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S1. Ans.(d)
Sol.
3 bowlers can be selected from the five players and 8 players can be selected from 12 players (17-5) = 12
is the number of ways of selecting the cricket team of 11 players.
P = C(5, 3) × C(12, 8)
S2. Ans.(b)
Sol.
log𝑎 27 + log8 32 log 27
log 9+
log 32
log 8
log 33
log 32 +log 25
log 23
3 log 3
2 log 3+
5 log 2
3 log 2
3
2+
5
3 =
9+10
6 =
19
6
S3. Ans.(a)
S4.Ans.(d)
Sol.
|a– x c b
c b– x ab a c– x
| = 0
c1 → c2 + c2 + c3
|a + b + c – x a + b + c – x a + b + c – x
c b– x ab a c– x
| = 0
(a + b + c – x) |1 1 1c b – x ab a c – x
| = 0
⇒ x = a + b + c = 0
[ ∵ a + b + c = 0]
S5. Ans.(b)
Sol.
Matrix A have an inverse iff |A| ≠ 0.
Consider |A| = 0.
⇒ 2x + 32 = 0
x =–32
2
x = –16
S6. Ans.(b)
Sol.
From the system of equations
2x + y – 3z = 5
3x – 2y + 2z = 5 and
5x – 3y – z = 16 𝑎1
𝑎2=
2
3≠
1
–2=
𝑏1
𝑏2
𝑎2
𝑎3=
3
5≠
2
3=
𝑏2
𝑏3
𝑎3
𝑎1=
2
5≠
1
–3=
𝑏3
𝑏1
Hence the given system of equations is consistent, with a unique solution.
S7. Ans.(d)
Sol.
Cube root of unity lie on the unit circle |z| = 1.
S8. Ans.(a)
Sol.
u = arp–1
v = arq–1
w = arr–1
consider |ln 𝑢 𝑝 1ln 𝑣 𝑞 1ln 𝑤 𝑟 1
|
= |
ln(arp–1) p 1
ln(𝑎𝑟𝑞−1) p 1
ln(arr–1) r 1
|
= |
(p– 1) ln(ar) p 1(q– 1) ln(ar) q 1(r– 1) ln(ar) r 1
| [∵ ln ab = b ln a]
= ln(ar) |p– 1 p 1q– 1 q 1r – 1 r 1
|
= ln(ar) ||p p 1q q 1r r 1
| – |1 p 11 q 11 r 1
||
ln(ar) [0 – 0] = 0 [∵ determinant have two columns are identical∴ its value is zero
]
S9. Ans.(c)
Sol.
The middle term of the expansion (1 + 𝑥)2𝑛
= (2𝑛
2+ 1)
𝑡ℎ
𝑡𝑒𝑟𝑚
The coefficient of the middle term of (1 + 𝑥)2𝑛
= 𝐶𝑛+1 =∝ 2𝑛
The middle terms of the expansion (1 + 𝑥)2𝑛−1
= (2𝑛−1+1
1)
𝑡ℎ
𝑡𝑒𝑟𝑚 𝑎𝑛𝑑 (2𝑛−1+1
2+ 1)
𝑡ℎ
𝑡𝑒𝑟𝑚
The coefficients of the middle term of the expansion (1 + 𝑥)2𝑛−1 = 𝐶𝑛 2𝑛−1 𝑎𝑛𝑑 𝐶𝑛+1
2𝑛−1
Given that 𝐶𝑛 = 𝛽 2𝑛−1
And 𝐶𝑛+1 2𝑛−1 = 𝛾
Consider,
α = β + γ
𝐶𝑛+1 2𝑛 = 𝐶𝑛
2𝑛−1 + 𝐶𝑛+1 2𝑛−1
(2𝑛)!
(𝑛+1)!(2𝑛−𝑛−1)!=
(2𝑛−1)!
(2𝑛−1−𝑛)!𝑛!+
(2𝑛−1)!
(𝑛+1)!(2𝑛−1−𝑛−1)!
(2𝑛)!
(𝑛+1)!(𝑛−1)!= (2𝑛 − 1)! [
1
(𝑛−1)!𝑛!+
1
(𝑛+1)!(𝑛−2)!]
= (2𝑛 − 1)! [(𝑛+1)!(𝑛−2)!+(𝑛−1)!𝑛!
(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]
= (2𝑛 − 1)! [(𝑛+1)𝑛!(𝑛−2)!+(𝑛−1)(𝑛−2)!𝑛!
(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]
= (2𝑛 − 1)! 𝑛! (𝑛 − 2)! [𝑛+1+𝑛−1
(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]
=2𝑛! 𝑛! (𝑛−2)!
(𝑛−1)! 𝑛! (𝑛+1)! (𝑛−2)!
=(2𝑛)!
(𝑛−1)!(𝑛+1)!
= 𝐿. 𝐻. 𝑆 = 𝑅. 𝐻. 𝑆.
Q10. Ans.(d)
Sol.
A function 𝑓(𝑥1 … . 𝑥𝑛) has the property, that for one set of values (𝑣1 … . 𝑣𝑛) there is at most one result. If
you compare. Your f(0)=1, but there are 2 values for y s.t 𝑦2 + 𝑥2 = 1 | 𝑥 = 0, namely {1, −1}.
S11. Ans.(c)
Sol.
Given Tm =1
n
a + (m– 1)d =1
n …(i)
and Tn =1
m
a + (n– 1)d =1
m …(ii)
solving (i) and (ii), we gets
d =1
mn and a =
1
mn
Now,
Tmn = a + (mn– 1)d
= 1
mn+ (mn– 1)
1
mn
= 1
S12. Ans.(b)
Sol.
𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 , 𝑓 > 0 ⇒ 𝑎 > 0.
𝑓(𝑥) + 𝑓′(𝑥) + 𝑓"(𝑥) = 𝑎𝑥2 + (2𝑎 + 𝑏)𝑥 + 𝑐 + 𝑏 + 2𝑎 = 𝑔(𝑥)
Reformulating g in terms of x + 1 gives
𝑎(𝑥 + 1)2 + (2𝑎 + 𝑏)(𝑥 + 1) + 𝑐 + 𝑏 + 2𝑎 − (2𝑎𝑥 + 𝑎) − (2𝑎 + 𝑏)
= 𝑎(𝑥 + 1)2 + (2𝑎 + 𝑏)(𝑥 + 1) + 𝑐 − 2𝑎(𝑥 + 1) + 𝑎
= 𝑎(𝑥 + 1)2 + 𝑏(𝑥 + 1) + 𝑐 + 𝑎 = 𝑔(𝑥)
So g(x) = f(x + 1) + a, so g(x) is f(x) translated by 1 to the left and by a upwards.
𝑓 > 0 ⇒ 𝑔 > 0.
S13. Ans.(c)
Sol.
1 and 3 only
S14. Ans.(b)
Sol.
First of all adding (11011)₂ & (10110110)₂ 1 1 0 1 1
1 0 1 1 0 1 1 0
1 1 0 1 0 0 0 1
Now by adding (11010001)₂ & (10011x 0y)₂
We get (101101101)₂
i.e. 1 1 0 1 0 0 0 1
1 0 0 1 1 𝑥 0 𝑦
1 0 1 1 0 1 1 0 1
⇒ y = 0 & x = 1
S15. Ans.(b)
Sol.
Given B = adj A
AB = A (adj A)
= |A|In where In is the identity matrix of A.
= kℓ [ ∵ |A| = k & In= ℓ]
S16. Ans.(c)
Sol.
Given that
ar = 2 …(i)
and 𝑎
1–𝑟= 8 …(ii) [∴ If number of terms is infinite then sum of the terms is S =
𝑎
1–𝑟, |r| < 1]
Solving (i) and (ii), we get
a = 4 and r = ½
∴ The G.P. is
4, 2, 1,1
2,
1
22 , …
Q17. Ans.(c)
Sol.
I. Let a, b, c are in A.P.
∴ 2b = a + c. ⇒ a = 2b – c
Consider 𝑎−𝑏
𝑏−𝑐=
2𝑏−𝑐−𝑏
𝑏−𝑐=
𝑏−𝑐
𝑏−𝑐= 1
II. Let a, b, c, are in G.P.
⇒ 𝑏2 = 𝑎𝑐. ⇒ 𝑐 =𝑏2
𝑎
Consider 𝑎−𝑏
𝑏−𝑐
=𝑎−𝑏
𝑏−𝑏2
𝑎
=(𝑎−𝑏)
𝑎𝑏−𝑏2
𝑎
=(𝑎−𝑏).𝑎
𝑏(𝑎−𝑏)
=𝑎
𝑏
III. Let a, b, c are in H.P
⇒ b = 2𝑎𝑐
𝑎+𝑐
Consider 𝑎−𝑏
𝑏−𝑐
= 𝑎−
2𝑎𝑐
𝑎+𝑐2𝑎𝑐
𝑎+𝑐−𝑐
= 𝑎2+𝑎𝑐−2𝑎𝑐
2𝑎𝑐−𝑎𝑐−𝑐2
=𝑎2−𝑎𝑐
𝑎𝑐−𝑐2 =𝑎(𝑎−𝑐)
𝑐(𝑎−𝑐)=
𝑎
𝑐
S18. Ans.(a)
Sol.
345 + 354 + 435 + 453 + 534 + 543 = 2664
S19. Ans.(b)
Sol.
