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Questions based on different algebraic expressions, equations (e.g. quadratic or higher order, square root, cube root, and inverse) or based on graphic representation of equations and the value of a variable is asked or an equation is required to be validated.
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Importance : Algebra based 2-3 questions are essentially asked in almost all competitive exams obviously this chapter should be given sufficient time and practice done. Scope of questions : Questions based on different algebraic expressions, equations (e.g. quadratic or higher order, square root, cube root and inverse) or based on graphic representation of equations and the value of a variable is asked or an equation is required to be validated. Way to success : Solution of questions of this chapter can be ensured by memorising the concerved formulae/rules and by regular practice.
Polynomials : An algebraic expression in which the variables involved have only non-negative integral powers is called a polynomial. General Form : p ( x ) = a 0 + a 1 x + a 2 x^2 + ... + anxn^ is a polynomial in variable x , where a 0 , a 1 , a 2 , a 3 ... an are real numbers and n is non-negative integer. Remainder Theorem : Let f ( x ) be a polynomial of degree n > 1, and let a be any real number. When f ( x ) is divided by ( x – a ), then the remainder is f ( a ). Proof : Suppose that when f ( x ) is divided by ( x – a ), the quotient is g ( x ) and the remainder is r ( x ). Then,degree r ( x ) < degree ( x – a )
Thus, when f ( x ) is divided by ( x – a ), then the quotient is g ( x ) and the remainder is r.
Putting x = a in (i), we get r = f ( a ). Thus, when f ( x ) is divided by ( x – a ), then the remainder is f ( a ). Remarks (i) If a polynomial p ( x ) is divided by ( x + a ), the remainder is the value of p ( x ) at x = – a i.e. p (– a )
(ii) If a polynomial p ( x ) is divided by ( ax – b ), the remainder
is the value of p ( x ) at x =
i.e. p
(iii) If a polynomial p ( x ) is divided by ( ax + b ), then
(iv) If a polynomial p ( x ) is divided by b – ax , the remainder
is the value of p ( x ) at x =
Factor Theorem Let p ( x ) be a polynomial of degree greater than or equal to 1 and a be a real number such that p ( a ) = 0, then ( x – a ) is a factor of p (x). Conversely, if ( x – a ) is a factor of p ( x ), then p ( a ) = 0
But by Remainder theorem, p ( x ) when divided by ( x – a ) gives the remainder equal to p ( a ).
Remarks (i) ( x + a ) is a factor of a polynomial iff (if and only if) p (– a ) = 0
(ii) ( ax – b ) is a factor of a polynomial if p
(iv) ( x – a ) ( x – b ) are factors of a polynomial p ( x ) if p ( a ) = 0 and p ( b ) = 0
ALGEBRAIC IDENTITIES
An algebraic identity is an algebraic equation which is true for all values of the variable (s).
= ( a + b + c )
1 2
(2 a^2 + 2 b^2 + 2 c^2 – 2 ab – 2 bc – 2 ac )
1 2
( a + b + c ) [( a – b )^2 + ( b – c )^2 + ( c – a )^2 ]
GRAPHIC REPRESENTATION OF STRAIGHT LINES
Ordered Pair : A pair of numbers a and b listed in a specific order with a at the first place and b at the second place is called an ordered pair ( a , b ).
Thus, (2, 3) is one ordered pair and (3, 2) is another ordered pair. CO-ORDINATE SYSTEM Co-ordinate Axes : The position of a point in a plane is determined with reference to two fixed mutually perpendicular lines, called the coordinate axes. Let us draw two lines X’OX and YOY’, which are perpendicular to each other and intersect at the point O. These lines are called the coordinate axes or the axes of reference. The horizontal line X’OX is called the x-axis. The vertical line YOY’ is called the y-axis. The point O is called the origin. The distance of a point from y-axis is called its x-co- ordinate or abscissa and the distance of the point from x-axis is called its y-co ordinate or ordinate. If x and y, denote respectively the abscissa and ordinate of a point P, then (x, y) are called the coordinates of the point P. The y–co-ordinate of every point on x - axis is zero. i.e. when a straight line intersects at x-axis, its y-co-ordinate is zero. So, the co-ordinates of any point on the x-axis are of the form (x, 0). The x- co-ordinate of every point on y-axis is zero. So, the co-ordinates of any point on y-axis are of the form (0, y). The co-ordinates of the origin are (0, 0). y = a where a is constant denotes a straight line parallel to x-axis. x = a where a is constant, denotes a straight line parallel to y-axis. x = 0 denotes y-axis. y = 0 denotes x-axis.
