SSC CGL MAINS 2017 -...
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SSC CGL MAINS 2017
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SSC CGL MAINS 2017
Important formulas:
( a + b)2 = a2 + 2ab + b2
( a − b)2 = a2 − 2ab + b2
( a + b)2 + ( a − b)2 = 2(a2 + b2)
( a + b)2 − ( a − b)2 = 4ab
( a + b)3 = a3 + b3 +3ab(a+ b)
( a − b)3 = a3 − b3 − 3ab(a − b)
( a2− b2) = (a + b)(a − b)
( a3 + b3) = (a + b) (a2 − ab + b2)
( a3 − b3) = (a − b)(a2+b2 +ab)
( a + b+ c)2 = a2 + b2 + c2+2(ab + bc + ca)
( a – b − c)2 = a2 + b2 +c2 − 2(ab + ac − bc)
( a + b + c)3 = a3 + b3 + c3 +3(a+b)(b+c)(c+a)
( a3 + b3 + c3 – 3abc) = (a+b+c)(a2+b2+c2 − ab−bc −ca)
If a + b + c = 0 then, a3 + b3 + c3 = 3abc
Previous year Question
EX-1: If a3+ b3 = 9 and a + b = 3 , then find 1
𝑎 +
1
𝑏 ?
Sol: (a + b)3 = a3+ b3+3ab(a+b)
33= 9 + 3ab × 3
ab = 2
1
𝑎 +
1
𝑏 =
a+𝑏
ab =
3
2
EX-2: If a = 4.965, b = 2.343 & c = 2.622 then find the a3 - b3 - c3- 3abc
Sol: a – b – c = 4.965 – 2.343 – 2.622 = 0 a3 – b3 – c3 – 3abc = (a - b – c )(a2+b2+c2- ab +bc-ca)
a3 – b3 – c3 – 3abc = 0
EX-3: If x2 + 4y2 + z2 + 3 = 2x + 4y + 2z, then find the value of x + y + z .
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Sol: x2 + 4y2 + z2 + 3 = 2x + 4y + 2z
x2 – 2X+1 + 4y2 – 4y +1+ z2 − 2z +1 = 0
(x−1)2 + (2y − 1)2 + (z −1)2 = 0
( x − 1) = 0 ,
( 2y − 1) = 0 ,
x = 1
y =1
2
( z –1) = 0 ,z = 1
x + y + z = 1+ 1
2+ 1 = 2
1
2
EX-4: If a3 – b3 – c3 = 0 then find the value of a9 – b9 – c9 -3a3b3c3
Sol: a3 – b3 – c3 = 0
a3 – b3 = c3
Cubing both side
a9 – b9 -3a3b3(a3 – b 3 ) = c9
a9 – b9 - c9 – 3a3b3c3= 0
EX- 5 : If x2 = y + z, y2 = z + x and z2 = x + y, then the value of
1
1+𝑥+
1
1+𝑦+
1
1+𝑧 is
Sol: 𝑥2 = 𝑦 + 𝑥
𝑥2 + 𝑥 = 𝑥 + 𝑦 + 𝑧
𝑥(𝑥 + 1) = 𝑥 + 𝑦 + 𝑧
𝑥 + 1 =𝑥 + 𝑦 + 𝑧
𝑥
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1
𝑥 + 1=
𝑥
𝑥 + 𝑦 + 𝑧
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑦,1
𝑦 + 1=
𝑦
𝑥 + 𝑦 + 𝑧
1
𝑧 + 1=
𝑧
𝑥 + 𝑦 + 𝑧
1
1 + 𝑥+
1
1 + 𝑦+
1
1 + 𝑧
=𝑥
𝑥 + 𝑦 + 𝑧+
𝑦
𝑥 + 𝑦 + 𝑧+
𝑧
𝑥 + 𝑦 + 𝑧
=𝑥 + 𝑦 + 𝑧
𝑥 + 𝑦 + 𝑧= 1
EX-6:If x2 + 1 = 2x, then the value of 𝑥4+
1
𝑥2
𝑥2 −3𝑥+1
Sol: 𝑥2 + 1 = 2𝑥 (𝐺𝑖𝑣𝑒𝑛)
𝑥 +1
𝑥= 2. . . . . . (i)
𝑥4 +1
𝑥2
𝑥2 − 3𝑥 + 1=
𝑥6 + 1𝑥2
(𝑥2 − 3𝑥 + 1)
=𝑥6 + 1
(𝑥2 + 1 − 3𝑥). 𝑥2
=𝑥6 + 1
(2𝑥 − 3𝑥)𝑥2=
𝑥6 + 1
−𝑥3
= − (𝑥6+1
𝑥3 ) = − (𝑥6
𝑥3 +1
𝑥3)
= − (𝑥3 +1
𝑥3)
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= − [(𝑥 +1
𝑥)
3− 3 (𝑥 +
1
𝑥)]
= −[23 − 3 × 2]
= −2
Ex: In the given figure PQ || RS, then find the value 𝛼 + 𝛽 + 𝛾 ?