Given, −𝑏+√𝑏2−4𝑎𝑐
−𝑏−√𝑏2−4𝑎𝑐=
−𝑞+√𝑞2−4𝑝𝑟
−𝑞−√𝑞2−4𝑝𝑟
−𝑏+√𝐷1
−𝑏−√𝐷1=
−𝑞+√𝐷2
−𝑞−√𝐷2
𝑏𝑞 + 𝑏√𝐷2 − 𝑞√𝐷1 − √𝐷1𝐷2 = 𝑏𝑞 − 𝑏√𝐷2 + 𝑞√𝐷1 − √𝐷1𝐷2
2𝑏√𝐷2 = 2𝑞√𝐷1
𝑏
𝑞=
√𝐷1
√𝐷2
⟹𝐷1
𝐷2=
𝑏2
𝑞2
S20. Ans.(b)
Sol.
𝐴 = sin2 𝜃 + cos4 𝜃
= sin2 𝜃(sin2 𝜃 + cos2 𝜃) + cos4 𝜃
= sin4 𝜃 + 2 sin2 𝜃 cos2 𝜃 + cos4 𝜃 − sin2 𝜃 cos2 𝜃
= (sin2 𝜃 + cos2 𝜃)2 − sin2 𝜃 cos2 𝜃
= 1 − sin2 𝜃 cos2 𝜃
= 1 −1
4(sin 2𝜃)2
∵ −1 ≤ sin 2𝜃 ≤ 1
0 ≤ sin2 2𝜃 ≤ 1
0 ≤1
4sin2 2𝜃 ≤
1
4
0 ≥ −1
4sin2 2𝜃 ≥ −
1
4
1 ≥ 1 −1
4sin2 2𝜃 ≥ 1 −
1
4
1 ≥ 1 −1
4sin2 2𝜃 ≥
3
4
S21. Ans.(d)
S22. Ans.(d)
S23. Ans.(c)
Sol.
Here, 𝑚1 = –ℓ
𝑚
𝑚2 =(–ℓ1)
𝑚1
Angle between two lines is
tan–1 θ = |𝑚1–m2
1+m1m2|
= |–ℓ
𝑚+
ℓ1
𝑚1
1+ℓ
𝑚
ℓ1
𝑚1
|
= |–ℓm1+ℓ1m
mm1+ℓℓ1|
= |ℓ𝑚1–ℓ1𝑚
𝑚𝑚1+ℓℓ1|
S24. Ans.(d)
S25. Ans.(a)
Sol.
Intersection of the lines 𝑥
2+
𝑦
3 = 1 and
𝑥
3+
𝑦
2 = 1 is (
6
5,
6
5)
Slope of the line 4x + 5y – 6 = 0 is –4
5
Slope of line passing through (6
5,
6
5) is same as the slope of the line 4x + 5y – 6 as they are same.
∴ the equation of line passing through (6
5,
6
5) and having slope is
–4
5
(𝑦–6
5) =
–4
5 (𝑥–
6
5)
⇒ 20x + 25y – 54 = 0
S26. Ans.(a)
Sol.
Distance of a point from the plane is
|𝑎𝑥1+𝑏𝑦1+𝑐𝑧1+𝑑
√𝑎2+𝑏2+𝑐2|
i.e. |3×2–6×3+2×4+11
√32+(–6)2+(2)2|
= |6–18+8+11
√9+36+4|
= |7
7|
= 1
S27. Ans.(c)
S28. Ans.(d)
Sol.
The line 𝑥−1
2=
𝑦−2
3=
𝑧−3
4 can be written
As 𝑥−1
3−1=
𝑦−2
5−2=
𝑧−3
7−3(𝑖. 𝑒.
𝑥−𝑥1
𝑥2−𝑥1=
𝑦−𝑦1
𝑦2−𝑦1=
𝑧−𝑧1
𝑧2−𝑧1)
This gives (𝑥1, 𝑦1, 𝑧1) = (1, 2, 3) 𝑎𝑛𝑑 (𝑥2, 𝑦2, 𝑧2) = (3, 5, 7)
As line is passing through (𝑥1, 𝑦1, 𝑧1) & (𝑥2, 𝑦2, 𝑧2)
∴ These two points must satisfy the equation of line.
Taking point (𝑥2, 𝑦2, 𝑧2) = (3, 5, 7)
Consider
L.H.S
3x + 2y – 3z
Put (x, y, z) = (3, 5, 7)
9 + 10 – 21 = – 2 = R. H. S
Again,
L.H.S
3x – 6y + 3z
Put (x, y, z) = (3, 5, 7)
9 – 30 + 21
= 0 = R. H. S.
S29. Ans.(c)
Sol.
I. cos 𝜃 = ±𝑎1𝑎2+𝑏1𝑏2+𝑐1𝑐2
√𝑎12+𝑏1
2+𝑐12√𝑎2
2+𝑏22+𝑐2
2
= ± (2–1+2
√(4+1+1)√1+1+4)
= ± (3
√6√6)
= 3
6=
1
2
∴ θ = 𝜋
3
II. 6x – 3y + 6z + 2 = 0
⇒ 2x – y + 2z + ⅔ = 0
Distance between the planes 2x – y + 2z + ⅔ = 0
And 2x – y + 2z + 4 = 0 is
|d1–d2
√a2+b2+c2|
= |2
3 – 4
√4+1+4|
= |2–12
3√9|
= |–10
9|
= 10
9
S30. Ans.(d)
Sol.
I.
Line OP = y – 0 = 𝑛−0
𝑚−0(𝑥 − 0)
𝑦 =𝑛
𝑚𝑥
nx – my = 0
and OQ = y – 0 = 𝑠−0
𝑟−0(𝑥 − 0)
𝑦 =𝑠
𝑟𝑥
sx - ry = 0
we know that if 𝑎1𝑥 + 𝑏1𝑦 + 𝑐1 = 0 𝑎𝑛𝑑 𝑎2𝑥 + 𝑏2𝑦 + 𝑐2 = 0 are two lines, the angle between them can be
calculated using the following formula
cos θ = 𝑎1𝑏2+𝑏1𝑏2
√(𝑎12+𝑏1
2)√𝑎22+𝑏2
2
∴ cosθ = 𝑛𝑠+𝑚𝑟
√(𝑛2+𝑚2√𝑠2+𝑟2
II. By the cosine rule,
cos A = 𝑏2+𝑐2−𝑎2
2𝑏𝑐
⇒ 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 𝑐𝑜𝑠𝐴
S31. Ans.(c)
Sol.
log7 log7√7√7√7
log7 log7√7√71+
1
2
log7 log7√7.7
3
2.1
2
log7 log7√71+
3
4
log7 log7 77
4.1
2
log7 log7 77
8
log7 (7
8log7 7)
= log7 (7
8)
= log7 7 – log7 8
= 1– log7 23
= 1– 3 log7 2
S32. Ans.(c)
Sol.
Sum of an infinite G.P. is
S =a
1–r, |r| < 1
i.e. 5 =x
1–r, |r| < 1.
⇒ 5 (1 – r) = x, |r| < 1
Given |r| < 1
i.e. –1 < r < 1
when r < 1
– 1 < – r
0 < 1 – r
⇒ 0 < 5 (1 – r) = x
⇒ x > 0 …(i)
When r > –1
1 > –r
1 + 1 > 1 – r
2 > 1 – r
⇒ 10 > 5 (1 – r) = x
i.e. x < 10 …(ii)
from (i) and (ii)
0 < x < 10.
S33. Ans.(a)
Sol.
As per the definition of Rational expressions
1, 4 & 5 are the rational functions
S34. Ans.(b)
Sol.
A square matrix A is called orthogonal if
ATA = I.
i.e. AT = A–1.
S35. Ans.(c)
Sol.
Let U = {1, 2, 3, 4, 5, 6, 7, 8}
A = {1, 2, 3, 4}
B = {3, 4, 5, 6}
C = {2, 3, 7, 8}
L.H.S.
A′ ∪ (B ∪ C) = (5, 6, 7, 8) ∪ [ 2, 3, 4, 5, 6, 7, 8]
= {2, 3, 4, 5, 6, 7, 8}
R.H.S. (C′ ∩ B)′ ∩ A′= [(1, 4, 5, 6) ∩ (3, 4, 5, 6)]′ ∩ (5, 6, 7, 8)
= (4, 5, 6)′ ∩ (5, 6, 7, 8)
= (1, 2, 3, 7, 8) ∩ (5, 6, 7, 8)
= (7, 8)
L.H.S. ≠ R.H.S.
S36. Ans.(b)
Sol.
Total numbers between 2999 to 8001= 5001
We find the numbers in which number doesn’t repeat than subtract it from total numbers
Numbers without repetition
5×9×8×7 = 2520
Required Answer = 5001 – 2520
= 2481
S37. Ans.(c)
Sol.
a = 3
r =–1
3
sum of infinite terms of G.P. is a
1 – r
= 3
1+1
3
= 34
3
= 3
1×
3
4 =
9
4
S38. Ans.(a) Sol.
Badminton = 125 Volleyball = 145 Tennis = 90 (B.+V.) + (V. +T.) + (T. + B.) = 32 Atq, 300 = B. + V. + T. – [common area] – 24 300 = 360 – 32 – 2x x = 14
S39. Ans.(c) Sol. Show figure in solution 3 Required value B + V + T – 2(32) – 3(14) = 360 – 64 – 42 = 254
S40. Ans.(d)
Sol.
