Category: Class 11

Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.2

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.2 Text Book Back Questions and Answers

Question 1.
Evaluate the following:

Solution:
(i) \(\lim _{x \rightarrow 2} \frac{x^{3}+2}{x+1}\)

(ii) \(\lim _{x \rightarrow \infty} \frac{2 x+5}{x^{2}+3 x+9}\)

[Takeout x from numerator and take x² from the denominator]

(iii) \(\lim _{x \rightarrow \infty} \frac{\Sigma n}{n^{2}}\)

(iv) \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{5 x}\)

(v) \(\lim _{x \rightarrow a} \frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{2}{3}}-a^{\frac{2}{3}}}\)

[Divide both numerator and denominator by x – a; \(\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n}\)]

(vi) \(\lim _{x \rightarrow 0} \frac{\sin ^{2} 3 x}{x^{2}}\)

Question 2.
If \(\lim _{x \rightarrow a} \frac{x^{9}-a^{9}}{x-a}=\lim _{x \rightarrow 3}(x+6)\), find the value of a.
Solution:
\(\lim _{x \rightarrow a} \frac{x^{9}-a^{9}}{x-a}=\lim _{x \rightarrow 3}(x+6)\)
9 . a^9-1 = 3 + 6
9 . a⁸ = 9
a⁸ = 1
Taking squareroot on bothsides, we get
\(\left(a^{8}\right)^{\frac{1}{2}}\) = ±1
a⁴ = ±1
But a⁴ = -1 is imposssible.
∴ a⁴ = 1
Again taking squareroot, we get
\(\left(a^{4}\right)^{\frac{1}{2}}\) = ±1
a² = ±1
a² = -1 is imposssible
∴ a² = 1
Again taking positive squareroot, a = ±1

Question 3.
If \(\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=448\), then find the least positive integer n.
Solution:

Question 4.
If f(x) = \(\frac{x^{7}-128}{x^{5}-32}\), then find \(\lim _{x \rightarrow 2} f(x)\)
Solution:
\(\lim _{x \rightarrow 2} f(x)\)

Question 5.
Let f(x) = \(\frac{a x+b}{x+1}\), if \(\lim _{x \rightarrow 0} f(x)=2\) and \(\lim _{x \rightarrow \infty} f(x)=1\), then show that f(-2) = 0
Solution:


Hence Proved.

Students can download 11th Business Maths Chapter 5 Differential Calculus Ex 5.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.1

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.1 Text Book Back Questions and Answers

Question 1.
Determine whether the following functions are odd or even?
(i) f(x) = \(\left(\frac{a^{x}-1}{a^{x}+1}\right)\)
(ii) f(x) = log(x² + \(\sqrt{x^{2}+1}\))
(iii) f(x) = sin x + cos x
(iv) f(x) = x² – |x|
(v) f(x) = x + x²
Solution:
(i) f(x) = \(\left(\frac{a^{x}-1}{a^{x}+1}\right)\)

Thus f(-x) = -f(x)
∴ f(x) is an odd function.

(ii) f(x) = log(x² + \(\sqrt{x^{2}+1}\))
f(-x) = log((-x)² + \(\sqrt{(-x)^{2}+1}\))
= log(x² + \(\sqrt{x^{2}+1}\))
Thus f(-x) = f(x)
∴ f(x) is an even function.

(iii) f(x) = sin x + cos x
f(-x) = sin(-x) + cos(-x)
= -sin x + cos x
= -[sin x – cos x]
Since f(-x) ≠ -f(x) (or) f(x) ≠ -f(x)
∴ f(x) is neither odd nor even function.

(iv) Given f(x) = x² – |x|
f(-x) = (-x)² – |-x|
= x² – |x|
= f(x)
∴ f(x) is an even function.

(v) f(x) = x + x²
f(-x) = (-x) + (-x)² = -x + x²
Since f(-x) ≠ f(x), f(-x) ≠ -f(x).
∴ f(x) is neither odd nor even function.

Question 2.
Let f be defined by f(x) = x³ – kx² + 2x, x ∈ R. Find k, if ‘f’ is an odd function.
Solution:
For a polynomial function to be an odd function each term should have odd powers pf x. Therefore there should not be an even power of x term.
∴ k = 0.

Question 3.
If f(x) = \(x^{3}-\frac{1}{x^{3}}\), then show that f(x) + f(\(\frac{1}{x}\)) = 0
Solution:
f(x) = \(x^{3}-\frac{1}{x^{3}}\) …….. (1)
\(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}\) = \(\frac{1}{x^{3}}-x^{3}\) …….. (2)
(1) + (2) gives \(f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}=0\)
Hence Proved.

Question 4.
If f(x) = \(\frac{x+1}{x-1}\), then prove that f(f(x)) = x.
Solution:

Hence proved.

Question 5.
For f(x) = \(\frac{x-1}{3 x+1}\), write the expressions of f(\(\frac{1}{x}\)) and \(\frac{1}{f(x)}\).
Solution:

Question 6.
If f(x) = e^x and g(x) = log_e x then find
(i) (f + g) (1)
(ii) (fg) (1)
(iii) (3f) (1)
(iv) (5g) (1)
Solution:
(i) (f+g) (1) = e¹ + log_e 1 = e + 0 = e
(ii) (fg) (1) = f(1) g(1) = e¹ \(\log _{e}^{1}\) = e × 0 = 0
(iii) (3f) (1) = 3 f(1) = 3 e¹ = 3e
(iv) (5g) (1) = 5 (g) (1) = 5 \(\log _{e}^{1}\) = 5 × 0 = 0

Question 7.
Draw the graph of the following functions:
(i) f(x) = 16 – x²
(ii) f(x) = |x – 2|
(iii) f(x) = x|x|
(iv) f(x) = e^2x
(v) f(x) = e^-2x
(vi) f(x) = \(\frac{|x|}{x}\)
Solution:
(i) f(x) = 16 – x²
Let y = f(x) = 16 – x²
Choose suitable values for x and determine y. Thus we get the following table.

Plot the points (-4, 0), (-3, 7), (-2, 12), (-1, 15), (0, 16), (1, 15), (2, 12), (3, 7), (4, 0).
The graph is as shown in the figure.

(ii) Let y = f(x) = |x – 2|


Plot the points (2, 0), (3, 1) (4, 2), (5, 3), (0, 2), (-1, 3), (-2, 4), (-3, 5) and draw a line.
The graph is as shown in the figure.

(iii) Let y = f(x) = x|x|


Plot the points (0, 0), (1, 1) (2, 4), (3, 9), (-1, -1), (-2, -4), (-3, -9) and draw a smooth curve.
The graph is as shown in the figure.

(iv) For x = 0, f(x) becomes 1 i.e., the curve cuts the y-axis at y = 1.
For no real value of x, f(x) equals to 0. Thus it does not meet the x-axis for real values of x.

(v) For x = 0, f(x) becomes 1 i.e., the curve cuts the y-axis at y = 1.
For no real value of x, f(x) equal to 0. Thus it does not meet the x-axis for real values of x.

(vi) If f: R → R is defined by


The domain of the function is R and the range is {-1, 0, 1}.

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.5

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.5 Text Book Back Questions and Answers

Question 1.
The degree measure of \(\frac{\pi}{8}\) is:
(a) 20°60′
(b) 22°30′
(c) 22°60′
(d) 20°30′
Answer:
(b) 22°30′
Hint:
We know that, one radian = \(\frac{180^{\circ}}{\pi}\)
∴ \(\frac{\pi}{8}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{8}\) degrees
= \(\frac{45^{\circ}}{2}\)
= 22.5°
= 22°30′

Question 2.
The radian measure of 37°30′ is:
(a) \(\frac{5 \pi}{24}\)
(b) \(\frac{3 \pi}{24}\)
(c) \(\frac{7 \pi}{24}\)
(d) \(\frac{9 \pi}{24}\)
Answer:
(a) \(\frac{5 \pi}{24}\)
Hint:

Question 3.
If tan θ = \(\frac{1}{\sqrt{5}}\) and θ lies in the first quadrant then cos θ is:
(a) \(\frac{1}{\sqrt{6}}\)
(b) \(\frac{-1}{\sqrt{6}}\)
(c) \(\frac{\sqrt{5}}{\sqrt{6}}\)
(d) \(\frac{-\sqrt{5}}{\sqrt{6}}\)
Answer:
(c) \(\frac{\sqrt{5}}{\sqrt{6}}\)
Hint:

