Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Miscellaneous Exercise 6

(I) Select the correct answer from the given alternatives.

Question 1.

If log (5x – 9) – log (x + 3) = log 2, then x = ________

(A) 3

(B) 5

(C) 2

(D) 7

Answer:

(B) 5

Hint:

log (5x – 9) – log (x + 3) = log 2

∴ \(\frac{5 x-9}{x+3}\) = 2

∴ 3x = 9 + 6

∴ x = 5

Question 2.

If log_{10} (log_{10} (log_{10} x)) = 0, then x = ________

(A) 1000

(B) 10^{10}

(C) 10

(D) 0

Answer:

(B) 10^{10}

Hint:

log_{10} log_{10} log_{10} x = 0

∴ log_{10} (log_{10} (x)) = 10^{0} = 1

∴ log_{10} x = 10^{1} = 10

∴ x = 10^{10}

Question 3.

Find x, if 2 log_{2} x = 4

(A) 4, -4

(B) 4

(C) -4

(D) not defined

Answer:

(B) 4

Hint:

2 log_{2} x = 4, x > 0

∴ log_{2} (x^{2}) = 4

∴ x^{2} = 16

∴ x = ±4

∴ x = 4

Question 4.

The equation \(\log _{x^{2}} 16+\log _{2 x} 64=3\) has,

(A) one irrational solution

(B) no prime solution

(C) two real solutions

(D) one integral solution

Answer:

(A), (B), (C), (D)

Hint:

\(\log _{x^{2}} 16+\log _{2 x} 64=3\)

∴ \(\frac{\log 16}{\log x^{2}}+\frac{\log 64}{\log 2 x}=3\)

∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x) = 3 (2 log x) (log 2 + log x)

Let log 2 = a, log x = t. Then

∴ 4at + 4a^{2} + 12at = 6at + 6t^{2}

∴ 6t^{2} – 10at – 4a^{2} = 0

∴ 3t^{2} – 5at – 2a^{2} = 0

∴ (3t + a) (t – 2a) = 0

∴ t = \(-\frac{1}{3}\)a, 2a

∴ log x = \(\log (2)^{-\frac{1}{3}}\), log (2^{2})

∴ x = \(2^{-\frac{1}{3}}\), 4

∴ x = \(\frac{1}{\sqrt[3]{2}}\), 4

Question 5.

If f(x) = \(\frac{1}{1-x}\), then f(f(f(x))) is

(A) x – 1

(B) 1 – x

(C) x

(D) -x

Answer:

(C) x

Hint:

Question 6.

If f: R → R is defined by f(x) = x^{3}, then f^{-1} (8) is equal to:

(A) {2}

(B) {-2.2}

(C) {-2}

(D) (-2.2)

Answer:

(A) {2}

Hint:

Question 7.

Let the function f be defined by f(x) = \(\frac{2 x+1}{1-3 x}\) then f^{-1} (x) is:

(A) \(\frac{x-1}{3 x+2}\)

(B) \(\frac{x+1}{3 x-2}\)

(C) \(\frac{2 x+1}{1-3 x}\)

(D) \(\frac{3 x+2}{x-1}\)

Answer:

(A) \(\frac{x-1}{3 x+2}\)

Hint:

f(x) = \(\frac{2 x+1}{1-3 x}\) = y, say. then

2x + 1 = y(1 – 3x)

∴ y – 1 = x(2 + 3y)

∴ x = \(\frac{y-1}{2+3 y}\) = f^{-1} (y)

∴ f^{-1} (x) = \(\frac{x-1}{2+3 x}\)

Question 8.

If f(x) = 2x^{2} + bx + c and f(0) = 3 and f(2) = 1, then f(1) is equal to

(A) -2

(B) 0

(C) 1

(D) 2

Answer:

(B) 0

Hint:

f(x) = 2x^{2} + bx + c

f(0) = 3

∴ 2(0) + b(0) + c = 3

∴ c = 3 ……..(i)

∴ f(2) = 1

∴ 2(4) + 2b + c = 1

∴ 2b + c = -7

∴ 2b + 3 = -7 …..[From (i)]

∴ b = -5

∴ f(x) = 2x^{2} – 5x + 3

∴ f(1) = 2(1)^{2} – 5(1) + 3 = 0

Question 9.