We can fix a convenient unit of length and taking the origin as zero, mark equal distances on the x-axis as well as on the y-axis. Convention of Signs : The distances measured along OX and OY are taken as positive and those along OX’ and OY’ are taken as negative, as shown in the figure given above. CO-ORDINATES OF A POINT IN A PLANE Let P be a point in a plane. Let the distance of P from the y-axis = a units. And, the distance of P from the x-axis = b units. Then, we say that the co-ordinates of P are (a, b). a is called the x-co-ordinate, or abscissa of P. b is called the y co-ordinate, or ordinate of P.
X' O X
Y'
Y
a
M
P
b
( a , b )
Quadrants : Let X’ OX and YOY’ be the co-ordinate axes. These axes divide the plane of the paper into four regions, called quadrants. The regions XOY, YOX’,X’OY’ and Y’OX are respectively known as the first, second, third and fourth quadrants.
Rule 22. If a^3 + b^3 + c^3 = 3abc, then a + b + c = 0 or a = b = c.
Now, a^3 + b^3 + c^3 – 3abc =
(a + b + c) [(a – b)^2 +
(b – c)^2 + (c – a)^2 ]
(^2) + (b – c) (^2) + (c – a) (^2) ]
= 0, i.e., a – b = 0
Rule 23. If a^2 + b^2 + c^2 = ab + bc + ca, then a = b = c. Rule 24. Componendo and Dividendo Rule, If
Rule 25. If
then (^) x + x + x + ...¥ = (^) bn + (^1) g
Rule 28. (a + b + c)^3 = a^3 + b^3 + c^3 – 3(a + b) (b + c) (c + a) Rule 29. a^4 + a^2 b^2 + b^4 = (a^2 + ab + b^2 ) (a^2 – ab + b^2 )
Rule 32. Binomial theorem : (a + b)n^ = nC 0 anb^0 + nC 1 an – 1b^1 + nC 2 an – 2b^2 + ... + nC n – 1a
(^1) bn – 1 (^) + nC na
(^0) bn, where, n is a positive number and
n
! (^) b g! Permutation and Combination Permutation : It is used where we have to arrange things. Out of total n things, r things (taken at a time) can be arranged as npr or P ( n , r )
P ( n , r ) = n Pr = (^) ( n n - r^ !)! where n > r
Combination : It is used where we have to select things. It is written as nCr or C( n , r )
C ( n , r ) =
Some important results.
nCo = nCn =1 ; nCr = nCn – r = nC 1 = nCn –1 = n.
Ex. (^7 )
b g!
b g! n! (is called as n factorial) 5! = 5.4! = 5.4.3! = 5.4.3.2! = 5.4.3.2.1!
Importance : Coordinate geometry is separate and important filled in mathematics but very rarely asked in competitive exams. However in two-dimensional (2–D) geometry introductory/easy questions should be practised for improving marks. Scope of questions : Mostly questions are related to distance between two points, linear/non-linear these coplaner points, cutting a line a specific ratio by a given point. Way to success : The concept of coordinate geometry and practice of above mentioned questions is very important to solve questions.
