(A - B)\):

\[ = \frac{\sin(a+h-a)}{h\cos(a+h)\cos a} = \frac{\sin h}{h\cos(a+h)\cos a} \]
Answer: (i) \(4\) (ii) \(\frac{2\cos\left(a+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}\) (iii) \(3a^2 + 3ah + h^2 + 2a + h\) (iv) \(\frac{\sin h}{h\cos(a+h)\cos a}\)

Question 3: Express (a) the area \(A\) of a square as a function of its perimeter \(P\); (b) the circumference \(C\) of a circle as a function of its area \(A\); (c) the surface area \(S\) of a cube as a function of its volume \(V\).

Har part mein pehle standard geometry formula likhenge, phir us mein se beech wali variable (side ya radius) eliminate karke required relation banayenge.

(a) Perimeter of square: \(P = 4s \implies s = \frac{P}{4}\). Area: \(A = s^2\). Substituting:

\[ A = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16} \]

(b) Area of circle: \(A = \pi r^2 \implies r = \sqrt{\frac{A}{\pi}}\). Circumference: \(C = 2\pi r\). Substituting:

\[ C = 2\pi\sqrt{\frac{A}{\pi}} = 2\sqrt{\pi A} \]

(c) Volume of cube: \(V = s^3 \implies s = V^{1/3}\). Surface area: \(S = 6s^2\). Substituting:

\[ S = 6\left(V^{1/3}\right)^2 = 6V^{2/3} \]
Answer: (a) \(A = \frac{P^2}{16}\) (b) \(C = 2\sqrt{\pi A}\) (c) \(S = 6V^{2/3}\)

Question 4: Find the domain and the range of \(g\), where (i) \(g(x) = 5 - x\) (ii) \(g(x) = \sqrt{x+2}\) (iii) \(g(x) = \begin{cases} 6x+7, & x \le -2 \\ 4-3x, & x > -2 \end{cases}\) (iv) \(g(x) = |x-5|\) (v) \(g(x) = \frac{x+2}{3-x}\).

Domain ke liye restrictions dekhte hain: denominator zero nahi, square root ke andar negative nahi. Range ke liye har branch/piece ka behaviour check karte hain; rational function mein \(y\) ke liye solve karke forbidden \(y\) nikaalte hain.

(i) \(g(x) = 5 - x\) — linear, no restrictions.

Domain \(= \mathbb{R}\), Range \(= \mathbb{R}\).

(ii) \(g(x) = \sqrt{x+2}\) — need \(x + 2 \ge 0\).

Domain \(= [-2, \infty)\). A square root is never negative, so Range \(= [0, \infty)\).

(iii) Piecewise. Both pieces together cover every real \(x\), so Domain \(= \mathbb{R}\). For \(x \le -2\), \(g(x) = 6x + 7\) is increasing: at \(x = -2\), \(g = -5\); as \(x \to -\infty\), \(g \to -\infty\). This piece gives \((-\infty, -5]\). For \(x > -2\), \(g(x) = 4 - 3x\) is decreasing: just right of \(-2\), \(g\) is just below \(10\); as \(x \to \infty\), \(g \to -\infty\). This piece gives \((-\infty, 10)\). Union: \((-\infty, 10)\).

Domain \(= \mathbb{R}\), Range \(= (-\infty, 10)\).

(iv) \(g(x) = |x-5|\) — defined everywhere, never negative, minimum \(0\) at \(x = 5\).

Domain \(= \mathbb{R}\), Range \(= [0, \infty)\).

(v) \(g(x) = \frac{x+2}{3-x}\) — denominator \(3 - x \neq 0 \implies x \neq 3\).

Domain \(= \mathbb{R} - \{3\}\). For the range, set \(y = \frac{x+2}{3-x}\) and solve for \(x\):

\[ y(3-x) = x + 2 \implies 3y - xy = x + 2 \implies 3y - 2 = x(y+1) \implies x = \frac{3y-2}{y+1} \]

This requires \(y + 1 \neq 0\), i.e. \(y \neq -1\). Range \(= \mathbb{R} - \{-1\}\).