Question 4.
The value of sin 15° is:
(a) \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
(b) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
(c) \(\frac{\sqrt{3}}{\sqrt{2}}\)
(d) \(\frac{\sqrt{3}}{2 \sqrt{2}}\)
Answer:
(b) \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
Hint:
sin 15° = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Question 5.
The value of sin(-420°)
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(-\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{-1}{2}\)
Answer:
(b) \(-\frac{\sqrt{3}}{2}\)
Hint:
sin(-420°) = -sin(420°) [∵ sin(-θ) = -sin θ]
= -sin(360° + 60°)
= -sin 60°
= \(-\frac{\sqrt{3}}{2}\)

Question 6.
The value of cos(-480°) is:
(a) √3
(b) \(-\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{-1}{2}\)
Answer:
(d) \(\frac{-1}{2}\)
Hint:
cos(-480°) = cos 480° [∵ cos(-θ) = cos θ]
= cos(360° + 120°)
= cos 120°
= cos(180° – 60°)
= -cos 60°
= \(\frac{-1}{2}\)

Question 7.
The value of sin 28° cos 17° + cos 28° sin 17°
(a) \(\frac{1}{\sqrt{2}}\)
(b) 1
(c) \(\frac{-1}{\sqrt{2}}\)
(d) 0
Answer:
(a) \(\frac{1}{\sqrt{2}}\)
Hint:
sin 28° cos 17° + cos 28° sin 17° = sin(28° + 17°)
This is of the form sin(A + B), A = 28°, B = 17°
= sin 45°
= \(\frac{1}{\sqrt{2}}\)

Question 8.
The value of sin 15° cos 15° is:
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{\sqrt{3}}{2}\)
(d) \(\frac{1}{4}\)
Answer:
(d) \(\frac{1}{4}\)
Hint:
sin 15° cos 15° = \(\frac{1}{2}\) (2 sin 15° cos 15°)
= \(\frac{1}{2}\) (sin 30°)
= \(\frac{1}{2}\left(\frac{1}{2}\right)\)
= \(\frac{1}{4}\)

Question 9.
The value of sec A sin(270° + A) is:
(a) -1
(b) cos² A
(c) sec² A
(d) 1
Answer:
(a) -1
Hint:
sec A (sin(270° + A)) = \(\frac{1}{\cos A}\) (-cos A) = -1

Question 10.
If sin A + cos A = 1 then sin 2A is equal to:
(a) 1
(b) 2
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(c) 0
Hint:
Given sin A + cos A = 1
Squaring both sides we get
sin² A + cos² A + 2 sin A cos A = 1
1 + sin 2A = 1
sin 2A = 0

Question 11.
The value of cos² 45° – sin² 45° is:
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{1}{2}\)
(c) 0
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(c) 0
Hint:
cos² 45° – sin² 45°
= cos 2 × 45° (∵ cos² A – sin² A = cos 2A)
= cos 90°
= 0

Question 12.
The value of 1 – 2 sin² 45° is:
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{4}\)
(d) 0
Answer:
(d) 0
Hint:
1 – 2 sin² 45°
= cos(2 × 45°) [∵ cos 2A = 1 – 2 sin² A]
= cos 90°
= 0

Question 13.
The value of 4 cos³ 40° – 3 cos 40° is
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(-\frac{1}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) \(-\frac{1}{2}\)
Hint:
4 cos³ 40° – 3 cos 40°
= cos (3 × 40°) [∵ cos 3A = 4 cos³ A – 3 cos A]
= cos 120°
= cos (180° – 60°)
= -cos 60°
= \(-\frac{1}{2}\)

Question 14.
The value of \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{3}}\)
(c) \(\frac{\sqrt{3}}{2}\)
(d) √3
Answer:
(d) √3
Hint:
We know that sin 2A = \(\frac{2 \tan A}{1+\tan ^{2} A}\)
\(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) = sin(2 × 30°) = sin 60° = \(\frac{\sqrt{3}}{2}\)
= tan 2A
= tan 60°
= √3

Question 15.
If sin A = \(\frac{1}{2}\) then 4 cos³ A – 3 cos A is:
(a) 1
(b) 0
(c) \(\frac{\sqrt{3}}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) 0
Hint:
Given sin A = \(\frac{1}{2}\)
sin A = sin 30°
∴ A = 30°
[∵ 4 cos³ A – 3 cos A = cos 3A]
= cos(3 × 30°)
= cos 90°
= 0

Question 16.
The value of \(\frac{3 \tan 10^{\circ}-\tan ^{3} 10^{\circ}}{1-3 \tan ^{2} 10^{\circ}}\) is:
(a) \(\frac{1}{\sqrt{3}}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{\sqrt{3}}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(a) \(\frac{1}{\sqrt{3}}\)
Hint:
\(\frac{3 \tan 10^{\circ}-\tan ^{3} 10^{\circ}}{1-3 \tan ^{2} 10^{\circ}}\) = tan(3 × 10°) [∵ tan 3A = \(\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}\)]
= tan 30°
= \(\frac{1}{\sqrt{3}}\)

Question 17.
The value of cosec^-1 (\(\frac{2}{\sqrt{3}}\)) is:
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{6}\)
Answer:
(c) \(\frac{\pi}{3}\)
Hint:
Let cosec^-1 (\(\frac{2}{\sqrt{3}}\))
\(\frac{2}{\sqrt{3}}\) = cosec A
cosec A = \(\frac{2}{\sqrt{3}}\)
sin A = \(\frac{\sqrt{3}}{2}\) = sin 60°
∴ A = 60° = \(\frac{\pi}{3}\)

Question 18.
sec^-1 (\(\frac{2}{3}\)) + cosec^-1 (\(\frac{2}{3}\)) =
(a) \(\frac{-\pi}{2}\)
(b) \(\frac{\pi}{2}\)
(c) π
(d) -π
Answer:
(b) \(\frac{\pi}{2}\)
Hint:
We know that sec^-1 x + cosec^-1 x = \(\frac{\pi}{2}\)
∴ sec^-1 (\(\frac{2}{3}\)) + cosec^-1 (\(\frac{2}{3}\)) = \(\frac{\pi}{2}\)

Question 19.
If α and β be between 0 and \(\frac{\pi}{2}\) and if cos(α + β) = \(\frac{12}{13}\) and sin(α – β) = \(\frac{3}{5}\) then sin 2α is:
(a) \(\frac{16}{15}\)
(b) 0
(c) \(\frac{56}{65}\)
(d) \(\frac{64}{65}\)
Answer:
(c) \(\frac{56}{65}\)
Hint:

Given that cos(α + β) = \(\frac{12}{13}\)
∴ sin(α + β) = \(\frac{5}{13}\)

Also given that sin(α – β) = \(\frac{3}{5}\)
∴ cos(α – β) = \(\frac{4}{5}\)
sin 2α = sin[(α + β) + (α – β)]
= sin(α + β) cos(α – β) + cos(α + β) sin(α – β)
= \(\frac{5}{13} \times \frac{4}{5}+\frac{12}{13} \times \frac{3}{5}\)
= \(\frac{20}{65}+\frac{36}{65}\)
= \(\frac{56}{65}\)

Question 20.
If tan A = \(\frac{1}{2}\) and tan B = \(\frac{1}{3}\) then tan(2A + B) is equal to:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Hint:
Given tan A = \(\frac{1}{2}\), tan B = \(\frac{1}{3}\)

Question 21.
tan(\(\frac{\pi}{4}\) – x) is:
(a) \(\left(\frac{1+\tan x}{1-\tan x}\right)\)
(b) \(\left(\frac{1-\tan x}{1+\tan x}\right)\)
(c) 1 – tan x
(d) 1 + tan x
Answer:
(b) \(\left(\frac{1-\tan x}{1+\tan x}\right)\)
Hint:

[∵ tan \(\frac{\pi}{4}\) = 1]

Question 22.
\(\sin \left(\cos ^{-1} \frac{3}{5}\right)\) is:
(a) \(\frac{3}{5}\)
(b) \(\frac{5}{3}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{5}{4}\)
Answer:
(c) \(\frac{4}{5}\)
Hint:

Let cos^-1 (\(\frac{3}{5}\)) = A
\(\frac{3}{5}\) = cos A
sin A = \(\frac{4}{5}\)
Now sin(cos-1 (\(\frac{3}{5}\))) = sin A = \(\frac{4}{5}\)

Question 23.
The value of \(\frac{1}{cosec(-45^{\circ})}\) is:
(a) \(\frac{-1}{\sqrt{2}}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) √2
(d) -√2
Answer:
(a) \(\frac{-1}{\sqrt{2}}\)
Hint:
\(\frac{1}{cosec(-45^{\circ})}\) = sin(-45°)
= -sin 45°
= \(\frac{-1}{\sqrt{2}}\)