Important Points : x–coordinate is called the abscissa of P, where (x, y) are co-ordinates of any point P. y–co-ordinate is called the ordinate of P, where (x, y) are co–ordinates of any point P. Quadrants :
IInd quadrant (–x, y) x¢ x
y¢
y Ist quadrant (x, y)
IIIrd quadrant (–x, –y) IVth quadrant (x, –y)
COORDINATE GEOMETRY
Cartesian Co–ordinate System : y
x¢ x
y¢
P (x, y)
y-Co-ordinate (ordinate)
x-Co-ordinate (abscissa)
C0-ordinate
Polar Coordinate System : y
x¢ x
y¢
P (r, q)
x
q
r (^) y ^ r = x + y^2 2
RULE 1 : The distance between any two points in the plane is the length of the line segment joining them. The distance between two points P (x 1 , y 1 ) and Q (x 2 , y 2 ) is
PQ = (^) (differenc e of abscissa ) 2 +(differenc e of ordinates)^2
RULE 2 : The area of a triangle, the Co-ordinates of whose vertices are (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is
F HG
I KJ^ |x^1 (y^2 – y^3 ) + x^2 (y^3 – y^1 ) + x^3 (y^1 – y^2 )|
F H
G
I K
J
1 1 2 2 3 3
If all three points are collinear,
RULE 3 : The Co-ordinates of the point which divides the line segment joining the points (x 1 , y 1 ) and (x 2 , y 2 ) internally in the ratio m : n are given by
x =
RULE 4 : If P is the mid-point of AB, such that it divides AB in the ratio 1 : 1, then its Co-ordinates are (x,y) =
F^ +^ + H
G
I K
, (^) Jalso called mid point formula.
RULE 5 : The Co–ordinates of the point which divides the line segment joining the points (x 1 , y 1 ) and (x 2 , y 2 ) externally in the ratio m : n, are
F HG^
I KJ
RULE 6 : The Co-ordinates of the centroid of a triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is given by
F^ +^ +^ +^ + HG^
I KJ
RULE 7 : The Co–ordinates of the in–centre of a triangle whose vertices are A (x 1 , y 1 ), B(x 2 , y 2 ), C(x 3 , y 3 ) are given by
F HG^
I KJ
b = CA and c = AB. Equation of straight line. A straight line is a curve such that every point on the line segment joining any two points on it lies on it. RULE 8 : If (x 1 , y 1 ) and (x 2 , y 2 ) are the Co-ordinates of any two points on a line, then its slope is
2 1 2 1
2 1
RULE 10 : If two lines having slopes m 1 and m 2 are (i) parallel if m 1 = m 2 (ii) Perpendicular if m 1 x m 2 = – RULE 11 : (Slope–Intercept) The equation of a line with slope m and making an intercept c on y-axis is y = mx + c. RULE 12 : (Point-Slope form) The equation of a line which passes through the point (x 1 , y 1 ) and has the slope ‘m’ is (y – y 1 ) = m(x – x 1 ) RULE 13 : (Two-point form) The equation of a line passing through two points (x 1 , y 1 ) and (x 2 , y 2 ) is given by
1 2 1
y y y y
1 2 2 RULE 14 : (Intercept form) The equation of a line which cuts off intercepts a and b respectively on the x and y–axes is
RULE 15 : (i) The slope of a line whose general quation
is given by Ax + By + C = 0 is
(ii) The intercepts of a line on x and y axes respectively whose general equation is Ax + By + C = 0 is given by :-
x-intercept =
and y-intercept =
RULE 16 : General equation of straight line is ax+ by + c = 0
line and its intercepts is
F- HG^
I KJ
F- HG^
I KJ
qq q
, the value of x is
(1) 31 (2) 32 (3) 36 (4) 37 (SSC Section Officer (Commercial Audit) Exam.16.11.2003)
5 2 1 1
b g
n (^) n
n n
is
(1) 1 (2) 9 (3) 3 (4) 3n (SSC CGL Prelim Exam. 08.02. (First Sitting)
value of (^) 0 6. × (3 y ) x^ is equal to (1) 1.0 (2) 0. (3) 0.6 (4) 1. (SSC CGL Prelim Exam. 08.02. (Second Sitting)
x = (^) e j , then x equals
(SSC CPO S.I. Exam. 05.09.2004)
x (^) = (^) , then the value of x
is (1) 3 (2) –
(3)
(SSC CPO S.I. Exam. 05.09.2004)
equal to (1) 2 (2) 4 (3) 5 (4) 6 (SSC Data Entry Operator Exam. 31.08.2008)
is
(SSC CHSL DEO & LDC Exam. 04.12.2011 (Ist Sitting (East Zone)
(^2) a perfect square?