The domain of \(\frac{1}{[x]-x}\), where [x] is greatest integer function is

(A) R

(B) Z

(C) R – Z

(D) Q – {0}

Answer:

(C) R – Z

Hint:

f(x) = \(\frac{1}{[x]-x}=\frac{1}{-\{x\}}\)

For f to be defined, {x} ≠ 0

∴ x cannot be integer.

∴ Domain = R – Z

Question 10.

The domain and range of f(x) = 2 – |x – 5| are

(A) R+, (-∞, 1]

(B) R, (-∞, 2]

(C) R, (-∞, 2)

(D) R+, (-∞, 2]

Answer:

(B) R, (-∞, 2]

Hint:

f(x) = 2 – |x – 5|

= 2 – (5 – x), x < 5

= 2 – (x – 5), x ≥ 5

∴ f(x) = x – 3, x < 5

= 7 – x, x ≥ 5

Domain = R,

Range (from graph) = (-∞, 2]

(II) Answer the following:

Question 1.

Which of the following relations are functions? If it is a function determine its domain and range.

(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3), (16, 4), (16, -4)}

(iii) {(2, 1), (3, 1), (5, 2)}

Solution:

(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}

Every element of set A has been assigned a unique element in set B

∴ Given relation is a function

Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3) (16, 4), (16, -4)}

∵ (1, 1), (1, -1) ∈ the relation

∴ Given relation is not a function.

As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) {(2, 1), (3, 1), (5, 2)}

Every element of set A has been assigned a unique element in set B.

∴ Given relation is a function.

Domain = {2, 3, 5}, Range = {1, 2}

Question 2.

Find whether the following functions are one-one.

(i) f: R → R defined by f(x) = x^{2} + 5

(ii) f: R – {3} → R defined by f(x) = \(\frac{5 x+7}{x-3}\) for x ∈ R – {3}

Solution:

(i) f: R → R, defined by f(x) = x^{2} + 5

Note that f(-x) = f(x) = x^{2} + 5

∴ f is not one-one (i.e., many-one) function.

(ii) f: R – {3} → R, defined by f(x) = \(\frac{5 x+7}{x-3}\)

Let f(x_{1}) = f(x_{2})

∴ \(\frac{5 x_{1}+7}{x_{1}-3}=\frac{5 x_{2}+7}{x_{2}-3}\)

∴ 5x_{1} x_{2} – 15x_{1} + 7x_{2} – 21 = 5x_{1} x_{2} – 15x_{2} + 7x_{1} – 21

∴ 22(x_{1} – x_{2}) = 0

∴ x_{1} = x_{2}

∴ f is a one-one function.

Question 3.

Find whether the following functions are onto or not.

(i) f: Z → Z defined by f(x) = 6x – 7 for all x ∈ Z

(ii) f: R → R defined by f(x) = x^{2} + 3 for all x ∈ R

Solution:

(i) f(x) = 6x – 7 = y (say)

(x, y ∈ Z)

∴ x = \(\frac{7+y}{6}\)

Since every integer y does not give integer x, f is not onto.

(ii) f(x) = x^{2} + 3 = y (say)

(x, y ∈ R)

Clearly y ≥ 3 …..[x^{2} ≥ 0]

∴ All the real numbers less than 3 from codomain R, have not been pre-assigned any element from the domain R.

∴ f is not onto.

Question 4.

Let f: R → R be a function defined by f(x) = 5x^{3} – 8 for all x ∈ R. Show that f is one-one and onto. Hence find f^{-1}.

Solution:

Question 5.

A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence find f^{-1}.

Solution:

Question 6.

A function f is defined as f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.

Solution:

f(x) = 4x + 5, -4 ≤ x < 0

f(-1) = 4(-1) + 5 = -4 + 5 = 1

f(-2) = 4(-2) + 5 = -8 + 5 = -3

x = 0 ∉ domain of f

∴ f(0) does not exist.

Question 7.

A function f is defined as f(x) = 5 – x for 0 ≤ x ≤ 4. Find the values of x such that

(i) f(x) = 3

(ii) f(x) = 5

Solution:

(i) f(x) = 3

∴ 5 – x = 3

∴ x = 5 – 3 = 2

(ii) f(x) = 5

∴ 5 – x = 5

∴ x = 0

Question 8.