Given that α & β are the roots of the equation x² + αx – β = 0
⇒ α + β = – α & αβ = –β
⇒ β = – 2α ⇒ α = – 1
⇒ β = 2
∴ we can write quadratic expression – x2 + αx + β as f(x) =– x2– x + 2
f ′(x) = – 2x– 1 = 0
𝑥 = –1
2
𝑓′′(𝑥) = – 2 < 0
⇒ f(x) is an increasing function,
∴ Greatest value of f(x) is f (–1
2)
i.e. 9
4
S41. Ans.(d)
Sol.
Middle term of (2 + 3𝑥)4 = (4
2+ 1)
𝑡ℎ
𝑡𝑒𝑟𝑚 = (3)𝑟𝑑𝑡𝑒𝑟𝑚
𝑇3+1 = 𝐶3(2)4−3(3𝑥)3. 4
= 𝐶3 (2)33 𝑥3
4
=4!
3!1!× 2 × 27 × 𝑥3
=4×3!
3!× 2 × 27 × 𝑥3
= 4 × 2 × 27 × 𝑥3
= 216
S42. Ans.(a)
Sol.
(𝜆𝐴)−1 =𝑎𝑑𝑗(𝜆𝐴)
|𝜆𝐴|=
𝜆𝑛−1 𝑎𝑑𝑗 𝐴
𝜆𝑛|𝐴|
=𝜆𝑛(𝐴)−1
𝜆 𝜆𝑛
=𝐴−1
𝜆
S43. Ans.(a)
Sol.
|
𝑥 𝑦 3
𝑥2 5𝑦3 9
𝑥3 10𝑦5 27
|
𝐶1 → 𝐶1 − 𝐶3.
|
𝑥 − 3 𝑦 3
(𝑥2 − 9) 5𝑦3 9
(𝑥3 − 27) 10𝑦5 27
|
= |
𝑥 − 3 𝑦 3
(𝑥 + 3)(𝑥 − 3) 5𝑦3 9
(𝑥 − 3)(𝑥2 + 9 + 3𝑥) 10𝑦5 27
|
= (𝑥 − 3) |
1 𝑦 3
𝑥 + 3 5𝑦3 9
𝑥2 + 3𝑥 + 9 10𝑦5 27
|
S44. Ans.(a)
Sol.
𝐴 = (cos(−𝜃) − sin(−𝜃)
− sin(−𝜃) cos(−𝜃))
A = (𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃
)
Adjoint A = [𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃
−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃]
S45. Ans.(b)
Sol.
We know that
1 + 𝜔𝑟 + 𝜔2𝑟 = {0, 𝑖𝑓 𝑟 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3
3, 𝑖𝑓 𝑟 𝑖𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3.
& 𝜔 =−1+𝑖√3
2, 𝜔2 =
−1−𝑖√3
2
Given (𝜔)3𝑛 + (𝜔2)3𝑛
⇒ 1 + 𝜔3𝑛 + 𝜔2(3𝑛) − 1
⇒ 3 – 1 = 2.
S46. Ans.(c)
Sol.
Consider (0.2)𝑥 = 2
log10(0.2)𝑥 = log10 2.
𝑥 log10(0.2) = log10 2.
𝑥 log10 (2
10) = log10 2.
𝑥[log10 2 − log10 10] = log10 2.
𝑥 =log10 2
log10 2−log10 10
=0.3010
0.3010−1
=0.3010
−0.6990
= −0.4
S47. Ans.(c)
Sol.
There are 9 choices for the first digits, since 0 can’t be used. For the second digit, you can use any of the
remaining 9 digits. For the third digit you can use any of the 8 digits not already used. For the next digit,
there are 7 choices. And for the final digit there are 6 choices left. Multiplying the values together gives
the stated answer.
9 × 9 × 8 × 7 × 6 = 27216
S48. Ans.(d)
Sol.
(𝑥 𝑦 𝑦 + 𝑧𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦
)
𝑅1 → 𝑅1 + 𝑅2 + 𝑅3
(𝑥 + 𝑦 + 𝑧 𝑥 + 𝑦 + 𝑧 2(𝑥 + 𝑦 + 𝑧)
𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦
)
= (𝑥 + 𝑦 + 𝑧) (1 1 1𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦
)
= (𝑥 + 𝑦 + 𝑧)(𝑧 − 𝑥)2
S49. Ans.(b)
Sol.
|1 1 1
1 + 𝑆𝑖𝑛𝐴 1 + 𝑆𝑖𝑛𝐵 1 + 𝑆𝑖𝑛𝐶𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 𝑆𝑖𝑛𝐵 + 𝑆𝑖𝑛2𝐵 𝑆𝑖𝑛𝐶 + 𝑆𝑖𝑛2𝐶
|
𝐶2 → 𝐶2 − 𝐶1 𝑎𝑛𝑑 𝐶3 → 𝐶3 − 𝐶1
|1 0 0
1 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 − 𝑠𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐵 − sin2 𝐴 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴 + sin2 𝐶 − 𝑆𝑖𝑛2𝐴
|
= (𝑠𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴)
|1 0 0
1 + 𝑆𝑖𝑛𝐴 1 1𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 1 + 𝑠𝑖𝑛𝐵 + 𝑆𝑖𝑛𝐴 1 + 𝑆𝑖𝑛𝐶 + 𝑆𝑖𝑛𝐴
|
= (𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐵) = 0 ⟹ 𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴 = 0 , 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴 = 0 , 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐵 = 0
𝑆𝑖𝑛𝐵 = 𝑆𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 = 𝑆𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 = 𝑆𝑖𝑛𝐵𝐴 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐵
⇒ The triangle ABC is equilateral.
S50. Ans.(b)
S51. Ans.(c)
Sol.
=2 tan 𝜃
1+tan2 𝜃
=2 tan 𝜃
sec2 𝜃
=2 sin 𝜃
cos 𝜃cos2 𝜃
= 2 sin 𝜃 cos 𝜃 = sin 2 𝜃.
S52.Ans (a)
Sol.
2sec𝜃 =sec(𝜃 + 𝛼)+sec(𝜃 − 𝛼)
2
𝑐𝑜𝑠𝜃=
cos(𝜃+𝛼)+cos (𝜃−𝛼)
cos(𝜃+𝛼)cos (𝜃−𝛼)
2
𝑐𝑜𝑠𝜃=
2cos(𝜃)+cos (𝛼)
cos(𝜃+𝛼)cos (𝜃−𝛼)
𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝛼 = 𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝛼
𝑠𝑖𝑛2𝛼 = 𝑐𝑜𝑠2𝜃(1 − 𝑐𝑜𝑠𝛼)
1-𝑠𝑖𝑛2𝜃 = 1 + 𝑐𝑜𝑠𝛼
𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠𝛼 = 0
53. Ans.(d)
Sol.
sin 2𝐴 − sin 2𝐵 − sin 2𝐶.
sin (2(180 − (𝐵 + 𝐶)) − sin 2𝐵 − sin 2𝐶.
sin (360 − (2𝐵 + 2𝐶) − sin 2𝐵 − sin 2𝐶.
− sin(2𝐵 + 2𝐶) − sin 2𝐵 − sin 2𝐶.
− sin 2𝐵 cos 2𝐶 − cos 2𝐵 sin 2𝐶 − sin 2𝐵 − sin 2𝐶
− sin 2𝐵(cos 2𝐶 + 1) − sin 2𝐶(cos 2𝐵 + 1)
− sin 2𝐵(2 cos2 𝐶 − 1 + 1) − sin 2C(2 cos2 𝐵 − 1 + 1)
−2 cos2 𝐶 sin 2𝐵 − 2 cos2 𝐵 sin 2𝐶
−4 cos2 𝐶 sin 𝐵 cos 𝐵 − 4 cos2 𝐵 sin 𝐶 cos 𝐶
−4 cos 𝐶 cos 𝐵[cos 𝐶 sin 𝐵 + sin 𝐶 cos 𝐵)
−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 sin (𝐵 + 𝐶)
−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 sin (180 − 𝐴)
−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 𝑠𝑖𝑛𝐴
S54. Ans.(b)
Sol.
tan 48° = 122
𝑥
𝑥 =122
tan 48°
=122
cot 42°
= 122 tan 42°
S55. Ans.(a)
Sol.
By going through the options
Option (A)
300° = (360° – 60°)
∵ the angle lies in 4th quadrant
∴ 300° = – 60°
Putting – 60° in option A
3[3 − tan2(−60°) − cot (−60°)]2
= 3 [3 − 3 +1
√3]
2
= 3×1
3= 1 𝐴𝑛𝑠.
S56. Ans.(b)
Sol.
In ∆ ABC.
tan 30° = 𝑦
𝑥
1
√3=
𝑦
𝑥
𝑥 = √3 𝑦 __________(1)
In ∆ ADE
tan 60° = 𝑦+ℎ
𝑥
√3 =𝑦+ℎ
𝑥
√3 𝑥 = 𝑦 + ℎ
√3 × √3𝑦 = 𝑦 + ℎ
3𝑦 = 𝑦 + ℎ
2y = h
y = ℎ
2
∴ height of hill = y + h = ℎ
2+ ℎ =
3ℎ
2
S57. Ans.(b)
Sol.
𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥 = √3 1
sin 𝑥+
cos 𝑥
sin 𝑥= √3
1+cos 𝑥
sin 𝑥= √3
1+2 cos2𝑥
2−1
2 sin𝑥
2cos
𝑥
2
= √3
cot𝑥
2= cot
𝜋
6
𝑥
2=
𝜋
6
𝑥 =𝜋
3
S58. Ans.(b)
Sol. (2𝑐𝑜𝑠𝜃 + 1)10(2 cos 2𝜃 − 1)10(2 cos 𝜃 − 1)10(2 cos 4𝜃 − 1)10
[(2 cos 𝜃 + 1)(2 cos 𝜃 − 1)(2 cos 2 𝜃 − 1)(2 cos 4𝜃 − 1)]10
[(4 cos2 𝜃 − 1)(2 cos 2𝜃 − 1)(2 cos 4𝜃 − 1)]10
[(2 (2 𝑐𝑜𝑠2 𝜃 − 1) + 1)(2 cos 2 𝜃 − 1)(2 cos 4𝜃 − 1)]10
[(2 cos 2𝜃 + 1)(2 cos 2𝜃 − 1)(2 cos 4 𝜃 − 1)]10
[(4 cos2 2 𝜃 − 1)(2 cos 4 𝜃 − 1)]10
[(2(2 cos2 2𝜃 − 1) + 1)(2 cos 4𝜃 − 1)]10
[(2 cos 4𝜃 + 1)(2 cos 4 𝜃 − 1)]10 [4 cos2 4𝜃 − 1]10
𝑎𝑡 𝜃 =𝜋
8
[4 cos2 𝜋
2− 1]
10
(−1)10 = 1
S59. Ans.(a)
Sol.
Given that
𝑐𝑜𝑠𝛼 + 𝑐𝑜𝑠𝛽 = 0 ⇒ 𝑐𝑜𝑠𝛼 = −𝑐𝑜𝑠𝛽 _______(1)
& 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛽 =−3
4
⇒ − cos2 𝛽 =−3
4 [𝑓𝑟𝑜𝑚 (1)]
⇒ cos2 𝛽 =3
4
Consider
𝑠𝑒𝑐𝛼 × 𝑠𝑒𝑐𝛽 1
𝑐𝑜𝑠𝛼×
1
𝑐𝑜𝑠𝛽
−1
cos2 𝛽= −
4
3
S60. Ans.(a)
Sol.
tan−1(2𝑥) + tan−1 3𝑥 =𝜋
4
tan−1 (2𝑥+3𝑥
1−6𝑥2) =𝜋
4
5𝑥
1−6𝑥2 = 1
5𝑥 = 1 − 6𝑥2
6𝑥2 + 5𝑥 − 1 = 0
6𝑥2 + 6𝑥 − 𝑥 − 1 = 0
6𝑥(𝑥 + 1) − 1(𝑥 + 1) = 0 (6𝑥 − 1)(𝑥 + 1) = 0
⇒ 𝑥 =1
6𝑜𝑟 𝑥 = −1.
S61. Ans.(a)
Sol.
lim𝑥→1
𝑓(𝑥)−𝑓(1)
𝑥−1
= lim𝑥→1
√25−𝑥2−√24
𝑥−1= (
0
0) 𝑓𝑜𝑟𝑚
Applying L’Hospital rule
= lim𝑥→1
1 (−2𝑥)
2√25−𝑥2= lim
𝑥→1
(−𝑥)
√25−𝑥2=
−1
√24
S62. Ans.(a)
Sol.
y = tan−1 (5−2 tan √𝑥
2+5 tan √𝑥)
𝑙𝑒𝑡 𝑢 =5−2 tan √𝑥
2+5 tan √𝑥
y = tan−1 𝑢 𝑑𝑦
𝑑𝑢=
1
1+𝑢2 _______(1)
u = 5−2 tan √𝑥
2+5 tan √𝑥
𝑑𝑢
𝑑𝑥=
(2+5 tan √𝑥)𝑑
𝑑𝑥(5−2 tan √𝑥)−(5−2 tan √𝑥)
𝑑
𝑑𝑥(2+5 tan √𝑥)
(2+5 tan √𝑥)2
=(2+5 tan √𝑥) (−2 sec2 √𝑥.
1
2√𝑥)−(5−2 tan √𝑥).5 sec2 √𝑥 .
1
2√𝑥
(2+5 tan √𝑥)2
=
−2 sec2 √𝑥
√𝑥−
5 𝑡𝑎𝑛√𝑥 sec2 √𝑥
√𝑥−
25
2
sec2 √𝑥
√𝑥+
5 tan √𝑥 sec2 √𝑥
√𝑥
(2+5 tan √𝑥)2
𝑑𝑢
𝑑𝑥=
−29
2√𝑥
sec2 √𝑥
(2+5 tan √𝑥)2 ______(2)
𝑑𝑦
𝑑𝑥=
𝑑𝑦
𝑑𝑢×
𝑑𝑢
𝑑𝑥
=1
1+(5−2 tan √𝑥
2+5 tan √𝑥)
2(−29)
2√𝑥
sec2 √𝑥
(2+5 tan √𝑥)2
=1
(2+5 tan √𝑥)2+(5−2 𝑡𝑎𝑛√𝑥)2
(2+5 tan √𝑥)2
.(−29)
2√𝑥.
sec2 √𝑥
(2+5 tan √𝑥)2
=1
29 (1+tan2 √𝑥)
(−29)
2√𝑥 . sec2 √𝑥.
=1
29 sec2 √𝑥
(−29) sec2 √𝑥
2√𝑥
=−1
2√𝑥
S63. Ans.(a)
Sol.
𝑓(𝑥) = 𝑥 sin 𝑥 + cos 𝑥 +1
2cos2 𝑥
⇒ 𝐹′(𝑥) = 𝑥 cos 𝑥 + sin 𝑥 − sin 𝑥 − sin 𝑥 cos 𝑥
cos 𝑥(𝑥 − sin 𝑥) > 0 𝑖𝑛 [0,𝜋
2]
S64. Ans.(c)
Sol.
= lim𝜃→0
√1−cos 𝜃
𝜃
= lim𝜃→0
√1−1+2 sin2𝜃
2
𝜃
= lim𝜃→0
√2 sin𝜃
2
𝜃= (
0
0) 𝑓𝑜𝑟𝑚
Applying L’Hospital rule
lim𝜃→0
√2 cos𝜃
2.
1
2=
√2
2=
1
√2
S65. Ans.(c)
Sol.
𝑓(𝑥) = 𝑥2 − 4𝑥 + 5
The coordinates x and y of the vertex of the graph of f are given by
x = – b/2a
= 4/2 = 2.
& y = f(2) = 4 – 8 + 5 = 1
The leading coefficient a = 1 is positive & therefore the graph of f has a minimum point at (x, y) = (2, 1)
the range of f is given by the interval [1, ∞)
But A = (1, 4)
at x = 4, y = 5
Hence the range of function is [1, 5]
S66. Ans.(c)
Sol.
∫ [𝑥]𝑑𝑥 + ∫ [−𝑥]𝑑𝑥.𝑏
𝑎
𝑏
𝑎
∫ [𝑥]𝑑𝑥 + ∫ −[𝑥]𝑑𝑥𝑏
𝑎
𝑏
𝑎 [∵ [−𝑥] = {
−[𝑥], 𝑖𝑓 𝑥 𝜖 𝐼
−[𝑥] − 1, 𝑖𝑓 𝑥 ∉ 𝐼
𝑎 ∫ 𝑑𝑥 − 𝑎 ∫ 𝑑𝑥𝑏
𝑎
𝑏
𝑎
= 0
Q67. Ans.(d)
Sol.
∫ |𝑥 − 5|𝑑𝑥8
2
− ∫ (𝑥 − 5)𝑑𝑥 + ∫ (𝑥 − 5)𝑑𝑥8
5
5
2
− [𝑥2
2− 5𝑥]
2
5
+ [𝑥2
2− 5𝑥]
5
8
− [25
2− 25 − 2 + 10] + [
64
2− 40 −
25
2+ 25]
= 9
2+
9
2
= 9
S68. Ans.(d)
Sol.
∫ 𝑆𝑖𝑛3𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥
∫ sin2 𝑥 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥
∫(1 − cos2 𝑥)𝑐𝑜𝑠𝑥 𝑆𝑖𝑛𝑥 𝑑𝑥
Let cos x = t
–Sin x dx = dt
Sin x dx = – dt
− ∫(1 − 𝑡2)𝑡 𝑑𝑡
∫ 𝑡3 − 𝑡 𝑑𝑡 𝑡4
4−
𝑡2
2+ 𝐶′
𝑡4−2𝑡2
4+ 𝐶′
cos4 𝑥−2 cos2 𝑥+1
4−
1
4+ 𝐶′.
(1−cos2 𝑥)2
4+ 𝐶
Where C = 𝐶′ −1
4
S69. Ans.(b)
Sol.
∫ 𝑒𝐿𝑛(tan 𝑥)𝑑𝑥.
∫ tan 𝑥 𝑑𝑥.
𝑙𝑛 | sec 𝑥| + 𝑐
S70. Ans.(d)
Sol.
= ∫𝑑
𝑑𝑥(tan−1 1
𝑥)𝑑𝑥1
−1
= 2 ∫𝑑
𝑑𝑥(tan−1 1
𝑥) 𝑑𝑥
1
0
= 2 [tan−1 1
𝑥]
0
1
= 2[tan−1 1 − tan−1 0]
= 2 [𝜋
4− 0] =
𝜋
2.
S71. Ans.(a)
Sol.