(SSC CPO S.I. Exam. 03.09.2006)
(4) All of the above (SSC CPO S.I. Exam. 03.09.2006)
= 1, then
the value of
is
(1) 1 (2) 2 (3) 3 (4) 4 (SSC CHSL DEO & LDC Exam. 04.12.2011 (IInd Sitting (East Zone) & (SSC GL Tier-I Exam. 19.05.2013)
(1)
(SSC CGL Prelim Exam. 04.02. (Second Sitting)
value of a is
(SSC CPO S.I. Exam. 16.12.2007)
(1)
(SSC CGL Prelim Exam. 27.07. (First Sitting)
(SSC CGL Prelim Exam. 27.07. (First Sitting)
F^3 6 2 HG
I KJ
F HG
I KJ^
= F HG
I KJ
is equal to (1) –2 (2) 2 (3) –1 (4) 1 (SSC CGL Tier-I Exam. 16.05. (First Sitting)
(SSC CHSL DEO & LDC Exam. 11.12.2011 (IInd Sitting (Delhi Zone)
x x = x x
then x is
equal to
(SSC CGL Tier-I Exam. 16.05. (First Sitting)
= 5, then
x x - x + is equal to
(SSC CHSL DEO & LDC Exam. 11.12.2011 (Ist Sitting (East Zone)
, then the value of
F
H
G
I
K
J
(SSC SAS Exam. 26.06. (Paper-1)
and y^ =^
, then
value of x^2 + y^2 is : (1) 14 (2) 13 (3) 15 (4) 10 (SSC CGL Prelim Exam. 11.05. (First Sitting)
, then the value of
x is
(SSC CISF ASI Exam. 29.08. (Paper-1)
x x x x
(^2) then x is
equal to (1) 2.4 (2) 3. (3) 4 (4) 5 (SSC (South Zone) Investigator Exam.12.09.2010)
x is (1) 14 (2) 16 (3) 18 (4) 20 (SSC CPO S.I. Exam. 12.12.2010 (Paper-I)
7 5 e j ¸^ e j =^
p (^) , then the
value of p is (1) 5 (2) 2
Exam. 12.12.2010 (Paper-I)
3
-^ x^ = , then x equals
(1) 2 (2) 4
1 3 b g
/
(SSC CGL Tier-1 Exam. 19.06. (First Sitting)
+^ x^ = , (^) then the value of
x is
(1)
(SSC CGL Tier-1 Exam. 19.06. (Second Sitting)
= a + b , (^) then
the values of a and b are respectively
(1)
(SSC CGL Tier-1 Exam. 19.06. (Second Sitting)
z
(3)
(SSC Delhi Police S.I.(SI) Exam. 19.08.2012)
value of
is
(1) 0.009 (2) 0. (3) 0.9 (4) 0. (SSC CGL Tier-1 Exam 19.06. (Second Sitting)
x =? (1) 2 (2) 3 (3) 4 (4) 5 (SSC CGL Tier-1 Exam 26.06. (First Sitting)
b g b gb g
2
b g b gb g
2
b g b (^) gb (^) g
2 is :
(SSC CHSL DEO & LDC Exam. 11.12.2011 (IInd Sitting (Delhi Zone) & (SSC CHSL DEO & LDC Exam. 27.10.2013)
(SSC CHSL DEO & LDC Exam. 11.12.2011 (IInd Sitting (East Zone)
x^2 y^2 z^2 9 25 16
(1) 12 (2) 9 (3) 3 (4) 1 (SSC Graduate Level Tier-I Exam. 19.05.2013)
and b =
then the value of
2 2 2 2
is
(SSC CGL Prelim Exam. 13.11. (Second Sitting)
, then the value of
x is (1) 1 (2) 0
(3)
(SSC Assistant Grade-III Exam. 11.11.2012 (IInd Sitting)
3 F H
G
I K
J
F^3 HG
I KJ^
F^4 HG
I KJ (SSC Graduate Level Tier-I Exam. 11.11.2012, Ist Sitting)
(SSC Multi-Tasking Staff Exam. 17.03.2013, Kolkata Region)
, then
x x
x x
is equal to
(SSC CPO S.I. Exam. 03.09.2006)
