RESONANCE SOL140213 - 1

1. (A)

Sol. a > 0, Let f(x) = ax2 + bx + c

)1.(..........0c)0(f

0cba)1(f

a < 0

)2.(..........0c)0(f

0cba)1(f

(1) & (2) c (a + b + c) > 0

2. (B)

Sol. + + = 0 ; 3 = 1

2 + 2 + 2 + 2 = (+ + )2

2 = 0 2 = 6

2 2 2 = (1)2 = 12 2 + 22 + 22 = ( + + )2 2 (+ + )= 32 2.(1)(0) = 9Now equation x3 ( 6) x2 + 9x 1 = 0 x3 + 6x2 + 9x 1 = 0

3. (A)Sol. D = (4m + 1)

Roots are rational if D is a perfect squareLet 4 m + 1 = K2

m = 4

1K2

K = 1, 3, 5, 7, 9 ................... m = 0 , 2, 6, 12, 20, 30, 42 but m is 0 Number of integral values of m are 6.

4. (A)

Sol. (p 5) x2 2px + (p 4) = 0 or x2 5p4p

x5p

p2 = 0

f(x) = x2 5pp2

x + 5p4p

f(0) > 0, f(2) < 0, f(3) > 0

f(0) > 0 5p4p

> 0 ...... (1)

f(2) < 0 5p24p

< 0 ..........(2)

f(3) > 0 5p49p4

> 0 ..........(3)

Intersection of (1),(2) & (3) gives p

24,

449

5. (B)Sol. Conditions are a 2 > 0, D < 0.

a 2 > 0 a > 2 ...... (i)D < 0 82 4 . (a 2) (a + 4) < 0

16 (a2 + 2a 8) < 0

HINTS & SOLUTIONS

DATE : 14-02-2013COURSE NAME : VIKAAS (JA) & VIPUL (JB)

a2 + 2a 24 > 0a2 + 6a 4a 24 > 0(a + 6) (a 4) > 0

a (, 6) (4, ) .......(ii)Intersection of (i) and (ii) gives a (4, )

least integer value of a = 56. (C)Sol. |x4.3|x2|.5x1| = x43|x2|.5x1

x4.3|x2|.5x1 = 0 x4 = 0so x = 0

7. (C)Sol. Let the 4 numbers p,q,r,s are in A.P. i.e. p = a 3d, q = a d,

r = a + d, s = a + 3dp < q < r < sp + q = 4r + s = 20p + q + r + s = 244a = 24a = 6pq = A, rs = Bp + q = 4 = 12 4dr + s = 20 = 12 + 4d 8d = 16d = 2 the numbers are 0, 4 , 8 , 12pq = A = a2 4ad + 3d2

A = 36 48 + 12 = 0rs = B = a2 + 4ad + 3d2

= 36 + 48 + 12 = 96 (A, B) = (0, 96)

8. (A)Sol. x2 + 1 = y and y2 + 1 = x

Substracting x2 y2 = y x (x y) (x + y + 1) = 0 x = y or x + y + 1 = 0If x = y then x2 + 1 = x x2 x + 1 = 0 which has no real roots.If x + y + 1 = 0 then x + y = 1Adding, given equations, x2 + y2 + 2 = x + y x2 + y2 + 2 = 1 x2 + y2 + 3 = 0 which is not possible.

9. (C)Sol. For integral roots

D = (2(a + 1))2 4a(a + 2)= 4a2 + 8a + 4 4a2 8a

x = a2

2)1a(2 x = 1,

2aa

For 2a

a

to be integer possible values of a are 0, 1, 3

and 4 If a = 2 then x = 1 total number of integer values of x = 5

10. (D)Sol. 2{(x p) (x q) + (p q) (p x q + x} = (p q)2 + (x p)2 +

(x q)2

2(x p) (x q) + (p q)2 = (x p)2 + (x q)2

2x2 2x(p + q) + 2pq + p2 + q2 2pq = 2x2 + p2 + q2 2x(p + q)It is an identity so infinite values of x are possible.

