Q1. 𝐼𝑓 𝑓: 𝑅 − {35} → 𝑅 be defined by 𝑓(𝑥) = 3𝑥+2
, then
5𝑥−3
A. f−1(x)=f(x)
B. f−1(x)=−f(x)
C. fof(x)=−x
D. 𝑓 −1 (𝑥) =
1
20
𝑓(𝑥)
Answer: A
Explanation: Given that,
3𝑥 + 2
5𝑥 − 3
3𝑥 + 2
⇒𝑦=
5𝑥 − 3
⇒ 3𝑥 + 2 = 5𝑥𝑦 − 3𝑦
⇒ 𝑥(3 − 5𝑦) = −3𝑦 − 2
3𝑦 + 2
⇒𝑥=
5𝑦 − 3
3𝑥 + 2
⇒ 𝑓 −1 (𝑥) =
5𝑥 − 3
−1
⇒ 𝑓 (𝑥) = 𝑓(𝑥)
𝑓(𝑥) =
Q2.
𝐿𝑒𝑡 𝑓: [2, ∞) → 𝑅 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 5,
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑓 𝑖𝑠
A. R
B. [1, ∞)
C. [4, ∞)
D. [5, ∞)
Answer: B
Explanation:
𝑮𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕
𝒇(𝒙) = 𝒙𝟐 − 𝟒𝒙 + 𝟓
𝐋𝐞𝐭
𝟐
𝒚 = 𝒙 − 𝟒𝒙 + 𝟓
⇒ 𝒚 = 𝒙𝟐 − 𝟒𝒙 + 𝟒 − 𝟒 + 𝟓
⇒ 𝒚 = (𝒙 − 𝟐)𝟐 − 𝟒 + 𝟓
⇒ 𝒚 = (𝒙 − 𝟐)𝟐 + 𝟏
⇒ 𝒚 − 𝟏 = (𝒙 − 𝟐)𝟐
⇒ 𝒙 − 𝟐 = √𝒚 − 𝟏
⇒ 𝒙 = 𝟐 + √𝒚 − 𝟏
∴ 𝒚 − 𝟏 ≥ 𝟎, 𝒚 ≥ 𝟏
𝑻𝒉𝒊𝒔 𝒊𝒎𝒑𝒍𝒊𝒆𝒔 𝒓𝒂𝒏𝒈𝒆 𝒊𝒔 [𝟏, ∞)
Q3.The set A contains 5 elements and the set B contains 6 elements, then the
number of one-one and onto mappings from A to B is
A. 720
B. 120
C. 0
D. None of these
Answer: C
Explanation: We know that if A and B are two non-empty finite sets containing m and n
elements respectively , then the number of one-one and onto mapping from A to B is:
𝑛! 𝑖f 𝑚 = 𝑛
0, 𝑖f 𝑚 ≠ 𝑛
Given that
𝑚=5
and
𝑛=6
∵𝑚≠𝑛
So, number of mappings = 0
Q4.
𝐿𝑒𝑡 𝑓: 𝑅 → 𝑅 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓(𝑥) = 𝑥3 + 5.
–1
𝑇ℎ𝑒𝑛 𝑓 (𝑥) 𝑖𝑠
A. (x+5)1/3
B. (x−5)1/3
C. (5−x)1/3
D. 5−x
Answer: B
Explanation: Since
𝑓(𝑥) = 𝑥 3 + 5
⇒ 𝑦 = 𝑥3 + 5
⇒ 𝑥3 = 𝑦 − 5
⇒ 𝑥 = (𝑦 − 5)1/3
⇒ 𝑓 −1 (𝑥) = (𝑥 − 5)1/3
Q5.Which of the following functions from Z into Z are bisections’?
A. f(x)=x3
B. f(x)=x+ 2
C. f(x)= 2x+ 1
D. f(x)=x2+ 1
Answer: B
Explanation: For
𝑓(𝑥) = 𝑥 + 2
𝑓(𝑥1 ) = 𝑓(𝑥2 )
⇒ 𝑥1 + 2 = 𝑥2 + 2
⇒ 𝑥1 = 𝑥2
Therefore f is one,
Also ,
𝑦 =𝑥+2
⇒ 𝑥 = 𝑦 − 2∀𝑦 ∈ 𝑥
Thus f(x) is onto
this implies f(x) is one-one and onto or bijective function.
Q6.If a relation R on the set {1, 2, 3} be defined by R = {(1, 2), ),(2,3),(1,3)}, then R
is
A. reflexive
B. transitive
C. symmetric
D. none of these
Answer: B
Explanation: R on the set {1,2,3} be defined by R = {(1, 2), (2,3), (1,3)}
This implies that R is transitive.
Since for (1,2) R and (2,3) R , (1,3 ) also R.
Q7.The maximum number of equivalence relations on the set A = {1, 2, 3} are
A. 1
B. 2
C. 3
D. 5
Answer: D
Explanation: Given A = {1, 2, 3}
Number of equivalence relations are as follows:
𝑅1 = {(1,1), (2,2), (3,3)}
𝑅2 = {(1,1), (2,2), (3,3), (1,2), (2,1)}
𝑅3 = {(1,1), (2,2), (3,3), (1,3), (3,1)}
𝑅4 = {(1,1), (2,2), (3,3), (2,3), (3,2)}
𝑅5 = {(1,2,3) ⇔ 𝐴 × 𝐴 = 𝐴2 }
Maximum number of equivalence relations on the set A={1,2,3} =5
Q8.Set A has 3 elements and the set B has 4 elements. Then the number of
injective mappings that can be defined from A to B is
A. 144
B. 12
C. 24
D. 64
Answer: C
Explanation:
𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒊𝒏𝒋𝒆𝒄𝒕𝒊𝒗𝒆 𝒎𝒂𝒑𝒑𝒊𝒏𝒈𝒔 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆
𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒊𝒏𝒈 𝟑 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏𝒕𝒐 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒊𝒏𝒈 𝟒 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒔𝟒 𝑷𝟑 = 𝟒! = 𝟐𝟒.
Q9.Let N be the set of natural numbers and the function f : N → N be defined by
f(n) = 2n + 3. Then f is
A. surjective
B. injective
C. bijective
D. none of these
Answer: B
Explanation:
𝒇(𝒏𝟏 ) = 𝒇(𝒏𝟐 )
⇒ 𝟐(𝒏𝟏 ) + 𝟑 = 𝟐(𝒏𝟐 ) + 𝟑
⇒ 𝒏𝟏 = 𝒏𝟐
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝒇 𝒊𝒔 𝒊𝒏𝒋𝒆𝒄𝒕𝒊𝒗𝒆,
𝑩𝒖𝒕 𝒂𝒔 𝒇(𝒏) ≠ 𝟏 𝒐𝒓 𝟐 𝒐𝒓 𝟑 𝒐𝒓 𝟒 𝒘𝒉𝒊𝒄𝒉 𝒐𝒄𝒄𝒖𝒓𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏 𝒐𝒇 𝑵,
𝒔𝒐 𝒇(𝒏) = 𝟐𝒏 + 𝟑 𝒊𝒔 𝒏𝒐𝒕 𝒔𝒖𝒓𝒋𝒆𝒄𝒕𝒊𝒗𝒆.
Q10. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan(12 sin−1 34) 𝑖𝑠:
A.
3+√7
4
B. 3−4 7
√
C. 4+3 7
√
D.
4−√7
3
Answer: D
Explanation: Let,
1
3
tan( sin−1 ) = 𝑦
2
4
1 −1 3
⇒ sin
= tan−1 𝑦
2
4
3
⇒ sin−1 = 2tan−1 𝑦
4
3
2𝑦
⇒ sin−1 = tan−1 (
)
4
1 − 𝑦2
3
2𝑦
⇒ tan−1
= tan−1 (
)
1 − 𝑦2
√7
3
2𝑦
⇒
=(
)
1 − 𝑦2
√7
⇒ 3𝑦 2 + 2√7 − 3 = 0
2
2𝑎
−1 (1−𝑎 ) = tan−1 ( 2𝑥 ) 𝑤ℎ𝑒𝑟𝑒 𝑎, 𝑥 ∈]0,1[,
𝐼𝑓 sin−1 (1+𝑎
2 ) + cos
1+𝑎2
1−𝑥2
Q11.
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑠:
A. 0
C.
2𝑎
1+𝑎2
Answer: D
Explanation:
B.
D.
𝑎
2
2𝑎
1−𝑎2
Let 𝑎 = tan𝜃
Then
2𝑎
1 − 𝑎2
2𝑥
−1
−1
−1
sin (
)
+
cos
(
)
=
tan
(
)
1 − 𝑎2
1 + 𝑎2
1 − 𝑥2
2tan𝜃
1 − tan2 𝜃
2𝑥
−1
−1
−1
⇒ sin (
)
+
cos
(
)
=
tan
(
)
1 − tan2 𝜃
1 + tan2 𝜃
1 − 𝑥2
2𝑥
⇒ sin−1 sin2𝜃 + cos −1 cos2𝜃 = tan−1 (
)
1 − 𝑥2
2𝑥
⇒ 4𝜃 = tan−1 (
)
1 − 𝑥2
2𝑥
⇒ 4tan−1 𝑎 = tan−1 (
)
1 − 𝑥2
2𝑥
⇒ 2 ⋅ 2tan−1 𝑎 = tan−1 (
)
1 − 𝑥2
2𝑎
2𝑥
−1
⇒ 2 ⋅ tan−1 (
)
=
tan
(
)
1 − 𝑎2
1 − 𝑥2
2𝑎
2⋅
1 − 𝑎2 ) = tan−1 ( 2𝑥 )
⇒ tan−1 (
2𝑎 2
1 − 𝑥2
1−(
)
2
1−𝑎
4𝑎
2𝑥
1 − 𝑎2
⇒
=
2𝑎 2 1 − 𝑥 2
1−(
)
1 − 𝑎2
2𝑎
⇒𝑥=
1 − 𝑎2
Q12. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑓𝑜𝑟 tan−1 2𝑥 + tan−1 3𝑥 = 𝜋4 𝑖𝑠:
A.
1
6
C. -1
Answer: A
Explanation:
B. −
D. 1
1
6
𝐭𝐚𝐧−𝟏 𝟐𝒙 + 𝐭𝐚𝐧−𝟏 𝟑𝒙 =
𝝅
𝟒
𝒙+𝒚
𝑼𝒔𝒊𝒏𝒈 𝒊𝒅𝒆𝒏𝒕𝒊𝒕𝒚 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (
) 𝒂𝒔,
𝟏 − 𝒙𝒚
𝝅
𝐭𝐚𝐧−𝟏 𝟐𝒙 + 𝐭𝐚𝐧−𝟏 𝟑𝒙 =
𝟒
𝟐𝒙
+
𝟑𝒙
𝝅
⇒ 𝐭𝐚𝐧−𝟏 (
)=
𝟏 − 𝟐𝒙 ⋅ 𝟑𝒙
𝟒
𝟓𝒙
𝝅
⇒ 𝐭𝐚𝐧−𝟏 (
)=
𝟏 − 𝟔𝒙𝟐
𝟒
𝟓𝒙
𝝅
⇒(
) = 𝐭𝐚𝐧
𝟏 − 𝟔𝒙𝟐
𝟒
𝟓𝒙
⇒(
)=𝟏
𝟏 − 𝟔𝒙𝟐
⇒ 𝟏 − 𝟔𝒙𝟐 = 𝟓𝒙
⇒ 𝟔𝒙𝟐 + 𝟓𝒙 − 𝟏 = 𝟎
⇒ 𝟔𝒙𝟐 + (𝟔 − 𝟏)𝒙 − 𝟏 = 𝟎
⇒ 𝟔𝒙𝟐 + 𝟔𝒙 − 𝒙 − 𝟏 = 𝟎
⇒ 𝟔𝒙(𝒙 + 𝟏) − 𝟏(𝒙 + 𝟏) = 𝟎
⇒ (𝟔𝒙 − 𝟏)(𝒙 + 𝟏) = 𝟎
𝑬𝒊𝒕𝒉𝒆𝒓
(𝟔𝒙 − 𝟏) = 𝟎
𝟏
⇒𝒙=
𝟔
𝑶𝒓
(𝒙 + 𝟏) = 𝟎
⇒ 𝒙 = −𝟏
since −1 does not lie in the domain of tanx , so we ignore this value.
Therefore
𝑥=
1
6
Q13. 𝑇ℎ𝑒 𝑠𝑖𝑚𝑝𝑙𝑒𝑠𝑡 𝑓𝑜𝑟𝑚 𝑜𝑓 tan−1 (cos𝑥−sin𝑥
) 𝑖𝑠:
cos𝑥+sin𝑥
𝜋
𝜋
A. − 𝑥
B. + 𝑥
C. 𝜋4 + 2𝑥
D. − 2𝑥
4
4
𝜋
4
Answer: A
Explanation: We have given
tan−1 (
cos𝑥 − sin𝑥
)
cos𝑥 + sin𝑥
𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑎𝑛𝑑 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑏𝑦 cos𝑥 𝑎𝑠,
cos𝑥 − sin𝑥
𝑡𝑎𝑛−1 (
)
cos𝑥 + sin𝑥
cos𝑥 − sin𝑥
cos𝑥
−1
= tan (
)
cos𝑥 + sin𝑥
cos𝑥
1
−
tan𝑥
= tan−1 (
)
1 + tan𝑥
𝜋
tan − tan𝑥
4
−1
= tan (
)
𝜋
1 + tan 4 ⋅ tan𝑥
𝜋
= tan−1 (tan( − 𝑥))
4
𝜋
= −𝑥
4
Q14. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan−1 (13) + tan−1 (15) + tan−1 (17) + tan−1 (18) 𝑖𝑠:
A.
C.
𝜋
3
𝜋
6
Answer: D
Explanation:
B.
D.
𝜋
2
𝜋
4
𝐭𝐚𝐧
−𝟏
𝟏 𝟏
𝟏 𝟏
+
+
𝟏
𝟏
𝟏
𝟏
𝟑
𝟓
−𝟏
−𝟏
−𝟏
−𝟏
−𝟏
( ) + 𝐭𝐚𝐧 ( ) + 𝐭𝐚𝐧 ( ) + 𝐭𝐚𝐧 ( ) = 𝐭𝐚𝐧 (
) + 𝐭𝐚𝐧 ( 𝟕 𝟖 )
𝟏 𝟏
𝟏 𝟏
𝟑
𝟓
𝟕
𝟖
𝟏−𝟑⋅
𝟏−𝟕⋅𝟖
𝟓
𝟖
𝟏𝟓
= 𝐭𝐚𝐧−𝟏 ( ) + 𝐭𝐚𝐧−𝟏 ( )
𝟏𝟒
𝟓𝟓
𝟒
𝟑
= 𝐭𝐚𝐧−𝟏 ( ) + 𝐭𝐚𝐧−𝟏 ( )
𝟕
𝟏𝟏
𝟒 𝟑
+
= 𝐭𝐚𝐧−𝟏 ( 𝟕 𝟏𝟏 )
𝟒 𝟑
𝟏 − 𝟕 ⋅ 𝟏𝟏
𝟒𝟒 + 𝟐𝟏
−𝟏
= 𝐭𝐚𝐧 ( 𝟕𝟕 )
𝟕𝟕 − 𝟏𝟐
𝟕𝟕
𝟒𝟒
+ 𝟐𝟏
= 𝐭𝐚𝐧−𝟏 (
)
𝟕𝟕 − 𝟏𝟐
𝟔𝟓
= 𝐭𝐚𝐧−𝟏 ( )
𝟔𝟓
−𝟏
= 𝐭𝐚𝐧 (𝟏)
𝝅
=
𝟒
Q15. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan(2tan−1 15) 𝑖𝑠:
A.
C.
7
12
9
12
Answer: B
Explanation:
B.
