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Punjab Board Class 11 Mathematics

Exercise 11.3 Solutions — Trigonometric Functions and their Graphs

Complete worked solutions for Exercise 11.3 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 11.3 Solutions

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

Question 1: Find the maximum and minimum values of the following functions:

Is sawal mein hum \(-1 \le \sin u \le 1\) aur \(-1 \le \cos u \le 1\) ka use karenge: maximum ke liye \(\sin u\) ya \(\cos u\) ki woh value rakhenge jo coefficient ke sign ke saath bara result de. Reciprocal wale parts mein denominator sab se chhota ho to fraction sab se bara hota hai.

(i) \(3 - \sin 3x\)

\[ \max = 3 - (-1) = 4 \quad (\sin 3x = -1), \qquad \min = 3 - (1) = 2 \quad (\sin 3x = 1) \]

(ii) \(3 + \sin 2x\)

\[ \max = 3 + 1 = 4, \qquad \min = 3 - 1 = 2 \]

(iii) \(\frac{1}{2} + \sin(5x + \pi)\)

\[ \max = \frac{1}{2} + 1 = \frac{3}{2}, \qquad \min = \frac{1}{2} - 1 = -\frac{1}{2} \]

(iv) \(\frac{3}{2} + \cos\left(x - \frac{\pi}{4}\right)\)

\[ \max = \frac{3}{2} + 1 = \frac{5}{2}, \qquad \min = \frac{3}{2} - 1 = \frac{1}{2} \]

(v) \(1 - 3\cos 2x\)

\[ \max = 1 - 3(-1) = 4 \quad (\cos 2x = -1), \qquad \min = 1 - 3(1) = -2 \quad (\cos 2x = 1) \]

(vi) \(1 + 2\sin\left(x + \frac{\pi}{6}\right)\)

\[ \max = 1 + 2 = 3, \qquad \min = 1 - 2 = -1 \]

(vii) \(\dfrac{1}{10 - 2\sin 3x}\)

The denominator ranges from \(10 - 2(1) = 8\) to \(10 - 2(-1) = 12\), always positive. The fraction is largest when the denominator is smallest:

\[ \max = \frac{1}{8}, \qquad \min = \frac{1}{12} \]

(viii) \(\dfrac{1}{7 + 3\cos(-2x)} = \dfrac{1}{7 + 3\cos 2x}\)

The denominator ranges from \(7 - 3 = 4\) to \(7 + 3 = 10\), always positive:

\[ \max = \frac{1}{4}, \qquad \min = \frac{1}{10} \]

(ix) \(\dfrac{1}{5 - 3\cos(3x - 1)}\)

The denominator ranges from \(5 - 3 = 2\) to \(5 + 3 = 8\), always positive:

\[ \max = \frac{1}{2}, \qquad \min = \frac{1}{8} \]

Common mistake: In (v), writing max \(= 1 - 3 = -2\) — the minus sign flips things: the maximum occurs when \(\cos 2x = -1\), giving \(1 + 3 = 4\).

Answer: (i) max \(4\), min \(2\); (ii) max \(4\), min \(2\); (iii) max \(\frac{3}{2}\), min \(-\frac{1}{2}\); (iv) max \(\frac{5}{2}\), min \(\frac{1}{2}\); (v) max \(4\), min \(-2\); (vi) max \(3\), min \(-1\); (vii) max \(\frac{1}{8}\), min \(\frac{1}{12}\); (viii) max \(\frac{1}{4}\), min \(\frac{1}{10}\); (ix) max \(\frac{1}{2}\), min \(\frac{1}{8}\)

Question 2: The temperature \(T\) in degrees Celsius of a certain city varies throughout the day according to the equation \(T(t) = \frac{13}{2} \sin\left(\frac{\pi}{6}t - \frac{\pi}{9}\right) + 15\), where \(t\) is the time in hours, with \(t = 0\) corresponding to midnight.