If f(x) = 3x^{4} – 5x^{2} + 7, find f(x – 1).

Solution:

f(x) = 3x^{4} – 5x^{2} + 7

∴ f(x – 1) = 3(x – 1)^{4} – 5(x – 1)^{2} + 7

= 3(x^{4} – ^{4}C_{1} x^{3} + ^{4}C_{2} x^{2} – ^{4}C_{3} x + ^{4}C_{4}) – 5(x^{2} – 2x + 1) + 7

= 3(x^{4} – 4x^{3} + 6x^{2} – 4x + 1) – 5(x^{2} – 2x + 1) + 7

= 3x^{4} – 12x^{3} + 18x^{2} – 12x + 3 – 5x^{2} + 10x – 5 + 7

= 3x^{4} – 12x^{3} + 13x^{2} – 2x + 5

Question 9.

If f(x) = 3x + a and f(1) = 7, find a and f(4).

Solution:

f(x) = 3x + a, f(1) = 7

∴ 3(1) + a = 7

∴ a = 7 – 3 = 4

∴ f(x) = 3x + 4

∴ f(4) = 3(4) + 4 = 12 + 4 = 16

Question 10.

If f(x) = ax^{2} + bx + 2 and f(1) = 3, f(4) = 42, find a and b.

Solution:

f(x) = ax^{2} + bx + 2

f(1) = 3

∴ a(1)^{2} + b(1) + 2 = 3

∴ a + b = 1 ….(i)

f(4) = 42

∴ a(4)^{2} + b(4) + 2 = 42

∴ 16a + 4b = 40

Dividing by 4, we get

4a + b = 10 …..(ii)

Solving (i) and (ii), we get

a = 3, b = -2

Question 11.

Find composite of f and g:

(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}

g = {(3, 6), (4, 8), (5, 10), (6, 12)}

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

g = {(1, 1), (3, 27), (4, 64)}

Solution:

(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}

g = {(3, 6), (4, 8), (5, 10), (6, 12)}

∴ f(1) = 3, g(3) = 6

f(2) = 4, g(4) = 8

f(3) = 5, g(5)=10

f(4) = 6, g(6) = 12

(gof) (x) = g (f(x))

(gof)(1) = g(f(1)) = g(3) = 6

(gof)(2) – g(f(2)) = g(4) = 8

(gof)(3) = g(f(3)) = g(5) = 10

(gof)(4) = g(f(4)) = g(6) = 12

∴ gof = {(1, 6), (2, 8), (3, 10), (4, 12)}

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

g = {(1, 1), (3, 27), (4, 64)}

f(1) = 1, g(1) = 1

f(2) = 4, g(3) = 27

f(3) = 4, g(4) = 64

f(4) = 3

(gof) (x) = g(f(x))

(gof) (1) = g(f(1)) = g(1) = 1

(gof) (2) = g(f(2)) = g(4) = 64

(gof) (3) = g(f(3)) = g(4) = 64

(gof) (4) = g(f(4)) = g(3) = 27

∴ gof = {(1, 1), (2, 64), (3, 64), (4, 27)}

Question 12.

Find fog and gof:

(i) f(x) = x^{2} + 5, g(x) = x – 8

(ii) f(x) = 3x – 2, g(x) = x^{2}

(iii) f(x) = 256x^{4}, g(x) = √x

Solution:

(i) f(x) = x^{2} + 5, g(x) = x – 8

(fog) (x) = f(g(x))

= f(x – 8)

= (x – 8)^{2} + 5

= x^{2} – 16x + 64 + 5

= x^{2} – 16x + 69

(gof) (x) = g(f(x))

= g(x^{2} + 5)

= x^{2} + 5 – 8

= x – 3

(ii) f(x) = 3x – 2, g(x) = x^{2}

(fog) (x) = f(g(x)) = f(x^{2}) = 3x^{2} – 2

(gof) (x) = g(f(x))

= g(3x – 2)

= (3x – 2)^{2}

= 9x^{2} – 12x + 4

(iii) f(x) = 256x^{4}, g(x) = √x

(fog) (x) = f(g(x)) = f (√x) = 256 (√x)^{4} = 256x^{2}

(gof) (x) = g(f(x)) = g(256x^{4}) = \(\sqrt{256 x^{4}}\) = 16x^{2}

Question 13.

