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

⇔ (x^{2} – 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 – x^{2} ≥ 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 – x^{2} > 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

⇔ x^{2} – 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 – x^{2}|

Solution:

Let y = f(x) = log|4 – x^{2}|

f(x) ∈ R

⇔ 4 – x^{2} ≠ 0

⇔ x ≠ ±2

∵ y = log|4 – x^{2}|

⇒ |4 – x^{2}| = 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) = x^{2} 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) = x^{2}

(3f – 2g) (x) = 3f(x) – 2g(x)

= 3(2x – 1) – 2x^{2}

= -2x^{2} + 6x – 3

(ii) (fg) (x)

= f(x) . g(x)

= (2x – 1) (x^{2})

= 2x^{3} – x^{2}

(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) + x^{2} + 2

= x^{2} + 2x + 1

= (x + 1)^{2}

Question 4.

If f = {(1, 2), (2, -3), (3, -1)} then find

(i) 2f

(ii) 2 + f

(iii) f^{2}

(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) f^{2} = {(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

⇔ x^{2} – 3x + 2 ≥ 0

⇔ (x- 1) (x – 2) ≥ 0

⇔ x ∈ (-∞, 1 ] ∪ [2, ∞]

∴ The domain of f is R – (1, 2)

(ii) f(x) = log(x^{2} – 4x + 3)

Solution:

f(x) = log(x^{2} – 4x + 3) ∈ R

⇔ x^{2} – 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 – x^{2} ≥ 0 and [x] + 2 > 0 (or) Case (ii) 4 – x^{2} ≤ 0 and [x] + 2 < 0

Case (i): 4 – x^{2} ≥ 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 – x^{2} ≤ 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 – x^{2}) ≥ 0

⇒ x – x^{2} ≤ (0.3)^{0}

⇒ x – x^{2} ≤ 1

⇒ -x^{2} + x – 1 ≤ 0

⇒ x^{2} – x + 1 ≥ 0

This is true for all x ∈ R ……..(1)

and x – x^{2} ≥ 0

⇒ x^{2} – 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, ∞]