5
12
D. 59
𝟐𝒙
𝑼𝒔𝒊𝒏𝒈 𝒊𝒅𝒆𝒏𝒕𝒊𝒕𝒚 𝟐𝐭𝐚𝐧−𝟏 𝒙 = 𝐭𝐚𝐧−𝟏 (
) 𝒂𝒔,
𝟏 − 𝒙𝟐
𝟏
𝐭𝐚𝐧(𝟐𝐭𝐚𝐧−𝟏 )
𝟓
𝟏
𝟐×
𝟓 ))
−𝟏
= 𝐭𝐚𝐧(𝐭𝐚𝐧 (
𝟏 𝟐
𝟏−( )
𝟓
𝟐
= 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 ( 𝟓 ))
𝟏
𝟏−
𝟐𝟓
𝟐
−𝟏 𝟓
= 𝐭𝐚𝐧(𝐭𝐚𝐧 ( ))
𝟐𝟒
𝟐𝟓
𝟏
= 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 ( ))
𝟏𝟐
𝟓
𝟓
= 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 ( ))
𝟏𝟐
𝟓
=
𝟏𝟐
Q16. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 sin(sec−1 17
) 𝑖𝑠:
15
A.
C.
9
17
17
8
B.
D.
8
17
8
15
Answer: B
𝟏𝟕
𝒉
Explanation: 𝑺𝒊𝒏𝒄𝒆, 𝐬𝐞𝐜 −𝟏 𝟏𝟓 = 𝐬𝐞𝐜 −𝟏 𝒃,
So,In right angle triangle
Use Pythagoras theorem to find unknown side as,
𝑝2 + 𝑏 2 = ℎ2
⇒ 𝑝2 = ℎ2 − 𝑏 2
⇒ 𝑝2 = 172 − 152
⇒ 𝑝2 = 289 − 225
⇒ 𝑝2 = 64
⇒𝑝=8
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 , 𝑢𝑠𝑖𝑛𝑔 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 sin−1
17
)
15
8
= sin(sin−1 )
17
8
=
17
𝑝
ℎ
= sec −1
ℎ
𝑏
sin(sec −1
Q17. 𝑇ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 cos−1 (cos 7𝜋
) 𝑖𝑠:
6
A.
C.
7𝜋
B. cos
6
5𝜋
D.
6
𝜋
6
𝜋
3
Answer: C
Explanation: Since 7π/6 does not lie between 0 and π.
Therefore
cos−1 (cos
7𝜋
7𝜋
)≠
6
6
To covert 7π/6 into a value that lies between 0 and π.
Let us proceed as,
cos −1 (cos
𝜋
⇒ cos −1 (cos(𝜋 + ))
6
7𝜋
)
6
(𝑆𝑖𝑛𝑐𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓cos(𝜋 + 𝜃) = cos(𝜋 − 𝜃))
𝜋
⇒ cos −1 (cos(𝜋 − ))
6
5𝜋
⇒
6
Q18.
𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 √1 + cos2𝑥 = √2cos−1 (cos𝑥)
𝑖𝑛 [𝜋2 , 𝜋] 𝑖𝑠:
A. 0
B. 1
C. 2
D. infinity
Answer: A
Explanation: This can be solved as,
√1 + cos2𝑥 = √2cos −1 (cos𝑥)
√1 + 2cos 2 𝑥 − 1 = √2cos −1 (cos𝑥)
√2cos𝑥 = √2cos −1 (cos𝑥)
cos𝑥 = 𝑥
Which is not true for any real value of x.
Q19. 𝐼𝑓 tan−1 𝑥 + tan−1 𝑦 = 4𝜋
𝑡ℎ𝑒𝑛 cot−1 𝑥 + cot−1 𝑦 =?
5
A.
C.
𝜋
5
3𝜋
5
B.
2𝜋
5
D. π
Answer: A
Explanation: This can be solved as,
tan−1 𝑥 + tan−1 𝑦 =
4𝜋
5
𝜋
𝜋
4𝜋
− cot −1 𝑥 + − cot −1 𝑦 =
2
2
5
4𝜋
⇒ −(cot −1 𝑥 + cot −1 𝑦) =
−𝜋
5
𝜋
⇒ −(cot −1 𝑥 + cot −1 𝑦) = −
5
𝜋
⇒ cot −1 𝑥 + cot −1 𝑦 =
5
⇒
Q20. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 cot(sin−1 𝑥) 𝑖𝑠:
A.
C.
√1+𝑥 2
B.
𝑥
1
D.
𝑥
𝑥
√1+𝑥 2
√1−𝑥 2
𝑥
Answer: D
Explanation: This can be solved as,
Let sin−1 𝑥 = 𝜃
⇒ 𝑥 = sin𝜃
1
⇒ = cosec𝜃
𝑥
Since
2
1 + cot 𝜃 = cosec 2 𝜃
1
⇒ cot 2 𝜃 = 2 − 1
𝑥
√1 − 𝑥 2
⇒ cot𝜃 =
𝑥
Q21. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan−1 tan 9𝜋
𝑖𝑠:
8
A.
C.
9𝜋
8
7𝜋
8
B.
D.
𝜋
8
3𝜋
8
Answer: B
Explanation: This can be solved as,
tan−1 tan
9𝜋
8
𝜋
⇒ tan−1 tan(𝜋 + )
8
𝜋
⇒ tan−1 tan( )
8
𝜋
⇒
8
Q22. 𝑇ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 cos−1 (2𝑥 − 1) 𝑖𝑠:
A. [0,1]
B. [-1,1]
C. (-1,1)
D. (0, π)
Answer: A
Explanation: This can be solved as,
𝑓(𝑥) = cos −1 (2𝑥 − 1)
⇒ −1 ≤ 2𝑥 − 1 ≤ 1
⇒ 0 ≤ 2𝑥 ≤ 2
⇒0≤𝑥≤1
⇒ 𝑥 ∈ [0,1]
Q23. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2sec−1 2 + sin−1 (12) 𝑖𝑠:
A.
C.
𝜋
6
7𝜋
6
B.
5𝜋
6
D. 1
Answer: B
Explanation: This can be solved as,
1
2sec −1 2 + sin−1 ( )
2
𝜋
𝜋
= 2sec −1 sec( ) + sin−1 sin
3
6
𝜋
𝜋
= 2( ) +
3
6
5𝜋
=
6
1 3 2 1
Q24. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ [1 𝑥 1][ 2 5 1][2] = O 𝑖𝑠:
15 3 2 𝑥
A. x=−2, −14
B. x=2,14x
C. x=−2, 14
D. x=2, −14
Answer: A
Explanation: This can be solved as,
3 2 1
5 1][2] = O
3 2 𝑥
1
[16 + 2𝑥 5𝑥 + 6 4 + 𝑥][2] = O
𝑥
1
[1 + 2𝑥 + 15 3 + 5𝑥 + 3 2 + 𝑥 + 2][2] = O
𝑥
2
[16 + 2𝑥 + 10𝑥 + 12 + 𝑥 + 4𝑥] = O
[𝑥 2 + 16𝑥 + 28] = O
[𝑥 2 + 2𝑥 + 14𝑥 + 28] = O
(𝑥 + 2)(𝑥 + 14) = O
𝑥 = −2, −14
[1 𝑥
1
][
1 2
15
2 0 −1
Q25. 𝐼𝑓 𝐼A = [5 1 0 ], 𝑡ℎ𝑒𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑚𝑎𝑡𝑟𝑖𝑥 𝑖𝑠:
0 1 3
3 1
1
A. [5 6 −5]
5 −2 15
3 1
1
B. [3 6 −5]
5 −2 15
3
1
1
C. [15 6 −5]
5 −2 2
3
−1 1
D. [−15 6 −5]
5
−2 2
Answer: D
Explanation: We have given
2 0 −1
A = [5 1 0 ]
0 1 3
Rewrite it as,
2 0 −1
1 0 0
[5 1 0 ] = [0 1 0]A
0 1 3
0 0 1
R1
ApplyingR1 →
2
1
1
1 0 −
0 0
2
2
[
]=[
]A
0 1 0
5 1 0
0 0 1
0 1 3
ApplyingR 2 → R 2 − 5R1
1
1
1 0 −
0 0
2
2
[
]A
5 ]=[ 5
0 1
−
1 0
2
2
0 1 3
0 0 1
3
Q26. 𝐼𝑓 𝐴 𝑖𝑠 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐴2 = 𝐴, 𝑇ℎ𝑒𝑛 (𝐼 + 𝐴) =?
A. 7A+I
B. 7A−I
C. 5A+2I
D. 7A+2I
Answer: A
Explanation:
𝑺𝒊𝒏𝒄𝒆 𝐀𝟐 = 𝐀
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆,
(𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 𝟐 +𝟐𝐀𝐈 + 𝐀𝟐 )(𝐈 + 𝐀)
⇒ (𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 + 𝟐𝐀 + 𝐀)(𝐈 + 𝐀)
⇒ (𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 + 𝟑𝐀)(𝐈 + 𝐀)
⇒ (𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 𝟐 +𝟑𝐀𝐈 + 𝐀𝐈 + 𝟑𝐀𝟐 )
⇒ (𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 + 𝟒𝐀 + 𝟑𝐀)
⇒ (𝐈 + 𝐀)(𝐈 + 𝐀)(𝐈 + 𝐀) = (𝐈 + 𝟕𝐀)
0 𝑎 3
𝐼𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 [2 𝑏 −1] 𝑖𝑠 𝑎 𝑠𝑘𝑒𝑤 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥,
Q27.
𝑐 1 0
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒:
A. a = −2 ; b= 0 ; c = −3
B. a = 2 ; b= 0 ; c = 3
C. a = 0 ; b= 0 ; c = −3
D. a = −2 ; b= 0 ; c = 0]
Answer: A
Explanation:
𝟎 𝒂 𝟑
𝑳𝒆𝒕 𝐀 = [𝟐 𝒃 −𝟏]
𝒄 𝟏 𝟎
𝑺𝒊𝒏𝒄𝒆 𝑨 𝒊𝒔 𝒔𝒌𝒆𝒘 𝒔𝒚𝒎𝒎𝒆𝒕𝒓𝒊𝒄 𝒎𝒂𝒕𝒓𝒊𝒙
𝐀′ = −𝐀
𝟎 𝟐 𝒄
𝟎 𝒂 𝟑
[𝒂 𝒃 𝟏] = −[𝟐 𝒃 −𝟏]
𝟑 −𝟏 𝟎
𝒄 𝟏 𝟎
𝟎 𝟐 𝒄
𝟎 −𝒂 −𝟑
[𝒂 𝒃 𝟏] = [−𝟐 −𝒃 𝟏 ]
𝟑 −𝟏 𝟎
−𝒄 −𝟏 𝟎
By equality of matrices, we get:
a = −2 ; b = 0 ; c = −3
Q28. 𝐼𝑓 A = [1 2], 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴2 + 2𝐴 + 7𝐼 𝑖𝑠:
4 1
16 18
]
18 8
B. [
16 8
]
18 18
18 8
D. [
]
16 18
A. [
C. [
8
8
]
16 18
Answer: D
Explanation: Since
1 2 1 2
][
]
4 1 4 1
9 4
A2 = [
]
8 9
A2 + 2A + 7I
9 4
1 2
1
=[
] + 2[
] + 7[
8 9
0
4 1
9 4
2 4
7
=[
]+[
]+[
8 9
8 2
0
18 8
=[
]
16 18
A2 = [
Q29.
0
]
1
0
]
7
𝑇ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑎12 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝐴 = [𝑎𝑖𝑗 ]2×2 𝑤ℎ𝑜𝑠𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑎𝑖𝑗
𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑎𝑖𝑗 = 𝑒2𝑖𝑥 sin𝑗𝑥 𝑖𝑠:
A. exsin2x
B. e2xsin2x
C. e2xsin4x
D. e2xsinx
Answer: B
Explanation:
𝒂𝒊𝒋 = 𝒆𝟐𝒊𝒙 𝐬𝐢𝐧𝒋𝒙
𝑭𝒐𝒓 𝒂𝟏𝟐
𝒊=𝟏
𝒋=𝟐
𝒂𝒊𝒋 = 𝒆𝟐𝒊𝒙 𝐬𝐢𝐧𝒋𝒙
⇒ 𝒂𝟏𝟐 = 𝒆𝟐𝒙 𝐬𝐢𝐧𝟐𝒙
2 −3 5
Q30. 𝑇ℎ𝑒 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑎12 𝑖𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 |6 0
4 | 𝑖𝑠:
1 5 −7
A. 46
B. -46
C. 12
D. 18
Answer: A
Explanation:
𝑪𝒐𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒂𝟏𝟐 𝒊𝒔:
𝟔 𝟒
𝑨𝟏𝟐 = (−𝟏)𝟏+𝟐 |
|
𝟏 −𝟕
𝑨𝟏𝟐 = −(−𝟒𝟐 − 𝟒)
𝑨𝟏𝟐 = −(−𝟒𝟔)
𝑨𝟏𝟐 = 𝟒𝟔
Q31.If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m
= n, then the order of matrix (5A – 2B) is:
A. m × 3
B. 3 × 3
C. m × n
D. 3 × n
Answer: D
𝑰𝒇 𝑨𝟑× 𝒎 𝒂𝒏𝒅 𝑩𝟑 × 𝒏 𝒂𝒓𝒆 𝒕𝒘𝒐 𝒎𝒂𝒕𝒓𝒊𝒄𝒆𝒔
𝑰𝒇 𝒎 = 𝒏 ,
Explanation:
𝑻𝒉𝒆𝒏 𝑨 𝒂𝒏𝒅 𝑩 𝒉𝒂𝒗𝒆 𝒔𝒂𝒎𝒆 𝒐𝒓𝒅𝒆𝒓 𝒂𝒔 𝟑 × 𝒏 𝒆𝒂𝒄𝒉.
𝑺𝒐 𝒕𝒉𝒆 𝒐𝒓𝒅𝒆𝒓 𝒐𝒇 (𝟓𝑨 – 𝟐𝑩) 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒔𝒂𝒎𝒆 𝒂𝒔 𝟑 × 𝒏
Q32.Total number of possible matrices of order 3 × 3 with each entry 2 or 0, is:
A. 9
B. 27
C. 81
D. 512
Answer: D
Explanation: 𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒎𝒂𝒕𝒓𝒊𝒄𝒆𝒔 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑 𝒘𝒊𝒕𝒉 𝒆𝒂𝒄𝒉 𝒆𝒏𝒕𝒓𝒚 𝟐 𝒐𝒓 𝟎 𝒊𝒔 𝟐𝟗 =
𝟓𝟏𝟐
Q33.If A and B are symmetric matrices of the same order, then (AB′ – BA′) is a:
A. Skew symmetric matrix
B. Null matrix
C. Symmetric matrix
D. None of these
Answer: A
Explanation: Since
(AB′ –BA′)′ = (AB′)′ – (BA′)′
1 5
Q34. 𝐼𝑓 A = [1 3 5] 𝑎𝑛𝑑 B = [3 2], 𝑡ℎ𝑒𝑛
2 7 9
0 6
A. Only AB is defined
B. Only BA is defined
C. AB and BA both are defined
D. AB and BA both are not defined.
Answer: C
Explanation: 𝑯𝒆𝒓𝒆 𝑨 = [𝒂𝒊𝒋 ]𝟐×𝟑 𝒂𝒏𝒅 𝑩 = [𝒃𝒊𝒋 ]𝟑×𝟐
To multiply two marices, the necessary condition is the number of columns in first marix will be equal to
number of rows in second matrix. Since both AB and BA are fulfilling this condition., so both AB and BA
are defined.