(a) Maximum/minimum ke liye sine ki values \(1\) aur \(-1\) rakhenge. (b) \(t = 9\) ko equation mein rakh kar angle simplify karein aur sine ki value nikalein.

(a) Find the maximum and minimum temperature during the day.

This is of the form \(a + b\sin u\) with \(a = 15\) and \(b = \frac{13}{2}\):

\[ T_{\max} = 15 + \frac{13}{2} = 21.5^\circ\text{C}, \qquad T_{\min} = 15 - \frac{13}{2} = 8.5^\circ\text{C} \]

(b) Find the temperature at \(t = 9\) hours (9:00 a.m.).

\[ T(9) = \frac{13}{2}\sin\left(\frac{\pi}{6} \cdot 9 - \frac{\pi}{9}\right) + 15 = \frac{13}{2}\sin\left(\frac{3\pi}{2} - \frac{\pi}{9}\right) + 15 \]
\[ = \frac{13}{2}\sin\left(\frac{25\pi}{18}\right) + 15 \approx \frac{13}{2}(-0.9397) + 15 \approx -6.11 + 15 = 8.89 \]

Here \(\frac{25\pi}{18} = 250^\circ\), and \(\sin 250^\circ = -\sin 70^\circ \approx -0.9397\).

Answer: (a) max \(21.5^\circ\)C, min \(8.5^\circ\)C; (b) \(T(9) \approx 8.89^\circ\)C

Question 3: A man on the top of a 100 m high light-house is in line with two ships on the same side of it, whose angles of depression from the man are \(17^\circ\) and \(19^\circ\) respectively. Find the distance between the ships.

Angle of depression neeche dekhne ka angle hai jo horizontal se banta hai — yeh usi triangle ke base angle (angle of elevation) ke barabar hota hai. Har ship ke liye \(\tan = \frac{100}{\text{distance}}\) likhein; bara angle matlab qareeb ship.

Let the distances of the two ships from the foot of the light-house be \(x_1\) (angle \(19^\circ\), the nearer ship) and \(x_2\) (angle \(17^\circ\), the farther ship). The angle of depression equals the angle of elevation from the ship:

\[ \tan 19^\circ = \frac{100}{x_1} \implies x_1 = \frac{100}{\tan 19^\circ} \approx \frac{100}{0.3443} \approx 290.42 \text{ m} \]
\[ \tan 17^\circ = \frac{100}{x_2} \implies x_2 = \frac{100}{\tan 17^\circ} \approx \frac{100}{0.3057} \approx 327.09 \text{ m} \]

Distance between the ships:

\[ x_2 - x_1 \approx 327.09 - 290.42 = 36.67 \text{ m} \]
Answer: \(\approx 36.7\) m

Question 4: P and Q are two points in line with a tree. If the distance between P and Q be 30 m and the angles of elevation of the top of the tree at P and Q are \(12^\circ\) and \(15^\circ\) respectively, find the height of the tree.

Darakht ki height \(h\) aur Q se darakht tak ka fasla \(x\) maan lein (Q qareeb hai kyunke uska angle bara hai). Dono points se \(\tan\) ki do equations banayein aur \(x\) eliminate karke \(h\) nikalein.

Let the height of the tree be \(h\) m and the distance of Q from the tree be \(x\) m. Since Q has the larger angle of elevation (\(15^\circ\)), Q is nearer; the distance of P from the tree is \(x + 30\).

\[ \tan 15^\circ = \frac{h}{x} \implies x = \frac{h}{\tan 15^\circ}, \qquad \tan 12^\circ = \frac{h}{x + 30} \implies x + 30 = \frac{h}{\tan 12^\circ} \]

Subtracting:

\[ 30 = h\left(\frac{1}{\tan 12^\circ} - \frac{1}{\tan 15^\circ}\right) = h(\cot 12^\circ - \cot 15^\circ) \]
\[ h = \frac{30}{4.7046 - 3.7321} = \frac{30}{0.9725} \approx 30.85 \text{ m} \]
Answer: \(\approx 30.8\) m

Question 5: A giant Ferris wheel has a diameter of 60 feet. The lowest point of the wheel is located 6 feet above the ground. The wheel completes one full revolution every 80 seconds.

