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Samacheer Kalvi 10th Maths Exercise 1.5 Solutions

10th Maths Exercise 1.5 solutions (Chapter 1 Relations and Functions) with step-by-step answers for all 18 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance13 answers · full steps below
  1. Q1 (i)f∘g=x2−6f\circ g=x^2-6, g∘f=x2−12x+36g\circ f=x^2-12x+36; f∘g≠g∘ff\circ g\neq g\circ f
  2. Q1 (ii)f∘g=22x2−1f\circ g=\dfrac{2}{2x^2-1}, g∘f=8−x2x2g\circ f=\dfrac{8-x^2}{x^2}; f∘g≠g∘ff\circ g\neq g\circ f
  3. Q1 (iii)f∘g=9−x3f\circ g=\dfrac{9-x}{3}, g∘f=3−x3g\circ f=\dfrac{3-x}{3}; f∘g≠g∘ff\circ g\neq g\circ f
  4. Q1 (iv)f∘g=g∘f=x−1f\circ g=g\circ f=x-1; so f∘g=g∘ff\circ g=g\circ f
  5. Q1 (v)f∘g=4x2+8x+3f\circ g=4x^2+8x+3, g∘f=4x2g\circ f=4x^2; f∘g≠g∘ff\circ g\neq g\circ f
  6. Q2 (i)k=−5k=-5
  7. Q2 (ii)k=−53k=-\dfrac{5}{3}
  8. Q4a=2a=2 or a=−2a=-2
  9. Q5Range of f∘g={3,9,19,33,… }f\circ g=\{3,9,19,33,\dots\}; Range of g∘f={9,25,49,81,… }g\circ f=\{9,25,49,81,\dots\}
  10. Q6 (i)f∘f(x)=x4−2x2f\circ f(x)=x^4-2x^2
  11. Q6 (ii)f∘f∘f(x)=x8−4x6+4x4−1f\circ f\circ f(x)=x^8-4x^6+4x^4-1
  12. Q7ff is one-one; gg is not one-one; f∘g(x)=x20f\circ g(x)=x^{20} is not one-one
  13. Q9f(x)=−4x−1f(x)=-4x-1

Q1. Using the functions ff and gg given below, find f∘gf \circ g and g∘fg \circ f …

Using the functions ff and gg given below, find f∘gf \circ g and g∘fg \circ f. Check whether f∘g=g∘ff \circ g = g \circ f.

(i)

f(x)=x−6f(x) = x - 6, g(x)=x2g(x) = x^2

  1. f∘g(x)=f(g(x))=f(x2)=x2−6f\circ g(x)=f(g(x))=f(x^2)=x^2-6

  2. g∘f(x)=g(f(x))=g(x−6)=(x−6)2=x2−12x+36g\circ f(x)=g(f(x))=g(x-6)=(x-6)^2=x^2-12x+36

  3. The two results are different.

Answer

f∘g=x2−6f\circ g=x^2-6, g∘f=x2−12x+36g\circ f=x^2-12x+36; f∘g≠g∘ff\circ g\neq g\circ f

(ii)

f(x)=2xf(x) = \dfrac{2}{x}, g(x)=2x2−1g(x) = 2x^2 - 1

  1. f∘g(x)=f(2x2−1)=22x2−1f\circ g(x)=f(2x^2-1)=\dfrac{2}{2x^2-1}

  2. g∘f(x)=g ⁣(2x)=2(2x)2−1=8x2−1=8−x2x2g\circ f(x)=g\!\left(\dfrac{2}{x}\right)=2\left(\dfrac{2}{x}\right)^2-1=\dfrac{8}{x^2}-1=\dfrac{8-x^2}{x^2}

  3. The two results are different.

Answer

f∘g=22x2−1f\circ g=\dfrac{2}{2x^2-1}, g∘f=8−x2x2g\circ f=\dfrac{8-x^2}{x^2}; f∘g≠g∘ff\circ g\neq g\circ f

