1.Algebra Booster

4.76 Algebra Booster
1È Ê3pˆ Ê5pˆ Ê7pˆ
=- cot cot cot
2
Í Á ˜ + Á ˜ + Á ˜
Î Ë14¯ Ë14¯ Ë14¯
Ê5pˆ Ê3pˆ Ê p ˆ
-cot Á -cot -cot
Ë
˜
14 ¯
Á
Ë
˜ Á ˜
14 ¯ Ë14¯
= 0
Ê p ˆ˘
+ cot Á
Ë14
˜˙ ¯˚
32. Find all real values of the parameter ‘a’ or which the
equation (a – 1)z 4 – 4z 4 – a + 2 = 0 has only purely
imaginary roots.
33. Given equation is
4 3 2
x + ax + bx + cx + d = 0
Let its roots are
a + ib, a - ib, g + id,
g -id
Given a + ib + g + id
= 3+
4i
( a + g) + i( b + d) = 3+
4i
( a + g) = 3,( b + d) = 4
Again, ( a -ib)( g - id) = 13 + 11i
( ag - bd )- i( ad + bg ) = 13+
11i
( ag - bd ) = 13, ( ad + bg ) = -11
Now, b =S ( a + ib)( a -ib)
= ( a + ib)( a - ib) + ( a + ib)( g + id)
+ ( a + ib)( g - id) + ( a - ib)( g + id)
+ ( a -ib)( g - id) + ( g + id)( g -id)
2 2 2 2
= a + b + g + d + ( a + ib)(2 g)
+ ( a -ib)(2 g)
2 2 2 2
= a + b + g + d + 4ag
2 2 2 2
= ( a + g ) + ( b + d ) + 4ag
2 2
=( a + g) + ( b + d) + 2( ag – bd)
= 9 + 16 + 2 ¥ 13
= 25 + 26
= 51
34. We have Z 2m + Z 2m – 1 + Z 2m – 2 + … + Z + 1 = 0
fi
2m
1
z
+ - 1 = 0
z - 1
fi z 2m + 1 – 1 = 0
Again Z r
, r = 1, 2, 3,..., 2m, m ΠN are the roots of the
equation (z 2m + 1 – 1) = 0
So,
2m+
1
( z - 1) = ( z -1)( z - z1)( z - z2) º ( z - zm
)
È
2m+
1
z - 1˘
fi Í ˙ = ( z - z1)( z - z2) º ( z - zm
)
Î ( z - 1) ˚
fi (z – z 1
)(z – z 2
)…(z – z m
)
= 1 + z + z 2 + … + z 2m
Taking log of both the sides, we get
log{( z - z1)( z - z2) …( z - zm
)}
2 2m
= log(1 + z + z + … + z )
Differentiating both sides w. r. t. z, we get
1 1 1
fi + +º+
z - z1 z - z2
z - zm
2 2m-1
1+ 2z + 3 z + … + 2mz
=
2 3 2m
1 + z + z + z + … + z
Put z = 1,
1 1 1 1
+ + +º+
1- z1 1- z2 1- z3 1-
z2m
1+ 2+ 3+º+
2m
=
1 + 1 + 1 +º+ 1(2m
times)
2 m(1+
2 m) 1
= ¥
2 (2m
+ 1)
= m
1 1 1
Thus, + + … + = m
z - z z - z z - z
fi
2m
r = 1
1 2
Ê 1 ˆ
ÂÁ
=- m
Ë Z - 1˜
¯
r
Integer Type Questions
1. Let the equation be
1 1 1 2
+ + =
( a + x) ( b + x) ( c + x)
x
Clearly, x = w, w 2 satifying the given equation
Put x = 1, we get,
1 1 1
2
a + 1 + b+ 1 + c+
1
=
2. Given expression is
È
1 cos {(1 )(1 2 ) ... (10 )(10 2 p ˘
+ Í -w - w + + -w -w
)}
900 ˙
Î
˚
ÈÏ
2 2 2
Ô(1 + 2 + ... + 10 ) ¸Ô
p ˘
= 1+ cosÍÌ
˝ ˙
ÍÎÔÓ
+ (1 + 2 + ... + 10) + 10Ô˛900˙˚
È
p ˘
= 1 + cos Í{ 385 + 55 + 10}
900 ˙
Î
˚
Ê p ˆ
= 1 + cosÁ450
¥
Ë
˜
900¯
Êp
ˆ
= 1+ cosÁ Ë
˜ 2 ¯
= 1 + 0
= 1
m