Q35.If A and B are square matrices of the same order, then (A + B) (A – B) is equal
to:
A. A2− B2
B. A2− BA− AB−B2
C. A2− B2+BA−AB
D. A2−BA+B2+AB
Answer: C
Explanation: This can be solved as,
(A + B) (A – B) = A (A – B) + B (A – B)
= A2− B2+BA−AB
1
Q36.Using properties of determinants, calculate the value of: |𝑥 2
𝑥
A. (1−x3)2
B. (1−x3)
C. (1−x3)3
D. (1−x2)3
Answer: A
Explanation:
𝑥
1
𝑥2
𝑥2
𝑥|
1
1 𝑥 𝑥2
Given, |𝑥 2 1 𝑥 |
𝑥 𝑥2 1
𝑅1 → 𝑅1 + 𝑅2 + 𝑅3
1 + 𝑥 + 𝑥2 1 + 𝑥 + 𝑥2 1 + 𝑥 + 𝑥2
=|
|
𝑥2
1
𝑥
2
𝑥
𝑥
1
1 1 1
= (1 + 𝑥 + 𝑥 2 )|𝑥 2 1 𝑥 |
𝑥 𝑥2 1
𝐶2 → 𝐶2 − 𝐶1
𝐶3 → 𝐶3 − 𝐶1
1
0
0
= (1 + 𝑥 + 𝑥 2 )|𝑥 2 1 − 𝑥 2 𝑥 − 𝑥 2 |
𝑥 𝑥2 − 𝑥 1 − 𝑥
1
0
0
2 𝑥 2 (1 − 𝑥)(1 + 𝑥) 𝑥(1 − 𝑥)
= (1 + 𝑥 + 𝑥 )|
|
𝑥
−𝑥(1 − 𝑥)
1−𝑥
1
0
0
= (1 − 𝑥)2 (1 + 𝑥 + 𝑥 2 )|𝑥 2 (1 + 𝑥) 𝑥 |
𝑥
−𝑥
1
= (1 − 𝑥)2 (1 + 𝑥 + 𝑥 2 )[1 + 𝑥 + 𝑥 2 ]
= (1 − 𝑥)2 (1 + 𝑥 + 𝑥 2 )2
= (1 − 𝑥 3 )2
𝑥
2
Q37.Using properties of determinants, calculate the value of: |𝑥
𝑥3
A. xyz(x−y)(y+z)(z−x)
B. xyz(x+y)(y−z)(z−x)
C. xyz(x−y)(y−z)(z−x)
D. 0
Answer: C
𝑦
𝑦2
𝑦3
𝑧
𝑧 2|
𝑧3
𝒙
𝟐
= |𝒙
𝒙𝟑
𝒚
𝒚𝟐
𝒚𝟑
𝒛
𝒛𝟐 |
𝒛𝟑
𝟏 𝟏 𝟏
= 𝒙𝒚𝒛| 𝒙 𝒚 𝒛 |
𝒙𝟐 𝒚𝟐 𝒛𝟐
𝑪𝟐 → 𝑪𝟐 − 𝑪𝟏
𝑪𝟑 → 𝑪𝟑 − 𝑪𝟏
Explanation:
𝟏
𝟎
𝟎
𝒙
𝒚
−
𝒙
𝒛
−
𝒙 |
= 𝒙𝒚𝒛|
𝟐
𝟐
𝟐
𝟐
𝒙 𝒚 − 𝒙 𝒛 − 𝒙𝟐
𝟏
𝟎
𝟎
𝟏
𝟏 |
= 𝒙𝒚𝒛(𝒚 − 𝒙)(𝒛 − 𝒙)| 𝒙
𝒙𝟐 𝒚 + 𝒙 𝒛 + 𝒙
= 𝒙𝒚𝒛(𝒚 − 𝒙)(𝒛 − 𝒙)(𝒛 + 𝒙 − 𝒚 − 𝒙)
= 𝒙𝒚𝒛(𝒚 − 𝒙)(𝒛 − 𝒙)(𝒛 − 𝒚)
= 𝒙𝒚𝒛(𝒙 − 𝒚)(𝒚 − 𝒛)(𝒛 − 𝒙)
Q38.Using properties of determinants, calculate the value
𝑦+𝑘
of: | 𝑦
𝑦
𝑦
𝑦+𝑘
𝑦
𝑦
𝑦 |
𝑦+𝑘
A. k(3y+k)
B. k2(y+k)
C. k2(3y−k)
D. k2(3y+k)
Answer: D
𝒚+𝒌
𝒚
𝒚
𝒚+𝒌
𝒚 |
=| 𝒚
𝒚
𝒚
𝒚+𝒌
𝑹𝟏 → 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑
𝟑𝒚 + 𝒌 𝟑𝒚 + 𝒌 𝟑𝒚 + 𝒌
𝒚+𝒌
𝒚 |
=| 𝒚
𝒚
𝒚
𝒚+𝒌
𝟏
𝟏
𝟏
𝒚 |
Explanation: = (𝟑𝒚 + 𝒌)|𝒚 𝒚 + 𝒌
𝒚
𝒚
𝒚+𝒌
𝑪𝟐 → 𝑪𝟐 − 𝑪𝟏
𝑪𝟑 → 𝑪𝟑 − 𝑪𝟏
𝟏 𝟎 𝟎
= (𝟑𝒚 + 𝒌)|𝒚 𝒌 𝟎|
𝒚 𝟎 𝒌
= (𝟑𝒚 + 𝒌)(𝒌𝟐 − 𝟎)
= 𝒌𝟐 (𝟑𝒚 + 𝒌)
Q39.Using properties of determinants, calculate the value of:
𝑏+𝑐
| 𝑏
𝑐
𝑎
𝑐+𝑎
𝑐
𝑎
𝑏 |
𝑎+𝑏
A. 2abc
B. 4abc
C. 3abc
D. 5abc
Answer: B
𝒃+𝒄
𝒂
𝒂
| 𝒃
𝒄+𝒂
𝒃 |
𝒄
𝒄
𝒂+𝒃
𝑹𝟏 → 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑
𝟐(𝒃 + 𝒄) 𝟐(𝒂 + 𝒄) 𝟐(𝒂 + 𝒃)
= |
|
𝒃
𝒄+𝒂
𝒃
𝒄
𝒄
𝒂+𝒃
𝐓𝐚𝐤𝐢𝐧𝐠 𝟐 𝐚𝐬 𝐜𝐨𝐦𝐦𝐨𝐧
𝒃+𝒄 𝒂+𝒄 𝒂+𝒃
Explanation:
= 𝟐| 𝒃
𝒄+𝒂
𝒃 |
𝒄
𝒄
𝒂+𝒃
𝑹𝟏 → 𝑹𝟏 − 𝑹𝟐
𝒄
𝟎
𝒂
= 𝟐 |𝒃 𝒄 + 𝒂
𝒃 |
𝒄
𝒄
𝒂+𝒃
= 𝟐[𝒄{(𝒃 + 𝒂)(𝒄 + 𝒂) − 𝒃𝒄} − 𝟎 + 𝒂{𝒃𝒄 − 𝒄(𝒄 + 𝒂)}]
= 𝟐(𝒂𝒃𝒄 + 𝒂𝒃𝒄)
= 𝟒𝒂𝒃𝒄
Q40.Calculate the area between the given points:
A (a,b+c),B (b,c+a),C (c,a+b)A (a,b+c),B (b,c+a),C (c,a+b)
A. 0
B. 1
C. 2
D. 3
Answer: A
Explanation:
Here, coordinates of points, 𝐴: (𝑎, 𝑏 + 𝑐); 𝐵: (𝑏, 𝑐 + 𝑎); 𝐶: (𝑐, 𝑎 + 𝑏)
𝐴𝑟𝑒𝑎 𝑜𝑓 𝛥𝐴𝐵𝐶 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑠
1 𝑎 𝑏+𝑐 1
𝛥 = |𝑏 𝑐 + 𝑎 1|
2
𝑐 𝑎+𝑏 1
𝑅2 → 𝑅2 − 𝑅1
𝑅3 → 𝑅3 − 𝑅1
𝑏+𝑐 1
1 𝑎
= |𝑏 − 𝑎 𝑎 − 𝑏 0|
2
𝑐−𝑎 𝑎−𝑐 0
𝑎 𝑏+𝑐 1
1
= (𝑎 − 𝑏)(𝑎 − 𝑐)|−1
1
0|
2
−1
1
0
1
= (𝑎 − 𝑏)(𝑎 − 𝑐)[−1 + 1]
2
=0
Hence, the points A, B, and C are collinear. "
Q41.
𝑏2 − 𝑎𝑏
𝑇ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 |𝑎𝑏 − 𝑎2
𝑏𝑐 − 𝑎𝑐
A. abc (b – c) (c – a) (a – b)
B. (b – c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. 0
Answer: D
Explanation: This can be solved as,
𝑏−𝑐
𝑎−𝑏
𝑐−𝑎
𝑏𝑐 − 𝑎𝑐
𝑏2 − 𝑎𝑏| 𝑒𝑞𝑢𝑎𝑙𝑠 𝑡𝑜:
𝑎𝑏 − 𝑎2
𝑏 2 − 𝑎𝑏 𝑏 − 𝑐 𝑏𝑐 − 𝑎𝑐
|𝑎𝑏 − 𝑎2 𝑎 − 𝑏 𝑏 2 − 𝑎𝑏|
𝑏𝑐 − 𝑎𝑐 𝑐 − 𝑎 𝑎𝑏 − 𝑎2
𝑏(𝑏 − 𝑎) 𝑏 − 𝑐 𝑐(𝑏 − 𝑎)
= |𝑎(𝑏 − 𝑎) 𝑎 − 𝑏 𝑏(𝑏 − 𝑎)|
𝑐(𝑏 − 𝑎) 𝑐 − 𝑎 𝑎(𝑏 − 𝑎)
Taking out (𝑏 − 𝑎) common from C1 and C3 , we get:
𝑏 𝑏−𝑐 𝑐
2
= (𝑏 − 𝑎) |𝑎 𝑎 − 𝑏 𝑏 |
𝑐 𝑐−𝑎 𝑎
C1 → C1 − C3
𝑏−𝑐 𝑏−𝑐 𝑐
= (𝑏 − 𝑎)2 |𝑎 − 𝑏 𝑎 − 𝑏 𝑏 |
𝑐−𝑎 𝑐−𝑎 𝑎
= 0 (∵ C1 and C2 are identical)
1+𝑥
1
1
𝐼𝑓 𝑥, 𝑦, 𝑧 𝑎𝑟𝑒 𝑎𝑙𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑧𝑒𝑟𝑜 𝑎𝑛𝑑 | 1
1+𝑦
1 | = 0,
Q42.
1
1
1+𝑧
𝑡ℎ𝑒𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥–1 + 𝑦–1 + 𝑧–1 𝑖𝑠
A. x y z
B. x–1y–1z–1
C. –x–y–z
D. –1
Answer: D
Explanation: This can be solved as,
1+𝑥
1
1
| 1
1+𝑦
1 |=0
1
1
1+𝑧
Applying C1 → C1 − C3 and C2 → C2 − C3
𝑥
0
1
|0
𝑦
1 |=0
−𝑧 −𝑧 1 + 𝑧
Expanding along R1
𝑥[𝑦(1 + 𝑧) + 𝑧] − 0 + 1(𝑦𝑧) = 0
𝑥(𝑦 + 𝑦𝑧 + 𝑧) + 𝑦𝑧 = 0
𝑥𝑦 + 𝑥𝑦𝑧 + 𝑥𝑧 + 𝑦𝑧 = 0
𝑥𝑦 + 𝑥𝑦𝑧 + 𝑥𝑦 + 𝑦𝑧
=0
𝑥𝑦𝑧
1 1 1
+ + = −1
𝑥 𝑦 𝑧
𝑥 −1 + 𝑦 −1 + 𝑧 −1 = −1
0
𝑥−𝑎 𝑥−𝑏
Q43. 𝐼𝑓 𝑓(𝑥) = |𝑥 + 𝑎
0
𝑥 − 𝑐 |, 𝑡ℎ𝑒𝑛
𝑥+𝑏 𝑥+𝑐
0
A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0
Answer: C
𝟎
𝒙−𝒂 𝒙−𝒃
𝐆𝐢𝐯𝐞𝐧, 𝒇(𝒙) = |𝒙 + 𝒂
𝟎
𝒙 − 𝒄|
𝒙+𝒃 𝒙+𝒄
𝟎
𝟎
𝟎−𝒂 𝟎−𝒃
⇒ 𝒇(𝟎) = |𝟎 + 𝒂
𝟎
𝟎 − 𝒄|
Explanation:
𝟎+𝒃 𝟎+𝒄
𝟎
𝟎 −𝒂 −𝒃
⇒ 𝒇(𝟎) = |𝒂 𝟎 −𝒄 |
𝒃 𝒄
𝟎
⇒ 𝒇(𝟎) = 𝒂𝒃𝒄 − 𝒂𝒃𝒄 = 𝟎.
Q44. 𝑇ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝛥 = |
1
1
1
1
1 + sin𝜃 1| 𝑖𝑠 (𝜃 𝑖𝑠 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟):
1 + cos𝜃
1
1
1
3
A. 2
B. √2
C. √2
D. 2√4
3
Answer: A
𝟏
𝟏
𝐆𝐢𝐯𝐞𝐧, 𝜟 = |
𝟏
𝟏 + 𝐬𝐢𝐧𝜽
𝟏 + 𝐜𝐨𝐬𝜽
𝟏
[𝐂𝟏 → 𝐂𝟏 − 𝐂𝟑 𝐚𝐧𝐝 𝐂𝟐 → 𝐂𝟐 − 𝐂𝟑
𝟎
𝟎
𝟏
⇒| 𝟎
𝐬𝐢𝐧𝜽 𝟏|
𝐜𝐨𝐬𝜽
𝟎
𝟏
= −𝐬𝐢𝐧𝜽𝐜𝐨𝐬𝜽
−𝐬𝐢𝐧𝟐𝜽
= 𝟐
Explanation:
𝟏
𝟏|
𝟏
]
𝐍𝐨𝐰, 𝐬𝐢𝐧𝟐𝜽 𝐥𝐢𝐞𝐬 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 − 𝟏 𝐚𝐧𝐝 𝟏, 𝐡𝐞𝐧𝐜𝐞 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐚𝐥𝐮𝐞 =
𝟏
𝟐
Q45.The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The
value of k will be
A. 9
B. 3
C. -9
D. 6
Answer: B
Explanation: This can be solved as,
Area of triangle =
⇒9 =
1 −3 0
|3 0
2
0 𝑘
1
1|
1
1
(−3(0 − 𝑘) − 0 + 1(3𝑘 − 0))
2
⇒ 18 = 3𝑘 + 3𝑘
⇒𝑘=3
Q46.If A is a matrix of order 3 × 3, then |3A| =
A. 27|A|
B. 3|A|
C. 9|A|
D. |27A|
Answer: A
𝑭𝒐𝒓 𝒂 𝒔𝒒𝒖𝒂𝒓𝒆 𝒎𝒂𝒕𝒓𝒊𝒙 𝑨 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓𝒏, |𝒌𝑨| = 𝒌𝒏 |𝑨|
𝑰𝒇 𝑨 𝒊𝒔 𝒂 𝒎𝒂𝒕𝒓𝒊𝒙 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑
Explanation:
𝑻𝒉𝒆𝒏, |𝟑𝑨| = 𝟑𝟑 |𝑨| = 𝟐𝟕 | 𝑨|.
Q47. 𝐼𝑓 |2𝑥 5| = |6 −2|, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑠:
7 3
8 𝑥
A. +3
B. ± 3
C. ± 6
D. +6
Answer: C
Explanation: This can be solved as,
6 −2
2𝑥 5
|=|
|
7 3
8 𝑥
2𝑥 2 − 40 = 18 + 14
2𝑥 2 = 18 + 14 + 40
2𝑥 2 = 72
𝑥 2 = 36
𝑥 = ±6
|
Q48. 𝐼𝑓 𝑥m 𝑦n = (𝑥 + 𝑦)
A.
C.
d𝑦
d𝑥
d𝑦
d𝑥
=
=
−𝑦
𝑥
𝑥
𝑦
Answer: B
B.
d𝑦
d𝑥
m+n
=
, 𝑡ℎ𝑒𝑛 𝑤ℎ𝑎𝑡 𝑖𝑠
𝑦
𝑥
D. None of these
d𝑦
?
d𝑥
𝒙𝒎 𝒚𝒏 = (𝒙 + 𝒚)𝒎+𝒏
𝐓𝐚𝐤𝐢𝐧𝐠 𝐥𝐨𝐠 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬
𝒎𝐥𝐨𝐠𝒙 + 𝒏𝐥𝐨𝐠𝒚 = (𝒎 + 𝒏)𝐥𝐨𝐠(𝒙 + 𝒚)
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧
𝒎
𝒏 𝒅𝒚
𝟏
𝒅𝒚
+ 𝒚 . 𝒅𝒙 = (𝒎 + 𝒏)[𝒙+𝒚 . (𝟏 + 𝒅𝒙)]
𝒙
𝒎
𝒏 𝒅𝒚
Explanation:
𝒏
(𝒚 −
(
𝒎+𝒏
+ 𝒚 . 𝒅𝒙 =
𝒙
)
=
)
=
𝒙+𝒚 𝒅𝒙
𝒏𝒙+𝒏𝒚−𝒎𝒚−𝒏𝒚 𝒅𝒚
𝒚(𝒙+𝒚)
𝒅𝒙
𝒏𝒙−𝒎𝒚 𝒅𝒚
(
+
𝒎+𝒏 𝒅𝒚
.
