Category: Class 11

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.9

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.9 Text Book Back Questions and Answers

Question 1.
Find y₂ for the following functions:
(i) y = e^3x+2
(ii) y = log x + a^x
(iii) x = a cosθ, y = a sinθ
Solution:
(i) y = e^3x+2

(ii) y = log x + a^x

(iii) x = a cosθ, y = a sinθ
\(\frac{d x}{d \theta}\) = a(-sinθ) = -a sinθ …….. (i)
\(\frac{d y}{d \theta}\) = a(cosθ)

Question 2.
If y = 500e^7x + 600e^-7x, then show that y₂ – 49y = 0.
Solution:
y = 500e^7x + 600e^-7x

(or) y₂ – 49y = 0

Question 3.
If y = 2 + log x, then show that xy₂ + y₁ = 0.
Solution:
y = 2 + log x

Question 4.
If = a cos mx + b sin mx, then show that y₂ + m²y = 0.
Solution:
y = a cos mx + b sin mx
y₁ = a \(\frac{d}{d x}\) (cos mx) + b \(\frac{d}{d x}\) (sin mx)
[∵ \(\frac{d}{d x}\) (sin mx) = cos mx \(\frac{d}{d x}\) (mx) = (cos mx) . m]
= a(-sin mx) . m + b(cos mx) . m
= -am sin mx + bm cos mx
y₂ = -am(cos mx) . m + bm(-sin mx) . m
= -am² cos mx – bm² sin mx
= -m² [a cos mx + b sin mx]
= -m²y
∴ y₂ + m²y = 0

Question 5.
If y = \(\left(x+\sqrt{1+x^{2}}\right)^{m}\), then show that (1 + x²) y₂ + xy₁ – m²y = 0
Solution:


\(y_{1}=\frac{m y}{\sqrt{1+x^{2}}}\)
Squaring both sides we get,
\(y_{1}^{2}=\frac{m^{2} y^{2}}{\left(1+x^{2}\right)}\)
(1 + x²) (\(y_{1}^{2}\)) = m²y²
Differentiating with respect to x, we get
(1 + x²) . 2(y₁) (y₂) + (y₁)² (2x) = 2m²yy₁
Dividing both sides by 2y₁ we get,
(1 + x²) y₂ + xy₁ = m²y
⇒ (1 + x²) y₂ + xy₁ – m²y = 0

Question 6.
If y = sin(log x), then show that x²y₂ + xy₁ + y = 0.
Solution:
y = sin(log x)
y₁ = cos(log x) \(\frac{d}{d x}\) (log x)
y₁ = cos(log x) . \(\frac{1}{x}\)
∴ xy₁ = cos(log x)
Differentiating both sides with respect to x, we get
xy₂ + y₁(1) = -sin(log x) . \(\frac{1}{x}\)
⇒ x[xy₂ + y₁] = -sin(log x)
⇒ x²y₂ + xy₁ = -y
⇒ x²y₂ + xy₁ + y = 0

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.8

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.8 Text Book Back Questions and Answers

Question 1.
Find \(\frac{d y}{d x}\) of the following functions:
(i) x = ct, y = \(\frac{c}{t}\)
(ii) x = log t, y = sin t
(iii) x = a cos³θ, y = a sin³θ
(iv) x = a(θ – sin θ), y = a(1 – cos θ)
Solution:
(i) x = ct, y = \(\frac{c}{t}\)

(ii) x = log t, y = sin t

(iii) x = a cos³θ, y = a sin³θ

(iv) x = a(θ – sin θ), y = a(1 – cos θ)

Question 2.
Differentiate sin³x with respect to cos³x.
Solution:
Let u = sin³x = (sin x)³ ; v = cos³x = (cos x)³
\(\frac{d u}{d x}\) = 3(sin x) cos x ; \(\frac{d v}{d x}\) = 3(cos x)² (-sin x)

Question 3.
Differentiate sin²x with respect to x².
Solution:
Let u = (sin x)² ; v = x²
\(\frac{d u}{d x}\) = (2 sin x) (cos x) = sin 2x ; \(\frac{d v}{d x}\) = 2x