Question 24.
If p sec 50° = tan 50° then p is:
(a) cos 50°
(b) sin 50°
(c) tan 50°
(d) sec 50°
Answer:
(b) sin 50°
Hint:
p sec 50° = tan 50°
p(\(\frac{1}{\cos 50^{\circ}}\)) = \(\frac{\sin 50^{\circ}}{\cos 50^{\circ}}\)
∴ p = sin 50°

Question 25.
(\(\frac{cos x}{cosec x})-\sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x}\) is:
(a) cos² x – sin² x
(b) sin² x – cos² x
(c) 1
(d) 0
Answer:
(d) 0
Hint:
(\(\frac{cos x}{cosec x})-\sqrt{1-\sin ^{2} x} \sqrt{1-\cos ^{2} x}\)
= cos x × sin x – \(\sqrt{\cos ^{2} x} \sqrt{\sin ^{2} x}\)
= cos x × sin x – cos x × sin x
= 0

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.4 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.4

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.4 Text Book Back Questions and Answers

Question 1.
Find the principal value of the following:
(i) sin^-1 (\(-\frac{1}{2}\))
(ii) tan^-1 (-1)
(iii) cosec^-1 (2)
(iv) sec^-1 (-√2)
Solution:
(i) sin^-1 (\(-\frac{1}{2}\))
Let sin^-1 (\(-\frac{1}{2}\)) = y
[where \(\frac{-\pi}{2} \leq y \leq \frac{\pi}{2}\)]
\(-\frac{1}{2}\) = sin y
sin y = \(-\frac{1}{2}\) (∵ \(\sin \frac{\pi}{6}=\frac{1}{2}\))
sin y = sin(\(-\frac{\pi}{6}\)) [∵ \(\sin \left(-\frac{\pi}{6}\right)=-\sin \left(\frac{\pi}{6}\right)\)]
∴ y = \(-\frac{\pi}{6}\)
∴ The principal value of sin-1 (\(-\frac{1}{2}\)) is \(-\frac{\pi}{6}\)

(ii) tan^-1 (-1) = y
(-1) = tan y where \(\frac{-\pi}{2} \leq y \leq \frac{\pi}{2}\)
(or) tan y = – 1
tan y = tan(\(-\frac{\pi}{4}\)) (∵ \(\tan \frac{\pi}{4}\) = 1)
∴ y = \(-\frac{\pi}{4}\) [∵ \(\tan \left(-\frac{\pi}{4}\right)=-\tan \left(\frac{\pi}{4}\right)=-1\)]
∴ The principal value of tan^-1 (-1) is \(-\frac{\pi}{4}\).

(iii) Let cosec^-1 (2) = y
2 = cosec y
(or) cosec y = 2
⇒ \(\frac{1}{\sin y}\) = 2
⇒ sin y = \(\frac{1}{2}\) (Take reciprocal)
⇒ sin y = \(\sin \left(\frac{\pi}{6}\right)\)
⇒ y = \(\frac{\pi}{6}\)
The principal value of cosec^-1 (-1) is \(\frac{\pi}{6}\).

(iv) Let sec^-1 (-√2 ) = y
-√2 = sec y
sec y = -√2
\(\frac{1}{\cos y}\) = -√2
Taking reciprocal cos y = \(\frac{-1}{\sqrt{2}}\) [where 0 ≤ y ≤ π]

∴ The principal value of sec^-1 (-√2) is \(\frac{3 \pi}{4}\)

Question 2.
Prove that
(i) 2 tan^-1 (x) = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
(ii) \(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}\)
Solution:
(i) Let tan^-1 x = θ
x = tan θ

(ii) \(\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}\)

Question 3.
Show that \(\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{2}{11}\right)=\tan ^{-1}\left(\frac{3}{4}\right)\)
Solution:
We know that tan^-1 x + tan^-1 y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)
Now LHS = \(\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{2}{11}\right)\)

Question 4.
Solve: tan^-1 2x + tan^-1 3x = \(\frac{\pi}{4}\).
Solution:
Given tan^-1 (2x) + tan^-1 (3x) = \(\frac{\pi}{4}\)

⇒ 5x = 1(1 – 6x²)
⇒ 6x² + 5x – 1 = 0
⇒ (x + 1) (6x – 1) = 0
⇒ x + 1 = 0 (or) 6x – 1 = 0
⇒ x = -1 (or) x = \(\frac{1}{6}\)
x = -1 is rejected. It doesn’t satisfies the question.
Note: Put x = -1 in the given question.

So the question changes.

Question 5.
Solve: tan^-1 (x + 1) + tan^-1 (x – 1) = \(\tan ^{-1}\left(\frac{4}{7}\right)\)
Solution:


⇒ 7x = 2(2 – x²)
⇒ 7x = 4 – 2x²
⇒ 2x² + 7x – 4 = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x + 4 = 0 (or) 2x – 1 = 0
⇒ x = -4 (or) x = \(\frac{1}{2}\)
x = -4 is rejected, since does not satisfies the question.
∴ x = \(\frac{1}{2}\)

Question 6.
Evaluate
(i) cos[tan^-1(\(\frac{3}{4}\))]
(ii) \(\sin \left[\frac{1}{2} \cos ^{-1}\left(\frac{4}{5}\right)\right]\)
Solution:
(i) Let \(\tan ^{-1}\left(\frac{3}{4}\right)\) = θ
\(\frac{3}{4}\) = tan θ
tan θ = \(\frac{3}{4}\)
∴ cos θ = \(\frac{4}{5}\)

Now \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\) = cos θ = \(\frac{4}{5}\)

(ii) Let \(\cos ^{-1}\left(\frac{4}{5}\right)\) = A
Then \(\frac{4}{5}\) = cos A
cos A = \(\frac{4}{5}\)

We know that

Question 7.
Evaluate: \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right)\)
Solution:

Let \(\sin ^{-1}\left(\frac{4}{5}\right)\) = A
sin A = \(\frac{4}{5}\)
∴ cos A = \(\frac{3}{5}\)

Let \(\sin ^{-1}\left(\frac{12}{13}\right)\) = B
\(\frac{12}{13}\) = sin B
sin B = \(\frac{12}{13}\)
∴ cos B = \(\frac{5}{13}\)
Now \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)\right)\) = cos (A + B)
= cos A cos B – sin A sin B
= \(\frac{3}{5} \times \frac{5}{13}-\frac{4}{5} \times \frac{12}{13}\)
= \(\frac{15}{65}-\frac{48}{65}\)
= \(-\frac{33}{65}\)

Question 8.
Prove that \(\tan ^{-1}\left(\frac{m}{n}\right)-\tan ^{-1}\left(\frac{m-n}{m+n}\right)=\frac{\pi}{4}\)
Solution:

Question 9.
Show that \(\sin ^{-1}\left(-\frac{3}{5}\right)-\sin ^{-1}\left(-\frac{8}{17}\right)=\cos ^{-1}\left(\frac{84}{85}\right)\)
Solution:
\(\sin ^{-1}\left(-\frac{3}{5}\right)-\sin ^{-1}\left(-\frac{8}{17}\right)=-\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{8}{17}\right)\)
= \(\sin ^{-1}\left(\frac{8}{17}\right)-\sin ^{-1}\left(\frac{3}{5}\right)\)

Let \(\sin ^{-1}\left(\frac{8}{17}\right)\) = A
\(\frac{8}{17}\) = sin A
sin A = \(\frac{8}{17}\)
∴ cos A = \(\frac{15}{17}\)

Let \(\sin ^{-1}\left(\frac{3}{5}\right)\) = B
sin B = \(\frac{3}{5}\)
∴ cos B = \(\frac{4}{5}\)
Consider cos(A – B) = cos A cos B + sin A sin B
= \(\frac{15}{17} \times \frac{4}{5}+\frac{8}{17} \times \frac{3}{5}\)
= \(\frac{60}{85}+\frac{24}{85}\)
cos (A – B) = \(\frac{84}{85}\)
∴ A – B = \(\cos ^{-1}\left(\frac{84}{85}\right)\)

Question 10.
Express \(\tan ^{-1}\left[\frac{\cos x}{1-\sin x}\right]\), \(-\frac{\pi}{2}<x<\frac{3 \pi}{2}\) in the simplest form.
Solution:
\(\tan ^{-1}\left[\frac{\cos x}{1-\sin x}\right]\)


[∵ a² – b² = (a + b) (a – b)]

[∵ Divide each term by cos \(\frac{x}{2}\)]

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.3 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.3

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.3 Text Book Back Questions and Answers