Sol. PQ || OM
∠MOQ = 180 - 𝛽
RS || OM
∠MOS = 180 - 𝛼
∠MOS = ∠VOT = 180 - 𝛽 +180 - 𝛼
= 180 - 𝛽 +180 - 𝛼
= 𝛼 + 𝛽 + 𝛾 = 3600
Ex: In ∆ ABC, AB = AC, ∠BAC = 400 .Then the external angle at B is
Sol: In ∆ ABC, AB = AC , So ∠ ABC = ∠ ABC
400 + ∠ ABC + ∠ ACB = 1800
2 ∠ ABC = 180 – 40
= 1400
∠ ABC = 700
700 + ∠ ABD = 1800
∠ ABD = 1100
Ex: In ∆ ABC, M is the mid-point of BC. Length of AM is 9. N is a point on
AM such that MN = 1. Distance of N from the centroid of ∆ABC is equal to -
Sol:
T
S
P
R
𝜷
Q
𝜶 𝜸 M
O
V
A
D C B
400
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Given, AM = 9 and MN = 1
∴ AG : GM = 2 : 1 ( ∵ AM is the median)
x : (9 - x) = 2 : 1
⇒ x = 2 (9 - x) ⇒ 3x=18
⇒ x = 6
∴ GM = GN + NM
⇒ GN = 3 - 1 = 2
Ex: In a triangle, distances from centroid to vertices are respectively 4 cm,
6 cm and 8 cm. find the length of medians.
Sol: 𝐴𝑂=4 cm ,BO = 6 cm, CO = 8 cm
O is centroid "(intersection point of median)
Then , 𝐴𝑂
𝑂𝐹=
2
1
⇒ 𝐴𝑂
2= 𝑂𝐹
⇒ OF=4
2= 2𝑐𝑚
Similarly, OD = 3 cm
And OE = 4 cm
AF = (AO + OF
AF = (4 + 2) = 6 cm
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BD = (6 + 3) = 9 cm
And CE = (CO + OE) = (8+4) = 12 cm
Ex: ABC is an equilateral triangle. P and Q are two points on AB and AC
respectively such that PQ is parallel to BC. If PQ = 5 cm, then find area of
triangle APQ.
Sol:
2 2
APQ ABC AA Similarity
ABC is an equilateral triangle
APQ would also be equilateral
Hence
3 25 3Ar( APQ) 5 cm
4 4
Ex: Find the value of
0 0 0
0 0 0
cos 90 .sec 360 .tan 180?
sec 720 .sin 540 .cot 90
A A A
A A A
Sol:
0 0 0
0 0 0
cos 90 .sec 360 .tan 180?
sec 720 .sin 540 .cot 90
sin .sec . tan
sec sin tan
sin .sec . tan
sin .sec tan
A A A
A A A
A A A
A A A
A A A
A A A
Ex: Find the value of 0 0 0 0
sin720 cot 270 sin150 cos120
Sol:
0 0 0 0
0 0 0 0
sin 720 cot 270 sin150 cos120
sin 2 360 0 cot 3 90 0 sin 90 60 cos 90 30
1 10 0
2 2
1
4
Ex: Find the value of
0 0
0 0
sin 37 cos 20
cos53 sin 70
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Sol:
0 0
0 0
0 0
sin 37 cos 20
cos53 sin 70
sin 90 53 cos 90 70
cos53 sin 70
1 1 1
Ex: Find the value of 2 0 2 0 2 0 2 0 2 0 2 0
sin 60 cos 30 cot 45 sec 60 cos 30 cos 0ec
Sol:
2 0 2 0 2 0 2 0 2 0 2 0
2 2
2 2 2 2
sin 60 cos 30 cot 45 sec 60 cos 30 cos 0
3 31 2 2 1
2 2
7
2
ec
Ex: If 3tan 4 0, where , ,2
then the value of
Sol:
3 tan 4 0,
4tan
3
4 3 3sin ,cos ,cot
5 5 4
2cot 5cos sin
3 3 42 5
4 5 5
3 43
2 5
15 30 8 23
10 10
Ex-: If 0sin17 ,
x
y then the value of
0 0sec17 sin73 is :
Sol:
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0 0
0 0 0
0 0
0
0
2
2 0 2 0 2
0 0 2
2
2 2
2 2 2 22
2
sec17 sin 73
sec17 sin 90 17
sec17 cos17
1cos17
cos17
1 cos 17 sin 17
cos17 cos171
x
y
x
y
x x
y x y y xy
y
EX- : If sin( θ + 340) = cos θ and (θ + 340) is acute , then θ = ?
Sol:
sin (+340) = cos
sin (+34o) = sin(900 −)
+340 = 900 −
2 = 560
= 280
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