then the average of x and y is (1) 2 (2) 3 (3) 4 (4) 5 (SSC Graduate Level Tier-I Exam. 21.04.2013 IInd Sitting)
2 2
(SSC Section Officer (Commercial Audit) Exam. 26.11. (Second Sitting)
F H
G
I K
J (^) and (^) x^2 + y^2 is
(SSC Graduate Level Tier-I Exam. 21.04.2013)
of P, x = 6? (1) 6 (2) 4 (3) 2 (4) 1 (SSC Graduate Level Tier-I Exam. 19.05.2013)
is
(SSC Graduate Level Tier-I Exam. 19.05.2013)
Then point out the correct alternative among the fou r alternatives given below. (1) b < a < c (2) a < c < b (3) b < c < a (4) a < b < c (SSC CHSL DEO & LDC Exam. 20.10.2013)
, the value of
is
(1) 1 (2) 2
(SSC CHSL DEO & LDC Exam. 27.10.2013 IInd Sitting)
is
e 3 -^7 j
e 7 -^3 j
e 7 +^3 j
e^7 -^3 j
(SSC CHSL DEO & LDC Exam. 10.11.2013, Ist Sitting)
then 11 x + 18 y = (1) –15 (2) 51 (3) 33 (4) 15 (SSC CHSL DEO & LDC Exam. 10.11.2013, IInd Sitting)
(SSC CHSL DEO & LDC Exam. 10.11.2013, IInd Sitting)
, then the value of
x – x^2 is :
(1) – a (2)
(4) a
(SSC Graduate Level Tier-I Exam. 21.04.2013, Ist Sitting)
(SSC Graduate Level Tier-I Exam. 19.05.2013 Ist Sitting)
4 x (^3 4 3 4 3 ) x
y y
z z
then the value of
(SSC Graduate Level Tier-I Exam. 19.05.2013 Ist Sitting)
non-zero numbers, then x equals to
(SSC CHSL DEO & LDC Exam. 10.11.2013, IInd Sitting)
equal to (1) 38 (2) 36 (3) 34 (4) 30 (SSC CGL Prelim Exam. 04.02. (Ist Sitting) & (SSC CGL Prelim Exam. 27.07.2008 (IInd Sitting) & (SSC Investigator Exam. 12.09.2010) (South Zone)
F HG^
I KJ^
2
ble values of q are
(SSC Graduate Level Tier-I Exam. 21.04.2013 IInd Sitting)
2 2
(SSC Graduate Level Tier-I Exam. 21.04.2013 IInd Sitting)
= 1 and b +
is equal to
(SSC CGL Prelim Exam. 04.02. (First Sitting)
(SSC Graduate Level Tier-I Exam. 21.04.2013)
(SSC Graduate Level Tier-I Exam. 19.05.2013)
2 2
(SSC Graduate Level Tier-I Exam. 19.05.2013 Ist Sitting)
of
2 2
2
F
H
G
I
K
J is :
(SSC CAPFs SI & CISF ASI Exam. 23.06.2013)
F HG^
I KJ
(3) x + y + z (4) 0 (SSC CHSL DEO & LDC Exam. 20.10.2013)
(SSC CPO S.I. Exam.12.01.2003)
( a + c )( b + a )
( c + a )( c + b )is: (1) 1 (2) 0 (3) –1 (4) – (SSC CHSL DEO & LDC Exam. 11.12.2011 (IInd Sitting (East Zone)
2 2 2 2
is
(1) 0 (2) 1 (3) 2 (4) 3 (SSC Graduate Level Tier-II Exam. 16.09.2012)
F HG^
I KJ
is
(SSC Graduate Level Tier-II Exam. 16.09.2012)
F + HG^
I KJ
is
(SSC CHSL DEO & LDC Exam. 21.10.2012 (Ist Sitting)
numerical value of
p q
q p
(1) 16 (2) 20 (3) 22 (4) 18 (SSC CHSL DEO & LDC Exam. 21.10.2012 (Ist Sitting)
the value of
2 2 2 2
is
(SSC CHSL DEO & LDC Exam. 21.10.2012 (IInd Sitting)
(4) None of the above is true (SSC CHSL DEO & LDC Exam. 04.11.2012, Ist Sitting)