Rider neeche se start karta hai is liye \(-\cos\) model use hoga: amplitude = radius, midline = centre ki height, \(c = \frac{2\pi}{\text{period}}\).

(a) Model an equation that represents the height \(h(t)\) of a rider on the Ferris wheel at any given time \(t\).

Radius \(r = 30\) ft. Centre height \(= 6 + 30 = 36\) ft. Period \(T = 80\) s, so \(c = \frac{2\pi}{80} = \frac{\pi}{40}\). Starting at the lowest point, use the negative cosine model:

\[ h(t) = -30\cos\left(\frac{\pi}{40}t\right) + 36 \]

(b) Find the maximum height of the rider.

\[ h_{\max} = 36 + 30 = 66 \text{ ft} \quad \left(\text{when } \cos\frac{\pi t}{40} = -1\right) \]

(c) Find the height of the rider from the ground after 35 seconds.

\[ h(35) = -30\cos\left(\frac{35\pi}{40}\right) + 36 = -30\cos\left(\frac{7\pi}{8}\right) + 36 \]
\[ = -30(-0.9239) + 36 \approx 27.72 + 36 = 63.72 \text{ ft} \]
Answer: (a) \(h(t) = -30\cos\left(\frac{\pi t}{40}\right) + 36\); (b) \(66\) ft; (c) \(\approx 63.7\) ft

Question 6: A child is playing on a swing in a playground. The height \(h(t)\) of the swing seat above the ground (in metres) at time \(t\) (in seconds) is modeled by the function: \(h(t) = 1.5 + 1.2 \sin(3\pi t)\).

Yeh \(a + b\sin(ct)\) form ka model hai: max \(= a + b\), min \(= a - b\), period \(= \frac{2\pi}{c}\). Part (d) mein equation solve karke pehla time nikalein.

(a) What is the maximum height reached by the swing seat?

\[ h_{\max} = 1.5 + 1.2 = 2.7 \text{ m} \]

(b) What is the minimum height reached by the swing seat?

\[ h_{\min} = 1.5 - 1.2 = 0.3 \text{ m} \]

(c) How long does it take for the swing to complete one full back-and-forth motion (period)?

\[ T = \frac{2\pi}{3\pi} = \frac{2}{3} \text{ seconds} \]

(d) At what time(s) does the swing seat first reach a height of 2.12 metres?

\[ 1.5 + 1.2\sin(3\pi t) = 2.12 \implies \sin(3\pi t) = \frac{0.62}{1.2} \approx 0.5167 \]
\[ 3\pi t = \sin^{-1}(0.5167) \approx 0.5428 \text{ rad} \implies t \approx \frac{0.5428}{3\pi} \approx 0.058 \text{ s} \]
Answer: (a) \(2.7\) m; (b) \(0.3\) m; (c) \(\frac{2}{3}\) s; (d) \(\approx 0.058\) s

Question 7: A carnival ride consists of a vertical wheel with a diameter of 40 feet. The centre of the wheel is 28 feet above the ground. The wheel rotates at a constant speed and takes 120 seconds to make one complete revolution. Model an equation that describes the height \(h(t)\) of a rider on the wheel as a function of time \(t\). How high is the rider from the ground after 90 seconds? At what times will the rider be 36 feet above the ground?

Radius \(= 20\), centre \(= 28\), period \(= 120\). Neeche se start karne par \(-\cos\) model banta hai. Height 36 ft par do times milenge (upar jate hue aur neeche aate hue).

