Inter 1st Year Maths 1A Functions Solutions Ex 1(c)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(c)

I.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\)
Solution:
f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\) ∈ R
⇔ (x² – 1) (x + 3) ≠ 0
⇔ (x + 1) (x – 1) (x + 3) ≠ 0
⇔ x ≠ -1, 1, -3
∴ Domain of f is R – {-1, 1, -3}

(ii) f(x) = \(\frac{2 x^{2}-5 x+7}{(x-1)(x-2)(x-3)}\)
⇔ (x – 1) (x – 2) (x – 3) ≠ 0
⇔ x ≠ 1, x ≠ 2, x ≠ 3
∴ Domain of f is R – {1, 2, 3}

(iii) f(x) = \(\frac{1}{\log (2-x)}\)
Solution:
f(x) = \(\frac{1}{\log (2-x)}\)
⇔ log (2 – x) ≠ 0 and 2 – x > 0
⇔ (2 – x) ≠ 1 and 2 > x
⇔ x ≠ 1 and x < 2
x ∈ (-∞, 1) ∪ (1, 2) (or) x ∈ (-∞, 2) – {1}
∴ Domain of f is {(-∞, 2) – {1}}

(iv) f(x) = |x – 3|
Solution:
f(x) = |x – 3| ∈ R
⇔ x ∈ R
∴ The domain of f is R

(v) f(x) = \(\sqrt{4 x-x^{2}}\)
Solution:
f(x) = \(\sqrt{4 x-x^{2}}\) ∈ R
⇔ 4x – x² ≥ 0
⇔ x(4 – x) ≥ 0
⇔ x ∈ [0, 4]
∴ Domain of f is [0, 4]

(vi) f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\) ∈ R
⇔ 1 – x² > 0
⇔ (1 + x) (1 – x) > 0
⇔ x ∈ (-1, 1)
∴ Domain of f is {x/x ∈ (-1, 1)}

(vii) f(x) = \(\frac{3^{x}}{x+1}\)
Solution:
f(x) = \(\frac{3^{x}}{x+1}\) ∈ R
⇔ 3^x ∈ R, ∀ x ∈ R and x + 1 ≠ 0
⇔ x ≠ -1
∴ Domain of f is R – {-1}

(viii) f(x) = \(\sqrt{x^{2}-25}\)
Solution:
f(x) = \(\sqrt{x^{2}-25}\) ∈ R
⇔ x² – 25 ≥ 0
⇔ (x + 5) (x – 5) ≥ 0
⇔ x ∈ (-∞, -5] ∪ [5, ∞)
⇔ x ∈ R – (-5, 5)
∴ Domain of f is R – (- 5, 5)

(ix) f(x) = \(\sqrt{x-[x]}\)
Solution:
f(x) = \(\sqrt{x-[x]}\) ∈ R
⇔ x – [x] ≥ 0
⇔ x ≥ [x]
⇔ x ∈ R
∴ Domain of f is R.

(x) f(x) = \(\sqrt{[x]-x}\)
Solution:
f(x) = \(\sqrt{[x]-x}\) ∈ R
⇔ [x] – x ≥ 0
⇔ [x] ≥ x
⇔ x ≤ [x]
⇔ x ∈ Z
∴ The domain of f is z (Where z denotes a set of integers)

Question 2.
Find the ranges of the following real-valued functions.
(i) log|4 – x²|
Solution:
Let y = f(x) = log|4 – x²|
f(x) ∈ R
⇔ 4 – x² ≠ 0
⇔ x ≠ ±2
∵ y = log|4 – x²|
⇒ |4 – x²| = e^y
∵ e^y > 0 ∀ y ∈ R
∴ The range of f is R.

(ii) \(\sqrt{[x]-x}\)
Solution:
Let y = f(x) = \(\sqrt{[x]-x}\)
f(x) ∈ R
⇔ [x] – x ≥ 0
⇔ x ≤ [x]
⇔ x ∈ z
∴ Domain of f is z. Then range of f is {0}

(iii) \(\frac{\sin \pi[x]}{1+[x]^{2}}\)
Solution:
Let f(x) = \(\frac{\sin \pi[x]}{1+[x]^{2}}\) ∈ R
⇔ x ∈ R
∴ The domain of f is R
For x ∈ R, [x] is an integer,
sin π[x] = 0, ∀ x ∈ R [∵ sin nπ = 0, ∀ n ∈ z]
∴ Range of f is {0}

(iv) \(\frac{x^{2}-4}{x-2}\)
Solution:
Let y = f(x) = \(\frac{x^{2}-4}{x-2}\) ∈ R
⇔ y = \(\frac{(x+2)(x-2)}{x-2}\)
⇔ x ≠ 2
∴ The domain of f is R – {2}
Then y = x + 2, [∵ x ≠ 2 ⇒ y ≠ 4]
Then its range R – {4}

(v) \(\sqrt{9+x^{2}}\)
Solution:
Let y = f(x) = \(\sqrt{9+x^{2}}\) ∈ R
The domain of f is R
When x = 0, f(0) = √9 = 3
For all values of x ∈ R – {0}, f(x) > 3
∴ The range of f is [3, ∞)

Question 3.
If f and g are real-valued functions defined by f(x) = 2x – 1 and g(x) = x² then find
(i) (3f – 2g)(x)
(ii) (fg) (x)
(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
(iv) (f + g + 2) (x)
Solution:
(i) (3f – 2g)(x)
f(x) = 2x – 1, g(x) = x²
(3f – 2g) (x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2x²
= -2x² + 6x – 3

(ii) (fg) (x)
= f(x) . g(x)
= (2x – 1) (x²)
= 2x³ – x²

(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
\(\frac{\sqrt{f(x)}}{g(x)}=\frac{\sqrt{2 x-1}}{x^{2}}\)

