1.Algebra Booster

Sequence and Series 1.57
fi 64 + e 2 – 16e £ 4(16 – e 2 )
fi 5e 2 – 16e £ 0
fi e(5e – 16) £ 0
\
5
0 £ e £
16
È 5 ˘
Thus, the range of e is Í0, Î 16 ˙˚ .
201. As we know that,
AM ≥ GM
ab + bc + ca
fi ≥ 3 ( ab)( bc)( ca)
3
8
≥
3
fi 3
fi
fi
fi
3
( ab)( bc)( ca)
Ê8ˆ Á ≥ ( abc)
Ë
˜
3¯
( abc)
Ê ˆ
£ Á
Ë ˜
3¯
2 8
Ê8ˆ
( abc)
£ Á
Ë ˜
3¯
Ê8ˆ
Hence, the maximum value of abc is Á
Ë
˜
3¯
202. Given equation is
2
3
3/2
ax 4 + bx 3 + cx 2 + dx + 6 = 0
It is given that a, b, g and d be its roots.
Thus, a + b + g + d = –b
Â
Â
Â
ab = c
abg =-d
and abgd = 6
Applying AM ≥ GM, we have,
Êa + b + g + dˆÊabg + abd + agd + bgdˆ
Á
Ë
˜Á
4 ¯Ë
˜
4 ¯
4 4 3 3 3 3
≥ abgd a b g d = abgd = 6
Ê bˆÊ dˆ
fi Á- - ≥ 6
Ë
˜Á
4¯Ë ˜
4¯
fi bd ≥ 96
Hence, the minimum value of bd is 96.
203. As we know that AM ≥ GM,
where a + 2b + c = a + b + b + c
Ê( a + b) + ( b + c)
ˆ
Thus, Á
≥ ( a + b)( b + c)
Ë
˜
2 ¯
fi
Ê4ˆ
( a + b)( b+ c) £ Á = 4
Ë
˜
2¯
2
3/2
.
fi (ab + bc + ca + b 2 ) £ 4
fi (ab + bc + ca) £ 4 – b 2
Hence the maximum value of (ab + bc + ca) is 4.
204. As we know that AM ≥ GM
Thus,
Êa + b + c + dˆ Á
≥ ( a + b)( c + d)
Ë
˜
2 ¯
Ê2ˆ
Hence, ( a + b)( c + d) £ Á = 1
Ë
˜
2¯
Thus the maximum value of (a + b)(c + d) is 1.
Ê1
ˆ Ê1 - aˆ
205. Á - 1
Ë
˜ =
a ¯
Á
Ë
˜
a ¯
Êa + b + c -aˆ
= Á
Ë
˜
a ¯
Êb+
cˆ
= Á
Ë ˜
a ¯
Similarly 1 b – 1 = c + a and 1 b c – 1, a + b
c
multiplying, we get
Ê 1 1 1
Á 1 ˆÊ 1 ˆÊ 1
ˆ - - - =
Ë
˜Á
a ¯Ë
˜Á ˜
b ¯Ëc
¯
Êb + cˆÊc + aˆÊa + bˆ Ê8abcˆ
Á ≥ = 8
Ë
˜Á
a ¯Ë
˜Á ˜ Á ˜
b ¯Ë c ¯ Ë abc ¯
Hence the minimum value is 8.
206. As we know that AM ≥ HM
Ê( b + c) + ( c + a) + ( a + b)
ˆ
Thus, Á
Ë
˜
3 ¯
Ê
3
ˆ
≥ Á 1 1 1 ˜
Á + + ˜
Ë ( b + c) ( c + a) ( a + b)
¯
fi 2 ( ) 1 1 1 3
3 a b c Ê
ˆ
+ + Á + + ≥
Ëb + c c + a a + b˜
¯
fi
fi
fi
Ê 1 1 1 ˆ 9
( a + b + c)
Á + + ≥
Ëb + c c + a a + b˜
¯ 2
Êa + b + c a + b + c a + b + cˆ
Á + + ≥
Ë b + c c + a a + b ˜
¯
Ê a b c ˆ 9
Á + 1+ + 1+ + 1 ≥
Ëb+ c c + a a + b ˜
¯ 2
Ê a b c ˆ 9 3
fi Á + + ≥ - 3 =
Ëb + c c + a a + b˜
¯ 2 2
Hence, the result.
207. (1 + x)(1 + y) – 2(xy + 1)
= 1 + x + y + xy – 2xy – 2
= –1 + x + y – xy
= –(x – 1)(y – 1)
= –ve, since x > 1, y > 1
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