(iii)

f(x)=x+63f(x) = \dfrac{x+6}{3}, g(x)=3−xg(x) = 3 - x

  1. f∘g(x)=f(3−x)=(3−x)+63=9−x3f\circ g(x)=f(3-x)=\dfrac{(3-x)+6}{3}=\dfrac{9-x}{3}

  2. g∘f(x)=g ⁣(x+63)=3−x+63=9−x−63=3−x3g\circ f(x)=g\!\left(\dfrac{x+6}{3}\right)=3-\dfrac{x+6}{3}=\dfrac{9-x-6}{3}=\dfrac{3-x}{3}

  3. The two results are different.

Answer

f∘g=9−x3f\circ g=\dfrac{9-x}{3}, g∘f=3−x3g\circ f=\dfrac{3-x}{3}; f∘g≠g∘ff\circ g\neq g\circ f

(iv)

f(x)=3+xf(x) = 3 + x, g(x)=x−4g(x) = x - 4

  1. f∘g(x)=f(x−4)=3+(x−4)=x−1f\circ g(x)=f(x-4)=3+(x-4)=x-1

  2. g∘f(x)=g(3+x)=(3+x)−4=x−1g\circ f(x)=g(3+x)=(3+x)-4=x-1

  3. Both give the same function.

Answer

f∘g=g∘f=x−1f\circ g=g\circ f=x-1; so f∘g=g∘ff\circ g=g\circ f

(v)

f(x)=4x2−1f(x) = 4x^2 - 1, g(x)=1+xg(x) = 1 + x

  1. f∘g(x)=f(1+x)=4(1+x)2−1=4x2+8x+3f\circ g(x)=f(1+x)=4(1+x)^2-1=4x^2+8x+3

  2. g∘f(x)=g(4x2−1)=1+(4x2−1)=4x2g\circ f(x)=g(4x^2-1)=1+(4x^2-1)=4x^2

  3. The two results are different.

Answer

f∘g=4x2+8x+3f\circ g=4x^2+8x+3, g∘f=4x2g\circ f=4x^2; f∘g≠g∘ff\circ g\neq g\circ f

Q2. Find the value of kk , such that f∘g=g∘ff \circ g = g \circ f

Find the value of kk, such that f∘g=g∘ff \circ g = g \circ f

(i)

f(x)=3x+2f(x) = 3x + 2, g(x)=6x−kg(x) = 6x - k

Show steps (6)

  1. f∘g(x)=f(6x−k)=3(6x−k)+2=18x−3k+2f\circ g(x)=f(6x-k)=3(6x-k)+2=18x-3k+2

  2. g∘f(x)=g(3x+2)=6(3x+2)−k=18x+12−kg\circ f(x)=g(3x+2)=6(3x+2)-k=18x+12-k

  3. Given f∘g=g∘ff\circ g=g\circ f:

  4. 18x−3k+2=18x+12−k18x-3k+2=18x+12-k

  5. −3k+k=12−2⇒−2k=10-3k+k=12-2\Rightarrow -2k=10

  6. k=−5k=-5

Answer

k=−5k=-5

(ii)

f(x)=2x−kf(x) = 2x - k, g(x)=4x+5g(x) = 4x + 5

Show steps (6)

  1. f∘g(x)=f(4x+5)=2(4x+5)−k=8x+10−kf\circ g(x)=f(4x+5)=2(4x+5)-k=8x+10-k

  2. g∘f(x)=g(2x−k)=4(2x−k)+5=8x−4k+5g\circ f(x)=g(2x-k)=4(2x-k)+5=8x-4k+5

  3. Given f∘g=g∘ff\circ g=g\circ f:

  4. 10−k=−4k+510-k=-4k+5

  5. 3k=−53k=-5

  6. k=−53k=-\dfrac{5}{3}

Answer

k=−53k=-\dfrac{5}{3}

Q3. If f(x)=2x−1f(x) = 2x - 1 , g(x)=x+12g(x) = \dfrac{x+1}{2} , show that f∘g=g∘f=xf \circ g = g \circ f = x

  1. f∘g(x)=f ⁣(x+12)=2(x+12)−1=x+1−1=xf\circ g(x)=f\!\left(\dfrac{x+1}{2}\right)=2\left(\dfrac{x+1}{2}\right)-1=x+1-1=x