𝒙+𝒚
𝒙+𝒚 𝒅𝒙
𝒎+𝒏 𝒅𝒚
𝒎+𝒏
𝒎
) 𝒅𝒙 =
𝒚
𝒅𝒚
𝒅𝒙
−
𝒙+𝒚
𝒙
𝒎𝒙+𝒏𝒙−𝒎𝒙−𝒎𝒚
𝒙(𝒙+𝒚)
𝒏𝒙−𝒎𝒚
𝒙
𝒚
=𝒙
Q49. 𝐼𝑓 𝑥 = 𝑎(cos𝑡 + 𝑡sin𝑡), 𝑦 = 𝑎(sin𝑡 − 𝑡cos𝑡), 𝑡ℎ𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒
A.
C.
d2 𝑦
d𝑡 2
d2 𝑦
d𝑡 2
=
=
cos𝑡+sin𝑡
B.
cos𝑡−𝑡sin𝑡
𝑡cos𝑡+sin𝑡
D.
cos𝑡−sin𝑡
d2 𝑦
d𝑡 2
d2 𝑦
d𝑡 2
=
=
𝑡cos𝑡+sin𝑡
cos𝑡−𝑡sin𝑡
𝑡cos𝑡+𝑡sin𝑡
cos𝑡−𝑡sin𝑡
Answer: B
𝒙 = 𝒂(𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕), 𝒚 = 𝒂(𝐬𝐢𝐧𝒕 − 𝒕𝐜𝐨𝐬𝒕)
𝒙 = 𝒂(𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕)
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧
𝒅𝒙
= 𝒂[−𝐬𝐢𝐧𝒕 + 𝒕𝐜𝐨𝐬𝒕 + 𝐬𝐢𝐧𝒕]
𝒅𝒕
𝒅𝒙
𝒅𝒕
𝒅𝟐 𝒙
𝒅𝒕𝟐
Explanation:
= 𝒂(𝒕𝐜𝐨𝐬𝒕)
= 𝒂[𝐜𝐨𝐬𝒕 − 𝒕𝐬𝐢𝐧𝒕] . . . (𝟏)
𝒚 = 𝒂(𝐬𝐢𝐧𝒕 − 𝒕𝐜𝐨𝐬𝒕)
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧
𝒅𝒚
= 𝒂[𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕 − 𝐜𝐨𝐬𝒕]
𝒅𝒕
𝒅𝒚
𝒅𝒕
𝒅𝟐 𝒚
𝒅𝒕𝟐
= 𝒂(𝒕𝐬𝐢𝐧𝒕)
= 𝒂[𝒕𝐜𝐨𝐬𝒕 + 𝐬𝐢𝐧𝒕]
. . . . (𝟐)
𝐎𝐧 𝐝𝐢𝐯𝐢𝐝𝐢𝐧𝐠 . . . (𝟐) 𝐚𝐧𝐝 . . . (𝟏)
𝒅𝟐 𝒚
𝒅𝒕𝟐
𝒕𝐜𝐨𝐬𝒕+𝐬𝐢𝐧𝒕
= 𝐜𝐨𝐬𝒕−𝒕𝐬𝐢𝐧𝒕
2
d 𝑦
d𝑡2
Q50. 𝐼𝑓 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥 = 0, 𝑡ℎ𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒
A.
C.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= −(
= −(
1+𝑦
)
B.
)
D.
1+𝑥
1−𝑦
1+𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=(
1+𝑦
1+𝑥
=(
𝑑𝑦
𝑑𝑥
)
1+𝑦
1−𝑥
)
Answer: A
𝒙√𝟏 + 𝒚 + 𝒚√𝟏 + 𝒙 = 𝟎
𝒙√𝟏 + 𝒚 = −𝒚√𝟏 + 𝒙
𝐎𝐧 𝐬𝐪𝐮𝐚𝐫𝐢𝐧𝐠 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬
𝒙𝟐 (𝟏 + 𝒚) = 𝒚𝟐 (𝟏 + 𝒙)
𝒙𝟐 + 𝒙𝟐 𝒚 = 𝒚𝟐 + 𝒚𝟐 𝒙
𝒙𝟐 − 𝒚𝟐 = 𝒚𝟐 𝒙 − 𝒙𝟐 𝒚
Explanation:
(𝒙 − 𝒚)(𝒙 + 𝒚) = −𝒙𝒚(𝒙 − 𝒚)
𝒙 + 𝒚 + 𝒙𝒚 = 𝟎
𝒙 = −𝒚(𝟏 + 𝒙)
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧
𝒅𝒚
𝟏 = −(𝟏 + 𝒙) 𝒅𝒙 − 𝒚
−1
Q51. 𝐼𝑓 𝑥 = √𝑎sin
A.
C.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
𝑦
𝑥
2𝑦
𝑥
Answer: D
𝒅𝒚
𝒅𝒙
= −(𝟏+𝒙)
𝑡,
𝑦 = √𝑎cos
B.
D.
𝟏+𝒚
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
−𝑥
𝑦
−𝑦
𝑥
−1 𝑡
, 𝑡ℎ𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒
𝑑𝑦
𝑑𝑥
−𝟏 𝒕
𝒙 = √𝒂𝐬𝐢𝐧
𝒚 = √𝒂𝐜𝐨𝐬
,
−𝟏 𝒕
−𝟏
𝐓𝐚𝐤𝐢𝐧𝐠, 𝒙 = √𝒂𝐬𝐢𝐧 𝒕
𝐎𝐧 𝐬𝐪𝐮𝐚𝐫𝐢𝐧𝐠 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬
−𝟏
𝒙𝟐 = 𝒂𝐬𝐢𝐧 𝒕
𝐓𝐚𝐤𝐢𝐧𝐠 𝐥𝐨𝐠 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬
𝟐𝐥𝐨𝐠𝒙 = 𝐬𝐢𝐧−𝟏 𝒕. 𝐥𝐨𝐠𝒂
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧 𝐰𝐢𝐭𝐡 𝐫𝐞𝐬𝐩𝐞𝐜𝐭 𝐭𝐨 𝒕
𝟐 𝒅𝒙
𝟏
. =
. 𝐥𝐨𝐠𝒂
𝒙 𝒅𝒕
√𝟏−𝒕𝟐
𝒅𝒙
𝒙.𝐥𝐨𝐠𝒂
=
. . . . (𝟏)
𝒅𝒕
𝟐√𝟏−𝒕𝟐
Explanation:
𝐒𝐢𝐦𝐢𝐥𝐚𝐫𝐥𝐲,
𝒚 = √𝒂𝐜𝐨𝐬
𝒅𝒚
𝒅𝒕
=
−𝟏 𝒕
−𝒚.𝐥𝐨𝐠𝒂
. . . . (𝟐)
𝟐√𝟏−𝒕𝟐
𝐃𝐢𝐯𝐢𝐝𝐢𝐧𝐠 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 . . . . (𝟏) 𝐚𝐧𝐝 . . . . (𝟐), 𝐰𝐞 𝐡𝐚𝐯𝐞
𝒅𝒚
𝒅𝒙
=
−𝒚
𝒙
Q52. 𝐼𝑓 𝑓(2) = 4 𝑎𝑛𝑑 𝑓’(2) = 1, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 lim 𝑥𝑓(2)−2𝑓(𝑥)
𝑖𝑠
𝑥−2
𝑥→2
A. 0
B. 2
C. 4
D. 8
Answer: B
𝑯𝒆𝒓𝒆, 𝐥𝐢𝐦
= 𝐥𝐢𝐦
𝒙−𝟐
𝒙→𝟐
𝒙𝒇(𝟐)−𝟐𝒇(𝒙)−𝟐𝒇(𝟐)+𝟐𝒇(𝟐)
𝒙−𝟐
(𝒙−𝟐)𝒇(𝟐)−𝟐[𝒇(𝒙)−𝒇(𝟐)]
𝒙→𝟐
= 𝐥𝐢𝐦
Explanation:
𝒙→𝟐
(𝒙−𝟐)𝒇(𝟐)
= 𝐥𝐢𝐦
𝒙→𝟐
𝒙−𝟐
= 𝐥𝐢𝐦𝒇(𝟐) − 𝟐𝐥𝐢𝐦𝒇′(𝟐)
𝒙→𝟐
𝒙𝒇(𝟐)−𝟐𝒇(𝒙)
𝒙→𝟐
𝒙−𝟐
− 𝟐𝐥𝐢𝐦
𝒙→𝟐
𝒇(𝒙)−𝒇(𝟐)
𝒙−𝟐
[𝒇′(𝟐) = 𝐥𝐢𝐦
𝒙→𝟐
𝒇(𝒙)−𝒇(𝟐)
]
𝒙−𝟐
= 𝟒 − (𝟐 × 𝟏) = 𝟐.
Q53. If 𝑥 = 𝑎 cos3 θ and𝑦 = 𝑎 sin3 θ, then the value of
2
d 𝑦
d𝑥2
at θ = π6 is
A.
C.
31
27a
32
27a
32a
B.
27
D.
32
5a
Answer: C
𝒙 = 𝒂𝐜𝐨𝐬𝟑 𝜽
𝒅𝒙
= 𝟑𝒂𝐜𝐨𝐬𝟐 𝜽(−𝐬𝐢𝐧𝜽)
𝒅𝜽
𝒅𝒙
𝒅𝜽
= −𝟑𝒂𝐜𝐨𝐬𝟐 𝜽𝐬𝐢𝐧𝜽
𝒚 = 𝒂𝐬𝐢𝐧𝟑 𝜽
𝒅𝒚
= 𝟑𝒂𝐬𝐢𝐧𝟐 𝜽𝐜𝐨𝐬𝜽
𝒅𝜽
𝒅𝒚
=
𝒅𝒙
𝒅𝒚
𝒅𝜽
𝒅𝒙
𝒅𝜽
𝒅𝒚
Explanation:
𝒅𝒙
𝒅𝟐 𝒚
𝒅𝒙𝟐
𝒅𝟐 𝒚
𝒅𝒙𝟐
𝒅𝟐 𝒚
𝒅𝒙𝟐
𝟑𝒂𝐬𝐢𝐧𝟐 𝜽𝐜𝐨𝐬𝜽
= −𝟑𝒂𝐜𝐨𝐬𝟐 𝜽𝐬𝐢𝐧𝜽
= −𝐭𝐚𝐧𝜽
𝒅𝜽
= −𝐬𝐞𝐜 𝟐 𝜽. 𝒅𝒙
𝐬𝐞𝐜 𝟐 𝜽
= 𝟑𝒂𝐜𝐨𝐬𝟐 𝜽𝐬𝐢𝐧𝜽
𝟏
= 𝟑𝒂𝐜𝐨𝐬𝟒 𝜽𝐬𝐢𝐧𝜽
𝒅𝟐 𝒚
(𝒅𝒙𝟐 )𝜽 = 𝝅 =
𝟔
𝒅𝟐 𝒚
𝒅𝒙𝟐
=
𝒅𝟐 𝒚
𝒅𝒙𝟐
𝟏
𝝅
𝟔
𝟑𝒂𝐬𝐢𝐧 𝐜𝐨𝐬 𝟒
𝝅
𝟔
𝟏
𝟏 √𝟑
𝟑𝒂( )( )𝟒
𝟐 𝟐
𝟑𝟐
= 𝟐𝟕𝒂
Q54.If f and g are two continuous functions on their common domain D, then
Choose the incorrect or incomplete options from the statements given below.
A. f + g is a continuous on D
B. f ‒ g is a continuous on D
C. f × g is a continuous on D
D. f / g is a continuous on D
Answer: D
Explanation: If f and g are two continuous functions on their common domain D, then
f + g is a continuous on D
f ‒ g is a continuous on D
f × g is a continuous on D
f / g is continuous on D ‒ {x: g (x) ≠ 0}.
Q55.Which of the following statements is false?
A. |x| is continuous at x = 0
B. |x| is differentiable at x = 0
C. |x| is not continuous but not differentiable at x = 0.
D. |x| is continuous at x = 1 and ‒1
Answer: C
Explanation: The graph of |x| is shown below
the graph is continuous at x = 0. But, the graph has a kink at x = 0.
Therefore, the function f (x) = |x| is not continuous but not differentiable at x = 0.
𝐼𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓(𝑥) = 𝑥3 + 𝑏𝑥2 + 𝑎𝑥 𝑜𝑛 [1, 3], 𝑅𝑜𝑙𝑙𝑒’𝑠
Q56.
𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ℎ𝑜𝑙𝑑𝑠 𝑤𝑖𝑡ℎ 𝑐 = 2 + √1 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒:
3
A. a = 11, b = − 6
B. a = −11, b = 6
C. a = −11, b = − 6
D. a =11, b = 6
Answer: A
𝑮𝒊𝒗𝒆𝒏, 𝒇(𝒙) = 𝒙𝟑 + 𝒃𝒙𝟐 + 𝒂𝒙 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 [𝟏, 𝟑],
⇒ 𝒇′ (𝒙) = 𝟑𝒙𝟐 + 𝟐𝒃𝒙 + 𝒂
⇒ 𝒇′ (𝒄) = 𝟑𝒄𝟐 + 𝟐𝒃𝒄 + 𝒂
𝑵𝒐𝒘, 𝑹𝒐𝒍𝒍𝒆’𝒔 𝒕𝒉𝒆𝒐𝒓𝒆𝒎 𝒉𝒐𝒍𝒅 𝒇𝒐𝒓 𝒇(𝒙) 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 [𝟏, 𝟑] 𝒘𝒊𝒕𝒉 𝒄 = 𝟐 +
𝟏
√𝟑
𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞, 𝒇(𝟏) = 𝒇(𝟑) = 𝒇′(𝐜) = 𝟎
𝒇′ (𝒄) = 𝟎
⇒ 𝟑𝒄𝟐 + 𝟐𝒃𝒄 + 𝒂 = 𝟎 . . . (𝒊)
𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒏𝒈 𝒄 = 𝟐 +
Explanation:
𝟑(𝟐 +
𝟏 𝟐
)
√𝟑
𝟏
√𝟑
𝐢𝐧 . . . (𝒊), 𝐰𝐞 𝐡𝐚𝐯𝐞
+ 𝟐𝒃(𝟐 +
⇒ 𝒂 + 𝟒𝒃 + 𝟏𝟑 +
𝟐
√𝟑
𝟏
)+𝒂 =𝟎
√𝟑
(𝒃 + 𝟔) = 𝟎. . . . (𝒊)
𝑨𝒍𝒔𝒐,
𝒇(𝟏) = 𝒇(𝟑)
⇒ 𝟏 + 𝒃 + 𝒂 = 𝟐𝟕 + 𝟗𝒃 + 𝟑𝒂
⇒ 𝒂 + 𝟒𝒃 + 𝟏𝟑 = 𝟎 . . . (𝒊𝒊)
𝑺𝒐𝒍𝒗𝒊𝒏𝒈 𝒃𝒐𝒕𝒉 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 … (𝒊) 𝒂𝒏𝒅 … (𝒊𝒊), 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒂 = 𝟏𝟏, 𝒃 = −𝟔.
Q57.Find a real number c between (1, 2) such that Lagrange’s mean value
theorem holds true for the function f (x) = x (x – 2) on the interval [1, 2].