Question 1.
Express each of the following as the sum or difference of sine or cosine:
(i) sin\(\frac{A}{8}\) sin\(\frac{3A}{8}\)
(ii) cos(60° + A) sin(120° + A)
(iii) cos\(\frac{7 A}{3}\) sin\(\frac{5 A}{3}\)
(iv) cos 7θ sin 3θ
Solution:
(i) sin\(\frac{A}{8}\) sin\(\frac{3A}{8}\)

[∵ 2 sin A sin B = cos(A – B) – cos(A + B)

[∵ cos(-θ) = cos θ]

(ii) cos(60° + A) sin(120° + A) = \(\frac{1}{2}\) [2 cos(60° + A) sin(120° + A)] [Multiply and divide by 2]
= \(\frac{1}{2}\) [sin((60° + A) + (120° + A))] – sin((60° + A) – (120° + A))]
[∵ 2 cos A sin B = sin(A + B) – sin(A – B)]
= \(\frac{1}{2}\) [sin(180° + 2A) – sin(60° + A – 120° – A)]
= \(\frac{1}{2}\) [(-sin 2A) – sin(-60°)]
= \(\frac{1}{2}\) [-sin 2A + sin 60°]
= \(\frac{1}{2}\) [-sin 2A + \(\frac{\sqrt{3}}{2}\)]

(iii) cos\(\frac{7 A}{3}\) sin\(\frac{5 A}{3}\)

(iv) cos 7θ sin 3θ = \(\frac{1}{2}\) [sin(7θ + 3θ) – sin(7θ – 3θ)]
= \(\frac{1}{2}\) (sin 10θ – sin 4θ)

Question 2.
Express each of the following as the product of sine and cosine
(i) sin A + sin 2A
(ii) cos 2A + cos 4A
(iii) sin 6θ – sin 2θ
(iv) cos 2θ – cos θ
Solution:
(i) sin A + sin 2A = 2 sin(\(\frac{A+2 A}{2}\)) cos(\(\frac{A-2 A}{2}\))
[∵ sin C + sin D = sin(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 sin \(\frac{3A}{2}\) cos \(\frac{A}{2}\) [∵ cos(-θ) = cos θ]

(ii) cos 2A + cos 4A = 2 cos(\(\frac{2 \mathrm{A}+4 \mathrm{A}}{2}\)) cos(\(\frac{2 \mathrm{A}-4 \mathrm{A}}{2}\))
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))
= 2 cos(\(\frac{6 \mathrm{A}}{2}\)) cos(\(\frac{6 \mathrm{-2A}}{2}\))
= 2 cos(3A) cos (-A) [∵ cos(-θ) = cos θ]
= 2 cos 3A cos A

(iii) sin 6θ – sin 2θ = 2 cos(\(\frac{6 \theta+2 \theta}{2}\)) cos(\(\frac{6 \theta-2 \theta}{2}\))
[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))
= 2 cos(\(\frac{8 \theta}{2}\)) sin(\(\frac{4 \theta}{2}\))
= 2 cos 4θ sin 2θ

(iv) cos 2θ – cos θ = -2 sin(\(\frac{2 \theta+\theta}{2}\)) sin(\(\frac{2 \theta-\theta}{2}\))
[∵ cos C – cos D = -2 sin(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))
= -2 sin(\(\frac{3 \theta}{2}\)) sin(\(\frac{\theta}{2}\))

Question 3.
Prove that
(i) cos 20° cos 40° cos 80° = \(\frac{1}{8}\)
(ii) tan 20° tan 40° tan 80° = √3.
Solution:
(i) cos 20° cos 40° cos 80° = \(\left(\frac{2 \sin 20^{\circ}}{2 \sin 20^{\circ}}\right)\) cos 20° cos 40° cos 80°
[multiply and divide by 2 sin 20°]

(Multiply and divide by 2)


[∵ sin(180° – θ) = sin θ]

(ii) tan 20° tan 40° tan 80°

Consider sin 20° × sin 40° sin 80°
= sin 20° sin (60° – 20°) sin (60° + 20°)
= sin 20° [sin² 60° – sin² 20°]

cos 20° × cos 40° cos 80° = \(\frac{1}{8}\) [∵ from (i)] …… (2)
divide (1) by (2) we get, tan 20° tan 40° tan 80° = \(\frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}}\) = √3

Question 4.
Prove that
(i) (cos α – cos β)² + (sin α – sin β)² = 4 \(\sin ^{2}\left(\frac{\alpha-\beta}{2}\right)\)
(ii) sin A sin(60° + A) sin(60° – A) = sin 3A
Solution:
(i) LHS = (cos α – cos β)² + (sin α – sin β)²

(ii) LHS = 4 sin A sin (60° + A) . sin (60° – A)
= 4 sin A {sin (60° + A) . sin (60° – A)}
= 4 sin A {sin² 60° – sin² A}
= 4 sin A {\(\frac{3}{4}\) – sin² A}
= 3 sin A – 4 sin³ A
= sin 3A
= RHS

Question 5.
Prove that
(i) sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
(ii) \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)
Solution:
Consider sin (A – B) sin C
= (sin A cos B – cos A sin B) sin C
= sin A cos B sin C – cos A sin B sin C …….. (1)
Similarly sin(B – C) sin A = sin B cos C sin A – cos B sin C sin A …….. (2)
[Replace A by B, B by C, C by A in (1)]
and sin(C – A) sin B [Replace A by B, B by C, C by A in (2)]
= sin C cos A sin B – cos C sin A sin B …….. (3)
Adding (1), (2) and (3) we get
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0

(ii) \(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)

[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]

[∵ cos(-θ) = cos θ]

[take 2 cos \(\frac{\pi}{3}\) as commom)

Hence proved.

Question 6.
Prove that
(i) \(\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2}\)
(ii) \(\frac{\cos 7 \mathbf{A}+\cos 5 \mathbf{A}}{\sin 7 \mathbf{A}-\sin 5 \mathbf{A}}=\cot \mathbf{A}\)
Solution:
(i) \(\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}=\tan \frac{A}{2}\)

(ii) \(\frac{\cos 7 \mathbf{A}+\cos 5 \mathbf{A}}{\sin 7 \mathbf{A}-\sin 5 \mathbf{A}}=\cot \mathbf{A}\)



Hence proved.

Question 7.
Prove that cos 20° cos 40° cos 60° cos 80° = \(\frac{3}{16}\).
Solution:
LHS = cos 20° cos 40° cos 60° cos 80°
= cos 20° cos 40° (\(\frac{1}{2}\)) cos 80° [∵ cos 60° = \(\frac{1}{2}\)]
= \(\frac{1}{2}\) (cos 20° cos 40° cos 80°)
= \(\frac{1}{2}\left(\frac{2 \sin 20^{\circ}}{2 \sin 20^{\circ}}\right)\) (cos 20° cos 40° cos 80°)
[multiply and divide by 2 sin 20°]

[multiply and divide by 2]

Hence Proved.

Question 8.
Evaluate:
(i) cos 20° + cos 100° + cos 140°
(ii) sin 50° – sin 70° + sin 10°
Solution:
(i) LHS = (cos 20° + cos 100°) + cos 140°
= 2 \(\cos \left(\frac{20^{\circ}+100^{\circ}}{2}\right) \cos \left(\frac{20^{\circ}-100^{\circ}}{2}\right)\) + cos 140°
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
= 2 cos 60° cos(-40°) + cos 140°
= 2 × \(\frac{1}{2}\) × cos(-40°) + cos(180° – 140°)
[∵ cos(-θ) = cos θ, cos 60° = \(\frac{1}{2}\)
= cos 40° – cos 40°
= 0
Hence Proved.

(ii) LHS = (sin 50° – sin 70°) + sin 10°
= 2 \(\cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right) \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right)\) + sin 10°
[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]
= 2 cos 60° sin(-10°) + sin 10°
= 2 × \(\frac{1}{2}\) (-sin 10°) + sin 10° [∵ sin(-θ) = -sin θ]
= -sin 10° + sin 10°
= 0
= RHS

Question 9.
If cos A + cos B = \(\frac{1}{2}\) and sin A + sin B = \(\frac{1}{4}\), prove that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\frac{\mathbf{1}}{2}\)
Solution:
Given that cos A + cos B = \(\frac{1}{2}\)
\(2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{2}\) …….. (1)
Also given that sin A + sin B = \(\frac{1}{4}\)
\(2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{1}{4}\) ……. (2)
\(\frac{(2)}{(1)}\) gives

Question 10.
If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.
Solution:
In A.P. commom difference are equal, namely t₂ – t₁ = t₃ – t₂
sin(z + x – y) – sin(y + z – x) = sin(x + y – z) – sin(z + x – y)


cos z sin (x – y) = cos x sin (y – z)
cos z (sin x cos y – cos x sin y) = cos x (sin y cos z – cos y sin z)
Divide bothsides by cos x cos y cos z we get

tan x – tan y = tan y – tan z
Multiply both sides by (-1) we get,
tan y – tan x = tan z – tan y
This means tan x, tan y, and tan z are in A.P.
Hence proved.

