1.Algebra Booster

Quadratic Equations and Expressions 2.67
fi
3 2
dk
+ dk
dk
+ d
=
1
4
2 1
fi k =
4
1
fi k =
2
Again g + d = 12
fi ak + d = 12
fi d(k + 1) = 12
3
fi d Ê Á
ˆ = 12
Ë
˜
2¯
fi d = 8
1 1 1
Thus, g = 8¥ = 4, b = 8¥ = 2, a = 8¥ = 1
2 4 8
Thus, A = ab = 2 and B = gd = 32
Hence, the value of
Ê B ˆ
+ 1
Ê 32 ˆ
Á 2
Ë A + A+
2
˜ = 1 4 1 5
¯
Á + = + =
Ë4+ 2+
2
˜
¯
Previous Years’ JEE-Advanced Examinations
1. Now,
2
(5 2 - 38 + 5 3 )
=50- 10 76+ 10 3 + (38+
5 3)
= 88 + 5 3 - 10 76 + 10 3
= 88 + 5 3 - 10 76 + 2 75
= 88 + 5 3 - 10 ( 75 + 1)
= 88 + 5 3 - 10( 75 + 1)
= 88 + 75 - 10( 75 + 1)
= 78 -9 75
= 3(26 -3 75)
= 3(26 -15 3)
26 - 15 3
Thus, the square of
5 2 - 38+
5 3
= 1 = Rational number.
3
2. Given a + b = –p, ab = 1
and g + d = –q, gd = 1
Now, (a – g)(b – g)(a + d)(b + d)
= (g – a)(g – b)(d + a)(d + b)
= (g 2 – (a + b)g + ab)(d 2 + (a + b)d + ab)
= (g 2 + pg + 1)(d 2 – pd + 1)
= (g 2 + 1 + pg)(d 2 + 1 – pd)
= (–qg + pg)(–qd – pd)
2
= (q – p)(q + p)gd
= (q – p)(q + p).1
= (q 2 – p 2 )
3. Given a + b = –p, ab = q
g + d = –r, gd = s
Now, (a – g)(a – d)(b – g)(b – d)
= (a – g)(a – d)(b – g)(b – d)
= (a 2 – (g + d)a + gd)(b 2 – (g + d)b + gd)
= (a 2 + ra + s)(b 2 + rb + s)
= (–pa – q + ra + s)(–pb – q + rb + s)
= ((r – p)a + (s – q))((r – p)b + (s – q)
= (r – p) 2 ab + (s – q) 2 + (r – p)(s – q)(a + b)
= (r – p) 2 q + (s – q) 2 – (r – q)p
4. Given a, b, c are the sides of a triangle.
2 2 2
b + c -a
So,
= cos A £ 1
2bc
fi (b 2 + c 2 – a 2 ) £ 2bc
Similarly, (a 2 + c 2 – b 2 ) £ 2ac
and (a 2 + b 2 + c 2 ) £ 2ab
Thus, (a 2 + b 2 + c 2 ) £ 2(ab + bc + ca)
fi (a 2 + b 2 + c 2 ) +2(ab + bc + ca) £ 4(ab + bc + ca)
fi (a + b + c) 2 £ 4(ab + bc + ca) …(i)
Again, a 2 + b 2 + c 2 – ab – bc – ca
1 ( ) 2 ( ) 2 ( ) 2
= È
0
2 a - b + b - c + c - a ˘
Î
˚
≥
(a 2 + b 2 + c 2 ) ≥ (ab + bc + ca)
fi (a 2 + b 2 + c 2 ) + 2(ab + bc + ca) ≥ 3(ab + bc + ca)
fi (a + b + c) 2 ≥ 3(ab + bc + ca) …(ii)
From Relations (i) and (ii), we get,
3(ab + bc + ca) £ (a + b + c) 2 £ 4(ab + bc + ca)
5. (i) x = 3; (ii) no integral solution
6. We have
U = x 2 + 4y 2 + 9z 2 – 6yz – 3zx – 2xy
= (x) 2 + (2y) 2 + (3z) 2 – (2y)(3z) – (x)(3z) – (x)(2y)
= 1 [( 2 )
2
( 2 3 ) 2 (3 )
2 ]
2 x - y + y - z + z - x
≥ 0
Thus, U always non-negative.
7. Now, D = b 2 – 4ac
If D ≥ 0 , then D – b 2 < 0
So, D < b 2 and the roots of
(- b±
D)
are negative
2a
When D < 0, the roots are ( - b±
i D ) , which have
negative real parts.
2a
8. Solving, we get,
25m
x =
2m
+ 15
2m
- 60
and y =
2m
+ 15