F + H
G
I K
J
F HG^
I KJ^ is : (1) 8 (2) – (3) 9 (4) 0 (SSC Graduate Level Tier-I Exam. 21.04.2013)
a +
value of abc is : (1) – 1 (2) 3 (3) – 3 (4) 1 (SSC Graduate Level Tier-I Exam. 21.04.2013)
s a s b s c s a b c
b g b g b g
(^2 2 2 ) 2 2 2
is equal to (1) a^2 + b^2 + c^2 (2) 0 (3) 1 (4) 2 (SSC Graduate Level Tier-I Exam. 21.04.2013)
2
is
(1) 36 (2) 30 (3) 32 (4) 34 (SSC Graduate Level Tier-I Exam. 19.05.2013 Ist Sitting)
F – HG^
I KJ^
2 2
2 2 2 2 2 2
is
(SSC CPO S.I. Exam. 12.01.2003)
is (1) 1 (2) 2 (3) 3 (4) 0 (SSC CHSL DEO & LDC Exam. 11.12.2011 (Ist Sitting (Delhi Zone) & (SSC CHSL DEO & LDC Exam. 10.11.2013)
x y
tx y
2 2
2 4
a perfect square, then the val- ues of t is (1) + 1 (2) + 2 (3) 0 (4) + 3 (SSC CHSL DEO & LDC Exam. 28.10.2012 (Ist Sitting)
is
(1) 2 (2) 1 (3) 0 (4) – (SSC CHSL DEO & LDC Exam. 28.10.2012, Ist Sitting)
(SSC Graduate Level Tier-I Exam. 19.05.2013 Ist Sitting)
x a b c
x b c a
x c a b
2 2 2
= 4( a + b + c ), then x is equal to (1) ( a + b + c )^2 (2) a^2 + b^2 + c^2 (3) ab + bc + ca (4) a^2 + b^2 + c^2 – ab – bc – ca (SSC Graduate Level Tier-II Exam. 29.09.2013)
and
2 2 2 2
is equal to
(SSC Graduate Level Tier-I Exam. 21.04.2013 IInd Sitting)
then the relation among x , y , z is (1) x + y + z = 0 (2) x + y = z
(4) x = y = z (SSC Graduate Level Tier-I Exam. 21.04.2013)
2 2
F HG^
I KJ^ is (1) 12 (2) 14 (3) 16 (4) 10 (SSC CGL Tier-I Re-Exam. (2013) 27.04.2014)
(^2) is a
perfect square?
(SSC CGL Tier-I Re-Exam. (2013) 27.04.2014)
the following relations is true?
(1)
(SSC CGL Tier-I Re-Exam. (2013) 27.04.2014)
a – b =
, then the value of c^2 – d^2 is
(1)
(SSC CGL Tier-I Re-Exam. (2013) 20.07.2014 (Ist Sitting)
and
b , c are all non-zero, then ‘ abc ’ is equal to
(1)
(SSC CGL Tier-I Exam. 19.10. TF No. 022 MH 3)
the value of a^2 + 3 a +
is
(1) 24 (2) 26 (3) 28 (4) 30 (SSC CGL Tier-I Exam. 19.10. TF No. 022 MH 3)
equal to (1) 10 (2) 12 (3) –12 (4) 14 (SSC CGL Tier-I Exam. 19.10. TF No. 022 MH 3)
–1 <
< 1 then the num- ber of values of x is (1) 4 (2) 3 (3) 2 (4) 5 (SSC CHSL (10+2) DEO & LDC Exam. 16.11.2014, IInd Sitting TF No. 545 QP 6)
0, the value of
(SSC CGL Tier-II Exam. 12.04. TF No. 567 TL 9)
(SSC CGL Tier-II Exam. 12.04. TF No. 567 TL 9)
F HG^
I KJ^ is
(SSC CGL Tier-II Exam. 12.04. TF No. 567 TL 9)
equals (1) 1 (2) 27
(SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
the value of
2 2
is
( a + 1)
( a – 1) (4) a – 1
(SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
+ _b_^2 + _c_^2 = 2 ( _a_ – _b_ – _c_ ) – 3, then the value of _a + b + c_ is (1) – 1 (2) 1 (3) 3 (4) 0 (SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
(SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
value of
(SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
2 2
2 2
2 2
= 1, then the value of
2
2
2
(SSC CGL Tier-II Exam, 2014 12.04.2015 (Kolkata Region) TF No. 789 TH 7)
, then
the value of ( a^2 – ax ) is (1) 1 (2) 2 (3) – 1 (4) 0 (SSC CAPFs SI, CISF ASI & Delhi Police SI Exam, 21.06. IInd Sitting)
then the value of 8 xy ( x^2 + y^2 ) is (1) 196 (2) 290 (3) 112 (4) 194 (SSC CAPFs SI, CISF ASI & Delhi Police SI Exam, 21.06. IInd Sitting)
then the value of
is
(1) 3 (2) 2 (3) 0 (4) 1 (SSC CAPFs SI, CISF ASI & Delhi Police SI Exam, 21.06. IInd Sitting)
2
m is (1) a^2 + b^2 – c^2 (2) a^2 + b^2 (3) a^2 + b^2 + c^2 (4) a^2 – b^2 – c^2 (SSC CGL Tier-I Exam, 09.08. (Ist Sitting) TF No. 1443088)
2 2
is
(SSC CGL Tier-I Exam, 09.08. (IInd Sitting) TF No. 4239378)
p ( p^2 + 3 p + 3) is : (1) 989898 (2) 988899 (3) 999999 (4) 998889 (SSC CGL Tier-I Exam, 16.08. (Ist Sitting) TF No. 3196279)
and y
then the value of
2 2 2 2
(SSC CGL Tier-I Exam, 16.08. (IInd Sitting) TF No. 2176783)
= 1 then the value of
(SSC CGL Tier-I Exam, 16.08. (IInd Sitting) TF No. 2176783)
, then
is equal to (1) 1 (2) 0
(SSC CGL Tier-I Re-Exam, 30.08.2015)
and y =
then the value of 3 x^2 – 5 xy + 3 y^2 is (1) 1717 (2) 1177 (3) 1771 (4) 1171 (SSC CGL Tier-II Exam, 25.10.2015, TF No. 1099685)
217. If a +
= b +
= c +
where a ¹ b ¹ c ¹ 0, then the value of a^2 b^2 c^2 is (1) –1 (2) abc (3) 0 (4) 1 (SSC CGL Tier-II Exam, 25.10.2015, TF No. 1099685)
ab + bc + ca , the value of
is (1) 3 (2) 1 (3) 2 (4) 0 (SSC CHSL (10+2) LDC, DEO & PA/SA Exam, 01.11.2015, IInd Sitting)
x^3 is (1) 1 (2) 27 (3) 0 (4) – (SSC CHSL (10+2) LDC, DEO & PA/SA Exam, 01.11.2015, IInd Sitting)
Exam, 01.11.2015, IInd Sitting)
and
then the value of
2 2
2 2
2 2
& PA/SA Exam, 15.11. (Ist Sitting) TF No. 6636838)
= 2, then the value of
( a – b ) is : (1) 1 (2) 2 (3) –1 (4) 0 (SSC CHSL (10+2) LDC, DEO & PA/SA Exam, 15.11. (IInd Sitting) TF No. 7203752)
2 is :
& PA/SA Exam, 15.11. (IInd Sitting) TF No. 7203752)
2 2
(SSC CHSL (10+2) Tier-I (CBE) Exam. 08.09.2016) (Ist Sitting)
is
a + f a – f (2) – (3) 2 ab (4) 1 (SSC CGL Tier-I (CBE) Exam. 09.09.2016) (Ist Sitting)
then c +
(SSC CAPFs (CPO) SI & ASI, Delhi Police Exam. 05.06.2016) (Ist Sitting)
expression
b – g b + g
(1) –32 (2) 11
(SSC CAPFs (CPO) SI & ASI, Delhi Police Exam. 05.06.2016) (Ist Sitting)
2
2
us of circle P and
which of the following best ap- proximates the circumference of circle P?