(iv) (f + g + 2) (x)
= f(x) + g(x) + 2
= (2x – 1) + x² + 2
= x² + 2x + 1
= (x + 1)²

Question 4.
If f = {(1, 2), (2, -3), (3, -1)} then find
(i) 2f
(ii) 2 + f
(iii) f²
(iv) √f
Solution:
Given f = {(1, 2), (2, -3), (3, -1)}
(i) 2f = {(1, 2 × 2), (2, 2(-3), (3, 2(-1))}
= {(1, 4), (2, -6), (3, -2)}

(ii) 2 + f = {(1, 2 + 2), (2, 2 + (-3)), (3, 2 + (-1)}
= {(1, 4), (2, -1), (3, 1)}

(iii) f² = {(1, 22), (2, (-3)2), (3, (-1)2)}
= {(1, 4), (2, 9), (3, 1)}

(iv) √f = {(1, √2)}
∵ √-3 and √-1 are not real

II.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\sqrt{x^{2}-3 x+2}\)
Solution:
f(x) = \(\sqrt{x^{2}-3 x+2}\) ∈ R
⇔ x² – 3x + 2 ≥ 0
⇔ (x- 1) (x – 2) ≥ 0
⇔ x ∈ (-∞, 1 ] ∪ [2, ∞]
∴ The domain of f is R – (1, 2)

(ii) f(x) = log(x² – 4x + 3)
Solution:
f(x) = log(x² – 4x + 3) ∈ R
⇔ x² – 4x + 3 > 0
⇔ (x – 1) (x – 3) > 0
⇔ x ∈ (-∞, 1) ∪ (3, ∞)
∴ Domain of f is R – [1, 3]

(iii) f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\)
Solution:
f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\) ∈ R
⇔ 2 + x ≥ 0, 2 – x ≥ 0, x ≠ 0
⇔ x ≥ -2, x ≤ 2, x ≠ 0
⇔ -2 ≤ x ≤ 2, x ≠ 0
⇔ x ∈ [-2, 2] – {0}
Domain of f is [-2, 2] – {0}

(iv) f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\)
Solution:
f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\) ∈ R
⇔ 4 – x > 0, 4 – x ≠ 1 and x – 2 ≠ 0
⇔ x < 4, x ≠ 3, x ≠ 2
∴ Domain of f is (-∞, 4) – {2, 3}

(v) f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\)
Solution:
f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\) ∈ R
Case (i) 4 – x² ≥ 0 and [x] + 2 > 0 (or) Case (ii) 4 – x² ≤ 0 and [x] + 2 < 0
Case (i): 4 – x² ≥ 0 and [x] + 2 > 0
⇔ (2 – x) (2 + x) ≥ 0 and [x] > -2
⇔ x ∈ [-2, 2] and x ∈ [-1, ∞]
⇔ x ∈ [-1, 2] ……..(1)
Case (ii): 4 – x² ≤ 0 and [x] + 2 < 0
⇔ (2 + x) (2 – x) ≤ 0 and [x] < – 2
⇔ x ∈ (-∞, -2] ∪ [2, ∞] and x ∈ (-∞, -2)
⇔ x ∈ (-∞, -2) ……(2)
from (1) and (2),
Domain of f is (-∞, -2) ∪ [-1, 2]

(vi) f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\)
Solution:
f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\) ∈ R
Then log_0.3(x – x²) ≥ 0
⇒ x – x² ≤ (0.3)⁰
⇒ x – x² ≤ 1
⇒ -x² + x – 1 ≤ 0
⇒ x² – x + 1 ≥ 0
This is true for all x ∈ R ……..(1)
and x – x² ≥ 0
⇒ x² – x ≤ 0
⇒ x(x – 1) ≤ 0
⇒ x ∈ (0, 1) …….(2)
From (1) and (2)
Domain of f is R ^ (0, 1) = (0, 1)
∴ The domain of f is (0, 1)

(vii) f(x) = \(\frac{1}{x+|x|}\)
Solution:
f(x) = \(\frac{1}{x+|x|}\) ∈ R
⇔ x + |x| ≠ 0
⇔ x ∈ (0, ∞)
∵ |x| = x, if x ≥ 0
|x| = -x, if x < 0
∴ The domain of f is (0, ∞)

Question 2.
Prove that the real valued function f(x) = \(\frac{x}{e^{x}-1}+\frac{x}{2}+1\) is an even function on R \ {0}.
Solution:
f(x) ∈ R, e^x – 1 ≠ 0
⇒ e^x ≠ 1
⇒ x ≠ 0

⇒ f(x) is an even function on R – {0}

Question 3.
Find the domain and range of the following functions.
(i) f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\)
Solution:
f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\) ∈ R
⇔ x ∈ R, since [x] is an integer tan π[x] and sin π[x] each is zero for ∀ x ∈ R and f(x) ∈ R
Domain of f is R
Its range = {0}

(ii) f(x) = \(\frac{x}{2-3 x}\)
Solution:

(iii) f(x) = |x| + |1 + x|
Solution:
f(x) = |x| + |1 + x| ∈ R
⇔ x ∈ R
∴ Domain of f is R
∵ |x| = x, if x ≥ 0
= -x, if x < 0
|1 + x| = 1 + x, if x ≥ -1
= -(1 + x) if x < -1
For x = 0, f(0) = |0| + |1 + 0| = 1
x = 1, f(1) = |1| + |1 + 1| = 1 + 2 = 3
x = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
x = -2, f(-2) = |-2| + |1 + (-2)| = 2 + 1 = 3
x = -1, f(-1) = |-1| + |1 +(-1)| = 1 + 0 = 1
∴ The range of f is [1, ∞]