Question 11.
If cosec A + sec A = cosec B + sec B prove that cot(\(\frac{A+B}{2}\)) = tan A tan B.
Solution:
Given that cosec A + sec A = cosec B + sec B

Arrange T-ratios of the sine and cosine in the separate side

[∵ sin C – sin D = 2 cos(\(\frac{C+D}{2}\)) sin(\(\frac{C-D}{2}\))]

Students can download 11th Business Maths Chapter 4 Trigonometry Ex 4.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.2

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.2 Text Book Back Questions and Answers

Question 1.
Find the values of the following:
(i) cosec 15°
(ii) sin (-105°)
(iii) cot 75°
Solution:
(i) cosec 15° = \(\frac{1}{\sin 15^{\circ}}\)
Consider sin 15° = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°

cosec 15° = \(\frac{1}{\sin 15^{\circ}}\) = \(\frac{2 \sqrt{2}}{\sqrt{3}-1}\)

(ii) sin (-105°) = -sin (105°) (∵ sin (-θ) = – sin θ)
= -[sin(60° + 45°)]
= -[sin 60° cos 45° + cos 60° sin 45°]

(iii) cot 75° = \(\frac{1}{\tan 75^{\circ}}\)
Consider tan 75° = tan (30° + 45°)

cot 75° = \(\frac{1}{\tan 75^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

Question 2.
Find the values of the following:
(i) sin 76° cos 16° – cos 76° sin 16°
(ii) \(\sin \frac{\pi}{4} \cos \frac{\pi}{12}+\cos \frac{\pi}{4} \sin \frac{\pi}{12}\)
(iii) cos 70° cos 10° – sin 70° sin 10°
(iv) cos² 15° – sin² 15°
Solution:
(i) Given that, sin 76° cos 16° – cos 76° sin 16° (∴ This is of the form sin(A – B))
= sin(76° – 16°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)

(ii) This is of the form sin(A + B) = \(\sin \left(\frac{\pi}{4}+\frac{\pi}{12}\right)\)
= \(\sin \left(\frac{3 \pi+\pi}{12}\right)\)
= \(\sin \frac{4 \pi}{12}\)
= \(\sin \frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\) (∵ sin 60° = \(\frac{\sqrt{3}}{2}\))

(iii) Given that cos 70° cos 10° – sin 70° sin 10°
(This is of the form of cos (A + B), A = 70°, B = 10°)
= cos (70° + 10°)
= cos 80°

(iv) cos² 15° – sin² 15°
[∵ cos 2A = cos² A – sin² A, Here A = 15°]
= cos (2 × 15°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

Question 3.
If sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) and cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\), find the values of the following:
(i) cos(A + B)
(ii) sin(A – B)
(iii) tan(A – B)
Solution:
Given that sin A = \(\frac{3}{5}\), 0 < A < \(\frac{\pi}{2}\) (i.e., A lies in first quadrant)
Since A lies in first quadrant cos A is positive.

cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
tan A = \(\frac{3}{4}\)
AB = \(\sqrt{5^{2}-3^{2}}\) = 4
Also given that cos B = \(\frac{-12}{13}\), π < B < \(\frac{3 \pi}{2}\) (i.e., B lies in third quadrant)
Now sin B lies in third quadrant. sin B is negative.

CA = \(\sqrt{13^{2}-12^{2}}\) = 5
sin B = \(\frac{-\text { Opposite side }}{\text { Hypotenuse }}=\frac{-5}{13}\)
tan B = \(\frac{-\text { Opposite side }}{\text { Adjacent }}=\frac{5}{12}\) [B lies in 3rd quadrant. tan B is positive.]
(i) cos(A + B) = cos A cos B – sin A sin B

(ii) sin(A – B) = sin A cos B – cos A sin B

(iii) tan(A – B)

Question 4.
If cos A = \(\frac{13}{14}\) and cos B = \(\frac{1}{7}\) where A, B are acute angles prove that A – B = \(\frac{\pi}{3}\)
Solution:
cos A = \(\frac{13}{14}\), cos B = \(\frac{1}{7}\)

cos(A – B) = cos A cos B + sin A sin B

cos(A – B) = cos 60°
A – B = 60° = \(\frac{\pi}{3}\)

Question 5.
Prove that 2 tan 80° = tan 85° – tan 5°.
Solution:
Consider tan 80° = tan(85° – 5°)


∴ 2 tan 80° = tan 85° – tan 5°
Hence Proved.

Question 6.
If cot α = \(\frac{1}{2}\), sec β = \(\frac{-5}{3}\), where π < α < \(\frac{3 \pi}{2}\) and \(\frac{\pi}{2}\) < β < π, find the value of tan(α + β). State the quadrant in which α + β terminates.
Solution:
Given that cot α = \(\frac{1}{2}\) where π < α < \(\frac{3 \pi}{2}\) (i.e,. α lies in third quadrant)
tan α = \(\frac{1}{\frac{1}{2}}\) = 2 [∵ In 3rd quadrant tan α is positive]
Also given that sec β = \(\frac{-5}{3}\) where \(\frac{\pi}{2}\) < β < π (i.e., β lies in second quadrant cos β and tan β are negative)


tan (α + β) = \(\frac{2}{11}\) which is positive.
α + β terminates in first quandrant.

Question 7.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22\(\frac{1}{2}\).
Solution:
Given A + B = 45°
tan (A + B) = tan 45°
\(\frac{\tan A+\tan B}{1-\tan A \tan B}=1\)
tan A + tan B = 1 – tan A . tan B
tan A + tan B + tan A tan B = 1
Add 1 on both sides we get,
(1 + tan A) + tan B + tan A tan B = 2
1(1+ tan A) + tan B (1 + tan A) = 2
(1 + tan A) (1 + tan B) = 2 ……. (1)
Put A = B = 22\(\frac{1}{2}\) in (1) we get
(1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2
⇒ (1 + tan22\(\frac{1}{2}\))² = 2
⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2
⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1
Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and therefore tan 22\(\frac{1}{2}\) = √2 – 1

Question 8.
Prove that
(i) sin(A + 60°) + sin(A – 60°) = sin A.
(ii) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Solution:
(i) LHS = sin (A + 60°) + sin (A – 60°)
= sin A cos 60° + cos A sin 60° + sin A cos 60° – cos A sin 60°
= 2 sin A cos 60°
= 2 sin A \(\left(\frac{1}{2}\right)\)
= sin A
= RHS

(ii) 4A = 3A + A
tan 4A = tan (3A + A)
tan 4A = \(\frac{\tan 3 \mathrm{A}+\tan \mathrm{A}}{1-\tan 3 \mathrm{A} \tan \mathrm{A}}\)
on cross multiplication we get,
tan 3A + tan A = tan 4A (1 – tan 3A tan A) = tan 4A – tan 4A tan 3A tanA
i.e., tan 4A tan 3A tan A + tan 3A + tan A = tan 4A
(or) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0

Question 9.
(i) If tan θ = 3 find tan 3θ
(ii) If sin A = \(\frac{12}{13}\), find sin 3A.
Solution:
(i) tan θ = 3

(ii) If sin A = \(\frac{12}{13}\)
We know that sin 3A = 3 sin A – 4 sin³ A

Question 10.
If sin A = \(\frac{3}{5}\), find the values of cos 3A and tan 3A.
Solution:

Given sin A = \(\frac{3}{5}\)
cos A = \(\frac{\text { Adjacent side }}{\text { Hypotenuse }}=\frac{4}{5}\)
and tan A = \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{3}{4}\)
We know that cos 3A = 4 cos³ A – 3 cos A

Question 11.
Prove that \(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)
Solution:
Consider \(\frac{\sin (B-C)}{\cos B \cos C}\)
= \(\frac{\sin \mathrm{B} \cos \mathrm{C}-\cos \mathrm{B} \sin \mathrm{C}}{\cos \mathrm{B} \cos \mathrm{C}}\)
= \(\frac{\sin B \cos C}{\cos B \cos C}-\frac{\cos B \sin C}{\cos B \cos C}\)
= tan B – tan C ……… (1)
Similarly we can prove \(\frac{\sin (C-A)}{\cos C \cos A}\) = tan C – tan A …….(2)
and \(\frac{\sin (A-B)}{\cos A \cos B}\) = tan A – tan B …….. (3)
Add (1), (2) and (3) we get
\(\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}+\frac{\sin (A-B)}{\cos A \cos B}=0\)

Question 12.
If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) = \(\frac{1}{x}+\frac{1}{y}\).
Solution:


Hence proved.