A. 3/2
B. 2/3
C. ½
D. 5/4
Answer: A
Explanation:
𝑺𝒊𝒏𝒄𝒆𝒇(𝒙) = 𝒙(𝒙– 𝟐), 𝒙 ∈ [𝟏, 𝟐] 𝒉𝒐𝒍𝒅𝒔𝑳𝒂𝒈𝒓𝒂𝒏𝒈𝒆’𝒔𝒎𝒆𝒂𝒏 𝒗𝒂𝒍𝒖𝒆 𝒕𝒉𝒆𝒐𝒓𝒆𝒎, 𝒕𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆
𝒇′ (𝒄) =
𝒇(𝒃)−𝒇(𝒂)
. . . . (𝒊)
𝒃−𝒂
𝒇(𝒄) = 𝒄(𝒄 − 𝟐) ⇒ 𝒇′(𝒄) = 𝟐𝒄 − 𝟐
𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 . . . (𝒊) 𝐰𝐢𝐥𝐥 𝐛𝐞 𝐫𝐞𝐝𝐮𝐜𝐞𝐝 𝐚𝐬,
𝒇(𝟐)−𝒇(𝟏)
𝟐−𝟏
𝟎−𝟏(−𝟏)
= 𝟐−𝟏
⇒ 𝟐𝒄 − 𝟐 =
⇒ 𝟐𝒄 − 𝟐
⇒ 𝟐𝒄 − 𝟐 = 𝟏
⇒ 𝟐𝒄 = 𝟑
𝟑
⇒ 𝒄 = 𝟐.
Q58.If 𝑓(𝑥) = 2𝑥 𝑎𝑛𝑑 𝑔(𝑥) =
𝑥2
2
+ 1,
then which of the following can be the
discontinuous function?
A. f(x)+g(x)
B. f(x)–g(x)
C. f(x).g(x)
D. f(x)/g(x)
Answer: D
Explanation: If f and g are continuous functions, then
(a) f + g is continuous
(b) f − g is continuous
(c) fg is continuous
(d) f /g is continuous at these points where g(x) ≠ 0
Here,
𝑓(𝑥)
2𝑥
4𝑥
= 2
= 2
𝑔(𝑥) 𝑥
𝑥 −2
2 −1
𝑓(𝑥)/𝑔(𝑥)𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡,
𝑥2 − 2 = 0
⇒ 𝑥2 = 2
⇒ 𝑥 = ±√2.
Q59.The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute. At, x = 8 cm and y = 6 cm, which
of the following options is correct?
A. Perimeter is decreasing at the rate of 2 cm/min
B. Area of the rectangle is increasing at the rate of 2 cm2/min
C. Both A & B
D. None of these
Answer: C
Explanation: We have,
𝑑𝑥
= −5 𝑐𝑚/min
𝑑𝑡
𝑑𝑦
= 4 𝑐𝑚/min
𝑑𝑡
The perimeter of a rectangle, P = 2(x + y)
𝑷 = 𝟐(𝒙 + 𝒚)
𝒅𝑷
𝒅𝒙 𝒅𝒚
= 𝟐( + )
𝒅𝒕
𝒅𝒕 𝒅𝒕
= 𝟐(−𝟓 + 𝟒)
= −𝟐 𝒄𝒎/𝐦𝐢𝐧
Hence, the perimeter is decreasing at the rate of 2 cm/min.
𝑻𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒂 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒍𝒆, 𝑨 = 𝒙 × 𝒚
𝑨 = 𝒙. 𝒚
𝒅𝑨 𝒅𝒙
𝒅𝒚
=
. 𝒚 + 𝒙.
𝒅𝒕
𝒅𝒕
𝒅𝒕
= −𝟓𝒚 + 𝟒𝒙
𝑾𝒉𝒆𝒏 𝒙 = 𝟖 𝒄𝒎 𝒂𝒏𝒅 𝒚 = 𝟔 𝒄𝒎,
𝒅𝑨
= −𝟓𝒚 + 𝟒𝒙
𝒅𝒕
= −𝟓(𝟔) + 𝟒(𝟖)
= −𝟑𝟎 + 𝟑𝟐
= 𝟐 𝒄𝒎𝟐 /𝐦𝐢𝐧
𝑯𝒆𝒏𝒄𝒆, 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒍𝒆 𝒊𝒔 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈 𝒂𝒕 𝒕𝒉𝒆 𝒓𝒂𝒕𝒆 𝒐𝒇 𝟐 𝒄𝒎𝟐 /𝒎𝒊𝒏.
Q60.
𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙– 𝟑 = 𝟎
𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒕𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒔 𝒂𝒓𝒆 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒙 − 𝒂𝒙𝒊𝒔.
A. (1, 2) and (1, −2)
B. (1, −2) and (1, 2)
C. (−1, 2) and (1, −2)
D. (−1, 2) and (1, 2)
Answer: A
𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙 − 𝟑 = 𝟎
𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧
Explanation: 𝟐𝒙 + 𝟐𝒚. 𝒅𝒚 − 𝟐 = 𝟎
𝒅𝒙
𝒅𝒚
𝒅𝒙
=
𝟏−𝒙
𝒚
Now, the tangents are parallel to the x-axis if the slope of the tangent = 0
𝑑𝑦
=0
𝑑𝑥
1−𝑥
⇒
=0
𝑦
⇒1−𝑥 =0
∴𝑥=1
𝐴𝑙𝑠𝑜, 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 1 𝑖𝑛 𝑥 2 + 𝑦 2 − 2𝑥 − 3 = 0
𝑊𝑒 ℎ𝑎𝑣𝑒, 𝑦 2 = 4
𝑦 = ±2
⇒
Hence, (1, 2) and (1, −2) are the points at which the tangents are parallel to the x-axis.
Q61.
𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒚 = 𝒙𝟑 𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒕𝒉𝒆 𝒔𝒍𝒐𝒑𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒊𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐
𝒕𝒉𝒆 𝒚 − 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕.
A. (2, 27)
B. (3, 27)
C. (3, 25)
D. (3, 26)
Answer: Option B
Explanation: y=x3
On differentiation
𝑑𝑦
= 3𝑥 2
𝑑𝑥
The slope of the tangent at the point (x, y) is,
𝒅𝒚
](𝒙,𝒚) = 𝟑(𝒙)𝟐
𝒅𝒙
When the slope of the tangent = equal to the y-coordinate of the point, then
𝒚 = 𝟑𝒙𝟐
𝑨𝒍𝒔𝒐, 𝒚 = 𝒙𝟑
∴ 𝟑𝒙𝟐 = 𝒙𝟑
𝒙𝟐 (𝒙 – 𝟑) = 𝟎
𝒙 = 𝟎, 𝒙 = 𝟑
𝑾𝒉𝒆𝒏 𝒙 = 𝟎, 𝒕𝒉𝒆𝒏 𝒚 = 𝟎 𝒂𝒏𝒅
𝑾𝒉𝒆𝒏 𝒙 = 𝟑, 𝒕𝒉𝒆𝒏 𝒚 = 𝟑(𝟑)𝟐 = 𝟐𝟕
𝑯𝒆𝒏𝒄𝒆, 𝒕𝒉𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒑𝒐𝒊𝒏𝒕𝒔 𝒂𝒓𝒆 (𝟎, 𝟎) 𝒂𝒏𝒅 (𝟑, 𝟐𝟕).
Q62.A circular disc of radius 3 cm is being heated. Due to expansion, its radius
increases at the rate of 0.05 cm/s. The rate at which its area is increasing when
radius is 3.2 cm is
A. 32.0π cm2/s
B. 3.20π cm2/s
C. 0.320π cm2/s
D. None of these
Answer: C
Explanation: Let r be the radius of the given disc and A be its area. Then,
𝐴 = 𝜋𝑟 2
𝑑𝐴
𝑑𝑟
⇒
= 2𝜋𝑟
𝑑𝑡
𝑑𝑡
Now approximate rate of increase of radius
𝑑𝑟 =
𝑑𝑟
𝑑𝑡
𝛥𝑡 = 0.05 cm/s
∴ the approximate rate of increase in area is given by
𝑑A
𝑑𝑟
(𝛥𝑡) = 2𝜋𝑟( 𝛥𝑡)
𝑑𝑡
𝑑𝑡
= 2𝜋(3.2)(0.05) = 0.320𝜋 𝑐𝑚2 /𝑠.
𝑑A =
Q63.
𝑰𝒕 𝒊𝒔 𝒈𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝒂𝒕 𝒙 = 𝟏, 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒙𝟒 − 𝟔𝟐𝒙𝟐 + 𝒂𝒙 + 𝟗 𝒂𝒕𝒕𝒂𝒊𝒏𝒔 𝒊𝒕𝒔 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆,
𝒐𝒏 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 [𝟎, 𝟐]. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒂.
A. 120
B. 110
C. -110
D. -120
Answer: A
Explanation:
𝒇(𝒙) = 𝒙𝟒 − 𝟔𝟐𝒙𝟐 + 𝒂𝒙 + 𝟗
𝒇′(𝒙) = 𝟒𝒙𝟑 − 𝟏𝟐𝟒𝒙 + 𝒂
𝒇′(𝟏) = 𝟒 − 𝟏𝟐𝟒 + 𝒂
= −𝟏𝟐𝟎 + 𝒂
𝒇′(𝟏) = 𝟎
−𝟏𝟐𝟎 + 𝒂 = 𝟎
𝒂 = 𝟏𝟐𝟎
Q64.Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm
A. 5π
B. 4π
C. 8π
D. 2π
Answer: C
Explanation: The area of a circle, A=πr2
On differentiating both sides
On differentiating both sides
𝑑𝐴
𝑑
=
(𝜋𝑟 2 )
𝑑𝑟 𝑑𝑟
= 2𝜋𝑟
𝑑𝐴
𝐻𝑒𝑟𝑒
𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑎𝑟𝑒𝑎
𝑑𝑟
𝑟 = 4 𝑐𝑚
𝑑𝐴
= 2𝜋𝑟
𝑑𝑟
= 2𝜋(4)
= 8𝜋
𝑻𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 𝒐𝒏 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒚𝟐 = 𝒙, 𝒘𝒉𝒆𝒓𝒆 𝒕𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒎𝒂𝒌𝒆𝒔
Q65.
𝝅
𝒂𝒏 𝒂𝒏𝒈𝒍𝒆 𝒐𝒇 𝟒 𝒘𝒊𝒕𝒉 𝒙 − 𝒂𝒙𝒊𝒔 𝒊𝒔:
1 1
1 1
A. (2 , 4)
B. (4 , 2)
1 1
C. (4 , 5)
D. (2, 4)
Answer: B
Explanation:
Given, 𝑦 2 = 𝑥
𝑑𝑦
⇒ 2𝑦
=1
𝑑𝑥
𝑑𝑦
1
⇒
=
𝑑𝑥 2𝑦
As, tangent makes an angle of
𝜋
4
𝑑𝑦
1
𝜋
=
= tan
𝑑𝑥 2𝑦
4
1
⇒
=1
2𝑦
1
⇒𝑦=
2
1
1
Putting 𝑦 = in 𝑦 2 = 𝑥, we have, 𝑥 =
2
4
1 1
So the required point is( , ).
4 2
∴
Q66.The tangent to the curve y=x1/5(0, 0) is:
A. parallel to Y axis
B. parallel to X axis
C. at an angle of 45° to the x-axis
D. not possible
Answer: A
𝒚 = 𝒙𝟏/𝟓
𝒅𝒚
𝒅𝒙
𝟏
𝟓
⇒ ( ) = 𝒙−𝟒/𝟓 =
Explanation:
𝟏
𝟓𝒙𝟒/𝟓
𝒅𝒚
⇒ (𝒅𝒙)(𝟎,𝟎) = ∞.
𝑯𝒆𝒏𝒄𝒆, 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒚 = 𝒙𝟏/𝟓 𝒂𝒕 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒂𝒕 (𝟎, 𝟎)
𝒘𝒉𝒊𝒄𝒉 𝒊𝒔 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒀 − 𝒂𝒙𝒊𝒔.
𝟓
Q67. 𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: ∫𝟐 [|𝒙 − 𝟐| + |𝒙 − 𝟑| + |𝒙 − 𝟓|]𝒅𝒙
A. 12
C.
B. 15
23
D. 23
5
2
Answer: C
𝟓
∫𝟐 [|𝒙 − 𝟐| + |𝒙 − 𝟑| + |𝒙 − 𝟓|]𝒅𝒙
𝟑
𝟓
= ∫𝟐 [𝒙 − 𝟐 + 𝟑 − 𝒙 + 𝟓 − 𝒙]𝒅𝒙 + ∫𝟑 [𝒙 − 𝟐 + 𝒙 − 𝟑 + 𝟓 − 𝒙]𝒅𝒙
𝟑
𝟓
= ∫𝟐 [𝟔 − 𝒙]𝒅𝒙 + ∫𝟑 [𝒙]𝒅𝒙
= [𝟔𝒙 −
Explanation:
𝒙𝟐 𝟑
]
𝟐 𝟐
𝒙𝟐
𝟐
+ [ ]𝟓𝟑
𝟗
𝟐𝟓
𝟗
= (𝟏𝟖 − 𝟐 − 𝟏𝟐 + 𝟐) + ( 𝟐 − 𝟐)
𝟗
𝟐
𝟐𝟑
𝟐
= (𝟖 − ) + 𝟖
=
√𝟑
Q68. 𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: ∫
𝟏
A. 𝜋6
C.
𝜋
3
Answer: D
𝒅𝒙
𝟏+𝒙𝟐
B.
D.
𝜋
4
𝜋
12
√𝟑
𝒅𝒙
𝟏+𝒙𝟐
∫
𝟏
√𝟑
𝑳𝒆𝒕 𝑰 = ∫
Explanation:
= [𝐭𝐚𝐧
𝟏
−𝟏
𝒅𝒙
𝟏+𝒙𝟐
𝒙]√𝟑
𝟏
= 𝐭𝐚𝐧−𝟏 (√𝟑) − 𝐭𝐚𝐧−𝟏 (𝟏)
𝝅
𝝅
=𝟑−𝟒
=
Q69. 𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: ∫
A.
C.
(1+log𝑥)3
3
(1+log𝑥)2
2
𝝅
𝟏𝟐
(𝟏+𝐥𝐨𝐠𝒙)𝟐
𝒅𝒙
𝒙
+𝐶
B.
+𝐶
D.
(1+log𝑥)3
2
(1−log𝑥)3
3
+𝐶
+𝐶
Answer: A
∫
(𝟏+𝐥𝐨𝐠𝒙)𝟐
𝒅𝒙
𝒙
𝑳𝒆𝒕 𝟏 + 𝐥𝐨𝐠𝒙 = 𝒕
Explanation: 𝑷𝒖𝒕 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇
𝟏
𝒅𝒙 = 𝒅𝒕
𝒙
𝟏
𝒅𝒙 𝒊𝒏 𝒕𝒉𝒆 𝐞𝐱𝐩𝒓𝒆𝒔𝒔𝒊𝒐𝒏
𝒙
𝟐
= ∫ 𝒕 𝒅𝒕
𝒕𝟑
+𝒄
𝟑
(𝟏+𝐥𝐨𝐠𝒙)𝟑
+
𝟑
=
=
𝒂
𝒄
Q70. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 ‘𝒂’ 𝒊𝒇 ∫𝟎 𝟑𝒙𝟐 𝒅𝒙 = 𝟖
A. 1
B. 3
C. 2
D. 4
Answer: C
𝒂
∫𝟎 𝟑𝒙𝟐 𝒅𝒙 = 𝟖
𝒙𝟑
𝟑. [ 𝟑 ]𝒂𝟎 = 𝟖
Explanation:
[𝒙𝟑 ]𝒂𝟎 = 𝟖
𝒂𝟑 − 𝟎 = 𝟖
𝒂=𝟐
Q71. ∫𝐥𝐨𝐠(𝟏 + 𝒙𝟐 )𝒅𝒙 =?