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.7

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.7 Text Book Back Questions and Answers

Question 1.
Differentiate the following with respect to x.
(i) x^{sin x}
(ii) (sin x)^x
(iii) (sin x)^{tan x}
(iv) \(\sqrt{\frac{(x-1)(x-2)}{(x-3)\left(x^{2}+x+1\right)}}\)
Solution:
(i) Let y = x^{sin x}
Taking logarithm on both sides we get,
log y = log(x^{sin x})
log y = sin x log x
Differentiating with respect to x,

(ii) Let y = (sin x)^x
Taking logarithm on both sides we get,
log y = x log(sin x)
Differentiating with respect to x,
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = x \(\frac{d}{d x}\) log(sin x) + log(sin x) \(\frac{d}{d x}\) (x)
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = x \(\frac{1}{\sin x}\) (cos x) + log(sin x) (1)
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = x cot x + log(sin x)
\(\frac{d y}{d x}\) = y[x cot x + log(sin x)]
\(\frac{d y}{d x}\) = (sin x)^x [x cot x + log(sin x)]

(iii) Let y = (sin x)^{tan x}
Taking logarithm on both sides we get,
log y = tan x log(sin x)
Differentiating with respect to x,

(iv) Let y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)\left(x^{2}+x+1\right)}}\)
y = \(\left(\frac{(x-1)(x-2)}{(x-3)\left(x^{2}+x+1\right)}\right)^{\frac{1}{2}}\)
Taking logarithm on both sides we get,
log y = \(\frac{1}{2}\) {[log(x – 1) + log(x – 2)] – [(log(x – 3) + log(x² + x + 1)]}
log y = \(\frac{1}{2}\) [log(x – 1) + log(x – 2) – log(x – 3) – log(x² + x + 1)]
Differentiating with respect to x,

Question 2.
If x^m . yⁿ = (x + y)^m+n, then show that \(\frac{d y}{d x}=\frac{y}{x}\)
Solution:
x^m . yⁿ = (x + y)^m+n
Taking logarithm on both sides we get,
m log x + n log y = (m + n) log(x + y)
Differentiating with respect to x,

Hence proved.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.6

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.6 Text Book Back Questions and Answers

Question 1.
Find \(\frac{d y}{d x}\) for the following functions:
(i) xy – tan(xy)
(ii) x² – xy + y² = 1
(iii) x³ + y³ + 3axy = 1
Solution:
(i) Given xy = tan(xy)
Differentiating both sides with respect to x, we get

(ii) x² – xy + y² = 7
Differentiating both side with respect to x,

(iii) x³ + y³ + 3axy = 1
Differentiating both sides with respect to x,

Question 2.
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\) and x ≠ y, then prove that \(\frac{d y}{d x}=-\frac{1}{(x+1)^{2}}\).
Solution:
Given \(x \sqrt{1+y}+y \sqrt{1+x}=0\)
\(x \sqrt{1+y}=-y \sqrt{1+x}\)
Squaring both sides we get
⇒ x² (1 + y) = y² (1 + x)
⇒ x² + x²y = y² + y²x
⇒ x² – y² + x²y – y²x = 0
⇒ (x + y) (x – y) + xy(x – y) = 0
⇒ (x – y) [(x + y) + xy] = 0
∴ x – y = 0 (or) x + y + xy = 0
x = y (or) x + y + xy = 0
Given that x ≠ y
x + y + xy = 0
⇒ y + xy = -x
⇒ y(1 + x) = -x

Hence proved.

