Daily Practice Problems Problems No-1 (II T JEE JEE 2013 - VCC) T o pi pi c : N um u m be b e r S ys ys t em em & Qu Qu ad ad r at at ic ic E q u at at io io ns ns
MATHEMATICS
D at a t e: e:
A Quadratic Equation with Real Coefficients 1.
The number of values of a for which (a2 – 3a + 2) x x2 + (a2 – 5a + 6) x x + a2 – 4 = 0 is an identity in x, is (a) 0
(b) 2
(c) 1
(d ) 3
Sol: Condition for identify of ax2 + bx + c = 0 is a = b = c = 0
i.e.,
a2 – 3a 2 0 ,
a 2 – 5a 6 0
a2 – 4 0
2 a – 2a – a 2 0
2 a – 3a – 2a 6 0
(a – 2)( a 2) 0
a(a – 2) – 1( a – 2) 0
a( a – 3) – 2( a – 3) 0
a = –2, 2
a = 1, 2 2.
The equation x
2 x 1
1
(a) n o root Sol: (a) x –
2
2 x 1
has :
(b) one r oot
1–
2
x – 1 x –1 For the validity validi ty of (1), x 1
(c) t wo equal r oots – 6
(d ) infinitely many roots
(c) – 1
(d ) none of these
.................... (1)
cancelling – 2 on both sides, we get x = 1 (a contradiction) x – 1 The given equation has no n o roots. 3.
The sum of the real roots of the equation x2 + | x | – 6 = 0 is (a) 4
(b) 0
Sol: x 2 | x | –6 0 | x |2 | x | –6 0
Let | x| y
y2 y – 6 0 y2 3 y– 2 y– 6 0
y( y 3) – 2( y 3) 0 | x | –3, 4.
| x | 2 ,
y 2,–3
x = 2, –2
The smallest value of k for for which both th e roots of the equation x2 8 kx 16 k2 k 0 are real, distinct and have values at least 4 is (a) 2
(b) 3
(c) 4
(d ) None
Sol: (a) x2 – 8 kx16( k2 – k 1) 0 2 2 D 64( k – ( k – k1) 64( k– 1) 0
f( 4) 0
16 – 32 k16( k2 – k1) 0
k 1 k1 or k 2
k 2
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Daily Practice Problems N o-1 (II T JEE 2013 - VCC) Topic: Num ber System & Quadratic Equations
MATHEMATICS 5.
Date:
If a R, b R then the equation x2 – abx – a2 = 0 has (a) one positive root and one negative roots
(b) both positive roots
(c) both negative roots
(d ) non-real roots
Sol: Hints : Use the conditions for +ve roots, –ve roots, –ve roots, non-real roots.
6.
If a and b are rational and b is not a perfect square, then the quadratic equation with rational coefficients whose one root is
1 a b
is (a) x2 – 2ax (a2 – b) = 0
1
Sol: (b)
(b) (a2 – b) x2 – 2ax + 1 = 0
a b
a b
a
a b a b a b a b Quadratic ecuation thus formed. 2
2
b a b 2
;
(c) (a2 – b) x2 – 2bx + 1 = 0
a a b 2
(d ) none of these
b a b 2
a b a b 2a x 2 2 x 2 0 2 a b a b a b
x
2a
2
a b 2
x
a2 b
a b 2
2
0
(a 2 b) x 2 2ax 1 0 7.
If x R, the number of solutions of x 1 (a) 1
Sol:
(b) 2
x 1
2
1
x 1
x 1 1 x 1 2 2
1 2 x 1 8.
x 1 1 is
2
(c) 4
(d ) infinite
2
x1
1 4( x 1)
4 x 5
x
5 4
If l, m, n are real, l m, then th e roots of the equation (l – m) x2 – 5(l + m) x – 2(l – m) = 0 are (a) real and equal
(b) complex
(c) real and unequal
(d ) none of these
Sol: (l m) x 2 5(l m) x 2(l m) 0 D b2 4 ac 25( l m) 2 8( l m) 2 25(l 2 m 2 2lm) 8(l 2 m 2 2 ml ) 33(l 2 m 2 34ml)
9.
If b < 0, then the roots x1 and x2 of the equation 2 x2 + 6 x + b = 0, satisfy the condition
x1 x2 k where k is equal to x x 2
(a) – 3
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(b) – 5
1
(c) – 6
(d ) – 2
2
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Daily Practice Problems No-1 (II T JEE 2013 - VCC) To pic: N um be r S yst em & Qu adr at ic Equ at io ns
MATHEMATICS
D at e:
Sol: 2 x2 6 x b 0 x1 x2 3 x1 .x2
b
2
x1 x2 x2 x1
x12 x22 x1x 2
K
K 2
x1 x2 2 x1 x2 x1 x2
9b
3
K ,
K
Proved
10. The condition that the equation
(a) b2 = m2 Sol:
1 x
1 x b
x b x x( x b )
1 x
1 x b
(b) b2 = 2m2
1 m
1 m
1 m b
has real roots that are equal in magnitude but opposite in sign is (c) 2b2 = m2
(d ) none of these
1 m b
m b m m (m b)
(2 x b)( m2 bm) (2 m b)( x2 bx)
2m2 x 2bmx bm2 b2 m
2mx2 2mbx bx2 b 2 x (2 m b) x 2 (b 2 2 m 2 x ) bm 2 b 2 0
Take one root then try your self ***End of the Paper***
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