A. xlog(1+x2)−2x+2tan−1x+C
B. −xlog(1+x2)−2x−2tan−1x+C
C. xlog(1+x2)+2x+2tan−1x+C
D. xlog(1+x2)−2x+2tan−1(1+x2)+C
Answer: A
𝐋𝐞𝐭, 𝐈 = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐 )𝒅𝒙
⇒ 𝐈 = ∫ 𝟏 ⋅ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐 )𝒅𝒙
𝟏
⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐 ) − ∫ (𝒙 ⋅ 𝟏+𝒙𝟐 ⋅ 𝟐𝒙)𝒅𝒙
𝒙𝟐
⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐 ) − 𝟐∫ (𝟏+𝒙𝟐)𝒅𝒙
Explanation:
𝒙𝟐 +𝟏−𝟏
)𝒅𝒙
𝟏+𝒙𝟐
𝟏
𝟐∫ 𝒅𝒙 + 𝟐∫ 𝟏+𝒙𝟐 𝒅𝒙
−𝟏
⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐 ) − 𝟐∫ (
⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐 ) −
⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐 ) − 𝟐𝒙 + 𝟐𝐭𝐚𝐧
𝟏
Q72. 𝑰𝒇 ∫
𝟎
𝒆𝒕
𝒅𝒕
𝟏+𝒕
A. 𝑎 − 1 +
C. a
Answer: B
𝑒
2
𝟏
= 𝒂, 𝒕𝒉𝒆𝒏 ∫
𝒆𝒕
𝟐
𝟎 (𝟏+𝒕)
𝒅𝒕 =?
B. 𝑎 + 1 −
D. a2
𝑒
2
𝒙 + 𝐂.
Explanation: According to question
1
∫
𝑒𝑡
𝑑𝑡 = 𝑎
1+𝑡
0
1
1
𝑒𝑡
𝑡 1
⇒|
𝑒 | − ∫[−
]𝑑𝑡 = 𝑎
1+𝑡 0
(1 + 𝑡)2
0
1
⇒
𝑒
𝑒𝑡
−1+∫
𝑑𝑡 = 𝑎
1+1
(1 + 𝑡)2
0
1
⇒
𝑒
𝑒𝑡
−1+∫
𝑑𝑡 = 𝑎
2
(1 + 𝑡)2
0
1
⇒∫
𝑒𝑡
𝑒
𝑑𝑡
=
𝑎
+
1
−
(1 + 𝑡)2
2
0
Q73. ∫
𝟏
𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐬𝟐 𝒙
𝒅𝒙 =?
A. tanx + cotx + C
B. (tanx + cotx)2 + C
C. tanx − cotx + C
D. (tanx − cotx )2+ C
Answer: C
Explanation: This can be solved as,
∫
1
𝑑𝑥
sin2 𝑥cos2 𝑥
2
2
sin 𝑥 + cos 𝑥
𝑑𝑥
sin2 𝑥cos 2 𝑥
sin2 𝑥
cos 2 𝑥
=∫
+
𝑑𝑥
sin2 𝑥cos 2 𝑥 sin2 𝑥cos 2 𝑥
= ∫ (sec 2 𝑥 + cosec 2 𝑥)𝑑𝑥
= tan𝑥 − cot𝑥 + C
=∫
Q74. 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒈𝒊𝒐𝒏 𝒃𝒐𝒖𝒏𝒅𝒆𝒅 𝒃𝒚 𝒚 = √𝟓 − 𝒙𝟐 𝒂𝒏𝒅 𝒊𝒔:
5
2
2
√5
5
2
2
√5
5
2
2
√5
5
2
2
√5
A. [sin−1
B. [sin−1
C. [sin−1
D. [sin−1
−1
1
√5
2
−1
1
√5
2
−1
1
√5
2
−1
1
√5
2
− sin−1 ( )] −
− sin−1 ( )] +
+ sin−1 ( )] −
+ sin−1 ( )] +
𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
Answer: Option A
Explanation: The given equations are
1−𝑥𝑥 <1
𝑥 2 + 𝑦 2 = 5 and 𝑦 = {
}
𝑥−1𝑥 > 1
Figure is as follows:
Point of intersection are:
C(2, 1) and D(-1, 2)
Therefore,
2
1
2
𝐴𝑟𝑒𝑎 = ∫ √5 − 𝑥 2 𝑑𝑥 − ∫(1 − 𝑥) 𝑑𝑥 − ∫(𝑥 − 1) 𝑑𝑥
−1
−1
1
2
𝑥
5 −1 𝑥 2
𝑥 1
𝑥2
2
√
𝐴𝑟𝑒𝑎 = [ 5 − 𝑥 + sin
]−1 − [𝑥 − ]−1 − [ − 𝑥]12
2
2
2
2
√5
5
2
1
5
−1
1
1
1
𝐴𝑟𝑒𝑎 = [1 + sin−1
+ ⋅ 2 − sin−1
] − [1 − + 1 + ] − [2 − 2 − + 1]
2
2
2
2
2
√5 2
√5
5
2
−1
1
𝐴𝑟𝑒𝑎 = 2 + [sin−1
− sin−1 ( )] − 2 −
2
2
√5
√5
5
2
−1
1
𝐴𝑟𝑒𝑎 = [sin−1
− sin−1 ( )] −
2
2
√5
√5
Q75.The area of the region in the first quadrant enclosed by the x-axis , the line
and the circle x2+y2=32.
A. 8 sq units
B. 4 sq units
C. 2 sq units
D. sq units
Answer: B
Explanation: The given equations are
2𝑦 = 𝑥 . . . . . . . . . . . . . . . . . . . . . (1)
𝑎𝑛𝑑 𝑥 2 + 𝑦 2 = 32 . . . . . . . . . . . (2)
Solving (1) and (2) as
we find that the line and the circle meet at B(4, 4) in the first quadrant (Fig).
Draw perpendicular BM to the x-axis.
Area of the region bounded by a circle and a line can be drawn as,
Since general equation of the circle passing through origin is :
𝑥2 + 𝑦2 = 𝑟2
𝐶𝑜𝑚𝑝𝑎𝑖𝑟𝑖𝑛𝑔 𝑥 2 + 𝑦 2 = 32 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑥 2 + 𝑦 2 = 𝑟 2 , 𝑤𝑒 𝑔𝑒𝑡
𝑟 2 = 32
⇒ 𝑟 = 4√2
For region OMBO, limits will be from 0 to intersecting point i.e 4 and for the region BMAB,
limits will be 4 to 4√2
Therefore, the required area = area of the region OBMO + area of the region BMAB.
𝟒√𝟐
𝟒
= ∫ 𝒚𝟏 𝒅𝒙 + ∫ 𝒚𝟐 𝒅𝒙
𝟎
=∫
𝟒
𝟒√𝟐
𝟒
𝒙𝒅𝒙 + ∫ √𝟑𝟐 − 𝒙𝟐 𝒅𝒙
𝟎
𝟒
𝟐
𝒙 𝟒
𝟏
𝟑𝟐
𝒙 𝟒√𝟐
]𝟎 + [ 𝒙√𝟑𝟐 − 𝒙𝟐 +
𝐬𝐢𝐧−𝟏
]𝟒
𝟐
𝟐
𝟐
√𝟑𝟐
𝟏𝟔
𝟏
𝟒√𝟐
𝟏
𝟒
= [ − 𝟎] + [ 𝟒√𝟐√𝟑𝟐 − 𝟑𝟐 + 𝟏𝟔𝐬𝐢𝐧−𝟏
] − [ 𝟒√𝟑𝟐 − 𝟏𝟔 + 𝟏𝟔𝐬𝐢𝐧−𝟏
]
𝟐
𝟐
𝟐
𝟒√𝟐
𝟒√𝟐
𝟏
= 𝟖 + [𝟎 + 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟏] − [𝟐 ⋅ 𝟒 + 𝟏𝟔𝐬𝐢𝐧−𝟏 ]
√𝟐
𝟏
= 𝟖 + 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟏 − 𝟖 − 𝟏𝟔𝐬𝐢𝐧−𝟏
√𝟐
𝝅 𝝅
= 𝟏𝟔( − )
𝟐 𝟒
𝟐𝝅 − 𝝅
= 𝟏𝟔(
)
𝟒
= 𝟒𝝅
=[
Hence the required area is 4π Square units.
Q76. 𝑻𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒈𝒊𝒐𝒏 {(𝒙, 𝒚): 𝟎 ≤ 𝒚 ≤ 𝒙𝟐 + 𝟏, 𝟎 ≤ 𝒚 ≤ 𝒙 + 𝟏; 𝟎 ≤ 𝒙 ≤ 𝟐} 𝒊𝒔:
A. 16/3 sq units
B. 13/3 sq units
C. 11/3 sq units
D. 7/3 sq units
Answer: C
𝑻𝒉𝒆 𝒄𝒖𝒓𝒗𝒆𝒔 𝒂𝒓𝒆𝒙 = 𝟐, 𝒚 = 𝒙 + 𝟏, 𝒚 = 𝒙𝟐 + 𝟏
𝑻𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆𝒔 𝒚 = 𝒙 + 𝟏, 𝒚 = 𝒙𝟐 + 𝟏
Explanation:
𝒙𝟐 + 𝟏 = 𝒙 + 𝟏
𝒙(𝒙 − 𝟏) = 𝟎
𝒙 = 𝟎, 𝟏
The shaded area is the required area.
1
1
2
∫ 𝑦2 𝑑𝑥 − ∫ 𝑦1 𝑑𝑥 + ∫ 𝑦2 𝑑𝑥
0
1
0
1
1
2
= ∫(𝑥 + 1) 𝑑𝑥 − ∫(𝑥 2 + 1) 𝑑𝑥 + ∫(𝑥 + 1) 𝑑𝑥
0
0
2
3
1
2
𝑥
𝑥
𝑥
+ 𝑥]10 − [ + 𝑥]10 + [ + 𝑥]12
2
3
2
1
1
4
1
= [ + 1] − [ + 1] + [ + 2 − − 1]
2
3
2
2
1 1
1
= − +4−
2 3
2
1
=− +4
3
−1 + 12
=
6
11
=
𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
3
=[
Q77.The area of the region bounded by the y-axis, y = cosx and, 0 ≤ x ≤π/2 is:
A. √2 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
B. (√2 + 1) 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
C. (√2 − 1) 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
D. (2√2 − 1) 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
Answer: Option C
Explanation: Intersection points can be calculated as,
cos𝑥 = sin𝑥
⇒ tan𝑥 = 1
𝜋
⇒𝑥=
4
So,
𝜋/4
𝑎𝑟𝑒𝑎 = ∫ (cos𝑥 − sin𝑥)𝑑𝑥
0
𝜋/4
⇒ 𝐴𝑟𝑒𝑎 = [sin𝑥 + cos𝑥]0
𝜋
𝜋
⇒ 𝐴𝑟𝑒𝑎 = [sin + cos − sin0 − cos0]
4
4
1
1
⇒ 𝐴𝑟𝑒𝑎 = [ +
− 0 − 1]
√2 √2
2
⇒ 𝐴𝑟𝑒𝑎 = [ − 1]
√2
⇒ 𝐴𝑟𝑒𝑎 = (√2 − 1) 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
Q78. 𝑻𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒈𝒊𝒐𝒏 𝒃𝒐𝒖𝒏𝒅𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒚 = √𝟏𝟔 − 𝒙𝟐 𝒂𝒏𝒅 𝒙 − 𝒂𝒙𝒊𝒔 𝒊𝒔
A. 8π sq units
B. 20πsq units
C. 16π sq units
D. 256π sq units
Answer: A
Explanation: This can be simplified as
𝑦 = √16 − 𝑥 2
𝐹𝑜𝑟 𝑦 = 0
𝑥 = ±4
4
𝑆𝑜 𝑎𝑟𝑒𝑎 = 2 × ∫ 𝑦𝑑𝑥
0
4
⇒ 𝐴𝑟𝑒𝑎 = 2 × ∫ √16 − 𝑥 2 𝑑𝑥
0
𝑥
16
𝑥
⇒ 𝐴𝑟𝑒𝑎 = 2 × [ √16 − 𝑥 2 + sin−1 ]40
2
2
4
⇒ 𝐴𝑟𝑒𝑎 = 2 × 4𝜋 = 8𝜋 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
Q79.The area of the region bounded by the curve y = x + 1 and the lines x = 2 and
x = 3 is
A. 7/2 sq units
B. 9/2 sq units
C. 11/2 sq units
D. 3/2 sq units
Answer: A
Explanation: This can be solved as,
3
𝐴𝑟𝑒𝑎 = ∫(𝑥 + 1)𝑑𝑥
2
𝑥2
⇒ 𝐴𝑟𝑒𝑎 = [ + 𝑥]32
2
9
4
⇒ 𝐴𝑟𝑒𝑎 = [ + 3 − − 2]
2
2
7
⇒ 𝐴𝑟𝑒𝑎 = 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
2
Q80.Calculate the area under the curve y = 2√x included between the lines x = 0
and x = 1.
A. 2/3 sq units
B. 1 sq units
C. 4/3 sq units
D. 5/3 sq units
Answer: C
Explanation: Since
1
𝐴𝑟𝑒𝑎 = ∫ 2√𝑥𝑑𝑥
0
𝑥 3/2
⇒ 𝐴𝑟𝑒𝑎 = 2[
⋅ 2]10
3
2
4
⇒ 𝐴𝑟𝑒𝑎 = 2[ ⋅ 1 − 0] = 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
3
3
Q81.
𝑾𝒉𝒂𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝒕𝒂𝒏−𝟏 𝒚 − 𝒙) 𝒅𝒚 = (𝟏 + 𝒚𝟐 ) 𝒅𝒙?
(𝑮𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕, 𝒂𝒕 𝒙 = 𝟎 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒚 = 𝟎)
A. 𝑥 = tan−1 𝑦 + 𝑒 −tan
−1 𝑦
B. 𝑥 = tan−1 𝑦 − 1 + 𝑒 tan
−1 𝑦
C. 𝑥 = tan−1 𝑦 − 5 + 𝑒 −tan
−1 𝑦
D. 𝑥 = tan−1 𝑦 − 1 + 𝑒 −tan
Answer: D
Explanation:
−1 𝑦
(tan−1 𝑦 − 𝑥)𝑑𝑦 = (1 + 𝑦 2 )𝑑𝑥
𝑑𝑥 (tan−1 𝑦 − 𝑥)
=
𝑑𝑦
(1 + 𝑦 2 )
𝑑𝑥 tan−1 𝑦
𝑥
=
−
2
𝑑𝑦 1 + 𝑦
1 + 𝑦2
𝑑𝑥
𝑥
tan−1 𝑦
+
=
𝑑𝑦 1 + 𝑦 2 1 + 𝑦 2
∫
𝑑𝑦
1+𝑦 2
−1
Integrating factor = 𝑒
= 𝑒 tan 𝑦
𝑁𝑜𝑤, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟
𝑑𝑥
𝑥
tan−1 𝑦
tan−1 𝑦
tan−1 𝑦
tan−1 𝑦
𝑒
+𝑒
.
=𝑒
.
𝑑𝑦
1 + 𝑦2
1 + 𝑦2
tan−1 𝑦
−1
−1
𝑥. 𝑒 tan 𝑦 = ∫ 𝑒 tan 𝑦 .
. 𝑑𝑦
1 + 𝑦2
Let 𝑡 = tan−1 𝑦
𝑑𝑦
𝑑𝑡 =
1 + 𝑦2
−1
𝑥. 𝑒 tan 𝑦 = ∫ 𝑒 t . 𝑡𝑑𝑡
Applying by parts
𝑑
−1
𝑥. 𝑒 tan 𝑦 = 𝑡(𝑒 𝑡 ) − ∫ 𝑒 t . { (𝑡)}𝑑𝑡
𝑑𝑡
tan−1 𝑦
𝑡
𝑥. 𝑒
= 𝑡(𝑒 ) − ∫ 𝑒 t 𝑑𝑡
−1
𝑥. 𝑒 tan 𝑦 = 𝑡𝑒 𝑡 − 𝑒 𝑡 + 𝐶
−1
−1
tan−1 𝑦
𝑥. 𝑒
= tan−1 𝑦(𝑒 tan 𝑦 ) − 𝑒 tan 𝑦 + 𝐶
When 𝑥 = 0, 𝑦 = 0
−1
0 = tan−1 0(𝑒 tan 0 ) − 𝑒 tan
0 = −𝑒 0 + 𝐶
𝐶=1
−1 0
+𝐶
Q82.The differential equation of the family of parabolas having vertex at the
origin and axis along positive y-axis is
A. xy'−y=0
B. xy'+2y=0
C. xy'−7y=0
D. xy'−2y=0
Answer: D
Explanation:
Vertex = (0, 0)
The equation of the parabola
𝑥 2 = 4𝑎𝑦 . . . . (1)
On differentiation
2𝑥 = 4𝑎𝑦′
𝑥 = 2𝑎𝑦′
𝑥
𝑎=
2𝑦′
Put value of a in equation (1)
𝑥
𝑥2 = 4
𝑦
2𝑦′
𝑦′𝑥 2 = 2𝑥𝑦
𝑦′𝑥 = 2𝑦
𝑥𝑦′ − 2𝑦 = 0
This is the required differential equation.