Question 13.
If sin α + sin β = a and cos α + cos β = b, then prove that cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)
Solution:
Consider a² + b² = sin²α + sin²β + 2 sin α sin β + cos²α + cos²β + 2 cos α cos β
a² + b² = (sin²α + cos²α) + (sin²β + cos²β) + 2[cos α cos β + sin α sin β]
a² + b² = 1 + 1 + 2 cos(α – β)
∴ cos(α – β) = \(\frac{a^{2}+b^{2}-2}{2}\)

Question 14.
Find the value of tan\(\frac{\pi}{8}\).
Solution:
Method 1:
\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=\frac{45^{\circ}}{2}=22 \frac{1}{2}\)
We know that tan 2A = \(\frac{2 \tan A}{1-\tan ^{2} A}\)
Put A = 22\(\frac{1}{2}\) in the above formula

On cross multiplication we get


Here a = 1, b = 2, c = -1

Since 22\(\frac{1}{2}\) is acute tan 22\(\frac{1}{2}\) is positive tan 22\(\frac{1}{2}\) = tan \(\frac{\pi}{8}\)
= -1 + √2
= √2 – 1

Method 2:



∴ \(\tan ^{2} 22 \frac{1}{2}=(\sqrt{2}-1)^{2}\)
Taking square root, \(\tan ^{2} 22 \frac{1}{2}\) = ±(√2 – 1)
But \(22 \frac{1}{2}\) lies in first quadrant, tan \(22 \frac{1}{2}\) is positive.
∴ tan 22\(\frac{1}{2}\) = √2 – 1

Method 3:
consider tan A = \(\frac{\sin 2 A}{1+\cos 2 A}\)
Put A = \(22 \frac{1}{2}\)



tan 22\(\frac{1}{2}\) = √2 – 1

Question 15.
If tan α = \(\frac{1}{7}\), sin β = \(\frac{1}{\sqrt{10}}\). Prove that α + 2β = \(\frac{\pi}{4}\) where 0 < α < \(\frac{\pi}{2}\) and 0 < β < \(\frac{\pi}{2}\).
Solution:
Given that tan α = \(\frac{1}{7}\)
We wish to find tan(α + 2β)

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 4 Trigonometry Ex 4.1

Samacheer Kalvi 11th Business Maths Trigonometry Ex 4.1 Text Book Back Questions and Answers

Question 1.
Convert the following degree measure into radian measure
(i) 60°
(ii) 150°
(iii) 240°
(iv) -320°
Solutions:
(i) 1°= \(\frac{\pi}{180}\) radians
∴ 60° = \(\frac{\pi}{180}\) × 60 radians = \(\frac{\pi}{3}\) radians.

(ii) 150° = \(\frac{\pi}{180}\) × 150 radians = \(\frac{5 \pi}{6}\) radians.

(iii) 240° = \(\frac{\pi}{180}\) × 240 radians = \(\frac{4 \pi}{3}\) radians.

(iv) -320° = \(\frac{\pi}{180}\) × -320 = \(\frac{-16 \pi}{9}\) radians

Question 2.
Find the degree measure corresponding to the following radian measure.
(i) \(\frac{\pi}{8}\)
(ii) \(\frac{9 \pi}{5}\)
(iii) -3
(iv) \(\frac{11 \pi}{18}\)
Solution:
We know that, one radian = \(\frac{180^{\circ}}{\pi}\)
(i) \(\frac{\pi}{8}\)
\(\frac{\pi}{8}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{8}\) degrees
= \(\frac{45}{2}\)
= 22.5°
= 22°30′ [∵ 0.5° = (0.5 × 60)’ = 30′]

(ii) \(\frac{9 \pi}{5}\)
\(\frac{9 \pi}{5}=\frac{180^{\circ}}{\pi} \times \frac{9 \pi}{5}\) degrees
= 36 × 9 degrees
= 324°

(iii) -3

= -171.81°
= -171°48′ (∵ 0.8° = (0.8 × 60)’ = 48′)

(iv) \(\frac{11 \pi}{18}\)
\(\frac{11 \pi}{18}=\frac{180}{\pi} \times \frac{11 \pi}{18}\)
= 10 × 11°
= 110°

Question 3.
Determine the quadrants in which the following degree lie.
(i) 380°
(ii) -140°
(iii) 1195°
Solution:
(i) 380° = 360°+ 20°
This is of the form 360° + θ
∴ After one completion of the round, the angle is 20°, 380° lies in the I quadrant.

(ii) -140° = -90° + (-50°)
The angle is negative it moves in the anti-clockwise direction.
-140° lies in the III quadrants.

(iii) 1195° = (3 × 360°) + 90° + 25°
∴ After three completion round, the angle will lie in the II quadrant.
1195° lies in the II quadrant.

Question 4.
Find the values of each of the following trigonometric ratios.
(i) sin 300°
(ii) cos(-210°)
(iii) sec 390°
(iv) tan(-855°)
(v) cosec 1125°
Solution:
(i) sin 300° = sin(360° – 60°)
[For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative]
= -sin 60°
= \(-\frac{\sqrt{3}}{2}\)

(ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ)
[∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative]
= cos(180° + 30°)
= -cos 30°
= \(-\frac{\sqrt{3}}{2}\)

(iii) sec 390° = sec(360° + 30°)
= sec 30°
= \(\frac{1}{\cos 30^{\circ}}\)
= \(\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}\)
= \(\frac{2}{\sqrt{3}}\)

(iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ)
[∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative]
= -tan(2 × 360° + 135°)
= -tan 135°
= -tan(180° – 45°)
= -(-tan 45°)
= -(-1)
= 1

(v) cosec 1125° = cosec(3 × 360°+ 45°)
= cosec 45°
= \(\frac{1}{\sin 45^{\circ}}\)
= \(\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}\)
= √2

Question 5.
Prove that:
(i) tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0.
(ii) 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\) = \(\frac{3}{2}\)
(iii) \(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)=-1\)
Solution:
(i) tan(-225°) = -(tan 225°)
= -(tan(180° + 45°))
= – tan 45°
= – 1
cot(-405°) = -(cot 405°)
= – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.]
= -cot 45°
= -1
tan(-765°) = -tan 765°
= -tan(2 × 360° + 45°)
= -tan 45°
= -1
cot 675° = cot (360°+ 315°)
= cot 315°
= cot(360° – 45°)
= -cot 45°
= -1
LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°)
= (-1) (-1) – (-1) (-1)
= 1 – 1
= 0
= RHS.
Hence proved.

(ii) 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\) = \(\frac{3}{2}\)
LHS = 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\)
[∵ \(\frac{7 \pi}{6}\) = 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio.
210° lies in 3rd quadrant, cosec θ is negative.]
= 2\(\left(\sin \frac{\pi}{6}\right)^{2}\) + (cosec (180° + 30°))² \(\left(\cos \frac{\pi}{3}\right)^{2}\)
= 2 \(\left(\frac{1}{2}\right)^{2}\) + (-cosec 30°)² . \(\left(\frac{1}{2}\right)^{2}\)
= \(2 \times \frac{1}{4}+(-2)^{2} \frac{1}{4}\)
= \(\frac{2}{4}+\frac{4}{4}=\frac{6}{4}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
= RHS

(iii) sec(\(\frac{3 \pi}{2}\) – θ) = sec (270° – θ) = -cosec θ
[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’]
sec(θ – \(\frac{5 \pi}{2}\)) = \(\sec \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)\)
= sec(\(\frac{5 \pi}{2}\) – θ) [∵ sec(-θ) = θ]
= sec(450° – θ)
= sec (360° + (90° – θ))
= sec (90° – θ)
= cosec θ
[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’]
tan(\(\frac{5 \pi}{2}\) + θ) = tan(450° + θ)
[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’]
= tan (360° + (90° + θ))
= tan (90° + θ)
= -cot θ
\(\tan \left(\theta-\frac{5 \pi}{2}\right)=\tan \left(-\left(\frac{5 \pi}{2}-\theta\right)\right)\)
= \(-\tan \left(\frac{5 \pi}{2}-\theta\right)\) [∵ tan(-θ) = -tan θ]
= -tan(450° – θ)
= -tan(360° + (90° – θ))
= -tan(90° – θ)
= -cot θ
LHS = \(\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{5 \pi}{2}\right)\)
= -cosec θ (cosec θ) + (-cot θ) (-cot θ)
= -cosec² θ + cot² θ
= -(1 + cot² θ) + cot² θ [∵ 1 + cot² θ = cosec² θ]
= -1
= RHS

Question 6.
If A, B, C, D are angles of a cyclic quadrilateral, prove that: cos A + cos B + cos C + cos D = 0.
Solution:
Note: If the vertices of a quadrilateral lie on the circle then the quadrilateral is called a cyclic quadrilateral.
In a cyclic quadrilateral sum of opposite angles are 180°.