  2. g∘f(x)=g(2x−1)=(2x−1)+12=2x2=xg\circ f(x)=g(2x-1)=\dfrac{(2x-1)+1}{2}=\dfrac{2x}{2}=x

  3. Hence both compositions give xx.

Answer

f∘g=g∘f=xf\circ g=g\circ f=x (shown)

Q4. If f(x)=x2−1f(x) = x^2 - 1 , g(x)=x−2g(x) = x - 2 find aa , if g∘f(a)=1g \circ f(a) = 1 .

  1. g∘f(a)=g(f(a))=g(a2−1)=(a2−1)−2=a2−3g\circ f(a)=g(f(a))=g(a^2-1)=(a^2-1)-2=a^2-3

  2. Given g∘f(a)=1g\circ f(a)=1:

  3. a2−3=1⇒a2=4a^2-3=1\Rightarrow a^2=4

  4. a=±2a=\pm 2

Answer

a=2a=2 or a=−2a=-2

Q5. Let A,B,C⊆NA, B, C \subseteq \mathbb{N} and a function f:A→Bf : A \to B be defined by f(x)=2x+1f(x) = 2x + 1 and g:B→Cg : B \to C …

Let A,B,C⊆NA, B, C \subseteq \mathbb{N} and a function f:A→Bf : A \to B be defined by f(x)=2x+1f(x) = 2x + 1 and g:B→Cg : B \to C be defined by g(x)=x2g(x) = x^2. Find the range of f∘gf \circ g and g∘fg \circ f.

  1. f∘g(x)=f(g(x))=f(x2)=2x2+1f\circ g(x)=f(g(x))=f(x^2)=2x^2+1

  2. Putting x=1,2,3,4,…x=1,2,3,4,\dots gives 3,9,19,33,…3,9,19,33,\dots

  3. g∘f(x)=g(f(x))=g(2x+1)=(2x+1)2g\circ f(x)=g(f(x))=g(2x+1)=(2x+1)^2

  4. Putting x=1,2,3,4,…x=1,2,3,4,\dots gives 9,25,49,81,…9,25,49,81,\dots

Answer

Range of f∘g={3,9,19,33,… }f\circ g=\{3,9,19,33,\dots\}; Range of g∘f={9,25,49,81,… }g\circ f=\{9,25,49,81,\dots\}

Q6. Let f(x)=x2−1f(x) = x^2 - 1 . Find

Let f(x)=x2−1f(x) = x^2 - 1. Find

(i)

f∘ff \circ f

  1. f∘f(x)=f(f(x))=f(x2−1)f\circ f(x)=f(f(x))=f(x^2-1)

  2. =(x2−1)2−1=(x^2-1)^2-1

  3. =x4−2x2+1−1=x4−2x2=x^4-2x^2+1-1=x^4-2x^2

Answer

f∘f(x)=x4−2x2f\circ f(x)=x^4-2x^2

(ii)

f∘f∘ff \circ f \circ f

  1. f∘f(x)=x4−2x2f\circ f(x)=x^4-2x^2

  2. f∘f∘f(x)=f(x4−2x2)=(x4−2x2)2−1f\circ f\circ f(x)=f(x^4-2x^2)=(x^4-2x^2)^2-1

  3. =x8−4x6+4x4−1=x^8-4x^6+4x^4-1

Answer

f∘f∘f(x)=x8−4x6+4x4−1f\circ f\circ f(x)=x^8-4x^6+4x^4-1

Q7. If f:R→Rf : \mathbb{R} \to \mathbb{R} and g:R→Rg : \mathbb{R} \to \mathbb{R} are defined by f(x)=x5f(x) = x^5 and g(x)=x4g(x) = x^4 then check …

If f:R→Rf : \mathbb{R} \to \mathbb{R} and g:R→Rg : \mathbb{R} \to \mathbb{R} are defined by f(x)=x5f(x) = x^5 and g(x)=x4g(x) = x^4 then check if ff, gg are one-one and f∘gf \circ g is one-one?