𝑓(𝑥) = 𝑥2 − 5𝑥 + 6
𝑓′(𝑥) = 2𝑥 − 5
As function is decreasing
∴ 𝑓′(𝑥) < 0
⇒ 2x – 5 < 0
⇒ 𝑥 <5
2
𝑖. 𝑒. 𝑥 < 2.5
Hence f(x) is decreasing in the interval (−∞, 2]
S72. Ans.(d)
Sol.
y = pcos (ax) + q sin (ax) 𝑑𝑦
𝑑𝑥= −𝑎𝑝 sin 𝑎𝑥 + 𝑎 𝑞 cos 𝑎𝑥
𝑑2𝑦
𝑑𝑥2 = −𝑎2𝑝 cos 𝑎 𝑥 − 𝑎2𝑞 sin 𝑎 𝑥.
𝑑2𝑦
𝑑𝑥2 = −𝑎2(𝑝 cos 𝑎 𝑥 + 𝑞 sin 𝑎𝑥)
𝑑2𝑦
𝑑𝑥2 = −𝑎2𝑦
𝑑2𝑦
𝑑𝑥2 + 𝑎2𝑦 = 0
Q73. Ans.(b)
Sol.
Given 𝑑𝑦
𝑑𝑥= −𝑥2 −
1
𝑥3
Integrate both sides w.r.t x.
∫𝑑𝑦
𝑑𝑥= ∫ (−𝑥2 −
1
𝑥3) 𝑑𝑥
= y = −𝑥3
3+
1
2𝑥2 + 𝐶
At (–1, –2)
−2 =−(−1)3
3+
1
2(−1)2 + 𝐶
−2 =1
3+
1
2+ 𝐶
−2 −5
6= 𝐶
−17
6= 𝐶
∴ 𝑦 = −𝑥3
3+
1
2𝑥2 −17
6
6𝑥2𝑦 = −2𝑥5 + 3 − 17𝑥2
6𝑥2𝑦 + 17𝑥2 + 2𝑥5 − 3 = 0
S74. Ans.(d)
Sol.
y = a cos x + b sin x + 𝑐𝑒−𝑥 + 𝑑
Order = 4.
As the order of differential equation is equal to the number of arbitrary constants in the given relation.
S75. Ans.(d)
Sol.
𝑙𝑛 (𝑑𝑦
𝑑𝑥) = 𝑎𝑥 + 𝑏𝑦
𝑑𝑦
𝑑𝑥= 𝑒𝑎𝑥+𝑏𝑦
𝑑𝑦
𝑑𝑥= 𝑒𝑎𝑥 . 𝑒𝑏𝑦
𝑒−𝑏𝑦𝑑𝑦 = 𝑒𝑎𝑥𝑑𝑥.
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠
∫ 𝑒−𝑏𝑦𝑑𝑦 = ∫ 𝑒𝑎𝑥𝑑𝑥
𝐶 +𝑒−𝑏𝑦
−𝑏=
𝑒𝑎𝑥
𝑎
𝐶 =𝑒𝑎𝑥
𝑎+
𝑒−𝑏𝑦
𝑏
S76. Ans.(b)
Sol.
Let 𝐸1, 𝐸2, 𝐸3 & 𝐴 be the events defined as follows:
𝐸1 = 𝐵𝑜𝑙𝑡 is manufactured by machine X.
𝐸2 = 𝐵𝑜𝑙𝑡 is manufactured by machine Y.
𝐸3 = Bolt is manufactured by machine Z.
A = Bolt is defective.
𝑃(𝐸1) =25
100
𝑃(𝐸2) =35
100
𝑃(𝐸3) =40
100
𝑃 (𝐴
𝐸1) = Probability that the bolt is defective given that it is manufactured by machine X =
2
100
𝑃 (𝐴
𝐸2) =
4
100
𝑃 (𝐴
𝐸3) =
5
100
Required Probability = Probability that the bolt is manufactured by machine X given that the bolt is
defective.
= 𝑃(𝐸2
𝐴)
=𝑃(𝐸2)𝑃(
𝐴
𝐸2)
𝑃(𝐸1)𝑃(𝐴
𝐸1)+𝑃(𝐸2)𝑃(
𝐴
𝐸2)+𝑃(𝐸3)𝑃(
𝐴
𝐸3)
=35
100×
4
40025
100×
2
100+
35
100×
4
100+
40
100×
5
100
=35×4
50+140+200
=35×4
390
=14
39
S77. Ans.(c)
Sol.
Let P denote the probability of getting head in a single toss of a coin. Then
P = 1
2 𝑎𝑛𝑑 𝑠𝑜, 𝑞 =
1
2
Let X denote the number of heads in a toss of 8 coins. Then X is a binomial variate with parameters n = 8
and p = 1
2 such that
𝑃(𝑋 = 𝑟) = 𝐶𝑟 8 (
1
2)
8−𝑟
(1
2)
𝑟
= 𝐶𝑟 8 (
1
2)
8
Where r = 0, 1, 2, ….8.
Probability of at least 6 heads = 𝑃(𝑋 ≥ 6)
= 𝑝(𝑋 = 6) + 𝑝(𝑋 = 7) + 𝑝(𝑋 = 8)
= 𝐶6 (1
2)
8
+ 𝐶7 8 (
1
2)
8
+ 𝐶8 8 (
1
2)
8
8
= (1
2)
8
[ 𝐶6 + 𝐶7 + 𝐶8 8 ]
8
8
= (1
2)
8[28 + 8 + 1]
=37
256
S78. Ans.(a)
Sol.
One girl and 2 boys can be selected in the following mutually exclusive ways:
Group 1 Group2 Group3
(I) Girl Boy Boy
(II) Boy Girl Boy
(III) Boy Boy Girl
Thus if we define 𝐺1, 𝐺2, 𝐺3 as the events of selecting a girl from first, second & third group respectively
and 𝐵1, 𝐵2, 𝐵3, as the events of selecting a boy from first, second and third group respectively. Then
𝐵1, 𝐵2, 𝐵3, 𝐺1, 𝐺2, 𝐺3 are independent events such that
𝑃(𝐺1) =3
4, 𝑃(𝐺2) =
2
4, 𝑃(𝐺3) =
1
4
𝑃(𝐵1) =1
4, 𝑃(𝐵2) =
2
4, 𝑃(𝐵3) =
3
4
Required Probability = P (selecting 1 girl and 2 boys)
= (I or II or III)
= P(I ∪ II ∪ III)
= 𝑃[(𝐺1 ∩ 𝐵2 ∩ 𝐵3) ∪ (𝐵1 ∩ 𝐺2 ∩ 𝐵3) ∪ (𝐵1 ∩ 𝐵2 ∩ 𝐺3)]
= 𝑃(𝐺1 ∩ 𝐵2 ∩ 𝐵3) + 𝑃(𝐵1 ∩ 𝐺2 ∩ 𝐵3) + 𝑃(𝐵1 ∩ 𝐵2 ∩ 𝐺3)
= 𝑃(𝐺1) 𝑃(𝐵2) 𝑃(𝐵3) + 𝑃((𝐵1)𝑃(𝐺2)𝑃(𝐵3) + 𝑃(𝐵1)𝑃(𝐵2)𝑃(𝐺3)
=3
4×
2
4×
3
4+
1
4×
2
4×
3
4+
1
4×
2
4×
1
4
=9
32+
3
32+
1
32=
13
32
S79. Ans.(c)
Sol.
1 and 3 only.
S80. Ans.(b)
Sol.
𝜎2 =1
𝑛Σ(𝑥𝑖 − 𝑥)2
Given that 𝜎2 = 4. 𝑎𝑛𝑑 𝑛 = 25
⇒ 4 = 1
25Σ(𝑥𝑖 − 𝑥)2 _________(1)
If 2 is added to each observation
Then the variance of 25 observation is
𝜎2 =1
25Σ(𝑥𝑖 + 2 − (𝑥 + 2))
2
=1
25Σ(𝑥𝑖 + 2 − 𝑥 − 2)2
1
25Σ(𝑥𝑖 − 𝑥)2 = 4 [form(1)]
S81. Ans.(a)
Sol.
Given that
P = (𝑥1. 𝑥2 … . 𝑥𝑛)1/𝑛
𝐿𝑜𝑔 𝑃 =1
𝑛 𝐿𝑜𝑔(𝑥1. 𝑥2, … . . 𝑥𝑛)
𝑛 𝐿𝑜𝑔 𝑃 = log 𝑥1 + log 𝑥2 + ⋯ + log 𝑥𝑛 _______(1)
Also,
Q = (𝑦1. 𝑦2. … … 𝑦𝑛)1/𝑛.
Log Q = 1
𝑛log(𝑦1. 𝑦2. … . . 𝑦𝑛)
𝑛 log 𝑄 = log 𝑦1 + log 𝑦2 + ⋯ + log 𝑦𝑛 ________(2)
Now,
G. M of 𝑥
4= (
𝑥1
𝑦1,
𝑥2
𝑦2. … .
𝑥𝑛
𝑦𝑛)
1/𝑛
Log (𝐺. 𝑀) =1
𝑛 𝐿𝑜𝑔 (
𝑥1
𝑦1.
𝑥2
𝑦2. … .
𝑥𝑛
𝑦𝑛).
n log (G. M) = 𝐿𝑜𝑔𝑥1
𝑦1+ log
𝑥2
𝑦2+ ⋯ + log
𝑥𝑛
𝑦𝑛
n log (G. M) = 𝐿𝑜𝑔 𝑥1 + 𝐿𝑜𝑔 𝑥2 + ⋯ + 𝐿𝑜𝑔 𝑥𝑛 − (log 𝑦1 + log 𝑦2 + ⋯ + log 𝑦𝑛).