Since A, B, C, D are angles of cyclic quadrilateral
A + C = 180° and B + D = 180°
LHS = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
= RHS

Question 7.
Prove that

(ii) sin θ . cos{sin(\(\frac{\pi}{2}\) – θ) . cosec θ + cos(\(\frac{\pi}{2}\) – θ) . sec θ} = 1
Solution:

(ii) sin θ . cos{sin(\(\frac{\pi}{2}\) – θ) . cosec θ + cos(\(\frac{\pi}{2}\) – θ) . sec θ} = 1

Question 8.
Prove that: cos 510° cos 330° + sin 390° cos 120° = -1.
Solution:
LHS = cos 510° cos 330° + sin 390° cos 120°
= cos(360° + 150°) cos(360° – 30°) + sin(360° + 30°) × cos(180° – 60°)
= cos 150° cos 30° + sin 30° (-cos 60°)
= cos(180° – 30°) cos 30° + sin 30° cos 60°
= -cos 30° cos 30° + \(\frac{1}{2} \times\left(\frac{-1}{2}\right)\)
= \(-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}\)
= \(-\frac{3}{4}-\frac{1}{4}\)
= \(\frac{-3-1}{4}\)
= -1

Question 9.
Prove that:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = sin² x.

Solution:
(i) tan(π + x) cot(x – π) – cos(2π – x) cos(2π + x) = (tan x) (-cot(π – x) – cos x cos x
[∵ cot(x – π) = cot(-(π – x)) = -cot(π – x) = cot x]
= tan x cot x – cos² x
= 1 – cos² x
= sin² x [∵ sin² x + cos² x = 1 ⇒ sin² x = (1 – cos² x)]

Question 10.
If sin θ = \(\frac{3}{5}\), tan φ = \(\frac{1}{2}\) and \(\frac{\pi}{2}\) < θ < π < φ < \(\frac{3 \pi}{2}\), then find the value of 8 tan θ – √5 sec φ.
Solution:

Given that sin θ = \(\frac{3}{5}=\frac{\text { Opposite side }}{\text { Hypotenuse }}\)
∵ AB = \(\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}\) = 4
Here θ lies in second quadrant [∵ \(\frac{\pi}{2}\) < θ < π]
∵ tan θ is negative.
tan θ = \(-\frac{3}{4}\)
Also given that tan Φ = \(\frac{1}{2}=\frac{\text { Opposite side }}{\text { Adjacent side }}\)
∴ PR = \(\sqrt{\mathrm{PQ}^{2}+\mathrm{QP}^{2}}=\sqrt{4+1}=\sqrt{5}\)
Here Φ lies in third quadrant (∵ π < Φ < \(\frac{3 \pi}{2}\))
∴ sec Φ is negative.
\(\sec \phi=\frac{1}{\cos \phi}=-\frac{1}{\left(\frac{2}{\sqrt{5}}\right)}=-\frac{\sqrt{5}}{2}\)
Now 8 tan θ – √5 sec Φ = \(8\left(-\frac{3}{4}\right)-\sqrt{5}\left(\frac{-\sqrt{5}}{2}\right)\)
= 2 × (-3) + \(\frac{5}{2}\)
= -6 + \(\frac{5}{2}\)
= \(\frac{-12+5}{2}\)
= \(\frac{-7}{2}\)

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.7

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.7 Text Book Back Questions and Answers

Question 1.
If m₁ and m₂ are the slopes of the pair of lines given by ax² + 2hxy + by² = 0, then the value of m₁ + m₂ is:
(a) \(\frac{2 h}{b}\)
(b) \(-\frac{2 h}{b}\)
(c) \(\frac{2 h}{a}\)
(d) \(-\frac{2 h}{a}\)
Answer:
(b) \(-\frac{2 h}{b}\)

Question 2.
The angle between the pair of straight lines x² – 7xy + 4y² = 0 is:
(a) \(\tan ^{-1}\left(\frac{1}{3}\right)\)
(b) \(\tan ^{-1}\left(\frac{1}{2}\right)\)
(c) \(\tan ^{-1}\left(\frac{\sqrt{33}}{5}\right)\)
(d) \(\tan ^{-1}\left(\frac{5}{\sqrt{33}}\right)\)
Answer:
(c) \(\tan ^{-1}\left(\frac{\sqrt{33}}{5}\right)\)
Hint:
x² – 7xy + 4y² = 0
Here 2h = -7, a = 1, b = 4

Question 3.
If the lines 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are the diameters of a circle, then its centre is:
(a) (-1, 1)
(b) (1, 1)
(c) ( 1, -1)
(d) (-1, -1)
Answer:
(c) ( 1, -1)
Hint:
To get centre we must solve the given equations
2x – 3y – 5 = 0 …….(1)
3x – 4y – 7 = 0 ………(2)
(1) × 3 ⇒ 6x – 9y = 15
(2) × 2 ⇒ 6x – 8y = 14
Subtracting, -y = 1 ⇒ y = -1
Using y = -1 in (1) we get
2x + 3 – 5 = 0
⇒ 2x = 2
⇒ x = 1

Question 4.
The x-intercept of the straight line 3x + 2y – 1 = 0 is
(a) 3
(b) 2
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{3}\)
Hint:
To get x-intercept put y = 0 in 3x + 2y – 1 = 0 we get
3x – 1 = 0
x = \(\frac{1}{3}\)

Question 5.
The slope of the line 7x + 5y – 8 = 0 is:
(a) \(\frac{7}{5}\)
(b) \(-\frac{7}{5}\)
(c) \(\frac{5}{7}\)
(d) \(-\frac{5}{7}\)
Answer:
(b) \(-\frac{7}{5}\)
Hint:
Slope of 7x + 5y – 8 = 0 is = \(\frac{-x \text { coefficient }}{y \text { coefficient }}\) = \(-\frac{7}{5}\)

Question 6.
The locus of the point P which moves such that P is at equidistance from their coordinate axes is:
(a) y = \(\frac{1}{x}\)
(b) y = -x
(c) y = x
(d) y = \(\frac{-1}{x}\)
Answer:
(c) y = x
Hint:

Given PA = PB
y₁ = x₁
∴ Locus is y = x

Question 7.
The locus of the point P which moves such that P is always at equidistance from the line x + 2y + 7 = 0:
(a) x + 2y + 2 = 0
(b) x – 2y + 1 = 0
(c) 2x – y + 2 = 0
(d) 3x + y + 1 = 0
Answer:
(a) x + 2y + 2 = 0
Hint:
Locus is line parallel to line x + 2y + 7 = 0 which is x + 2y + 2 = 0

Question 8.
If kx² + 3xy – 2y² = 0 represent a pair of lines which are perpendicular then k is equal to:
(a) \(\frac{1}{2}\)
(b) \(-\frac{1}{2}\)
(c) 2
(d) -2
Answer:
(c) 2
Hint:
Here a = k, b = -2
Condition for perpendicular is
a + b = 0
⇒ k – 2 = 0
⇒ k = 2

Question 9.
(1, -2) is the centre of the circle x² + y² + ax + by – 4 = 0, then its radius:
(a) 3
(b) 2
(c) 4
(d) 1
Answer:
(a) 3
Hint:
Given centre (-g, -f) = (1, -2)
From the given equation c = -4
Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-(-4)}=\sqrt{9}\) = 3

Question 10.
The length of the tangent from (4, 5) to the circle x² + y² = 16 is:
(a) 4
(b) 5
(c) 16
(d) 25
Answer:
(b) 5
Hint:
Length of the tangent from (x₁, y₁) to the circle x² + y² = 16 is \(\sqrt{x_{1}^{2}+y_{1}^{2}-16}=5\)

Question 11.
The focus of the parabola x² = 16y is:
(a) (4 , 0)
(b) (-4, 0)
(c) (0, 4)
(d) (0, -4)
Answer:
(c) (0, 4)
Hint:
x² = 16y
Here 4a = 16 ⇒ a = 4
Focus is (0, a) = (0, 4)

Question 12.
Length of the latus rectum of the parabola y² = -25x:
(a) 25
(b) -5
(c) 5
(d) -25
Answer:
(a) 25
Hint:
y² = -25a
Here 4a = 25 which is the length of the latus rectum.