If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), show that (fof) (x) = x.

Solution:

Question 14.

If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then show that (fog) (x) = x.

Solution:

f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\)

(fog)(x) = f(g(x))

= f(\(\frac{3+5 x}{4 x-1}\))

Question 15.

Let f: R – {2} → R be defined by f(x) = \(\frac{x^{2}-4}{x-2}\) and g: R → R be defined by g(x) = x + 2. Examine whether f = g or not.

Solution:

f(x) = \(\frac{x^{2}-4}{x-2}\), x ≠ 2

∴ f(x) = x + 2, x ≠ 2 and g(x) = x + 2,

The domain of f = R – {2}

The domain of g = R

Here, f and g have different domains.

∴ f ≠ g

Question 16.

Let f: R → R be given by f(x) = x + 5 for all x ∈ R. Draw its graph.

Solution:

f(x) = x + 5

Question 17.

Let f: R → R be given by f(x) = x^{3} + 1 for all x ∈ R. Draw its graph.

Solution:

Let y = f(x) = x^{3} + 1

Question 18.

For any base show that log(1 + 2 + 3) = log 1 + log 2 + log 3

Solution:

L.H.S. = log(1 + 2 + 3) = log 6

R.H.S. = log 1 + log 2 + log 3

= 0 + log (2 × 3)

= log 6

∴ L.H.S. = R.H.S.

Question 19.

Find x, if x = \(3^{3 \log _{3} 2}\).

Solution:

x = \(3^{3 \log _{3} 2}\)

= \(3^{\log 3\left(2^{3}\right)}\)

= 2^{3} ….[\(a^{\log _{a} b}\) = b]

= 8

Question 20.

Show that, log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x| = 0.

Solution:

L.H.S. = log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x|

= \(\log \left|\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)\right|\)

= log|x^{2} + 1 – x^{2}|

= log 1

= 0

= R.H.S.

Question 21.

Show that \(\log \frac{\mathrm{a}^{2}}{\mathrm{bc}}+\log \frac{\mathrm{b}^{2}}{\mathrm{ca}}+\log \frac{\mathrm{c}^{2}}{\mathrm{ab}}=0\).

Solution:

Question 22.

Simplify log (log x^{4}) – log(log x).

Solution:

log (log x^{4}) – log (log x)

= log (4 log x) – log (log x) …..[log m^{n} = n log m]

= log 4 + log (log x) – log (log x) …..[log (mn) = log m + log n]

= log 4

Question 23.

Simplify \(\log _{10} \frac{28}{45}-\log _{10} \frac{35}{324}+\log _{10} \frac{325}{432}-\log _{10} \frac{13}{15}\)

Solution:

Question 24.

If log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b), then show that a = b.

Solution:

log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b)

∴ 2 log (\(\frac{a+b}{2}\)) = log a + log b

∴ log \(\left(\frac{a+b}{2}\right)^{2}\) = log ab

∴ \(\frac{(a+b)^{2}}{4}\) = ab

∴ a^{2} + 2ab + b^{2} = 4ab

∴ a^{2} + 2ab – 4ab + b^{2} = 0

∴ a^{2} – 2ab + b^{2} = 0

∴ (a – b)^{2} = 0

∴ a – b = 0

∴ a = b

Question 25.

If b^{2} = ac. Prove that, log a + log c = 2 log b.

Solution:

b^{2} = ac

Taking log on both sides, we get

log b^{2} = log ac

∴ 2 log b = log a + log c

∴ log a + log c = 2 log b

Question 26.

Solve for x, log_{x} (8x – 3) – log_{x} 4 = 2.

Solution:

log_{x} (8x – 3) – log_{x} 4 = 2

∴ \(\log _{x}\left(\frac{8 x-3}{4}\right)\) = 2

∴ x^{2} = \(\frac{8 x-3}{4}\)

∴ 4x^{2} = 8x – 3

∴ 4x^{2} – 8x + 3 = 0

∴ 4x^{2} – 2x – 6x + 3 = 0

∴ 2x(2x – 1) – 3(2x – 1) = 0

∴ (2x – 1)(2x – 3) = 0

∴ 2x – 1 = 0 or 2x – 3 = 0

∴ x = \(\frac{1}{2}\) or x = \(\frac{3}{2}\)

Question 27.