Question 3.
If 4x + 3y = log(4x – 3y), then find \(\frac{d y}{d x}\)
Solution:
Given 4x + 3y = log(4x – 3y)
Differentiating both sides with respect to x,

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.5

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.5 Text Book Back Questions and Answers

Question 1.
Differentiate the following with respect to x.
(i) 3x⁴ – 2x³ + x + 8
(ii) \(\frac{5}{x^{4}}-\frac{2}{x^{3}}+\frac{5}{x}\)
(iii) \(\sqrt{x}+\frac{1}{\sqrt[3]{x}}+e^{x}\)
(iv) \(\frac{3+2 x-x^{2}}{x}\)
(v) x³ e^x
(vi) (x² – 3x + 2) (x + 1)
(vii) x⁴ – 3 sin x + cos x
(viii) \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)
Solution:
(i) Let y = 3x⁴ – 2x³ + x + 8

(ii) Let y = \(\frac{5}{x^{4}}-\frac{2}{x^{3}}+\frac{5}{x}\)

(iii) Let y = \(\sqrt{x}+\frac{1}{\sqrt[3]{x}}+e^{x}\)
\(y=x^{\frac{1}{2}}+x^{\frac{1}{3}}+e^{x}\)
[∵ \(\frac{1}{\sqrt[3]{x}}=\frac{1}{(x)^{\frac{1}{3}}}=x^{\frac{1}{3}}\)]

(iv) Let y = \(\frac{3+2 x-x^{2}}{x}\)

(v) Let y = x³ e^x
[∵ We have product of two functions, so use product rule]
= x³ e^x + e^x (3x²)
= e^x (x³ + 3x²)
= x² e^x (x + 3)

(vi) Let y = (x² – 3x + 2) (x + 1)
y = x³ – 3x² + 2x + x² – 3x + 2
y = x³ – 2x² – x + 2
\(\frac{dy}{dx}\) = 3x² – 4x – 1

(vii) Let y = x⁴ – 3 sin x + cos x
\(\frac{dy}{dx}\) = \(\frac{d}{d x}\) (x4) – 3 \(\frac{d}{d x}\) (sin x) + \(\frac{d}{d x}\) (cos x)
= 4x³ – 3 cos x – sin x

(viii) \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)

Question 2.
Differentiate the following with respect to x.
(i) \(\frac{e^{x}}{1+x}\)
(ii) \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
(iii) \(\frac{e^{x}}{1+e^{x}}\)
Solution:
(i) Let y = \(\frac{e^{x}}{1+x}\)

(ii) Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)

(iii) Let y = \(\frac{e^{x}}{1+e^{x}}\)

Question 3.
Differentiate the following with respect to x.
(i) x sin x
(ii) e^x sin x
(iii) e^x (x + log x)
(iv) sin x cos x
(v) x³ e^x
Solution:
(i) Let y = x sin x
\(\frac{d y}{d x}\) = x \(\frac{d}{d x}\) (sin x) + sin x \(\frac{d}{d x}\) (x)
= x cos x + sin x . 1
= x cos x + sin x

(ii) Let y = e^x sin x
\(\frac{d y}{d x}\) = e^x \(\frac{d}{d x}\) (sin x) + sin x \(\frac{d}{d x}\) (e^x)
= e^x cos x + sin x e^x
= e^x (cos x + sin x)

(iii) Let y = e^x (x + log x)

(iv) Let y = sin x cos x
\(\frac{d y}{d x}\) = sin x \(\frac{d}{d x}\) (cos x) + cos x \(\frac{d}{d x}\) (sin x)
= sin x (-sin x) + cos x cos x
= -sin² x + cos² x
= cos² x – sin² x
= cos 2x [∵ cos 2x = cos² x – sin² x]
(or) y = sin x cos x
y = \(\frac{1}{2}\) (2 sin x cos x)
y = \(\frac{1}{2}\) sin 2x
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) cos 2x . 2 = cos 2x

(v) Let y = x³ e^x
\(\frac{d y}{d x}\) = x³ \(\frac{d}{d x}\) (e^x) + e^x \(\frac{d}{d x}\) (x³)
= x³ (e^x) + e^x (3x²)
= x² e^x (x + 3)