  1. f(x)=x5f(x)=x^5: if f(a)=f(b)f(a)=f(b) then a5=b5⇒a=ba^5=b^5\Rightarrow a=b. So ff is one-one.

  2. g(x)=x4g(x)=x^4: g(1)=1g(1)=1 and g(−1)=1g(-1)=1 but 1≠−11\neq-1. So gg is not one-one.

  3. f∘g(x)=f(g(x))=f(x4)=(x4)5=x20f\circ g(x)=f(g(x))=f(x^4)=(x^4)^5=x^{20}

  4. f∘g(1)=1f\circ g(1)=1 and f∘g(−1)=1f\circ g(-1)=1 with 1≠−11\neq-1, so f∘gf\circ g is not one-one.

Answer

ff is one-one; gg is not one-one; f∘g(x)=x20f\circ g(x)=x^{20} is not one-one

Q8. Consider the functions f(x)f(x) , g(x)g(x) , h(x)h(x) as given below. Show …

Consider the functions f(x)f(x), g(x)g(x), h(x)h(x) as given below. Show that (f∘g)∘h=f∘(g∘h)(f \circ g) \circ h = f \circ (g \circ h) in each case.

(i)

f(x)=x−1f(x) = x - 1, g(x)=3x+1g(x) = 3x + 1 and h(x)=x2h(x) = x^2

  1. f∘g(x)=f(3x+1)=(3x+1)−1=3xf\circ g(x)=f(3x+1)=(3x+1)-1=3x

  2. (f∘g)∘h(x)=(f∘g)(x2)=3x2(f\circ g)\circ h(x)=(f\circ g)(x^2)=3x^2 ...(1)

  3. g∘h(x)=g(x2)=3x2+1g\circ h(x)=g(x^2)=3x^2+1

  4. f∘(g∘h)(x)=f(3x2+1)=(3x2+1)−1=3x2f\circ(g\circ h)(x)=f(3x^2+1)=(3x^2+1)-1=3x^2 ...(2)

  5. From (1) and (2) both are equal.

Answer

(f∘g)∘h=f∘(g∘h)=3x2(f\circ g)\circ h=f\circ(g\circ h)=3x^2 (shown)

(ii)

f(x)=x2f(x) = x^2, g(x)=2xg(x) = 2x and h(x)=x+4h(x) = x + 4

  1. f∘g(x)=f(2x)=(2x)2=4x2f\circ g(x)=f(2x)=(2x)^2=4x^2

  2. (f∘g)∘h(x)=(f∘g)(x+4)=4(x+4)2=4x2+32x+64(f\circ g)\circ h(x)=(f\circ g)(x+4)=4(x+4)^2=4x^2+32x+64 ...(1)

  3. g∘h(x)=g(x+4)=2(x+4)=2x+8g\circ h(x)=g(x+4)=2(x+4)=2x+8

  4. f∘(g∘h)(x)=f(2x+8)=(2x+8)2=4x2+32x+64f\circ(g\circ h)(x)=f(2x+8)=(2x+8)^2=4x^2+32x+64 ...(2)

  5. From (1) and (2) both are equal.

Answer

(f∘g)∘h=f∘(g∘h)=4x2+32x+64(f\circ g)\circ h=f\circ(g\circ h)=4x^2+32x+64 (shown)

(iii)

f(x)=x−4f(x) = x - 4, g(x)=x2g(x) = x^2 and h(x)=3x−5h(x) = 3x - 5

  1. f∘g(x)=f(x2)=x2−4f\circ g(x)=f(x^2)=x^2-4

  2. (f∘g)∘h(x)=(f∘g)(3x−5)=(3x−5)2−4=9x2−30x+21(f\circ g)\circ h(x)=(f\circ g)(3x-5)=(3x-5)^2-4=9x^2-30x+21 ...(1)

  3. g∘h(x)=g(3x−5)=(3x−5)2=9x2−30x+25g\circ h(x)=g(3x-5)=(3x-5)^2=9x^2-30x+25

  4. f∘(g∘h)(x)=9x2−30x+25−4=9x2−30x+21f\circ(g\circ h)(x)=9x^2-30x+25-4=9x^2-30x+21 ...(2)