If a^{2} + b^{2} = 7ab, show that \(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)

Solution:

a^{2} + b^{2} = 7ab

a^{2} + 2ab + b^{2} = 7ab + 2ab

(a + b)^{2} = 9ab

\(\frac{(a+b)^{2}}{9}\) = ab

\(\left(\frac{a+b}{3}\right)^{2}\) = ab

Taking log on both sides, we get

log \(\left(\frac{a+b}{3}\right)^{2}\) = log (ab)

2 log \(\left(\frac{a+b}{3}\right)\) = log a + log b

Dividing throughout by 2, we get

\(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)

Question 28.

If \(\log \left(\frac{x-y}{5}\right)=\frac{1}{2} \log x+\frac{1}{2} \log y\), show that x^{2} + y^{2} = 27xy.

Solution:

Question 29.

If log_{3} [log_{2} (log_{3} x)] = 1, show that x = 6561.

Solution:

log_{3} [log_{2} (log_{3} x)] = 1

∴ log_{2} (log_{3} x) = 3^{1}

∴ log_{3} x = 2^{3}

∴ log_{3} x = 8

∴ x = 3^{8}

∴ x = 6561

Question 30.

If f(x) = log(1 – x), 0 ≤ x < 1, show that f(\(\frac{1}{1+x}\)) = f(1 – x) – f(-x).

Solution:

Question 31.

Without using log tables, prove that \(\frac{2}{5}\) < log_{10} 3 < \(\frac{1}{2}\).

Solution:

We have to prove that, \(\frac{2}{5}\) < log_{10} 3 < \(\frac{1}{2}\)

i.e., to prove that \(\frac{2}{5}\) < log_{10} 3 and log_{10} 3 < \(\frac{1}{2}\)

i.e., to prove that 2 < 5 log_{10} 3 and 2 log_{10} 3 < 1

i.e., to prove that 2 log_{10} 10 < 5 log_{10} 3 and 2 log_{10} 3 < log_{10} 10 ……[∵ log_{a} a = 1]

i.e., to prove that log_{10} 10^{2} < log_{10} 3^{5} and log_{10} 3^{2} < log_{10} 10

i.e., to prove that 10^{2} < 3^{5} and 3^{2} < 10

i.e., to prove that 100 < 243 and 9 < 10 which is true

∴ \(\frac{2}{5}\) < log_{10} 3 < \(\frac{1}{2}\)

Question 32.

Show that \(7 \log \left(\frac{15}{16}\right)+6 \log \left(\frac{8}{3}\right)+5 \log \left(\frac{2}{5}\right)+\log \left(\frac{32}{25}\right)\) = log 3

Solution:

Question 33.

Solve : \(\sqrt{\log _{2} x^{4}}+4 \log _{4} \sqrt{\frac{2}{x}}=2\)

Solution:

Question 34.

Find the value of \(\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}\)

Solution:

Question 35.

If \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\), show that abc = 1.

Solution:

Let \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\) = k

∴ log a = k(x + y – 2z), log b = k(y + z – 2x), log c = k(z + x – 2y)

log a + log b + log c = k(x + y – 2z) + k(y + z – 2x) + k(z + x – 2y)

= k(x + y – 2z + y + z – 2x + z + x – 2y)

= k(0)

= 0

∴ log (abc) = log 1 …….[∵ log 1 = 0]

∴ abc = 1

Question 36.

Show that, log_{y} x^{3} . log_{z} y^{4} . log_{x} z^{5} = 60.

Solution:

Question 37.

If \(\frac{\log _{2} \mathrm{a}}{4}=\frac{\log _{2} \mathrm{~b}}{6}=\frac{\log _{2} \mathrm{c}}{3 \mathrm{k}}\) and a^{3}b^{2}c = 1, find the value of k.

Solution:

Question 38.

If a^{2} = b^{3} = c^{4} = d^{5}, show that loga bcd = \(\frac{47}{30}\).

Solution:

Question 39.

Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.