Question 13.
The centre of the circle x² + y² – 2x + 2y – 9 = 0 is:
(a) (1, 1)
(b) (-1, 1)
(c) (-1, 1)
(d) (1, -1)
Answer:
(d) (1, -1)
Hint:
2g = -2, 2f = 2
g = -1, f = 1
Centre = (-g, -f) = (1, -1)

Question 14.
The equation of the circle with centre on the x axis and passing through the origin is:
(a) x² – 2ax + y² = 0
(b) y² – 2ay + x² = 0
(c) x² + y² = a²
(d) x² – 2ay + y² = 0
Answer:
(a) x² – 2ax + y² = 0
Hint:
Let the centre on the x-axis as (a, 0).
This circle passing through the origin so the radius

Now centre (h, k) = (a, 0)
Radius = a
Equation of the circle is (x – a)² + (y – 0)² = a²
⇒ x² – 2ax + a² + y² = a²
⇒ x² – 2ax + y² = 0

Question 15.
If the centre of the circle is (-a, -b) and radius is \(\sqrt{a^{2}-b^{2}}\) then the equation of circle is:
(a) x² + y² + 2ax + 2by + 2b² = 0
(b) x² + y² + 2ax + 2by – 2b² = 0
(c) x² + y² – 2ax – 2by – 2b² = 0
(d) x² + y² – 2ax – 2by + 2b² = 0
Answer:
(a) x² + y² + 2ax + 2by + 2b² = 0
Hint:
Equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + a)² + (y + b)² = a² – b²
⇒ x² + y² + 2ax + 2by + a² + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 16.
Combined equation of co-ordinate axes is:
(a) x² – y² = 0
(b) x² + y² = 0
(c) xy = c
(d) xy = 0
Answer:
(d) xy = 0
Hint:
Equation of x-axis is y = 0
Equation of y-axis is x = 0
Combine equation is xy = 0

Question 17.
ax² + 4xy + 2y² = 0 represents a pair of parallel lines then ‘a’ is:
(a) 2
(b) -2
(c) 4
(d) -4
Answer:
(a) 2
Hint:
Here a = 0, h = 2, b = 2
Condition for pair of parallel lines is b² – ab = 0
4 – a(2) = 0
⇒ -2a = -4
⇒ a = 2

Question 18.
In the equation of the circle x² + y² = 16 then v intercept is (are):
(a) 4
(b) 16
(c) ±4
(d) ±16
Answer:
(c) ±4
Hint:
To get y-intercept put x = 0 in the circle equation we get
0 + y² = 16
∴ y = ±4

Question 19.
If the perimeter of the circle is 8π units and centre is (2, 2) then the equation of the circle is:
(a) (x – 2)² + (y – 2)² = 4
(b) (x – 2)² + (y – 2)² = 16
(c) (x – 4)² + (y – 4)² = 16
(d) x² + y² = 4
Answer:
(c) (x – 2)² + (y – 2)² = 16
Hint:
Perimeter, 2πr = 8π
r = 4
Centre is (2, 2)
Equation of the circle is (x – 2)² + (y – 2)² = 4² = 16

Question 20.
The equation of the circle with centre (3, -4) and touches the x-axis is:
(a) (x – 3)² + (y – 4)² = 4
(b) (x – 3)² + (y + 4)² = 16
(c) (x – 3)² + (y – 4)² = 16
(d) x² + y² = 16
Answer:
(b) (x – 3)² + (y + 4)² = 16
Hint:
Centre (3, -4).
It touches the x-axis.
The absolute value of y-coordinate is the radius, i.e., radius = 4.
Equation is (x – 3)² + (y + 4)² = 16

Question 21.
If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
(a) 6
(b) 3
(c) 12
(d) 4
Answer:
(a) 6
Hint:

Question 22.
The eccentricity of the parabola is:
(a) 3
(b) 2
(c) 0
(d) 1
Answer:
(d) 1

Question 23.
The double ordinate passing through the focus is:
(a) focal chord
(b) latus rectum
(c) directrix
(d) axis
Answer:
(b) latus rectum
Hint:

Question 24.
The distance between directrix and focus of a parabola y² = 4ax is:
(a) a
(b) 2a
(c) 4a
(d) 3a
Answer:
(b) 2a

Question 25.
The equation of directrix of the parabola y² = -x is:
(a) 4x + 1 = 0
(b) 4x – 1 = 0
(c) x – 1 = 0
(d) x + 4 = 0
Answer:
(b) 4x – 1 = 0
Hint:
y² = -x.
It is a parabola open leftwards.
Here 4a = 1 ⇒ a = \(\frac{1}{4}\)
Equation of directrix is x = a.
i.e., x = \(\frac{1}{4}\) (or) 4x – 1 = 0

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.6

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.6 Text Book Back Questions and Answers

Question 1.
Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 0.
Solution:
F(-1, -2)
l : 4x – 3y + 2 = 0
Let P(x, y) be any point on the parabola.
FP = PM
⇒ FP² = PM²
⇒ (x + 1)² + (y + 2)² = \(\left[\frac{4 x-3 y+2}{\sqrt{4^{2}+(-3)^{2}}}\right]^{2}\)
⇒ x² + 2x + 1 + y² + 4y + 4 = \(\frac{16 x^{2}+9 y^{2}+4-24 x y+16 x-12 y}{(16+9)}\)
⇒ 25(x² + y² + 2x + 4y + 5) = 16x² + 9y² – 24xy + 16x – 12y + 4
⇒ (25 – 16)x² + (25 – 9)y² + 24xy + (50 – 16)x + (100 + 12)y + 125 – 4 = 0
⇒ 9x² + 16y² + 24xy + 34x + 112y + 121 = 0

Question 2.
The parabola y² = kx passes through the point (4, -2). Find its latus rectum and focus.
Solution:
y² = kx passes through (4, -2)
(-2)² = k(4)
⇒ 4 = 4k
⇒ k = 1
y² = x = 4(\(\frac{1}{4}\))x
a = \(\frac{1}{4}\)
Equation of LR is x = a or x – a = 0
i.e., x = \(\frac{1}{4}\)
⇒ 4x = 1
⇒ 4x – 1 = 0
Focus (a, 0) = (\(\frac{1}{4}\), 0)

Question 3.
Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola y² – 8y – 8x + 24 = 0.
Solution:
y² – 8y – 8x + 24 = 0
⇒ y² – 8y – 4² = 8x – 24 + 4²
⇒ (y – 4)² = 8x – 8
⇒ (y – 4)² = 8(x – 1)
⇒ (y – 4)² = 4(2) (x – 1)
∴ a = 2
Y² = 4(2)X where X = x – 1 and Y = y – 4

Question 4.
Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola (a) y² = 20x, (b) x² = 8y, (c) x² = -16y
Solution:
(a) y² = 20x
y² = 4(5)x
∴ a = 5

(b) x² = 8y = 4(2)y
∴ a = 2

(c) x² = -16y = -4(4)y
∴ a = 4

Question 5.
The average variable cost of the monthly output of x tonnes of a firm producing a valuable metal is ₹ \(\frac{1}{5}\) x² – 6x + 100. Show that the average variable cost curve is a parabola. Also, find the output and the average cost at the vertex of the parabola.
Solution:
Let output be x and average variable cost = y
y = \(\frac{1}{5}\) x² – 6x + 100
⇒ 5y = x² – 30x + 500
⇒ x² – 30x + 225 = 5y – 500 + 225
⇒ (x – 15)² = 5y – 275
⇒ (x – 15)² = 5(y – 55) which is of the form X² = 4(\(\frac{5}{4}\))Y
∴ Y average variable cost curve is a parabola
Vertex (0, 0)
x – 15 = 0; y – 55 = 0
x = 15; y = 55
At the vertex, output is 15 tonnes and average cost is ₹ 55.

Question 6.
The profit ₹ y accumulated in thousand in x months is given by y = -x² + 10x – 15. Find the best time to end the project.
Solution:
y = -x² + 10x – 15
⇒ y = -[x² – 10x + 5² – 5² + 15]
⇒ y = -[(x – 5)² – 10]
⇒ y = 10 – (x – 5)²
⇒ (x – 5)² = -(y – 10)
This is a parabola which is open downwards.
Vertex is the maximum point.
∴ Profit is maximum when x – 5 = 0 (or) x = 5 months.
After that profit gradually reduces.
∴ The best time to end the project is after 5 months.