1.Algebra Booster

2.74 Algebra Booster
73. Ans. (c)
74. Ans. (a)
Since roots are real, so
D > 0
fi 4(a + b + c) 2 – 12l(ab + bc + ca) > 0
fi (a + b + c) 2 – 3l(ab + bc + ca) > 0
fi (a + b + c) 2 + 3l(ab + bc + ca)
fi (a 2 + b 2 + c 2 ) > (3l – 2)(ab + bc + ca)
fi
2 2 2
( a + b + c )
(3 l – 2) <
( ab + bc + ca)
fi (3l – 2) < 2
fi
4
l <
2 2 2
b c a
Ísince cos A = + - £ 1
3 Î
2bc
fi
b 2 + c 2 – a 2 £ 2bc
Similarly, c 2 + a 2 – b 2 £ 2ac
and a 2 + b 2 – c 2 £ 2ab
Thus, a 2 + b 2 – c 2 £ 2
(ab + bc + ca)
fi
2 2 2
( a + b + c ) ˘
£ 2˙
( ab + bc + ca)
˚
75. Given a, b are the roots of x 2 – 10cx – 11d = 0 and c, d
are the roots of x 2 – 10ax – 11b = 0.
Thus, a + b = 10c, c + d = 10a
and ab = –11d, cd = –11b
So, a + b + c + d = 10(c + a)
…(i)
and (a + b) – (c + d) = 10(c – a)
fi (b – d) = 11(c – a) …(ii)
Also, a is a root of x 2 – 10cx – 11d = 0 and c is a root of
x 2 – 10ax – 11b = 0.
Thus, c 2 – 10ac = 11b, a 2 – 10ca = 11d
fi c 2 – a 2 = 11(b – d)
fi c 2 – a 2 = 11(b – d) = 11 ¥ 11(c – d)
fi c + a = 11 ¥ 11 = 121
From Eq. (i), we get,
a + b + c + d = 121 ¥ 10 = 1210.
76. Given a, b be the roots of x 2 – px + r = 0
Thus, a + b = p, ab = r
a
Also, , 2b be the roots of x
2
– qx + r = 0
2
a
a
Thus, + 2b
= qand ◊ 2b
= r
2 2
fi
a
+ 2 b = q, ab = r.
2
Since b is a roots of x 2 – px + r = 0 and 2b is a root of
x 2 – qx + r = 0, so we can write
pb – b 2 = 2qb – 4b 2
fi 3b 2 = (2q – p)b
fi 3b = (2q –p)
fi
(2 q - p)
b =
3
Therefore, r = pb – b 2
2
p(2 q - p) (2 q - p)
= -
3 9
(2 q - p) = (3 p - 2 q + p )
9
22 ( q - p)
(2 p -q)
=
9
77. (i) Æ A, C, D ; (ii) Æ B, D ;
(iii) Æ B , D ; (iv) Æ A, C, D.
Given
2
x - 6x+
5
f( x)
=
2
x - 5x+
6
( x -1)( x -5)
=
( x -2)( x -3)
+ – + – +
–1 2 3 5
Also , Range of f(x) is (– , 1)
(i) when –1 < x < 1, clearly 0 < f(x) < 1
(ii) when 1 < x < 2, clearly f(x) < 0
(iii) when 3 < x < 5, then also f(x) < 0
(iv) when x > 5 , then f(x) > 0.
79. Given equation is x 2 – 8kx + 16(k 2 – k + 1) = 0
Since the roots are real and distinct, so
64k 2 – 64(k 2 – k + 1) > 0
fi k 2 – (k 2 – k + 1) > 0
fi k – 1 > 0
fi k > 1
Thus, k = 2, 3
Hence the smallest integral value of k is 2.
80. Given a + b = –p, a 3 + b 3 = q
Now, a 3 + a 3 = q
fi (a + b) 3 – 3ab(a + a) = q
fi (–p) 3 – 3ab(–p) = q
fi –p 3 + 3ab.p = q
fi
3
p + q
ab =
3p
Now,
2 2
a b a + b
+ =
b a ab
2
( a + b) - 2ab
=
ab
2
( a + b)
= - 2
ab
2
p
= -2
Ê
3
p + qˆ
Á
3p
˜
Ë ¯
3
3p
= - 2
3
p + q
3
p - 2q
=
3
p + q