Question 4.
Differentiate the following with respect to x.
(i) sin² x
(ii) cos² x
(iii) cos³ x
(iv) \(\sqrt{1+x^{2}}\)
(v) (ax² + bx + c)ⁿ
(vi) sin(x²)
(vii) \(\frac{1}{\sqrt{1+x^{2}}}\)
Solution:
For the following problems chain rule to be used:
\(\frac{d}{d x}\) f(g(x)) = f'(g(x)) . g'(x)
\(\frac{d}{d x}\) [f(x)]ⁿ = n[f(x)]^n-1 × \(\frac{d}{d x}\) f(x)
(i) Let y = sin² x = (sin x)²
\(\frac{d y}{d x}\) = 2(sin x)^2-1 \(\frac{d}{d x}\) (sin x)
= 2 sin x (cos x)
= sin 2x

(ii) y = cos² x = (cos x)²
\(\frac{d y}{d x}\) = 2(cos x)^2-1 \(\frac{d}{d x}\) (cos x)
= 2 cos x (-sin x)
= -2 sin x cos x
= -sin 2x

(iii) y = cos³ x
y = (cos x)³
\(\frac{d y}{d x}\) = 3(cos x)^3-1 \(\frac{d}{d x}\) (cos x)
= 3 cos² x (-sin x)
= -3 cos² x sin x
= -3 cos x (sin x cos x) [Multiply and divide by 2]
= \(\frac{-3}{2}\) cos x (2 sin x cos x)
= \(\frac{-3}{2}\) cos x sin 2x

(iv) Let y = \(\sqrt{1+x^{2}}\)
y = \(\left(1+x^{2}\right)^{\frac{1}{2}}\)
Here f(x) = 1 + x²; n = \(\frac{1}{2}\)

(v) Let y = (ax² + bx + c)ⁿ
\(\frac{d y}{d x}\) = n(ax² + bx + c)^n-1 \(\frac{d}{d x}\) (ax² + bx + c)
= n(ax² + bx + c)^n-1 (2ax + b)

(vi) Let y = sin(x²)
\(\frac{d}{d x}\) f(g(x)) = f'(g(x)) . g'(x)
Here f = sin x, g = x²
\(\frac{d y}{d x}\) = cos(x²) \(\frac{d}{d x}\) (x²)
= cos(x²) (2x)
= 2x cos(x²)

(vii) Let y = \(\frac{1}{\sqrt{1+x^{2}}}\)
y = \(\left(1+x^{2}\right)^{-\frac{1}{2}}\)
Here n = \(-\frac{1}{2}\); f(x) = 1 + x²

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.4

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.4 Text Book Back Questions and Answers

Question 1.
Find the derivative of the following functions from first principle.
(i) x²
(ii) e^x
(iii) log(x + 1)
Solution:
Let f(x) = x² then f(x + h) = (x + h)²
Now \(\frac{d}{d x}\) f(x)

Thus \(\frac{d}{d x}\) (x²) = 2x

(ii) Let f(x) = e^-x then f(x + h) = \(e^{-(x+h)}\)
Now \(\frac{d}{d x}\) f(x)

(iii) Let f(x) = log(x + 1)
Then f(x + h) = log(x + h + 1) = log((x + 1) + h)
Now \(\frac{d}{d x}\) f(x)

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.3

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.3 Text Book Back Questions and Answers

Question 1.
Examine the following functions for continuity at indicated points.

Solution:
(a) f(x) = \(\frac{x^{2}-4}{x-2}\), also given that f(2) = 0

[∵ x = 2 – h, where h → 0, x → 2]


∴ The given function is not continuous at x = 2.

(b) Given that f(x) = \(\frac{x^{2}-9}{x-3}\) and f(3) = 6

[∵ x = 3 – h, where h → 0, x → 3]


[∵ x = 3 + h, where x → 3, h → 0]


= 0 + 6
= 6
Also given that f(3) = 6

∴ The given function f(x) is continuous at x = 3.

Question 2.
Show that f(x) = |x| is continuous at x = 0.
Solution:

[∵ x = 0 – h]



[∵ |x| = x if x > 0]

∴ f(x) is continuous at x = 0.

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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}.

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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 \(\frac{x}{2}\)]