(i) 1 < |x – 1| < 4

(ii) |x^{2} – x – 6| = x + 2

(iii) |x^{2} – 9| + |x^{2} – 4| = 5

(iv) -2 < [x] ≤ 7

(v) 2[2x – 5] – 1 = 7

(vi) [x]2 – 5 [x] + 6 = 0

(vii) [x – 2] + [x + 2] + {x} = 0

(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)

Solution:

(i) 1 < |x – 1| < 4

∴ -4 < x – 1 < -1 or 1 < x – 1 < 4

∴ -3 < x < 0 or 2 < x < 5

∴ Solution set = (-3, 0) ∪ (2, 5)

(ii) |x^{2} – x – 6| = x + 2 …..(i)

R.H.S. must be non-negative

∴ x ≥ -2 …..(ii)

|(x – 3) (x + 2)| = x + 2

∴ (x + 2) |x – 3| = x + 2 as x + 2 ≥ 0

∴ |x – 3| = 1 if x ≠ -2

∴ x – 3 = ±1

∴ x = 4 or 2

∴ x = -2 also satisfies the equation

∴ Solution set = {-2, 2, 4}

(iii) |x^{2} – 9| + |x^{2} – 4| = 5

∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i)

Case I: x < -3

Also, x < -2, x < 2, x < 3

∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0

Equation (i) reduces to

x^{2} – 9 + x^{2} – 4 = 5

∴ 2x^{2} = 18

∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2

As x < -2, x < 3

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0

Equation (i) reduces to

-(x^{2} – 9) + x^{2} – 4 = 5

∴ 5 = 5 (true)

-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3

∴ (x – 3) (x + 3) < 0,

(x – 2) (x + 2) < 0

Equation (i) reduces to

9 – x^{2} + 4 – x2 = 5

∴ 2x^{2} = 13 – 5

∴ x^{2} = 4

∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0

Equation (i) reduces to

9 – x^{2} + x^{2} – 4 = 5

∴ 5 = 5 (true)

∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2

∴ (x + 3) (x – 3) > 0,

(x – 2) (x + 2) > 0

Equation (i) reduces to

x^{2} – 9 + x^{2} – 4 = 5

∴ 2x^{2} = 18

∴ x^{2} = 9

∴ x = 3 …..(v)

(x = -3 rejected as x ≥ 3)

From (ii), (iii), (iv), (v), we get

∴ Solution set = [-3, -2] ∪ [2, 3]

(iv) -2 < [x] ≤ 7

∴ -2 < x < 8

∴ Solution set = (-2, 8)

(v) 2[2x – 5] – 1 = 7

∴ [2x – 5] = \(\frac{7+1}{2}\) = 4

∴ [2x] – 5 = 4

∴ [2x] = 9

∴ 9 ≤ 2x < 10

∴ \(\frac{9}{2}\) ≤ x < 5

∴ Solution set = [\(\frac{9}{2}\), 5)

(vi) [x]^{2} – 5[x] + 6 = 0

∴ ([x] – 3)([x] – 2) = 0

∴ [x] = 3 or 2

If [x] = 2, then 2 ≤ x < 3

If [x] = 3, then 3 ≤ x < 4

∴ Solution set = [2, 4)

(vii) [x – 2] + [x + 2] + {x} = 0

∴ [x] – 2 + [x] + 2 + {x} = 0

∴ [x] + x = 0 …..[{x} + [x] = x]

∴ x = 0

(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)

L.H.S. = an integer

R.H.S. = an integer

∴ x = 6k, where k is an integer

Question 40.

Find the domain of the following functions.

(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}\)

(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)

(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)

(iv) f(x) = x!

(v) f(x) = \({ }^{5-x} P_{x-1}\)

(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)

(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)

Solution:

(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}=\frac{x^{2}+4 x+4}{(x+3)(x-2)}\)

For f to be defined, x ≠ -3, 2

∴ Domain of f = (-∞, -3) ∪ (-3, 2) ∪ (2, ∞)

(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)

For f to be defined,

x – 3 ≥ 0, 5 – x > 0 and 5 – x ≠ 1

x ≥ 3, x < 5 and x ≠ 4

∴ Domain of f = [3, 4) ∪ (4, 5)

(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)

Equation (i) gives solution set = [-1, 1]

∴ Domain of f = [-1, 1]

(iv) f(x) = x!

