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Punjab Board Class 11 Mathematics

Exercise 11.1 Solutions — Trigonometric Functions and their Graphs

Complete worked solutions for Exercise 11.1 from Chapter 11 (Trigonometric Functions and their Graphs), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 11.1 Solutions

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

Question 1: Determine whether the following functions are even, odd or neither even nor odd.

Is sawal mein hum har function mein \(x\) ki jagah \(-x\) rakhenge aur result ko \(f(x)\) aur \(-f(x)\) se compare karenge: agar \(f(-x) = f(x)\) to even, agar \(f(-x) = -f(x)\) to odd, warna neither. Basic identities yaad rakhein: \(\sin(-x) = -\sin x\), \(\cos(-x) = \cos x\), \(\tan(-x) = -\tan x\), \(\sec(-x) = \sec x\), \(\cot(-x) = -\cot x\).

(i) \(f(x) = \sin^2 x\)

\[ f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x) \]

Since \(f(-x) = f(x)\), the function is even.

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

\[ f(-x) = \sin(-x) + \cos(-x) = -\sin x + \cos x \]
\[ f(x) = \sin x + \cos x, \qquad -f(x) = -\sin x - \cos x \]

\(f(-x)\) equals neither \(f(x)\) nor \(-f(x)\), so the function is neither even nor odd.

(iii) \(f(x) = \sin^4 x + \cos^4 x\)

\[ f(-x) = \sin^4(-x) + \cos^4(-x) = (-\sin x)^4 + (\cos x)^4 = \sin^4 x + \cos^4 x = f(x) \]

Since \(f(-x) = f(x)\), the function is even.

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

\[ f(-x) = \tan(-x) + \sec(-x) = -\tan x + \sec x \]
\[ f(x) = \tan x + \sec x, \qquad -f(x) = -\tan x - \sec x \]

\(f(-x)\) equals neither, so the function is neither even nor odd.

(v) \(f(x) = \dfrac{1}{\csc x}\)

\[ f(x) = \frac{1}{\csc x} = \sin x, \qquad f(-x) = \sin(-x) = -\sin x = -f(x) \]

Since \(f(-x) = -f(x)\), the function is odd.

(vi) \(f(x) = \dfrac{\sin x + \sin 3x}{\cos x + \cos 3x}\)

\[ f(-x) = \frac{\sin(-x) + \sin(-3x)}{\cos(-x) + \cos(-3x)} = \frac{-\sin x - \sin 3x}{\cos x + \cos 3x} = -f(x) \]

Since \(f(-x) = -f(x)\), the function is odd.

(vii) \(f(x) = \dfrac{1}{\sec x + \sec^3 x}\)

\[ f(-x) = \frac{1}{\sec(-x) + \sec^3(-x)} = \frac{1}{\sec x + \sec^3 x} = f(x) \]

Since \(\sec(-x) = \sec x\), the function is even.

(viii) \(f(x) = \dfrac{1}{\sec x + \cot^2 x}\)

\[ f(-x) = \frac{1}{\sec(-x) + \cot^2(-x)} = \frac{1}{\sec x + (-\cot x)^2} = \frac{1}{\sec x + \cot^2 x} = f(x) \]

Since \(f(-x) = f(x)\), the function is even.

Common mistake: Declaring \(\sin x + \cos x\) "odd" because it "looks like" a sum of odd-looking trig functions — always compute \(f(-x)\) and compare with both \(f(x)\) and \(-f(x)\) instead of guessing.

Answer: (i) even, (ii) neither, (iii) even, (iv) neither, (v) odd, (vi) odd, (vii) even, (viii) even

Question 2: Find the periods of the following functions.

Period ka formula: \(\sin bx, \cos bx, \sec bx, \csc bx\) ka period \(\frac{2\pi}{|b|}\) aur \(\tan bx, \cot bx\) ka period \(\frac{\pi}{|b|}\) hota hai, jahan \(b\) woh number hai jo \(x\) ke saath multiply ho raha hai. Constants jaise \(+12\) ya \(-5\) period ko affect nahi karte.

(i) \(\sin 5x\) — here \(b = 5\):

\[ \text{Period} = \frac{2\pi}{5} \]

(ii) \(\cos 7x\) — here \(b = 7\):

\[ \text{Period} = \frac{2\pi}{7} \]

(iii) \(\tan 3x\) — tangent, so \(\frac{\pi}{|b|}\) with \(b = 3\):

\[ \text{Period} = \frac{\pi}{3} \]

(iv) \(\cot \frac{x}{2}\) — cotangent, so \(\frac{\pi}{|b|}\) with \(b = \frac{1}{2}\):

\[ \text{Period} = \frac{\pi}{1/2} = 2\pi \]

(v) \(19\sin\left(\frac{\pi}{20}x\right)\) — \(b = \frac{\pi}{20}\):

\[ \text{Period} = \frac{2\pi}{\pi/20} = 2\pi \cdot \frac{20}{\pi} = 40 \]

(vi) \(\csc\left(\frac{2x}{5}\right)\) — \(b = \frac{2}{5}\):

\[ \text{Period} = \frac{2\pi}{2/5} = 2\pi \cdot \frac{5}{2} = 5\pi \]

(vii) \(\frac{1}{2}\sin\left(\frac{3x}{2} - \frac{\pi}{2}\right)\) — \(b = \frac{3}{2}\) (the phase shift \(-\frac{\pi}{2}\) does not affect the period):

\[ \text{Period} = \frac{2\pi}{3/2} = \frac{4\pi}{3} \]

(viii) \(-5 - 3\sec\left(7\pi x + \frac{\pi}{4}\right)\) — \(b = 7\pi\):

\[ \text{Period} = \frac{2\pi}{7\pi} = \frac{2}{7} \]

(ix) \(12 + 10\tan\left(\frac{\pi x}{30}\right)\) — \(b = \frac{\pi}{30}\):

\[ \text{Period} = \frac{\pi}{\pi/30} = 30 \]

(x) \(6 - 4\cot\left(\frac{7x}{4} + \frac{\pi}{4}\right)\) — cotangent with \(b = \frac{7}{4}\):

\[ \text{Period} = \frac{\pi}{7/4} = \frac{4\pi}{7} \]

Common mistake: Using \(2\pi/|b|\) for tangent and cotangent — those have period \(\pi/|b|\). Also, the phase shift inside the brackets never changes the period.

Answer: (i) \(\frac{2\pi}{5}\), (ii) \(\frac{2\pi}{7}\), (iii) \(\frac{\pi}{3}\), (iv) \(2\pi\), (v) \(40\), (vi) \(5\pi\), (vii) \(\frac{4\pi}{3}\), (viii) \(\frac{2}{7}\), (ix) \(30\), (x) \(\frac{4\pi}{7}\)