PM SHRI KENDRIYA VIDYALAYA PITAMPURA SHIFT-1

CHAPTER-1 RELATIONS AND FUNCTIONS

SHORT ANSWER TYPE QUESTIONS
1 Show that the function f : N → N given by f(x) = 2x is one-one
but not onto
One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain N.
⇒ 2 𝑥1 = 2 𝑥2 ⇒ 𝑥1 = 𝑥2
⇒ f is one-one.
Onto:
Let f(x) = 2x =y ∈ Codmain N
𝑦
⇒𝑥=
2
𝑦
But doesn’t belong to Domain N for all y ∈ Domain N
2
Hence, f is not onto.
2 𝑥−2
Let A = R- {3}, B =R-{1}. If f: A→ B be defined by f(x) = ,
𝑥−3
∀x ∈A. Then, show that f is bijective
One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain A.
𝑥1 − 2 𝑥2 − 2
=
𝑥1 − 3 𝑥2 − 3
After solving by cross multiplication, 𝑥1 = 𝑥2
⇒ f is one-one.
Onto:
𝑥−2
Let f(x) = =y ∈ Codmain B
𝑥−3
⇒ 𝑦𝑥 − 3𝑦 = 𝑥 − 2 ⇒ 𝑦𝑥 − 𝑥 = 3𝑦 − 2
𝑥−2
⇒ 𝑥(𝑦 − 3) = 𝑥 − 2⇒𝑥 = ∈ Domain A for all y ∈ Codomain B
𝑦−3
Hence, f is onto.
3 Show that the function f in 𝐴 = 𝑅 − {
2
} defined as (𝑥) =
4𝑥+3
is
3 6𝑥−4
one-one.
One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain A.
4𝑥1 + 3 4𝑥2 + 3
=
6𝑥1 − 4 6𝑥2 − 4