𝑛 (𝐿𝑜𝑔 (𝐺. 𝑀) = 𝑛 𝐿𝑜𝑔 𝑃 − 𝑛 𝐿𝑜𝑔 𝑄
𝑛 log(𝐺. 𝑀. ) = 𝑛(𝐿𝑜𝑔 𝑝 − log 𝜃)
𝐿𝑜𝑔 𝐺. 𝑀 = 𝐿𝑜𝑔𝑃
𝑄
G. M = 𝑃
𝑄
S82. Ans.(d)
Sol.
I. Since P(exactly one of A, B occurs) = q (given), we get
𝑃(𝐴 ∪ 𝐵) − 𝑃(𝐴 ∩ 𝐵) = 𝑞
p – 𝑃(𝐴 ∩ 𝐵) = 𝑞
= 𝑃(𝐴 ∩ 𝐵) = 𝑝 − 𝑞
1 − 𝑃(𝐴 ∪ 𝐵) = 𝑝 − 𝑞
𝑃(𝐴 ∪ 𝐵) = 1 − 𝑝 + 𝑞
𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 1 − 𝑝 + 𝑞
𝑃(𝐴) + 𝑃(𝐵) = (1 − 𝑝 + 𝑞) + 𝑃(𝐴 ∩ 𝐵)
= (1 − 𝑝 + 𝑞) + (1 − 𝑃(𝐴 ∪ 𝐵))
= (1 − 𝑝 + 𝑞) + (1 − 𝑝)
= 2 − 2𝑝 + 𝑞
II. P(𝐴 ∩ 𝐵) = 1 − 𝑃(𝐴 ∪ 𝐵)
= 1 – p
S83. Ans.(b)
Sol.
Correlation coefficient = √𝑟𝑥𝑦 + 𝑟𝑦𝑥
−1
2= √𝑟𝑥𝑦 × −6
1
4= 𝑟𝑥𝑦 × −6
−1
24= 𝑟𝑥𝑦
S84. Ans (d) by definition.
S85. Ans.(c)
Sol.
No. of pairs of numbers from the set {0, 1, ---10}
{
(0,0), (0,1) … . . (0, 10)(1, 0), (1, 1) … . . (1, 10)
(10, 0) (10, 1) … (10, 10)
}
= 121
Favorable Outcomes = {(6, 0), (7, 0), (8, 0),(9, 0), (10, 0), (10, 1), (9, 1), (8, 1), (7, 1), (10, 2),
(9, 2), (8, 2), (10, 3), (9, 3), (10, 4), (0,6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 10), (1, 9), (1, 8), (1, 7),
(2, 10), (2, 9), (2, 8), (3, 10), (3, 9), (4, 10)}
∴ Probability = 30
121
S86. Ans.(d)
Sol.
Average = 500×1860+600×1750
1100= 1800
Combined variance
=500(81+3600)+600(100+2500)
1100
=(5×3681)+(6×2600)
1100
≈ 3092
S87. Ans.(d)
Sol.
I II III
⑥, ⑤, ④, ③ ← 2 ← 1
⑥, ⑤, ④ ← 3
⑥, ⑤ ← 4
⑥ ← 5
I II III
⑥, ⑤, ④ ← 3 ←2
⑥, ⑤ ← 4
⑥ ← 5
I II III
⑥, ⑤ ← 4 ← 3
⑥ ←5
I II III
⑥ ← 5 ←4
Number of possible outcome =20
S88. Ans.(b)
S89. Ans.(b)
Sol.
For discrete series
σ = √1
𝑁Σ 𝑓𝑖(𝑥𝑖 − 𝑥)2
i.e. y = √𝑥
𝑦2 = 𝑥
⇒ 𝑥 ≥ 𝑦
S 90.Ans (c)
S91. Ans.(d)
Sol.
(𝑥1,1
𝑥1) , (𝑥2,
1
𝑥2) , (𝑥3,
1
𝑥3) are the vertices of the triangle.
Area of the triangle
=1
2|𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3(𝑦1 − 𝑦2)|
=1
2|𝑥1 (
1
𝑥2−
1
𝑥3) + 𝑥2 (
1
𝑥3−
1
𝑥1) + 𝑥3 (
1
𝑥1−
1
𝑥2)|
=1
2|𝑥1
(𝑥3−𝑥2)
𝑥2 𝑥3+
𝑥2(𝑥1−𝑥3)
𝑥1 𝑥3+
𝑥3(𝑥2−𝑥1)
𝑥1 𝑥2|
1
2|
𝑥12(𝑥3−𝑥2)+𝑥2
2(𝑥1−𝑥3)+𝑥32(𝑥2−𝑥1)
𝑥1𝑥2𝑥3|
=1
2|
𝑥12𝑥3−𝑥1
2𝑥2+𝑥1𝑥22−𝑥2
2𝑥3+𝑥2𝑥32−𝑥1𝑥3
2
𝑥1𝑥2𝑥3|
=1
2|
𝑥1𝑥2𝑥3+𝑥12𝑥3−𝑥1
2𝑥2+𝑥1𝑥22−𝑥2
2𝑥3+𝑥2𝑥32−𝑥1𝑥3
2−𝑥1𝑥2𝑥3
𝑥1𝑥2𝑥3|
=1
2|
(𝑥1𝑥2−𝑥1𝑥3−𝑥22+𝑥2𝑥3)(𝑥3−𝑥1)
𝑥1𝑥2𝑥3|
=1
2|
(𝑥1−𝑥2)(𝑥2−𝑥3)(𝑥3−𝑥1)
𝑥1𝑥2𝑥3|
S92. Ans.(a)
Sol.
Let the circle touches the Y-axis and has its centre C(h, k).
Then the equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = ℎ2
Or 𝑥2 + 𝑦2 − 2ℎ𝑥 − 2𝑘𝑦 + 𝑘2 = 0 _____(1)
Given equation is 𝑥2 + 𝑦2 + 𝑔𝑥 + 𝑓𝑦 +𝑐
4= 0 _____(2)
By comparison of (1) & (2), we get
–2h = g and –2k =f
h = −𝑔
2 k = –
𝑓
2
i.e. (−𝑔
2,
−𝑓
2)
S93. Ans.(d)
Sol.
(�⃗� + �⃗⃗�). (�⃗� + �⃗⃗�) = �⃗�. �⃗� + �⃗�. �⃗⃗� + �⃗⃗�. �⃗� + �⃗⃗� . �⃗⃗�
= |�⃗�|2 + �⃗�. �⃗⃗� + �⃗⃗�. �⃗� + |�⃗⃗�|2
We can conclude that (�⃗� + �⃗⃗�). (�⃗� + �⃗⃗�) = |�⃗�|2 + |�⃗⃗�|2
If and only if �⃗�. �⃗⃗� = −�⃗⃗�. �⃗�
i.e. �⃗� 𝑎𝑛𝑑 �⃗⃗� are anti parallel.
S94. Ans.(d)
Sol.
Given that 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧�̂�
Consider
�̂�. (𝑖̂ + 𝑗̂ + �̂�).
(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧�̂�)(𝑖̂ + 𝑗̂ + �̂�)
= (𝑥 + 𝑦 + 𝑧) [
∵ 𝑖̂. 𝑖̂ = 1 𝑖̂. 𝑗̂ = 0
𝑗̂. 𝑗̂ = 1 𝑎𝑛𝑑 𝑗̂. �̂� = 0
�̂�. �̂� = 1 �̂�. 𝑖̂ = 0
]
S95. Ans.(a)
Sol.
Let unit vector �̂� = 𝑥𝑖̂+y𝑗̂ + 𝑧�̂�
Let A = 2𝑖̂ − 𝑗̂ + �̂�
B = 3𝑖̂ − 4𝑗̂ − �̂�
Given that �̂�. 𝐴 = 0
⇒ 2𝑥 − 𝑦 + 𝑧 = 0 _______(1)
And �̂�. 𝐵 = 0
⇒ 3x – 4y – z = 0 _________(2)
From (1) & (2), we get
x = y = – z.
Hence required unit vector
=𝑥
√𝑥2+𝑦2+𝑧2𝑖̂ +
𝑦
√𝑥2+𝑦2+𝑧2𝑗̂ −
𝑧
√𝑥2+𝑦2+𝑧2�̂�
=1
√3𝑖̂ +
1
√3𝑗̂ −
1
√3�̂�
S96. Ans.(d)
Sol.
Given that,
|�⃗� − �⃗⃗�| = 5
|�⃗� − �⃗⃗�|2
= 25
|�⃗�|2 + |�⃗⃗�|2
− 2(�⃗�. �⃗⃗�) = 25
9 + 16 − 2(�⃗�. �⃗⃗�) = 25
�⃗�. �⃗⃗� = 0 _____(1)
Consider
|�⃗� + �⃗⃗�|2
= |�⃗�|2 + |�⃗⃗�|2
+ 2(�⃗�. �⃗⃗�)
= 9 + 16 + 0
= 25
⇒ |�⃗� + �⃗⃗�| = 5
S97. Ans.(c)
Sol.
Given that �⃗�, �⃗⃗�, 𝑐 are three mutually perpendicular vectors.
∴ �⃗�. �⃗� = �⃗⃗�. �⃗⃗� = 𝑐. 𝑐 = 1
And �⃗�. �⃗⃗� = �⃗⃗�. 𝑐 = 𝑐. �⃗� = 0
Also given that each of unit magnitude
∴ |�⃗�| = 1 = |�⃗⃗�| = |𝑐|
|𝐴| = √�⃗�2 + �⃗⃗�2 + 𝑐2 = √3
|�⃗⃗�| = √�⃗�2 + (−�⃗⃗�)2
+(𝑐)2 = √3
|𝑐| = √�⃗�2 + (−�⃗⃗�)2
+ (−𝑐)2 = √3
Hence, |𝐴| = |�⃗⃗�| = |𝐶|
S98. Ans.(c)
Sol.
