1.Algebra Booster

1.80 Algebra Booster
Hence, the value of 12(a + d)
Ê 1 1ˆ
=12Á
+
Ë
˜ 12 6 ¯
1+
2
= 12 ¥
12
= 3.
15. Given equation ab + bc + ca = 12 will provide us the
greatest value if
12
ab = bc = ca = = 4
3
Hence, the greatest value of
(abc) 2 = 4 3 = 8 2
fi abc = 8
Previous Years’ JEE-Advanced Examinations
1. Given x + y + z = 15
5
Now, x + y + z = ( a + b )
2
fi 5 ( ) 15
2 a b fi (a + b) = 6 …(i)
1 1 1 2
Also, + + =
x y z 1 1
+
a b
fi 2 ab 5
a + b = 3
fi 2 ab 5 =
6 3
fi ab = 5
Solving Eqs (i) and (ii), we get
a = 1, b = 5.
2. Given
log(x + z) + log(x + z – y)
= 2 log(x – z)
fi log{(x + z)(x + z – 2y)} = log{(x – z) 2 }
fi (x + z)(x + z – 2y) = (x – z) 2
fi (x + z) 2 – 2y(x + z) = (x – z) 2
fi 4xz = 2y(x + z)
2xz
fi y =
( y + z)
Thus, x, y, z are in HP.
3. Let a and d be the first term and the common difference,
respectively of the given AP whereas b and r be
the first term and common ratio, respectively of the
given GP.
Thus, x = a + (m – 1)d, x = br m – 1
y = a + (n – 1)d, y = br n – 1
z = a + (p – 1)d, z = br p – 1
Now, x – y = (m – n)d
y – z = (n – p)d
z – x = (p – m)d
We have x y – z ◊ y z – x ◊ x x – y
= (br m – 1 ) (n – p)d ◊ (br n – 1 ) (p – m)d ◊ (br p – 1 (m – n)d
)
= b (n – p + p – m + m – n)d ◊ r
= b 0 ◊ r 0
= 1
4. Let the number of sides be n.
Here, a = 120°, d = 5°
n[2a + {(n – 1)d}] = (2n – 4) ¥ 180°
fi n[2 ◊ 120° + {(n – 1)5°}] = (2n – 4) ¥ 180°
fi 240n + 5(n 2 – n) = 360n – 720
fi 5n 2 – 125n + 720 = 0
fi n 2 – 25n + 144 = 0
fi (n – 9)(n – 16) = 0
fi n = 9 and 16
fi n = 9 is the required solution.
5. Given A, B, C are in AP.
fi 2B = A + C
fi 3B = A + B + C = 180°
fi B = 60°
Given b: c = 3: 2
Now, by sine rule,
b c
=
sin B sin C
fi
b sin B
=
c sin C
fi
fi
fi
sin (60°) 3
sin (120° - A ) =
2
3
2 3
sin (120° - A ) =
2
1 1
2 sin (120° - A ) =
2
(m – 1)(n – p) + (n – 1)(p – m) + (p – 1)(m – n)
fi sin (120° - A)
=
1
2
fi sin(120° – A) = sin(45°)
fi A = 85°
6. Given a 2
– a 1
= a 3
– a 2
= … = a n
– a n – 1
= d
a n
= a 1
+ (n – 1)d
(n – 1)d = a n
– a 1
We have
1 1 1
+ +º+
a + a a + a a + a
1 2 2 3 n-1
1 Ê d d d ˆ
= + +º+
d Á a1 a2 a2 a3 an
1 a ˜
Ë + + - + n ¯
1 Ê a
a
2 - a1
a3-
a2
n - an-1
ˆ
=
d
Á + +º+
˜
Ë a1 + a2 a2 + a3 an-1
+ an
¯
1
= [( a2 - a1 ) + ( a3 - a2 ) +º+ ( an
- an–1
)]
d
n