∴ Domain of f = set of whole numbers (W)

(v) f(x) = \({ }^{5-x} P_{x-1}\)

5 – x > 0, x – 1 ≥ 0, x – 1 ≤ 5 – x

∴ x < 5, x ≥ 1 and 2x ≤ 6

∴ x ≤ 3

∴ Domain of f = {1, 2, 3}

(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)

x – x^{2} ≥ 0

∴ x^{2} – x ≤ 0

∴ x(x – 1) ≤ 0

∴ 0 ≤ x ≤ 1 …..(i)

5 – x ≥ 0

∴ x ≤ 5 …..(ii)

Intersection of intervals given in (i) and (ii) gives

Solution set = [0, 1]

∴ Domain of f = [0, 1]

(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)

For f to be defined,

log (x^{2} – 6x + 6) ≥ 0

∴ x^{2} – 6x + 6 ≥ 1

∴ x^{2} – 6x + 5 ≥ 0

∴ (x – 5)(x – 1) ≥ 0

∴ x ≤ 1 or x ≥ 5 …..(i)

[∵ The solution of (x – a) (x – b) ≥ 0 is x ≤ a or x ≥ b, for a < b]

and x^{2} – 6x + 6 > 0

∴ (x – 3)^{2} > -6 + 9

∴ (x – 3)^{2} > 3

∴ x < 3 – √3 0r x > 3 + √3 ……..(ii)

From (i) and (ii), we get

x ≤ 1 or x ≥ 5

Solution set = (-∞, 1] ∪ [5, ∞)

∴ Domain of f = (-∞, 1] ∪ [5, ∞)

Question 41.

(i) f(x) = |x – 5|

(ii) f(x) = \(\frac{x}{9+x^{2}}\)

(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\)

(iv) f(x) = [x] – x

(v) f(x) = 1 + 2x + 4x

Solution:

(i) f(x) = |x – 5|

∴ Range of f = [0, ∞)

(ii) f(x) = \(\frac{x}{9+x^{2}}\) = y (say)

∴ x^{2}y – x + 9y = 0

For real x, Discriminant > 0

∴ 1 – 4(y)(9y) ≥ 0

∴ y^{2} ≤ \(\frac{1}{36}\)

∴ \(\frac{-1}{6}\) ≤ y ≤ \(\frac{1}{6}\)

∴ Range of f = [\(\frac{-1}{6}\), \(\frac{1}{6}\)]

(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\) = y, (say)

∴ √x y + y = 1

∴ √x = \(\frac{1-y}{y}\) ≥ 0

∴ \(\frac{y-1}{y}\) ≤ 0

∴ o < y ≤ 1

∴ Range of f = (0, 1]

(iv) f(x) = [x] – x = -{x}

∴ Range of f = (-1, 0] …..[0 ≤ {x} < 1]

(v) f(x) = 1 + 2x + 4x

Since, 2x > 0, 4x > 0

∴ f(x) > 1

∴ Range of f = (1, ∞)

Question 42.

Find (fog) (x) and (gof) (x)

(i) f(x) = e^{x}, g(x) = log x

(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)

Solution:

(i) f(x) = e^{x}, g(x) = log x

(fog) (x) = f(g(x))

= f(log x)

= e^{log x}

= x

(gof) (x) = g(f(x))

= g(e^{x})

= log (e^{x})

= x log e

= x …..[∵ log e = 1]

(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)

Question 43.

Find f(x), if

(i) g(x) = x^{2} + x – 2 and (gof) (x) = 4x^{2} – 10x + 4

(ii) g(x) = 1 + √x and f [g(x)] = 3 + 2√x + x.

Solution:

(i) g(x) = x^{2} + x – 2

(gof) (x) = 4x^{2} – 10x + 4

= (2x – 3)^{2} + (2x – 3) – 2

= g(2x – 3)

= g(f(x))

∴ f(x) = 2x – 3

(gof) (x) = 4x^{2} – 10x + 4

= (-2x + 2)^{2} + (-2x + 2) – 2

= g(-2x + 2)

= g(f(x))

∴ f(x) = -2x + 2

(ii) g(x) = 1 + √x

f(g(x)) = 3 + 2√x + x

= x + 2√x + 1 + 2

= (√x + 1)^{2} + 2

f(√x + 1) = (√x + 1)^{2} + 2

∴ f(x) = x^{2} + 2

Question 44.

Find (fof) (x) if

(i) f(x) = \(\frac{x}{\sqrt{1+x^{2}}}\)

(ii) f(x) = \(\frac{2 x+1}{3 x-2}\)

Solution: