**Chapter 6 Functions Miscellaneous Exercise 6**

## 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) 7Answer:**

(B) 5

Hint:

**Question 3.Find x, if 2 log _{2} x = 4(A) 4, -4(B) 4(C) -4(D) not definedAnswer:**

(B) 4

Hint:

∴ x = ±4

∴ x = 4

Question 4.

**Question 5.(A) x – 1(B) 1 – x(C) x(D) -xAnswer:**

(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.**

**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) 2Answer:**

(B) 0

Hint:

**Question 9.(A) R(B) Z(C) R – Z(D) Q – {0}Answer:**

(C) R – Z

Hint:

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.**

**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 ∈ RSolution:

(i) f(x) = 6x – 7 = y (say)

(x, y ∈ Z)

∴ x = 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) = 5 x ^{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) = 3x/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) = 5Solution:**

(i) f(x) = 3

∴ 5 – x = 3

∴ x = 5 – 3 = 2

(ii) f(x) = 5

∴ 5 – x = 5

∴ x = 0

**Question 8.**

**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.**

Solution:

**Question 14.**

**Question 15.Solution:**

f(x) = , 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 3Solution:**
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.**

**Question 20.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.Solution:**

**Question 24.**

**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:**

**Question 27.**

**Question 28.Solution:**

**Question 29.**

**Question 30.Solution:**

**Question 31.Solution:**

**Question 33.Solution:**

**Question 34.Solution:**

**Question 35.**

∴ 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.Solution:**

**Question 37.Solution:**

**Question 38.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.**

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

∴ 2 x^{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)

(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

L.H.S. = an integer

R.H.S. = an integer

∴ x = 6k, where k is an integer

**Question 40.**

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)

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

∴ Domain of f = [-1, 1]

(iv) f(x) = x!

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

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}

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]

**Question 41.**

∴ Range of f = [0, ∞)

(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.**

**Question 43.**

(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.**

**Solution:**