(�⃗� − �⃗⃗�) × (�⃗� + �⃗⃗�)
|𝑖̂ 𝑗̂ �̂�
𝑎1 − 𝑏1 𝑎2 − 𝑏2 𝑎3 − 𝑏3
𝑎1 + 𝑏1 𝑎2 + 𝑏2 𝑎3 + 𝑏3
|
= 𝑖̂[(𝑎2 − 𝑏2)(𝑎3 + 𝑏3) − (𝑎2 + 𝑏2)(𝑎3 − 𝑏3)] − 𝑗̂[(𝑎1 − 𝑏1)(𝑎3 + 𝑏3) − (𝑎1 + 𝑏1)(𝑎3 − 𝑏3)]
+�̂�[(𝑎1 − 𝑏1)(𝑎2 + 𝑏2) − (𝑎1 + 𝑏1)(𝑎2 − 𝑏2)]
= 𝑖̂[�⃗�2 𝑎⃗⃗⃗ ⃗3 + �⃗�2 �⃗⃗�3 − �⃗�3 �⃗⃗�2 − �⃗⃗�2 �⃗⃗�2 − �⃗�2 �⃗�3 + �⃗�2 �⃗⃗�3 − �⃗⃗�2 �⃗�3 + �⃗⃗�2 �⃗⃗�3]
−𝑗̂[�⃗�1�⃗�3 + �⃗�1�⃗⃗�3 − �⃗⃗�1 �⃗�3 − �⃗⃗�1 �⃗⃗�3 − �⃗�1 �⃗�3 + �⃗�1 �⃗⃗�3 − �⃗⃗�1 �⃗�3 + �⃗⃗�1 �⃗⃗�3]
+�̂�[�⃗�1�⃗�2 + �⃗�1 �⃗⃗�2 − �⃗⃗�1 �⃗�2 − �⃗⃗�1 �⃗⃗�2 − �⃗�1 �⃗�2 + �⃗�1 �⃗⃗�2 − �⃗⃗�1 �⃗�2 + �⃗⃗�1 �⃗⃗�2]
= 𝑖̂(2�⃗�2�⃗⃗�3 − 2�⃗⃗�2�⃗�3) − 𝑗̂(2�⃗�1 �⃗⃗�3 − 2�⃗⃗�1 �⃗�3) + �⃗⃗� (2�⃗�1 �⃗⃗�2 − 2�⃗⃗�1 �⃗�2)
= 2[𝑖(�⃗�2 �⃗⃗�3 − �⃗⃗�2�⃗�3) − 𝑗(�⃗�1�⃗⃗�3 − �⃗⃗�1�⃗�3) + �̂�(�⃗�1�⃗⃗�2 − �̂�1�̂�2)
= 2(�⃗� × �⃗⃗�)
S99. Ans.(b)
Sol.
Moment = r × F, where r be the position vector.
𝑀 = |𝑖̂ 𝑗̂ �̂�1 2 30 0 𝜆
|
= 𝑖(2𝜆) − 𝑗(𝜆)
|𝑀| = √(2𝜆)2 + (−𝜆)2 = √5 𝜆
S100. Ans.(a)
Sol.
𝐴𝐶⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗ (Triangle law)
−𝐶𝐴⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗
𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗ + 𝐶𝐴⃗⃗⃗⃗⃗⃗ = �⃗⃗�
S101. Ans(b)
Sol.
𝑦 = cos−1(sin 𝑥) 𝑑𝑦
𝑑𝑥=
−1
√1−sin2 𝑥. 𝑐𝑜𝑠𝑥
=−1 .cos 𝑥
√cos2 𝑥
= −− cos 𝑥
cos 𝑥
= −1
Given that 𝑑𝑦
𝑑𝑥= 𝑡𝑎𝑛𝜃
⇒ tan 𝜃 = −1
⇒ 𝜃 =3𝜋
4
S102. Ans.(d)
Sol.
For 𝑓(𝑥) 𝑡𝑜 𝑏𝑒 𝑑𝑖𝑓𝑖𝑛𝑒𝑑,
𝑥 − 1 ≥ 0 ⇒ 𝑥 ≥ 1.
And 𝑥 − 4 > 0 ⇒ 𝑥 > 4
∴ 𝑥 𝜖 [1, 4) ∪ (4, ∞)
S103. Ans.(a)
Sol.
𝑓(𝑥) = {
𝑠𝑖𝑛2𝑥
5𝑥𝑖𝑓 𝑥 ≠ 0
2
15𝑖𝑓 𝑥 = 0
lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
sin 2𝑥
5𝑥
= limℎ→0
sin 2(0+ℎ)
5(0+ℎ)
= limℎ→0
sin 2 ℎ
5ℎ
= limℎ→0
2 cos 2 ℎ
5
=2
5
And 𝑓(0) =2
15
𝑓(0) ≠ lim𝑥→0+
𝑓(𝑥)
⇒ 𝑓(𝑥) is not continuous at x = 0
S104. Ans.(b)
Sol.
𝑓(𝑥) = |𝑥 − 3|
= {𝑥 − 3 𝑖𝑓 𝑥 ≥ 3
−(𝑥 − 3) 𝑖𝑓 𝑥 < 3
lim𝑥→3+
𝑓(𝑥)
= lim𝑥→3+
(𝑥 − 3)
= limℎ→0
(3 + ℎ − 3)
= limℎ→0
ℎ = 0
And
lim𝑥→3−
𝑓(𝑥)
lim𝑥→3−
−(𝑥 − 3)
limℎ→0
−(3 − ℎ − 3)
limℎ→0
ℎ
= 0
Hence, lim𝑥→3+
𝑓(𝑥) = lim𝑥→3−
𝑓(𝑥)
⇒ f is continuous at x = 3.
S105. Ans.(b)
Sol.
∵ f is continuous at each point in its domain
⇒ 𝑓(0) = lim𝑥→0
𝑓(𝑥)
= lim𝑥→0
2𝑥−sin−1 𝑥
2𝑥+tan−1 𝑥
= (0
0) form
Applying L’ Hospital rule
lim𝑥→0
2−1
√1−𝑥2
2+1
1+𝑥2
=1
3
S106. Ans.(a)
Sol. 𝑢 𝑑𝑢
𝑑𝑥+
𝑣 𝑑𝑣
𝑑𝑥
𝑒𝑎𝑥 sin 𝑏𝑥[𝑏. 𝑒𝑎𝑥 cos 𝑏𝑥 + sin 𝑏𝑥. 𝑎. 𝑒𝑎𝑥] + 𝑒𝑎𝑥 cos 𝑏𝑥[−𝑏𝑒𝑎𝑥 sin 𝑏𝑥 + 𝑎 cos 𝑏𝑥. 𝑒𝑎𝑥]
= 𝑏𝑒2𝑎𝑥 sin 𝑏𝑥 cos 𝑏𝑥 + 𝑎. 𝑒2𝑎𝑥 sin2 𝑏𝑥 − 𝑏𝑒2𝑎𝑥 sin 𝑏𝑥 𝑐𝑜𝑠𝑥 + 𝑎𝑒2𝑎𝑥 cos2 𝑥.
= 𝑎 𝑒2𝑎𝑥
S107. Ans.(c)
Sol.
𝑦 = sin (𝑙𝑛𝑥) 𝑑𝑦
𝑑𝑥=
cos (𝐿𝑛𝑥)
𝑥
𝑑2𝑦
𝑑𝑥2 =−𝑥 sin(𝐿𝑛𝑥).
1
𝑥−cos (𝐿𝑛𝑥)
𝑥2
=− sin(𝐿𝑛𝑥)−cos (𝐿𝑛𝑥)
𝑥2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟,
𝑥2 𝑑2𝑦
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥+ 𝑦
𝑥2 − (− sin(𝑙𝑛𝑥)−cos (𝑙𝑛𝑥)
𝑥2 ) +𝑥.cos (𝐿𝑛𝑥)
𝑥+ sin 𝐿𝑛𝑥
− sin(𝐿𝑛𝑥) − cos(𝐿𝑛𝑥) + cos(𝐿𝑛𝑥) + sin (𝐿𝑛𝑥)
= 0
S108. Ans.(c)
Sol.
Length of wire = arc length + r + r
40 = (𝜃
360× 2𝜋𝑟) + 2𝑟
40−2𝑟
2𝜋𝑟=
𝜃
360
20−𝑟
𝜋𝑟=
𝜃
360 ______(1)
Now the area of sector
A = 𝜃
360× 𝜋𝑟2
=(20−𝑟)
𝜋𝑟× 𝜋𝑟2[𝑓𝑟𝑜𝑚 (1)]
= (20 − 𝑟)𝑟
= 20 r – r²
Area is greatest
∴ A’ = 0
20 – 2r = 0
⇒ r = 10
S109. Ans.(a)
Sol.