  5. From (1) and (2) both are equal.

Answer

(f∘g)∘h=f∘(g∘h)=9x2−30x+21(f\circ g)\circ h=f\circ(g\circ h)=9x^2-30x+21 (shown)

Q9. Let f={(−1,3),(0,−1),(2,−9)}f = \{(-1,3),(0,-1),(2,-9)\} be a linear function from Z\mathbb{Z} into Z\mathbb{Z} . Find …

Let f={(−1,3),(0,−1),(2,−9)}f = \{(-1,3),(0,-1),(2,-9)\} be a linear function from Z\mathbb{Z} into Z\mathbb{Z}. Find f(x)f(x).

  1. A linear function is f(x)=ax+bf(x)=ax+b.

  2. f(0)=−1⇒b=−1f(0)=-1\Rightarrow b=-1

  3. f(−1)=3⇒−a+b=3⇒−a−1=3⇒a=−4f(-1)=3\Rightarrow -a+b=3\Rightarrow -a-1=3\Rightarrow a=-4

  4. So f(x)=−4x−1f(x)=-4x-1.

  5. Check: f(2)=−8−1=−9f(2)=-8-1=-9 ✓

Answer

f(x)=−4x−1f(x)=-4x-1

Q10. In electrical circuit theory, a circuit C(t)C(t) is called a linear circuit …

In electrical circuit theory, a circuit C(t)C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1+bt2)=aC(t1)+bC(t2)C(at_1 + bt_2) = aC(t_1) + bC(t_2), where a,ba, b are constants. Show that the circuit C(t)=3tC(t) = 3t is linear.

  1. C(t)=3tC(t)=3t

  2. LHS =C(at1+bt2)=3(at1+bt2)=3at1+3bt2=C(at_1+bt_2)=3(at_1+bt_2)=3at_1+3bt_2

  3. RHS =aC(t1)+bC(t2)=a(3t1)+b(3t2)=3at1+3bt2=aC(t_1)+bC(t_2)=a(3t_1)+b(3t_2)=3at_1+3bt_2

  4. LHS == RHS, so the superposition principle holds.

Answer

C(at1+bt2)=aC(t1)+bC(t2)C(at_1+bt_2)=aC(t_1)+bC(t_2), hence C(t)=3tC(t)=3t is a linear circuit

Quick check

  1. Let ff and gg be two functions given by f={(0,1),(2,0),(3,−4),(4,2),(5,7)}f = \{(0,1),(2,0),(3,-4),(4,2),(5,7)\}, g={(0,2),(1,0),(2,4),(−4,2),(7,0)}g = \{(0,2),(1,0),(2,4),(-4,2),(7,0)\} then the range of f∘gf \circ g is

    1. (1) {0,2,3,4,5}\{0,2,3,4,5\}
    2. (2) {−4,1,0,2,7}\{-4,1,0,2,7\}
    3. (3) {1,2,3,4,5}\{1,2,3,4,5\}
    4. (4) {0,1,2}\{0,1,2\}
    Show answer

    Answer: (4) {0,1,2}\{0,1,2\}

    1. (f∘g)(0)=f(2)=0(f\circ g)(0)=f(2)=0, (f∘g)(1)=f(0)=1(f\circ g)(1)=f(0)=1.

    2. (f∘g)(2)=f(4)=2(f\circ g)(2)=f(4)=2, (f∘g)(−4)=f(2)=0(f\circ g)(-4)=f(2)=0, (f∘g)(7)=f(0)=1(f\circ g)(7)=f(0)=1.

    3. Range ={0,1,2}=\{0,1,2\}.

    Question 1 of 5

  2. If there are 1024 relations from a set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} to a set BB, then the number of elements in BB is

    1. (1) 3
    2. (2) 2
    3. (3) 4
    4. (4) 8
    Show answer

    Answer: (2) 2

    1. Number of relations from AA to BB is 2n(A)⋅n(B)2^{n(A)\cdot n(B)}.

    2. 25 n(B)=1024=2102^{5\,n(B)}=1024=2^{10}.