After solving by cross multiplication

⇒ 𝑥1 = 𝑥2
⇒ f is one-one.
1
4 Let f: 𝑅+ → [-9,∞) be a function defined as : f(x) = 5𝑥 2 + 6x – 9.
Show than f(x) is bijective.
One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain 𝑅+ .
5 𝑥1 2 + 6 𝑥1 – 9= 5 𝑥2 2 + 6 𝑥2 – 9
After solving
⇒ 𝑥1 = 𝑥2
⇒ f is one-one.
Onto:
Let f(x) = 5𝑥 2 + 6x – 9 =y ∈ Codmain [-9,∞)
⇒ 5𝑥 2 + 6x – (9+y) =0
−6±√36+4×5×(9+𝑦)
⇒𝑥=
2×5
−6±√216+20𝑦
⇒𝑥 = ∈ Domain 𝑅+ for all y ∈ Codomain [-9,∞)
10
Hence, f is onto.
So f is bijective.
5 Show that the function 𝑓: 𝑅 → 𝑅 defined as f(𝑥) = 𝑥 2 is neither
one-one nor onto.
One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain R.
⇒ 𝑥1 2 = 𝑥2 2 ⇒ 𝑥1 =± 𝑥2
⇒ f is not one-one.
Onto:
Let f(𝑥) = 𝑥 2 =y ∈ Codmain R
⇒ 𝑥 = ±√𝑦
But ±√𝑦 doesn’t belong to Domain R for all y ∈ Codmain R
Hence, f is not onto.
6 In each of the following cases, state whether the function is one-
one, onto or bijective. Justify your answer.
(i) f : R → R defined by 𝑓 (𝑥) = 3 – 4𝑥
(ii) f : R → R defined by 𝑓 (𝑥) = 1 + 𝑥 2
(i) One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain R.
⇒ 3 − 4 𝑥1 =3 − 4 𝑥2 ⇒ 𝑥1 = 𝑥2
⇒ f is one-one.
Onto:
Let f(x) = 3-4x =y ∈ Codmain R
3−𝑦
⇒𝑥= belong to Domain R for all y ∈ Codomain R
4
Hence, f is onto.
2
 (ii) One-One:
Let f( 𝑥1 ) = f (𝑥2 ) where 𝑥1 ,𝑥2 ∈ Domain R.
⇒ 1 + 𝑥1 2 =1 + 𝑥2 2 ⇒ 𝑥1 =± 𝑥2
⇒ f is not one-one.
Onto:
Let f(x) = 1 + 𝑥 2 =y ∈ Codmain R
⇒ 𝑥 = ±√𝑦 − 1 doesn’t belong to Domain R for all y ∈ Codomain R
Hence, f is not onto.
LONG ANSWER TYPE QUESTIONS
7 Show that the relation R on R defined as R = {(a, b):a ≤ b}, is
reflexive and transitive but not symmetric.
R = {(a, b)} = a ≤ b}.
Reflexivity:
Since, a ≤ a ∀ a ∈ R,
∴ (a, a) ∈ R, Thus, R reflexive.
Symmetricity:
Let (a, b) ∈ R so, a ≤ b
But b ≤ a is incorrect for all a ,b ∈ R
so (b, a) ∉ R for all a ,b
OR
∵ (3, 5) ∈ R but (5, 3) ∉ R as 3 ≤ 5 but 5 > 3
Thus, R is not symmetric.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a ≤ b and b ≤ c
⇒a≤c
⇒ (a, c) ∈ R.
Thus, R is transitive
8 Check whether the relation 𝑹 in R defined by 𝑹={(𝒂, 𝒃) ∶𝒂 ≤ 𝑏 3 }
is reflexive, symmetric or transitive.
Show by taking example that R is not reflexive, not symmetric
and not transitive.
9 Let A = (x ∈Z : 0 ≤ x ≤ 12}. Show that R = {(a, b) : a, b ∈ A; |a
– b| is divisible by 4} is an equivalence relation. Find the set of
all elements related to 1. Also write the equivalence class [2].
R = {(a, b): a, b ∈ A; |a – b| is divisible by 4}.
Reflexivity:
|a – a| = 0, which is divisible by 4.
∴ (a, a) ∈ R. Thus, R is reflexive.
Symmetricity:
3
 Let (a, b) ∈ R
⇒ |a – b| is divisible by 4 ⇒ |b – a| is divisible by 4
So, (b, a) ∈ R, Thus, R is symmetric
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ |a – b| is divisible by 4 and |b – c| is divisible by 4
⇒ |a – c| is divisible by 4 , ⇒ (a, c) ∈ R.
Thus, R is transitive.
Now, R is reflexive, symmetric and transitive. Hence, R is an
equivalence relation.
Equivalence Class:
If first element is 1 second will be 1, 5, 9
⇒ [1]={1, 5, 9}. Similarly [2] = {2, 6, 10}.
10 Let N denote the set of all natural numbers and R be the relation
on N x N defined by : (a, b) R(c,d) is ad(b + c) = bc(a + d).
Show that R is an equivalence relation
(a, b) R (c, d) ⇒ ad (b + c) = bc (a + d) on N.
Reflexivity:
ab(b + a) = ba (a + b)
⇒ (a, b) R (a, b) ⇒. Thus R is reflexive.
Symmetricity:
Let (a, b) R (c, d)
⇒ ad(b + c) = bc (a + d)
⇒ bc(a + d) = ad (b + c)
⇒ cb(d + a) = da (c + b)
⇒ (c,d) R (a,b). Thus R is symmetric
Transitivity:
Let (a , b) R (c, d) and (c, d) R ( e, f)
∴ ad (b + c) = bc (a + d) and cf(d + e) = de(c + f)
⇒ be(a+f) = af(b+e)
⇒ af(b+e) = be(a+f)
⇒ (a, b) R (e,f).
Thus R is transitive. Hence, R is an equivalence relation.
11 Let A = N x N be the set of ail ordered pairs of natural numbers
and R be the relation on the set A defined by (a, b) R (c, d) iff
ad = bc. Show that R is an equivalence relation
12 Let R be the relation in N x N defined by (a, b) R (c , d) iff 𝑎2 +
𝑑 2 = 𝑏 2 + 𝑐 2 for all a, b, c ,d ∈ N .Prove that R is an equivalence
relation.
(a, b) R (c, d) ⇒ 𝑎2 + 𝑑 2 = 𝑏 2 + 𝑐 2 on N.
4
 Reflexivity:
𝑎2 + 𝑏 2 = 𝑏 2 + 𝑎2 ⇒ (a, b) R (a, b) ⇒. Thus R is reflexive.
Symmetricity:
Let (a, b) R (c, d)
⇒ 𝑎2 + 𝑑 2 = 𝑏 2 + 𝑐 2 ⇒ 𝑐 2 + 𝑏 2 = 𝑑 2 + 𝑎2
⇒ (c,d) R (a,b). Thus R is symmetric
Transitivity:
Let (a , b) R (c, d) and (c, d) R ( e, f)
∴ 𝑎2 + 𝑑 2 = 𝑏 2 + 𝑐 2 and 𝑐 2 + 𝑓 2 = 𝑑 2 + 𝑒 2
⇒ 𝑎2 + 𝑑 2 + 𝑐 2 + 𝑓 2 = 𝑏 2 + 𝑐 2 +𝑑 2 + 𝑒 2
⇒ 𝑎2 + 𝑓 2 = 𝑏 2 + 𝑒 2
⇒ (a, b) R (e,f).
Thus R is transitive. Hence, R is an equivalence relation.

PM SHRI KENDRIYA VIDYALAYA PITAMPURA SHIFT-1

CHAPTER-2 INVERSE TRIGONOMETRIC FUNCTIONS
VERY SHORT TYPE QUESTION ANSWERS
1 The value of 𝐜𝐨𝐬 −𝟏 (𝐜𝐨𝐬
𝟑𝛑
) is ..........
𝟐
3π π
cos −1 (cos ) = cos −1 (0) =
2 2
2 Write the range of 𝒔𝒊𝒏−𝟏 𝒙.
𝜋 𝜋
The range of 𝑠𝑖𝑛−1 𝑥 is [− , ]
2 2
3 What is the principal value of 𝐬𝐢𝐧−𝟏 (𝐬𝐢𝐧
𝟐𝝅
) + 𝒄𝒐𝒔−𝟏 (𝐜𝐨𝐬
𝟐𝝅
)
𝟑 𝟑
𝑠𝑖𝑛−1 (sin 2 /3) + cos −1 (cos 2 /3)=𝑠𝑖𝑛−1 (sin(π −  /3) + 2 /3
𝜋 2
=𝑠𝑖𝑛−1 (sin(  /3) + 2 /3=𝑠𝑖𝑛−1 (sin(  /3) + 2 /3= + = 𝜋
3 3
4 𝝅
Evaluate: 𝒔𝒊𝒏 [ − 𝒔𝒊𝒏 −𝟏
(− 𝟐)]
𝟏
𝟑
𝜋 1 𝜋 𝜋 𝜋
𝑠𝑖𝑛 [ − 𝑠𝑖𝑛−1 (− )]= 𝑠𝑖𝑛 [ + ] = s𝑖𝑛 [ ] = 1
3 2 3 6 2
5 The value of 𝒕𝒂𝒏𝟐 (𝐬𝐞𝐜 −𝟏 𝟐) + 𝒄𝒐𝒕𝟐 (𝐜𝐨𝐬𝐞𝐜 −𝟏 𝟑) is ____
tan2(sec–12) + cot2(cosec–13) = sec2(sec–12) – 1 + cosec2(cos 3) – 1
= (2)2 –1 + (3)2 – 1 = 4 – 1 + 9 – 1 = 11.
OR
tan (sec 2) + cot (cosec 3) = tan (tan–1√3) + cot2(cot–1√8)
2 –1 2 –1 2

= (√3)2 + (√8)2 = 3 + 8= 11
6 Find the value of the following expression: 𝐜𝐨𝐬−𝟏 (𝐜𝐨𝐬
𝟏𝟑𝝅
)
𝟔
13𝜋 𝜋 𝜋 𝜋
cos −1 (cos ) = cos −1 (cos(2π + )) = cos −1 (cos ) =
6 6 6 6
5
7 Find the principal value of 𝐜𝐨𝐭 −𝟏 (−√𝟑)
𝜋 5𝜋
𝜋− =
6 6
8 Write the principal values of 𝐬𝐞𝐜 −𝟏 (−𝟐)
𝜋 2𝜋
𝜋− =
3 3
9 Write the principal values of 𝐬𝐞𝐜 −𝟏 (− )
𝟐
√𝟑
𝜋 5𝜋
𝜋− =
6 6
10 𝟒𝝅
Write the principal value of the following:𝐬𝐢𝐧−𝟏 (𝐬𝐢𝐧 )
𝟓
4𝜋 𝜋 𝜋 𝜋
sin−1 (sin ) = sin−1 (sin(𝜋 − 5 )) = sin−1 (sin( 5 ))=5
5
SHORT TYPE QUESTION ANSWERS
√𝟑
11 Write the value of 𝐭𝐚𝐧−𝟏 [𝟐 𝐬𝐢𝐧 (𝟐 𝐜𝐨𝐬 −𝟏 )]
𝟐
√3 𝜋 𝜋
tan−1 [2 sin (2 cos −1 )] = tan−1 [2 sin (2 × )] = tan−1 [2 sin ( )]
2 6 3
√3 𝜋
= tan−1 [2 × ] = tan−1 [√3] =
2 3
12 Find the domain of 𝒚 = 𝐜𝐨𝐬−𝟏 (𝒙𝟐 − 𝟒)
As 𝒚 = 𝐜𝐨𝐬−𝟏 (𝒙𝟐 − 𝟒) ⇒ cos 𝑦= 𝒙𝟐 − 𝟒
Since – 1 ≤ cos y ≤ 1 i.e. – 1 ≤ 𝒙𝟐 − 𝟒 ≤ 1
⇒ 3 ≤ 𝒙𝟐 ≤ 5⇒ √3 ≤ |𝑥| ≤ √5 ⇒ x ∈ [−√5, −√3 ] ∪ [√3, √5, ]
13 Prove that: sec2(tan–1 2) + cosec2(cot–1 3) = 15.
sec2 (tan–1 2)+cosec2 (cot–1 3)= 1+tan2 (tan–1 2) +1+cot2 (cot–1 3)

= 1 + {tan (tan–1 2)}2 + 1 + {cot (cot–1 3)}2 = 1 + (2)2 + 1 + (3)2 = 15.

OR
LHS= sec2 (tan–1 2) + cosec2 (cot–1 3)

= sec2 (sec–1 √5) + cosec2 (cosec–1 √10) = (√5)2 + (√10)2 = 5+ 10= 10

√3 𝜋
14 Prove that tan−1 [2𝑠𝑖𝑛 (2 cos −1 )] =
2 3
√ 3 𝜋 √3 𝜋
L.H.S. = tan−1 [2𝑠𝑖𝑛 (2 cos −1 )] =tan−1 [2𝑠𝑖𝑛 (2. )] = tan−1 [2. ] =
2 6 2 3
𝟑
15 −𝟏 𝟑𝒙−𝒙
Express 𝐭𝐚𝐧 ( ) in simplest form.
𝟏−𝟑𝒙𝟐
3𝑥−𝑥 3
Let 𝑥 = tan 𝜃 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠 𝜃 = tan−1 𝑥 , 𝑆𝑜 tan−1 ( ) =
1−3𝑥 2
3𝑡𝑎𝑛𝜃−tan3 𝜃
tan−1 ( ) = tan−1 (tan 3𝜃) = 3𝜃 = 3 tan−1 𝑥
1−3 tan2 𝜃