(SSC CPO SI & ASI, Online Exam. 06.06.2016) (IInd Sitting)
= 0 if one of its roots is
(SSC CPO SI & ASI, Online Exam. 06.06.2016) (IInd Sitting)
2 2
then the value of a^2 + b^2 + 3 ab is (1) 115 (2) 121 (3) 125 (4) 127 (SSC CGL Tier-I (CBE) Exam. 27.08.2016) (Ist Sitting)
of ( a – 2)^2 +
2
F HG^
I KJ^ is : (1) 0 (2) 2 (3) –2 (4) 4 (SSC CGL Tier-I (CBE) Exam. 27.08.2016) (IInd Sitting)
, then the value of
is
(1) 6 (2) 8 (3) 2 (4) 3 (SSC CGL Tier-I (CBE) Exam. 27.08.2016) (IInd Sitting)
2
(SSC CGL Tier-I (CBE) Exam. 28.08.2016) (IInd Sitting)
F – HG^
I KJ^
2 2
(SSC CGL Tier-I (CBE) Exam. 28.08.2016) (IInd Sitting)
2 2
is
(1) –2 (2) 0 (3) 1 (4) – (SSC CGL Tier-I (CBE) Exam. 29.08.2016) (IInd Sitting)
of
(SSC CGL Tier-I (CBE) Exam. 30.08.2016) (Ist Sitting)
will be
(1) 8 (2) 10 (3) 12 (4) None of these (SSC CGL Tier-I (CBE) Exam. 31.08.2016) (Ist Sitting)
2 2
is ( a ¹ 0)
(1) 1 (2) – (3) 0 (4) 2 (SSC CGL Tier-I (CBE) Exam. 02.09.2016) (Ist Sitting)
value of x^2 +
(SSC CGL Tier-I (CBE) Exam. 02.09.2016) (IInd Sitting)
(SSC CGL Tier-I (CBE) Exam. 04.09.2016) (Ist Sitting)
= 5, then the value of
is
(SSC CGL Tier-I (CBE) Exam. 04.09.2016) (Ist Sitting)
2
2
2
= 1 then find the value of
(SSC CGL Tier-I (CBE) Exam. 04.09.2016) (Ist Sitting)
is equal to
(SSC CGL Tier-I (CBE) Exam. 06.09.2016) (Ist Sitting)
(SSC CGL Tier-I (CBE) Exam. 06.09.2016) (Ist Sitting)
then x is equal to (1) 2 a – b (2) a + b (3) a – b (4) 2 a + b (SSC CGL Tier-I (CBE) Exam. 07.09.2016) (Ist Sitting)
the value of
(SSC CGL Tier-I (CBE) Exam. 30.08.2016) (IInd Sitting)
F HG^
I KJ is (1) 1 (2) 0
(3) p + q + n (4)
(SSC CGL Tier-I (CBE) Exam. 01.09.2016) (IInd Sitting)
2
2
2
then
is
(1) 1 (2) 2 (3) 3 (4) 4 (SSC CGL Tier-I (CBE) Exam. 01.09.2016) (IInd Sitting)
value of ( a )^2 +
a f a^2 is (1) 81 (2) 18 (3) 79 (4) 83 (SSC CGL Tier-I (CBE) Exam. 02.09.2016) (IInd Sitting)
+ 3 _p_ + 3) is (1) 9999 (2) 999999 (3) 99999 (4) 9999999 (SSC CGL Tier-II (CBE) Exam. 30.11.2016) then the value of x is
(1) c
(2) c , c^2 (3) c , 2 c (4) 0, 1 (SSC CGL Tier-II (CBE) Exam. 30.11.2016)
F HG^
I KJ^ =^
, the value of 3
F H
G I K
J (^) is :
(1) –1 (2) 1 (3) –2 (4) 2 (SSC CGL Tier-I (CBE) Exam. 28.08.2016 (IST Sitting)
, find
the value of ( pa + qb + rc ). (1) 0 (2) 1 (3) 2 (4) – (SSC CGL Tier-I (CBE) Exam. 29.08.2016 (IST Sitting)
(SSC CGL Tier-I (CBE) Exam. 30.08.2016 (IIIrd Sitting)
F HG^
I KJ
2 2
F + HG^
I KJ is equal to (1) 0 (2) 2 (3) 4 (4) 8 (SSC CGL Tier-I (CBE) Exam. 31.08.2016 (IIIrd Sitting)
then the value of 8 xy ( x^2 + y^2 ) is :
(SSC CGL Tier-I (CBE) Exam. 31.08.2016 (IIIrd Sitting)