    3. 5 n(B)=10⇒n(B)=25\,n(B)=10\Rightarrow n(B)=2.

    Question 2 of 5

  3. Let A={1,2,3,4}A = \{1,2,3,4\} and B={4,8,9,10}B = \{4,8,9,10\}. A function f:A→Bf : A \to B given by f={(1,4),(2,8),(3,9),(4,10)}f = \{(1,4),(2,8),(3,9),(4,10)\} is a

    1. (1) Many-one function
    2. (2) Identity function
    3. (3) One-to-one function
    4. (4) Into function
    Show answer

    Answer: (3) One-to-one function

    1. Images are 4,8,9,104,8,9,10 — all different.

    2. Different elements of AA have different images.

    3. So ff is one-to-one.

    Question 3 of 5

  4. If f(x)=2x2f(x) = 2x^2 and g(x)=13xg(x) = \dfrac{1}{3x}, then f∘gf \circ g is

    1. (1) 32x2\dfrac{3}{2x^2}
    2. (2) 23x2\dfrac{2}{3x^2}
    3. (3) 29x2\dfrac{2}{9x^2}
    4. (4) 16x2\dfrac{1}{6x^2}
    Show answer

    Answer: (3) 29x2\dfrac{2}{9x^2}

    1. (f∘g)(x)=f(g(x))=f(13x)(f\circ g)(x)=f\big(g(x)\big)=f\left(\dfrac{1}{3x}\right).

    2. =2(13x)2=2\left(\dfrac{1}{3x}\right)^{2}.

    3. =29x2=\dfrac{2}{9x^{2}}.

    Question 4 of 5

  5. A={a,b,p}A = \{a,b,p\}, B={2,3}B = \{2,3\}, C={p,q,r,s}C = \{p,q,r,s\} then n[(A∪C)×B]n[(A \cup C) \times B] is

    1. (1) 8
    2. (2) 20
    3. (3) 12
    4. (4) 16
    Show answer

    Answer: (3) 12

    1. A∪C={a,b,p,q,r,s}A\cup C=\{a,b,p,q,r,s\}, so n(A∪C)=6n(A\cup C)=6.

    2. n(B)=2n(B)=2.

    3. n[(A∪C)×B]=6×2=12n[(A\cup C)\times B]=6\times 2=12.

    Question 5 of 5
Chapter 1 formulas
  • Cartesian product of A with B

    A×B={(a,b)∣for all a∈A, b∈B}A \times B = \{(a,b) \mid \text{for all } a \in A,\ b \in B\}

  • A relation R from A to B is a subset of A × B

    R⊆A×BR \subseteq A \times B

  • Condition for a relation R from X to Y to be a function

    for every x∈X there exists only one y∈Y\text{for every } x \in X \text{ there exists only one } y \in Y

  • Identity function

    f(x)=xf(x) = x

  • Reciprocal function

    f(x)=1xf(x) = \dfrac{1}{x}

  • Constant function

    f(x)=cf(x) = c

  • Linear function

    f(x)=ax+b, a≠0f(x) = ax + b,\ a \neq 0

  • Quadratic function

    f(x)=ax2+bx+c, a≠0f(x) = ax^{2} + bx + c,\ a \neq 0

  • Cubic function

    f(x)=ax3+bx2+cx+d, a≠0f(x) = ax^{3} + bx^{2} + cx + d,\ a \neq 0

  • Composition of two functions f : A → B and g : B → C

    g∘f:A→C,(g∘f)(x)=g(f(x))  for all x∈Ag \circ f : A \to C,\quad (g \circ f)(x) = g(f(x)) \ \text{ for all } x \in A

  • Composition is not commutative in general

    f∘g≠g∘ff \circ g \neq g \circ f

  • Composition is associative

    f∘(g∘h)=(f∘g)∘hf \circ (g \circ h) = (f \circ g) \circ h

Answers checked against the Samacheer Kalvi textbook · Updated

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