Model: Radius \(r = 20\) ft, centre height \(d = 28\) ft, \(c = \frac{2\pi}{120} = \frac{\pi}{60}\). Assuming the ride starts at the lowest point:

\[ h(t) = -20\cos\left(\frac{\pi}{60}t\right) + 28 \]

Height after 90 seconds:

\[ h(90) = -20\cos\left(\frac{90\pi}{60}\right) + 28 = -20\cos\left(\frac{3\pi}{2}\right) + 28 = -20(0) + 28 = 28 \text{ ft} \]

Times when the rider is 36 feet above the ground:

\[ -20\cos\left(\frac{\pi t}{60}\right) + 28 = 36 \implies \cos\left(\frac{\pi t}{60}\right) = -0.4 \]
\[ \frac{\pi t}{60} = \cos^{-1}(-0.4) \approx 1.9823 \text{ rad} \quad \text{or} \quad 2\pi - 1.9823 \approx 4.3009 \text{ rad} \]
\[ t \approx \frac{60 \times 1.9823}{\pi} \approx 37.9 \text{ s}, \qquad t \approx \frac{60 \times 4.3009}{\pi} \approx 82.1 \text{ s} \]
Answer: \(h(t) = -20\cos\left(\frac{\pi t}{60}\right) + 28\); after 90 s \(= 28\) ft; rider is at 36 ft at \(t \approx 37.9\) s and \(t \approx 82.1\) s

Question 8: Suppose the temperature \(T\) in degrees Fahrenheit of Lahore city in a month of December throughout the day can be modeled by the equation: \(T = 64 + 8 \sin\left(\frac{\pi}{12}(t - 8)\right)\), where \(t\) is the time in hours. The temperature oscillates 8 degrees above and below an average temperature of 64 degrees.

Average \(= 64\), amplitude \(= 8\). Maximum tab hoga jab sine \(= 1\) ho — us waqt ka \(t\) nikal kar 12-hour format mein batayein.

(a) Find the temperature at \(t = 9\) hours?

\[ T(9) = 64 + 8\sin\left(\frac{\pi}{12}(9 - 8)\right) = 64 + 8\sin\left(\frac{\pi}{12}\right) \]
\[ = 64 + 8(0.2588) \approx 64 + 2.07 = 66.07^\circ\text{F} \]

(b) At what time will the temperature be maximum?

Maximum occurs when \(\sin\left(\frac{\pi}{12}(t-8)\right) = 1\):

\[ \frac{\pi}{12}(t - 8) = \frac{\pi}{2} \implies t - 8 = 6 \implies t = 14 \]

So the maximum occurs at \(t = 14\) hours, i.e. 2:00 p.m.

(c) Calculate the maximum temperature.

\[ T_{\max} = 64 + 8 = 72^\circ\text{F} \]
Answer: (a) \(\approx 66.07^\circ\)F; (b) \(t = 14\) hours (2 p.m.); (c) \(72^\circ\)F

Question 9: Suppose the population of a coastal city follows a sinusoidal pattern due to seasonal migration. The population of the city over the course of a year can be modeled by the equation: \(P(t) = 70000 + 10000 \cos\left(\frac{\pi}{6}t - \frac{\pi}{2}\right)\), where \(P(t)\) is the population at time \(t\) (\(t\) is the time in months, with \(t = 0\) corresponding to January 1st).

Average population \(= 70000\), amplitude \(= 10000\). \(t = 7\) rakh kar cosine ka angle simplify karein: \(\frac{7\pi}{6} - \frac{\pi}{2} = \frac{2\pi}{3}\).

(a) Find the population of the city at \(t = 7\) months.

\[ P(7) = 70000 + 10000\cos\left(\frac{\pi}{6} \cdot 7 - \frac{\pi}{2}\right) = 70000 + 10000\cos\left(\frac{7\pi}{6} - \frac{3\pi}{6}\right) \]
\[ = 70000 + 10000\cos\left(\frac{2\pi}{3}\right) = 70000 + 10000\left(-\frac{1}{2}\right) = 65000 \]

(b) Find the maximum population.

\[ P_{\max} = 70000 + 10000 = 80000 \]
Answer: (a) \(65000\); (b) \(80000\)