Answer: (i) Dom \(\mathbb{R}\), Ran \(\mathbb{R}\) (ii) Dom \([-2,\infty)\), Ran \([0,\infty)\) (iii) Dom \(\mathbb{R}\), Ran \((-\infty,10)\) (iv) Dom \(\mathbb{R}\), Ran \([0,\infty)\) (v) Dom \(\mathbb{R}-\{3\}\), Ran \(\mathbb{R}-\{-1\}\)

Common mistake: In part (iii), students include \(y = 10\) in the range — but at \(x = -2\) the second piece does not apply (it needs \(x > -2\)), so \(10\) is never attained.

Question 5: Given \(f(x) = x^3 - ax^2 + bx + 1\), if \(f(2) = -3\) and \(f(-1) = 0\), find the values of \(a\) and \(b\).

Dono conditions ko function mein rakh kar \(a\) aur \(b\) mein do linear equations banayenge, phir substitution se solve karenge.

From \(f(2) = -3\):

\[ (2)^3 - a(2)^2 + b(2) + 1 = -3 \implies 9 - 4a + 2b = -3 \implies 2a - b = 6 \quad \cdots (1) \]

From \(f(-1) = 0\):

\[ (-1)^3 - a(-1)^2 + b(-1) + 1 = 0 \implies -a - b = 0 \implies a = -b \quad \cdots (2) \]

Substitute (2) into (1): \(2(-b) - b = 6 \implies -3b = 6 \implies b = -2\). Then \(a = -(-2) = 2\).

Answer: \(a = 2\), \(b = -2\)

Question 6: A stone falls from a height; its height after \(x\) seconds is \(h(x) = 40 - 10x^2\) (metres). (i) Find the height when (a) \(x = 1\) s (b) \(x = 1.5\) s (c) \(x = 1.7\) s. (ii) When does the stone strike the ground?

Part (i) mein sirf values substitute karni hain. Part (ii) mein jab stone ground par lagega to height zero hogi — \(h(x) = 0\) rakh kar \(x\) solve karenge, aur negative time reject kar denge.

(i)

(a) \(h(1) = 40 - 10(1)^2 = 40 - 10 = 30\) m

(b) \(h(1.5) = 40 - 10(1.5)^2 = 40 - 10(2.25) = 40 - 22.5 = 17.5\) m

(c) \(h(1.7) = 40 - 10(1.7)^2 = 40 - 10(2.89) = 40 - 28.9 = 11.1\) m

(ii) On striking the ground, \(h(x) = 0\):

\[ 40 - 10x^2 = 0 \implies 10x^2 = 40 \implies x^2 = 4 \implies x = \pm 2 \]

Time cannot be negative, so \(x = 2\) seconds.

Answer: (i) (a) \(30\) m (b) \(17.5\) m (c) \(11.1\) m; (ii) after \(2\) seconds

Question 7: Consider \(f(x) = 3x - 5\). (i) Determine the domain and range. (ii) Is it one-to-one? Justify. (iii) Is it onto if the co-domain is all real numbers? Explain.

Linear function har real \(x\) par defined hai. One-to-one ke liye \(f(x_1) = f(x_2)\) se \(x_1 = x_2\) prove karenge. Onto ke liye range aur co-domain compare karenge.

(i) A linear polynomial is defined for every real number, and takes every real value. Domain \(= \mathbb{R}\), Range \(= \mathbb{R}\).

(ii) Let \(f(x_1) = f(x_2)\). Then:

\[ 3x_1 - 5 = 3x_2 - 5 \implies 3x_1 = 3x_2 \implies x_1 = x_2 \]

Hence \(f\) is one-to-one.

(iii) The co-domain is \(\mathbb{R}\) and the range is also \(\mathbb{R}\). Since range = co-domain, \(f\) is onto.

Answer: (i) Domain \(\mathbb{R}\), Range \(\mathbb{R}\) (ii) Yes, one-to-one (iii) Yes, onto

Question 8: Let \(f: \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = \frac{2x-3}{x+1}\). (i) Find the domain and range. (ii) Is \(f\) onto? (iii) Prove \(f\) is one-to-one.

Denominator zero nahi ho sakta — yahan se domain. Range ke liye \(y = f(x)\) se \(x\) solve karke dekhenge ke kaunsi \(y\) impossible hai. One-to-one proof mein cross-multiplication ke baad terms cancel honge.