Q83.Write the differential equation representing the family of curves y = m x,
where m is an arbitrary constant.
A. xdy−ydx=0
B. xdy−dx=0
C. xdy+ydx=0
D. dy−ydx=0
Answer: A
Explanation: We have,
𝑦 = 𝑚𝑥
On differentiation
𝑑𝑦
=𝑚
𝑑𝑥
𝑦
𝑚=
𝑥
𝑇ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑎𝑚𝑖𝑙𝑦 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒𝑠 𝑦 = 𝑚𝑥, 𝑖𝑠
𝑥𝑑𝑦 − 𝑦𝑑𝑥 = 0
Q84. 𝑻𝒉𝒆 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏
A. y=x(logx+1)+C
B. y+x(logx+1)=C
C. y=x(logx−1)+C
D. None of these
𝒅𝒚
𝒅𝒙
= 𝐥𝐨𝐠𝒙 𝒊𝒔:
Answer: C
Explanation: This can be solved by using “by parts” method of integration.
𝒅𝒚
= 𝐥𝐨𝐠𝒙
𝒅𝒙
⇒ ∫ 𝒅𝒚 = ∫ 𝟏 ⋅ 𝐥𝐨𝐠𝒙𝒅𝒙
𝟏
⇒ 𝒚 = 𝒙𝐥𝐨𝐠𝒙 − ∫ 𝒙 ⋅ 𝒅𝒙
𝒙
⇒ 𝒚 = 𝒙𝐥𝐨𝐠𝒙 − ∫ 𝟏𝒅𝒙
⇒ 𝒚 = 𝒙𝐥𝐨𝐠𝒙 − 𝒙 + 𝑪
⇒ 𝒚 = 𝒙(𝐥𝐨𝐠𝒙 − 𝟏) + 𝑪.
𝐆𝐢𝐯𝐞𝐧,
Q85.The order of the differential equation of all circle of radius r, having centre on
y-axis and passing through the origin is
A. 1
B. 2
C. 3
D. 4
Answer: Option A
Explanation: The equation of circle with centre on y-axis and radius r
(𝑥)2 + (𝑦 − ℎ)2 = 𝑟 2
This circle passes through origin i.e. (0, 0). Putting (0, 0) in above equation we have,
ℎ2 = 𝑟 2 ⇒ ℎ = 𝑟
Therefore, the equation of circle of radius r and centre on y axis and passing through origin will
be:
𝑥 2 + (𝑦 − 𝑟)2 = 𝑟 2
⇒ 𝑥 2 + 𝑦 2 − 2𝑟 + 𝑟 2 = 𝑟 2
⇒ 𝑥 2 + 𝑦 2 = 2𝑟
The equation has only one arbitrary constant that needs to be removed, so the order of
differential equation is 1.
→
→
→
→
→
→
Q86. (𝒂 × 𝒃)𝟐 + (𝒂 . 𝒃)𝟐 = 𝟏𝟒𝟒 𝒂𝒏𝒅 |𝒂| = 𝟒, 𝒕𝒉𝒆𝒏 |𝒃| =
A. 8
B. 5
C. 7
D. 3
Answer: D
Explanation: This can be solved as,
→
→
→ →
(𝑎 × 𝑏 )2 + (𝑎 . 𝑏 )2 = 144
→
→
→
→
|𝑎 |2 |𝑏 |2 sin2 𝜃 + |𝑎 |2 |𝑏 |2 cos 2 𝜃 = 144
→
→
|𝑎 |2 |𝑏 |2 (sin2 𝜃 + cos2 𝜃) = 144
→
42 |𝑏 |2 (1) = 144
→
144
|𝑏 |2 =
16
→
|𝑏 |2 = 9
→
|𝑏 | = 3
Q87.If a = 2i + j + k, b = 3i – 4j + 2k and c = i + 2j + 2k then the projection of a + b
on c isA.
C.
17
3
4
3
B.
5
3
D. none of these
Answer: B
Explanation: This can be solved as follows:
projection of a + b on c
(𝑎 + 𝑏) ⋅ 𝑐
|𝑐|
(5𝑖 − 3𝑗 + 3𝑘) ⋅ (𝑖 + 2𝑗 + 2𝑘)
=
=
√1 + 4 + 4
5
=
√9
5
=
3
→ → →
Q88.
𝑰𝒇 𝒂, 𝒃, 𝒄 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒎𝒖𝒕𝒖𝒂𝒍𝒍𝒚 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒆𝒂𝒄𝒉 𝒐𝒇
→
→
→
𝒎𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒖𝒏𝒊𝒕𝒚, 𝒕𝒉𝒆𝒏 |𝒂 + 𝒃 + 𝒄 | =
A. 1
B. 3
C. √3
D. 3√3
Answer: C
Explanation:
→ → →
𝑆𝑖𝑛𝑐𝑒 𝑎 , 𝑏 , 𝑐 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑒𝑎𝑐ℎ
𝑜𝑓 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑢𝑛𝑖𝑡𝑦, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,
|𝑎| = |𝑏| = |𝑐| = 1
𝑎⋅𝑏 =𝑎⋅𝑐 =𝑐⋅𝑏 =0
𝑁𝑜𝑤,
→
→
→
|𝑎 + 𝑏 + 𝑐 |2 = |𝑎|2 + |𝑏|2 + |𝑐|2 + 2𝑎 ⋅ 𝑏 + 2𝑎 ⋅ 𝑐 + 2𝑐 ⋅ 𝑏
→
→
→
⇒ |𝑎 + 𝑏 + 𝑐 |2 = 1 + 1 + 1 + 0 + 0 + 0
→
→
→
⇒ |𝑎 + 𝑏 + 𝑐 |2 = 3
→
→
→
⇒ |𝑎 + 𝑏 + 𝑐 |2 = √3
^
^
^
^
^
^
Q89. 𝑰𝒇 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒊 + 𝟐𝒋 + 𝟑𝒌 𝒂𝒏𝒅 𝟑𝒊 − 𝟐𝒋 + 𝒌 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝒕𝒉𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕
𝒔𝒊𝒅𝒆𝒔 𝒐𝒇 𝒂 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍𝒐𝒈𝒓𝒂𝒎, 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍𝒐𝒈𝒓𝒂𝒎 𝒊𝒔
A. 4√3
B. 6√3
C. 8√3
D. 16√3
Answer: C
Explanation: We have,
𝑖
𝑗 𝑘
(𝑎 × 𝑏) = |1 2 3|
3 −2 1
(𝑎 × 𝑏) = 𝑖(2 + 6) − 𝑗(1 − 9) + 𝑘(−2 − 6)
= 8𝑖 + 8𝑗 − 8𝑘
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎𝑟𝑒𝑎
|(𝑎 × 𝑏)| = |8𝑖 + 8𝑗 − 8𝑘|
= √|82 + 82 + 82 |
= 8√3
Q90.If the position vector of three consecutive vertices of any parallelogram are
respectively i + j + k, i + 3j + 5k, 7i + 9j + 11k, then the position vector of fourth
vertex is A. 6(i+j+k)
B. 7(i+j+k)
C. 2j−4k
D. 6i+8j+10k
Answer: B
Explanation: Let the position vector of 4 vertices are
𝐴(𝑖 + 𝑗 + 𝑘),
𝐵(𝑖 + 3𝑗 + 5𝑘),
𝐶(7𝑖 + 9𝑗 + 11𝑘)
𝐷(𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘)
Since the diagonals of parallelogram bisects, so the midpoint of AC and BD coincides, therefore,
7+1 1+𝑥
=
2
2
⇒𝑥=7
9+1 3+𝑦
=
2
2
⇒𝑦=7
11 + 1 5 + 𝑧
=
2
2
⇒𝑧=7
Therefore, fourth position vector is 7i + 7j + 7k
Q91.If vector 3j + 2j + 8k and 2i + xj + k are perpendicular then x is equal to
A. 7
B. -7
C. 5
D. -4
Answer: B
Explanation: For perpendicular vectors,
3⋅2+2⋅𝑥+8⋅1= 0
⇒ 6 + 2𝑥 + 8 = 0
⇒ 2𝑥 = −14
⇒ 𝑥 = −7
Q92.If i + 2j + 3k is parallel to sum of the vector 3i + λj + 2k and -2i + 3j + k, then
equals to:
A. 1
B. -1
C. 2
D. -2
Answer: B
Explanation: According to given condition
𝑖 + 2𝑗 + 3𝑘 = 3𝑖 + 𝜆𝑗 + 2𝑘 + −2𝑖 + 3𝑗 + 𝑘
⇒ 𝑖 + 2𝑗 + 3𝑘 = 𝑖 + (3 + 𝜆)𝑗 + 3𝑘
⇒ 𝑖 + 2𝑗 + 3𝑘 = 𝑖 + (3 + 𝜆)𝑗 + 3𝑘
For vectors to be parallel
(3+λ)=2⇒(3+λ)=−1
Q93. 𝑻𝒉𝒆 𝒊𝒎𝒂𝒈𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 (𝟏, 𝟔, 𝟑) 𝒊𝒏 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆
A. (0, 1,7)
B. (1,0,7)
C. (7, 0,1)
D. (7,1,0)
𝒙
𝟏
=
𝒚−𝟏
𝟐
=
𝒛−𝟐
𝟑
𝒊𝒔:
Answer: B
Explanation: :- Let P be the point whose image we have to find. Then point P is
(1,6,3)
𝑥
𝑦−1
1
2
Let Q be the image of point in the line =
=
𝑧−2
3
and M be the foot of perpendicular
drawn from P to this line .
Then PM = MQ
𝑥
1
Since M lies on line =
𝑦−1
2
=
𝑧−2
,
3
Then the coordinates of M can be obtained through line by
considering the given equation equal to a variable r as,
Then the coordinates of M will be
𝑥=𝑟
𝑦 = 2𝑟 + 1
𝑧 = 3𝑟 + 2
𝑆𝑖𝑛𝑐𝑒 𝑃 𝑖𝑠 (1,6,3)𝑎𝑛𝑑 𝑀 𝑖𝑠 (𝑟, 2𝑟 + 1,3𝑟 + 2).
The direction ratios of MP are proportional to
𝑟 − 1,2𝑟 + 1 − 6,3𝑟 + 2 − 3
⇒ 𝑟 − 1,2𝑟 − 5,3𝑟 − 1
Since MP is perpendicular to the given line.
Since for two lines,
𝑙1 𝑥 + 𝑚1 𝑦 + 𝑛1 𝑧 + 𝑑1 = 0
𝑙2 𝑥 + 𝑚2 𝑦 + 𝑛2 𝑧 + 𝑑2 = 0
Condition of perpendicularity is𝑙1 𝑙2 + 𝑚1 𝑚2 + 𝑛1 𝑛2 = 0
Therefore, using condition of perpendicularity
1(𝑟 − 1) + 2(2𝑟 − 5) + 3(3𝑟 − 1) = 0
⇒ 14𝑟 − 14 = 0
⇒𝑟=1
So the coordinates of M are
(𝑟, 2𝑟 + 1,3𝑟 + 2)
⇒ (1,2(1) + 1,3(1) + 2)
⇒ (1,3,5)
𝐿𝑒𝑡 (𝑥1 , 𝑦1 , 𝑧1 ) 𝑏𝑒 𝑡ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑄 .
Since M is the midpoint of PQ .
Therefore,
𝑥1 + 1
=1
2
⇒ 𝑥1 = 2 − 1
⇒ 𝑥1 = 1
𝑦1 + 6
=3
2
⇒ 𝑦1 = 6 − 6
⇒ 𝑦1 = 0
𝑧1 + 3
=5
2
𝑧1 = 10 − 3
⇒ 𝑧1 = 7
Thus the coordinates of Q are (1,0,7).
Q94. 𝑰𝒇 𝒍𝒊𝒏𝒆𝒔
𝒙−𝟏
𝟐
=
𝒚−𝟐
𝟑
=
𝒛−𝟑
𝟒
𝒂𝒏𝒅
𝒙−𝟒
𝟓
=
𝒚−𝟏
𝟐
= 𝒛, intersect each other ,then their
point of intersection are:
A. (1, −1, −1)
B. (1, 1, 1)
C. (−1, −1, −1)
D. (−1, 1,−1)
Answer: C
Explanation: The coordinates of any point on first line are given by
𝑥−1 𝑦−2 𝑧−3
=
=
=𝜆
2
3
4
𝑥 = 2𝜆 + 1
𝑦 = 3𝜆 + 2
𝑧 = 4𝜆 + 3
The coordinates of any point on second line are given by
𝑥−4 𝑦−1
=
=𝑧=𝜇
5
2
𝑥 = 5𝜇 + 4
𝑦 = 2𝜇 + 1
𝑧=𝜇
The point of intersection can be calculated by considering corresponding coordinates equal as,
2𝜆 + 1 = 5𝜇 + 4
⇒ 2𝜆 − 5𝜇 = 3
3𝜆 + 2 = 2𝜇 + 1
⇒ 3𝜆 − 2𝜇 = −1
4𝜆 + 3 = 𝜇
⇒ 4𝜆 − 𝜇 = −3
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 2𝜆 − 5𝜇 = 3 𝑏𝑦 2 𝑎𝑠,
2𝜆 − 5𝜇 = 3
⇒ 2(2𝜆 − 5𝜇 = 3)
⇒ 4𝜆 − 10𝜇 = 6
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 4𝜆 − 𝜇 = −3 𝑓𝑟𝑜𝑚 4𝜆 − 10𝜇 = 6 𝑎𝑠,
4𝜆 − 10𝜇 = 6
−(4𝜆 − 𝜇 = −3)
_
−9𝜇 = 9
⇒ 𝜇 = −1
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝜇 = −1 𝑖𝑛 4𝜆 − 𝜇 = −3 𝑤𝑒 𝑔𝑒𝑡
4𝜆 − (−1) = −3
⇒ 4𝜆 + 1 = −3
⇒ 4𝜆 = −4
⇒ 𝜆 = −1
𝑇ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒
𝑥 = 2(−1) + 1 = −1
𝑦 = 3(−1) + 2 = −1
𝑧 = 4(−1) + 3 = −1
𝐼. 𝑒 (−1, −1, −1)
Q95.The equation of the plane through the line of intersection of the planes x + y
+ z and 2x + 3y + 4z = 5 which is ⊥ of the plane x - y + z = 0 is:
A. x−z+2=0
B. x+z+2=0
C. y+z+2=0
D. y+z−2=0
Answer: A
Explanation: The equation of a plane passing through the line of intersection of the planes
𝑥 + 𝑦 + 𝑧 = 1 𝑎𝑛𝑑 2𝑥 + 3𝑦 + 4𝑧 = 5 𝑖𝑠:
(𝒙 + 𝒚 + 𝒛 − 𝟏) + 𝝀(𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎
⇒ (𝟏 + 𝟐𝝀)𝒙 + (𝟏 + 𝟑𝝀)𝒚 + (𝟏 + 𝟒𝝀)𝒛 − 𝟏 − 𝟓𝝀 = 𝟎. . . . . . . . . . (𝟏)
𝑺𝒊𝒏𝒄𝒆 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒕𝒐 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒙 − 𝒚 + 𝒛 = 𝟎 … … … … … … (𝟐)
𝑺𝒊𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒘𝒐 𝒑𝒍𝒂𝒏𝒆𝒔,
𝒂𝟏 𝒙 + 𝒃𝟏 𝒚 + 𝒄𝟏 𝒛 + 𝒅𝟏 = 𝟎
𝒂𝟐 𝒙 + 𝒃𝟐 𝒚 + 𝒄𝟐 𝒛 + 𝒅𝟐 = 𝟎
𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓𝒊𝒕𝒚 𝒊𝒔 𝒂𝟏 𝒂𝟐 + 𝒃𝟏 𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 𝟎
𝑼𝒔𝒊𝒏𝒈 𝒕𝒉𝒊𝒔 𝒊𝒏 𝒂𝒃𝒐𝒗𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝟏) 𝒂𝒏𝒅 (𝟐) 𝒂𝒔,
(𝟏 + 𝟐𝝀)𝟏 − (𝟏 + 𝟑𝝀)𝟏 + (𝟏 + 𝟒𝝀)𝟏 = 𝟎
𝟏 + 𝟐𝝀 − 𝟏 − 𝟑𝝀 + 𝟏 + 𝟒𝝀 = 𝟎
𝟑𝝀 = −𝟏
𝟏
𝝀=−
𝟑
𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒏𝒈 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝝀 𝒊𝒏 𝒂𝒃𝒐𝒗𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏
𝒐𝒇 𝒑𝒍𝒂𝒏𝒆, 𝑻𝒉𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 :
(𝒙 + 𝒚 + 𝒛 − 𝟏) + 𝝀(𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎
𝟏
⇒ (𝒙 + 𝒚 + 𝒛 − 𝟏) − (𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎
𝟑
𝟑𝒙 + 𝟑𝒚 + 𝟑𝒛 − 𝟑 − 𝟐𝒙 − 𝟑𝒚 − 𝟒𝒛 + 𝟓
⇒
=𝟎
𝟑
⇒ 𝟑𝒙 + 𝟑𝒚 + 𝟑𝒛 − 𝟑 − 𝟐𝒙 − 𝟑𝒚 − 𝟒𝒛 + 𝟓 = 𝟎
⇒𝒙−𝒛+𝟐=𝟎
Q96.If the points (1,1, p) and (-3,0,1)be equidistant from the plane
A. 5/3 or -5/3
B. 1/3 or -7/3
C. 2/3 or -7/3
D. 1/3 or -2/3
Answer: B
Q97.The position vector of the foot of perpendicular drawn from the point
P(1,8,4) to the line joining A(0,-1,3) and B(5,4,4) is:
A. (5, 5, 5)
B. (5, 4, 4)
C. (4, 4, 4)
D. (5,-4, 4)
Answer: B
Q98.The coordinates of the foot of the perpendicular drawn from the point (2, 5,
7) on the x-axis are given by
A. (2, 0, 0)
B. (0, 5, 0)
C. (0, 0, 7)
D. (0, 5, 7)
Answer: A
Explanation: Since on x-axis y and z coordinates are zero, therefore, coordinates of the foot of
the perpendicular drawn from the point (2, 5, 7) would be (2, 0, 0)
Q99.A line makes equal angles with co-ordinate axis. Direction cosines of this line
are:
A.