𝑓(𝑥) = [𝑥(𝑥 − 1) + 1)1
3
𝑓′(𝑥) =1
3[𝑥(𝑥 − 1) + 1]−
2
3[𝑥 + 𝑥 − 1]
𝑓′(𝑥) =1
3
(2𝑥−1)
[𝑥(𝑥−1)+1]2/3
𝑃𝑢𝑡 𝑓′(𝑥) = 0
0 =1
3
(2𝑥−1)
(𝑥(𝑥−1)+1]23
⇒ 𝑥 =1
2
Put the value, x = 1
2 in f(x)
𝑓 (1
2) = [
1
2(
1
2− 1) + 1]
1/3
= [1
2×
−1
2+ 1]
1/3
= [−1
4+ 1]
1/3
= [3
4]
1/3
S110. Ans.(c)
Sol.
𝑦 = |𝑠𝑖𝑛𝑥||𝑥|
𝐿𝑛𝑦 = |𝑥|𝐿𝑛|𝑠𝑖𝑛𝑥| 1
𝑦
𝑑𝑦
𝑑𝑥= [|𝑥|
1
|𝑠𝑖𝑛𝑥||𝑐𝑜𝑠𝑥| + log | sin 𝑥|]
𝑑𝑦
𝑑𝑥= |𝑠𝑖𝑛𝑥||𝑥| [|𝑥|
1
|sinx|| cos 𝑥| + log | sin 𝑥|]
𝐴𝑡 𝑥 = −𝜋
6
𝑑𝑦
𝑑𝑥= |sin (
−𝜋
6)|
|−𝜋
6|
[|−𝜋
6| .
|cos(−𝜋
6)|
|sin(−𝜋
6)|
+ |𝐿𝑛 (−𝜋
6)|]
= |− sin𝜋
6|
𝜋
6[
𝜋
6
|cos𝜋
6|
|− sin𝜋
6|+ 𝐿𝑛 |− sin
𝜋
6|]
= |−1
2|
𝜋
6[
𝜋
6.
√3
21
2
− 𝐿𝑛 (1
2)]
= (1
2)
𝜋
6[
𝜋√3
6+ 𝐿𝑛(2)]
(2)−𝜋/6 [√3𝜋+6𝐿𝑛(2)
6]
S111. Ans.(b)
Sol. 𝑑(√1−𝑠𝑖𝑛2𝑥)
𝑑𝑥
=𝑑
𝑑𝑥(√𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 − 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥)
=𝑑
𝑑𝑥√(cos 𝑥 − 𝑠𝑖𝑛𝑥)2
=𝑑
𝑑𝑥(cos x-sinx)
=-sinx -cosx = - (cos x+sinx)
S112. Ans.(a)
Sol.
= ∫𝑑𝑥
𝑎2 sin2 𝑥+𝑏2 cos2 𝑥
= ∫sec2 𝑥
𝑎2 tan2 𝑥+𝑏2 𝑑𝑥
Putting tan x = t and sec2 𝑥 𝑑𝑥 = 𝑑𝑡, we get
I = ∫𝑑𝑡
𝑎2𝑡2+𝑏2 =1
𝑎2 ∫𝑑𝑡
𝑡2+(𝑏
𝑎)
2 =1
𝑎2 ×1
𝑏/𝑎tan−1 (
𝑡
𝑏/𝑎) + 𝐶
=1
𝑎𝑏tan−1 (
𝑎𝑡
𝑏) + 𝐶 =
1
𝑎𝑏tan−1 (
𝑎 𝑡𝑎𝑛𝑥
𝑏) + 𝐶
S113. Ans.(d)
Sol.
𝑓′(𝑥) = limℎ→0
[𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ]
= limℎ→0
[𝑓(𝑥)𝑓(ℎ)−𝑓(𝑥)
ℎ] [∵ 𝑓 (𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦)]
= limℎ→0
𝑓(𝑥) [𝑓(ℎ)−1
ℎ]
= 𝑓(𝑥) limℎ→0
[𝑓(ℎ)−1
ℎ]
= 𝑓(𝑥) limℎ→0
[1+ℎ 𝑔 (ℎ) 𝜙 (ℎ)−1
ℎ]
= 𝑓(𝑥) limℎ→0
𝑔(ℎ) 𝜙 (ℎ)
= 𝑓(𝑥) (limℎ→0
𝑔(ℎ)) (limℎ→0
𝜙(ℎ))
= 𝑓(𝑥). 𝑎. 𝑏
= 𝑎𝑏 𝑓(𝑥)
S114. Ans.(c)
Sol. 𝑑𝑥
𝑑𝑦=
𝑥+𝑦+1
𝑥+𝑦−1
⇒𝑑𝑦
𝑑𝑥=
𝑥+𝑦−1
𝑥+𝑦+1
Let x + y = v and 𝑑𝑦
𝑑𝑥=
𝑑𝑣
𝑑𝑥− 1
∴𝑑𝑣
𝑑𝑥− 1 =
𝑣−1
𝑣+1⇒
𝑑𝑣
𝑑𝑥=
𝑣−1+𝑣+1
𝑣+1
⇒ 𝑣+1
2𝑣𝑑𝑣 = 𝑑𝑥 ⇒
1
2∫ 1 𝑑𝑣 +
1
2∫
1
𝑣𝑑𝑣 = ∫ 1 𝑑𝑥
⇒1
2𝑣 +
1
2log 𝑣 = 𝑥 + 𝑐1
⇒ 𝑥 + 𝑦 + log(𝑥 + 𝑦) = 2𝑥 + 𝑐 [∵ 𝑐 = 2𝑐1]
∴ (𝑦 − 𝑥) + log(𝑥 + 𝑦) = 𝑐
S115. Ans.(d)
Sol.
lim𝑥→
𝜋
6
2 sin2 𝑥+𝑠𝑖𝑛𝑥−1
2 sin2 𝑥−3𝑠𝑖𝑛𝑥+1
= (0
0) 𝑓𝑜𝑟𝑚
Applying L’ Hospital Rule
lim𝑥→
𝜋
6
4 sin 𝑥 cos 𝑥+cos 𝑥
4 sin 𝑥 cos 𝑥−3 cos 𝑥
= lim𝑥→
𝜋
6
4 sin 𝑥+1
4 sin 𝑥−3
=4×
1
2+1
4×1
2−3
=2+1
2−3= −3.
S116. Ans.(c)
Sol.
Sample space = {(1, 5), (2, 5), 3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
Favorable outcome = {(5, 5), (5, 6), (6, 5)}
∴ Probability = 3
11
S117.Ans(b)
S118. Ans.(d)
Sol.
Given that average of group of women = 21 years
⇒𝑥1+⋯+𝑥𝑛
𝑛= 21
𝑥1 + ⋯ + 𝑥𝑛 = 21𝑛 ________(1)
And average of group of men = 26 years. 𝑦1+⋯+𝑦𝑚
𝑚= 26
𝑦1 + ⋯ + 𝑦𝑚 = 26𝑚 ________(2)
and average of combined group = 25 𝑥1+⋯+𝑥𝑛+𝑦1+⋯+𝑦𝑛
𝑚+𝑛= 25
𝑥1 + ⋯ + 𝑥𝑛 + 𝑦1 + ⋯ 𝑦𝑛 = 25(𝑚 + 𝑛)
21n + 26m = 25 (m + n) [from (1) & (2)]
⇒ 𝑚 = 4𝑛
Percentage of men in group = 𝑚
𝑚+𝑛× 100
=4𝑛
4𝑛+𝑛× 100
=4𝑛
5𝑛× 100
= 80%
∴ women = 20%
S119. Ans.(c)
Sol.
Given that sinβ is the harmonic mean of sinα and cosα.
⇒ 1
𝑠𝑖𝑛𝛽 is the arithmetic mean of
1
𝑠𝑖𝑛𝛼 and
1
𝑐𝑜𝑠𝛼
⇒ 1
𝑠𝑖𝑛𝛽=
1
𝑠𝑖𝑛𝛼+
1
𝑐𝑜𝑠𝛼
2
1
sin 𝛽=
cos 𝛼+sin 𝛼
2 sin 𝛼 cos 𝛼
sin 2𝛼
sin 𝛽= cos 𝛼 + sin 𝛼 ______(1)
Consider
I. L.H.S √2 sin (𝛼 +𝜋
4) sin 𝛽
√2 (sin 𝛼 cos𝜋
4+ cos 𝛼 sin
𝜋
4) sin 𝛽
√2 (sin 𝛼1
√2+ cos 𝛼
1
√2) sin 𝛽
(𝑠𝑖𝑛𝛼 + 𝑐𝑜𝑠𝛼) sin 𝛽 sin 2𝛼
sin 𝛽sin 𝛽 [𝑓𝑟𝑜𝑚 (1)]
= sin 2 α
= R. H. S.
∵ Given that sin θ is the arithmetic mean of sin α and cos α.
∴ sin θ = sin 𝛼+cos 𝛼
2 ______(2)
Consider,
II. R.H.S
cos (𝛼 −𝜋
4)
= cos 𝛼 cos𝜋
4+ sin 𝛼 sin
𝜋
4
= cos 𝛼1
√2+ sin 𝛼
1
√2
=1
√2(cos 𝛼 + sin 𝛼)
=2
√2(
cos 𝛼+sin 𝛼
2)
= √2 sin 𝜃
= L.H.S.
S120. Ans.(b)
Sol.
Given that
= 𝑃(𝐵) = 1.5 𝑃(𝐴) 𝑃(𝐵)
𝑃(𝐴)= 1.5 =
3
2
𝑃(𝑐) = 0.5 𝑃(𝐵) 𝑃(𝐶)
𝑃(𝐵)=
1
2
𝑃(𝐴): 𝑃(𝐵): 𝑃(𝐶) = 4 ∶ 6 ∶ 3
∴ P(A) = 4
4+6+3=
4
13