(i) Domain: \(x + 1 \neq 0 \implies x \neq -1\), so Domain \(= \mathbb{R} - \{-1\}\).

Range: let \(y = \frac{2x-3}{x+1}\). Then:

\[ y(x+1) = 2x - 3 \implies yx + y = 2x - 3 \implies yx - 2x = -3 - y \implies x(y-2) = -(y+3) \]
\[ x = \frac{-(y+3)}{y-2}, \qquad y \neq 2 \]

So Range \(= \mathbb{R} - \{2\}\).

(ii) The co-domain is \(\mathbb{R}\) but the range is \(\mathbb{R} - \{2\}\). Since range \(\neq\) co-domain, \(f\) is not onto.

(iii) Let \(f(x_1) = f(x_2)\) with \(x_1, x_2 \neq -1\). Cross-multiplying:

\[ \frac{2x_1-3}{x_1+1} = \frac{2x_2-3}{x_2+1} \implies (2x_1-3)(x_2+1) = (2x_2-3)(x_1+1) \]
\[ 2x_1x_2 + 2x_1 - 3x_2 - 3 = 2x_1x_2 + 2x_2 - 3x_1 - 3 \]
\[ 2x_1 - 3x_2 = 2x_2 - 3x_1 \implies 5x_1 = 5x_2 \implies x_1 = x_2 \]

Hence \(f\) is one-to-one.

Answer: (i) Domain \(\mathbb{R}-\{-1\}\), Range \(\mathbb{R}-\{2\}\) (ii) Not onto (iii) One-to-one (proved)

Question 9: Consider \(f: \mathbb{R}^+ \to \mathbb{R}^+\) defined by \(f(x) = e^x\). Show whether \(f\) is bijective.

One-to-one to asani se prove ho jata hai (natural log le kar). Lekin onto check karte waqt dekhenge ke co-domain ka har element — masalan \(y = \frac{1}{2}\) — ka pre-image domain \(\mathbb{R}^+\) mein hai ya nahi.

One-to-one: Let \(f(x_1) = f(x_2)\), i.e. \(e^{x_1} = e^{x_2}\). Taking the natural logarithm of both sides: \(\ln(e^{x_1}) = \ln(e^{x_2}) \implies x_1 = x_2\). So \(f\) is one-to-one.

Onto check: Take \(y = \frac{1}{2} \in \mathbb{R}^+\) (the co-domain). Solving \(f(x) = y\) gives \(e^x = \frac{1}{2} \implies x = \ln\frac{1}{2} = -\ln 2 < 0\). This \(x\) is not in the domain \(\mathbb{R}^+\) (positive reals). So \(y = \frac{1}{2}\) has no pre-image, and \(f\) is not onto.

Since onto fails, \(f: \mathbb{R}^+ \to \mathbb{R}^+\) is not bijective as stated.

Answer: One-to-one but not onto, hence not bijective (e.g. \(y = \frac{1}{2}\) has no pre-image in \(\mathbb{R}^+\)). Note: with domain \(\mathbb{R}\) the function \(f(x) = e^x : \mathbb{R} \to \mathbb{R}^+\) is bijective.

Common mistake: Assuming \(e^x\) is automatically onto \(\mathbb{R}^+\) — for \(0 < x\) we only get \(e^x > 1\), so values like \(\frac{1}{2}\) are missed.

Question 10: Let \(g: \mathbb{R} \to \mathbb{R}\) be given by \(g(x) = x^2 - 3x\). Determine whether \(g\) is injective and/or surjective.

Injective (one-to-one) disprove karne ke liye ek counter-example kaafi hai. Surjective (onto) check karne ke liye completing square se range nikaal kar co-domain se compare karenge.

Injective: \(g(0) = 0 - 0 = 0\) and \(g(3) = 9 - 9 = 0\). Since \(0 \neq 3\) but \(g(0) = g(3)\), \(g\) is not injective.

Surjective: Complete the square to find the range:

\[ g(x) = x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} \ge -\frac{9}{4} \]

The range is \(\left[-\frac{9}{4}, \infty\right)\), which is not all of the co-domain \(\mathbb{R}\) (e.g. \(-3\) is never attained). So \(g\) is not surjective.

Answer: \(g\) is neither injective nor surjective.