1
,
1
,
1
√2 √2 √2
1 1 1
C. , ,
2 2 2
B.
1
,
1
,
1
√ 3 √3 √3
1 1 1
D. , ,
5 5 5
Answer: B
Explanation: Let the line makes angle α with each of the axis.
Then, its direction cosines are cos α, cos α, cos α
𝑆𝑖𝑛𝑐𝑒 𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛼 = 1. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,
𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛼 = 1
⇒ 3𝑐𝑜𝑠 2 𝛼 = 1
1
⇒ 𝑐𝑜𝑠 2 𝛼 =
3
1
⇒ 𝑐𝑜𝑠𝛼 =
√3
Q100.A cooperative society of farmers has 50 hectares of land to grow two crops
A and B. The profits from crops A and B per hectare are estimated as Rs 10,500
and Rs 9.000 respectively. To control weeds, a liquid herbicide has to be used for
crops A and B at the rate of 20 litres and 10 litres per hectare, respectively.
Further not more than 800 litres of herbicide should be used in order to protect
fish and wildlife using a pond which collects drainage from this land. Keeping in
mind that the protection of fish and other wildlife is more important than earning
profit, how much land should be allocated to each crop so as to maximize the
total profit? Form an LPP from the above and solve it graphically. Which of the
following is true?
A. Maximum profit = Rs. 4, 95, 000
B. Maximum profit = Rs. 4, 55, 000
C. Maximum profit = Rs. 3, 95, 000
D. Maximum profit = Rs. 5, 95, 000
Answer: A
Explanation: Let x = land allocated for crop A
y = land allocated for crop B
I31As per the question,
𝑥 + 𝑦 ≤ 50. . . . . . . . . . . . . . . (1)
20𝑥 + 10𝑦 ≤ 800
2𝑥 + 𝑦 ≤ 80. . . . . . . . . . . . . (2)
𝑥 ≥ 0. . . . . . . . . . . . . . . . . . . . . . (3)
𝑦 ≥ 0. . . . . . . . . . . . . . . . . . . . . . (4)
Total Profit=Rs.(10,500x+9,000y)Z=1500(7x+6y)
Maximum profit = Rs. 4, 95, 000
Hence, 30 hectares of land should be allocated for crop A and 20 hectares of land should be
allocated for crop B.
Q101.The feasible solution for a LPP is shown in Fig.. Let Z = 3x - 4y be the
objective function. Minimum of Z occurs at
A. (0, 0)
B. (0, 8)
C. (5, 0)
D. (4, 10)
Answer: B
Explanation:
Corner points
Z value for
(0,0)
0
(5,0)
15
(6,5)
-2
(6,8)
-14
(4,10)
-28
(0,8)
-32 (minimum)
Q102.The common region determined by all the linear constraints of a LPP is/are
called:
A. corner points
B. Feasible region
C. unbounded region
D. Bounded region
Answer: B
Explanation: Feasible region
Q103.Any point in the feasible region that gives the maximum or minimum value
of the objective function is also known as
A. optimal solution
B. infeasible solution
C. constraints
D. linear values
Answer: A
Explanation: Any point in the feasible region that gives the optimal value (maximum or minimum) of
the objective function is called an optimal solution.
Q104.The inequations or equations in the variables of linear programming
problems which describes the condition under which the optimization
(maximization or minimization) is to be accomplished are called
A. objective functions
B. objective variables
C. constraints
D. decision variables
Answer: C
Explanation: The in equations or equations in the variables of linear programming problems
which describes the condition under which the optimization (maximization or minimization) is
to be accomplished are called constraints.
Q11.Linear function Z = ax + by, where a, b are constants, which has to be
maximized or minimized is called a linear objective function. Here, x and y are
known as
A. Constraints
B. Decision variables
C. Objective variables
D. None of these
Answer: B
Explanation: Linear function Z = ax + by, where a, b are constants, which has to be maximised
or minimized is called a linear objective function. Variables x and y are known as decision
variables.
Q105.Let R be the feasible region for a linear programming problem, and let Z = ax
+ by be the objective function. If R is bounded, then the objective function Z
A. only has a minimum value on R
B. only has a minimum value on R
C. may have a maximum and a minimum value on R
D. must has both a maximum and a minimum value on R
Answer: D
Explanation: Let R be the feasible region for a linear programming problem, and let Z = ax + by
be the objective function. If R is bounded, then the objective function Z has both a maximum
and a minimum value on R and each of these occurs at a corner point (vertex) of R
Q106.How many times must a fair coin be tossed so that the probability of getting
at least one head is more than 80%.
A. 5
B. 4
C. 3
D. 2
Answer: C
Explanation: We have given
Probability of getting at least one head is 80%. This implies the value of head can be 1 or 2 but
not 0.
This implies
𝑃(𝑋 ≥ 1) > 80%
80
⇒ 𝑃(𝑋 ≥ 1) >
100
80
⇒ 1 − 𝑃(𝑋 = 0) >
100
80
⇒ 𝑃(𝑋 = 0) < 1 −
100
2
⇒ 𝑃(𝑋 = 0) <
10
1
1
⇒𝑛 𝐶0 ( )𝑛 <
2
5
1 𝑛 1
⇒( ) <
2
5
To make denominator bigger the value of n would be more than 2 as with n=1 and 2 we get
denominator as 2, and 4 which is less than 5 , so not satisfying the inequality.
To satisfy this inequality The value of n can be n = 3, 4, 5 .......
This implies man must toss at least 3 times
Q107.Two numbers are selected at random ( without replacement) from first six
positive integers. let X denote the larger of the two numbers obtained. Find the
probability distribution of X. Then the mean of distribution is.
A. 4.47
B. 1.97
C. 2.27
D. 8.67
Answer: A
Explanation: The greater of two numbers can be 1, 2,3,4,5 and 6.
Therefore X can be 1,2,3,4,5,6.
1
36
3
𝑃(2) =
36
5
𝑃(3) =
36
7
𝑃(4) =
36
9
𝑃(5) =
36
11
𝑃(6) =
36
𝑃(1) =
Probability distribution of X is as follows:
X
1
Probability 1/36
2
3
4
5
6
31/36
5/36
17/36
9/36
11/36
Expectation (mean) is calculated as:
=1⋅
1
3
5
7
9
11
+2⋅
+3⋅
+4⋅
+5⋅
+6⋅
36
36
36
36
36
36
1 + 6 + 15 + 28 + 45 + 66
=
36
161
=
36
= 4.47
Q108.The average age of S and G is 35 years. If K replaces S, the average age
becomes 32 years and if K replaces G, then the average age becomes 38 years. If
the average age of D and I be half of the average age of S, G and K, then the
average age of all the five people is:
A. 28 years
B. 32 years
C. 25 years
D. None of these
Answer: Option A
Explanation:
Average Age Total
S+G 35
70
K+G 32
64
S+K 38
76
S+K+G 35
105
D+I
35
∴
𝑺 + 𝑲 + 𝑮 + 𝑫 + 𝑰 𝟏𝟎𝟓 + 𝟑𝟓
=
= 𝟐𝟖
𝟓
𝟓
Q109.A cricket player has an average score of 40 runs for 52 innings played by
him. In an innings his highest score exceeds his lowest score by 100 runs. If these
two innings are excluded, his average of the remaining 50 innings is 38 runs. Find
his highest score in an innings.
A. 80
B. 40
C. 140
D. 60
Answer: C
Explanation: Let the lowest score of the cricketer be X.
Cricketer’s Highest score = X + 100
X + X + 100 = 40 × 52 -50 × 38
2X+100 = 2080 -1900
2X = 80
X = 40
Highest score is 140 runs.
Q110.Of the four numbers, whose average is 60, the first is one-fourth of the sum
of the last three. The second number is one-third of the sum of other three, and
the third is half of the other three. Find the fourth number.
A. 52
B. 48
C. 80
D. 60
Answer: A
𝑳𝒆𝒕 𝒕𝒉𝒆 𝒏𝒐𝒔. 𝒃𝒆 𝒂, 𝒃, 𝒄, 𝒅
𝟏
𝒂 = 𝟒 (𝒃 + 𝒄 + 𝒅)
⇒ 𝟒𝒂 = 𝒃 + 𝒄 + 𝒅
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 = 𝟔𝟎
𝒂+𝒃+𝒄+𝒅
𝟒
Explanation:
𝟓𝒂
𝟒
= 𝟔𝟎
= 𝟔𝟎 ⇒ 𝒂 = 𝟒𝟖
𝑺𝒊𝒎𝒊𝒍𝒂𝒓𝒍𝒚, 𝒃 = 𝟔𝟎 & 𝒄 = 𝟖𝟎
𝑨𝒍𝒔𝒐, 𝒊𝒕 𝒊𝒔 𝒌𝒏𝒐𝒘𝒏 𝒕𝒉𝒂𝒕 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 = 𝟔𝟎,
=> 𝒂 + 𝒃 + 𝒄 + 𝒅 = 𝟐𝟒𝟎 => 𝒅 = 𝟓𝟐
Q111.The average annual income (in Rs.) of certain group of illiterate workers is A
and that of other workers is W. The number of illiterate workers is 11 times that
of other workers. Then the average monthly income (in Rs.) of all the workers is :
A.
𝐴+𝑊
2
1
C. 11𝐴 + 𝑊
B.
D.
𝐴+11𝑊
2
11𝐴+𝑊
12
Answer: D
Explanation:
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑛𝑜. 𝑜𝑓 𝑜𝑡ℎ𝑒𝑟 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 𝑥
𝑁𝑜. 𝑜𝑓 𝑖𝑙𝑙𝑖𝑡𝑒𝑟𝑎𝑡𝑒 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑥
𝑇𝑜𝑡𝑎𝑙 𝑛𝑜 𝑜𝑓 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑥 + 𝑥 = 12𝑥
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑓 𝑖𝑙𝑙𝑖𝑡𝑒𝑟𝑎𝑡𝑒 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑊𝑥
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑓 𝑜𝑡ℎ𝑒𝑟 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 𝑥 × 𝑊 = 𝑥𝑊
𝐶𝑜𝑚𝑏𝑖𝑛𝑒𝑑 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 11𝐴𝑥 + 𝑊𝑥
11𝐴𝑥 + 𝑊𝑥 11𝐴 + 𝑊
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 =
=
12𝑥
12
Q112.The average age of a family of 5 members 4 year ago was 24 years. Mean
while a child was born in this family and still the average age of the whole family
is same today. The present age of the child is:
1
A. 2 years
B. 1 2 years
C. 4 years
D. data insufficient
Answer: C
Explanation: Total age of family of 5 members
= 24 × 5 + 4 × 5 = 140
𝑇𝑜𝑡𝑎𝑙 𝑎𝑔𝑒 𝑜𝑓 𝑓𝑎𝑚𝑖𝑙𝑦 𝑜𝑓 6 𝑚𝑒𝑚𝑏𝑒𝑟𝑠
= 24 𝑥 6 = 144
𝑆𝑜, 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑔𝑒 𝑜𝑓 𝑐ℎ𝑖𝑙𝑑 = 144— 140 = 4 𝑦𝑒𝑎𝑟𝑠
Q113.The average of five positive numbers is 99. The averages of the first two and
the last two numbers are 117 and 92 respectively. What is the third number?
A. 46
B. 54
C. 56
D. 77
Answer: D
Explanation: Average of 5 numbers = 99
So, the sum of all 5 numbers = 5× 99 = 495
The average of first two numbers = 117
So, sum of first two numbers = 117 × 2 = 234
The average of second two numbers = 92
So, sum of second two numbers = 92×2 = 184
The sum of 1st, 2nd, 4th and 5th numbers = 234+184 = 418
The fifth number = 495-418 = 77
Q114.The average weight of 16 boys in a class is 50.25 kgs of which 8 playing boys
is 45.15 kgs. Find the average weight of the non-playing boys in the class.
A. 47.55 kgs
B. 48 kgs
C. 49.25 kgs
D. 55.35 kgs
Answer: D
Explanation: Total Weight of 16 boys = 16 × 50.25
Weight of 8 playing boys = 8 × 45.15
Total Weight of remaining 8 non playing boys =16 × 50.25 - 8 × 45.15 = 804 - 361.2 = 442.8
Required Average = 442.8/8 = 55.35.
Q115.Of the four numbers, whose average is 60, the first is one-fourth of the sum
of the last three. The second number is one-third of the sum of other three, and
the third is half of the other three. Find the fourth number.
A. 52
B. 48
C. 80
D. 60
Answer: A
Explanation:
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑛𝑜𝑠. 𝑏𝑒 𝑎, 𝑏, 𝑐, 𝑑
1
𝑎 = (𝑏 + 𝑐 + 𝑑)
4
⇒ 4𝑎 = 𝑏 + 𝑐 + 𝑑
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 60
𝑎+𝑏+𝑐+𝑑
= 60
4
5𝑎
= 60 ⇒ 𝑎 = 48
4
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑏 = 60 & 𝑐 = 80
𝐴𝑙𝑠𝑜, 𝑖𝑡 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛 𝑡ℎ𝑎𝑡 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 60,
=> 𝑎 + 𝑏 + 𝑐 + 𝑑 = 240 => 𝑑 = 52