TEST - 1TARGET : JEE (IITs) 2014

COURSE CODE : CLINIC CLASSES Mathematics - Quadratic Equation

RESONANCE SOL140213 - 2

11. (B)Sol. f(1) < 0

(2 2k + k 4) < 0k > 2 ................(i)f(2) < 0(8 4k + k 4) < 0

k > 34

............(ii)

from (i), (ii) k > 34

12. (B)Sol. 2 2 + 3 = 0 and 2 2 + 3 = 0

now 3 32 + 5 2= (2 3 + 5) 2= (3 + 5) 2= 2 2 2= 3 2= 1and 3 2 + + 5= (2 + 1) + 5= (2 3 + 1) + 5 = 2 2 + 5 = 2 equation x2 3x + 2 = 0

13. (A)

Sol.2x25

1xx25.2 3x25.3 = 0

x2x2 25 xx2

5.10 3.53 = 0t2 10t 375 = 0(t 25) (t + 15) = 0 t = 25, 15

xx25 = 52 x2 x = 2

x2 x 2 = 0 (x 2)(x + 1) = 0

x = 2, 1 sum = 114. (B)Sol. Roots of (x a) (x a 1) = 0 are a, a + 1

Roots of (x + a) (x + a2 2) are a, a2 + 2Since the roots of (x a) (x a 1) = 0 lies between roots of(x + a) (x + a2 2) = 0 f(a) = (a + a) (a2 2 + a) < 0

= (2a) (a2 + a 2) < 0 = (2a) (a 1) (a + 2) < 0

a ( , 2) (0, 1)f(a + 1) = (2a + 1) [a + 1 2 + a2]

= (2a + 1) (a2 + a 1)

= (2a + 1)

251

a

251

a

a

251

,

25

21

,21

a ( , 2)

251

,0

p = 2 q = 0 q p = 2

15. (A)Sol. ax2 + bx + a2 + b2 ab bc ca = 0

= 2ax2 + 2bx + (a b)2 + (b c)2 + (c 1)2 = 0Let f(x) = 2ax2 + 2bx + (a b)2 + (b c)2 + (c a)2

f(0) = (a b)2 + (b c)2 + (c a)2 > 0and D < 0 f(x) > 0

x R f(1) > 02a 2b + (a b)2 + (b c)2 + (c a)2 > 0 2(a b) + (a b)2 + (b c)2 + (c a)2 > 0

16. (ABCD)Sol. (1 + )x2 (6 + 4) x + (8 + 3) = 0

Discriminant = 4(2 + 1) > 0, R {1}

17. (AB)Sol. a + b + c = 1 a2 + b2 + c2 + 2 (ab + bc + ca) = 1

ab + bc + ca = 21

a3 + b3 + c3 3abc= (a + b + c) (a2 + b2 + c2 ab bc ca)

3 3abc = (1)

21

2

3 2 21

= 3abc abc = 61

18. (ABCD)

Sol. Let y = (x a)(x b)

(x c) x2 (a + b + y)x + ab + cy = 0Since x is real D 0 (a + b + y)2 4 (ab + cy) y2 + 2 (a + b 2c)y + (a b)2 0Now D1 = 4(a + b 2c)

2 4 (a b)2 = 16(c a)(c b) Range of y = R if D1 0 i.e (c a) (c b) 0i.e a c b or b c aand Range of y R other wise

19. (AD)Sol. Adding x2 9 = 0 x = 3

If x = 3 9 3a 3 = 0 a = 2and9 + 3a 15 = 0 a = 2

if x = 3 9 + 3a 3 = 0 a = 2and9 3a 15 = 0 a = 2

20. (CD)Sol. Let f(x) = ax2 + 2bx 4

f(x) = 0 does not have two real & distinct roots and f(0)= 4 f(x) 0 x Rfurther f(2) = 4a 4b 4 0 a b 1f(1) = a 2b 4 0 2b a 4f(2) = 4a + 4b 4 0 a + b 1

21. (CD)

Sol. (a 1) (x2 + 3 x + 1)2 (a + 1) [(x2 + 1)2 (x 3 )2] 0

or (a 1) (x2 + 3 x + 1)2 (a + 1) (x2 + x 3 + 1)

(x2 x 3 1) 0

2(x2 + 1) + 2a 3 x 0

x2 a 3 x + 1 0 x R 3a2 4 0 (D 0)

a

3

2,

3

2

number of possible integral value of 'a' is {1,0,1}3 Ans.and sum of all integral values of 'a' is 1 + 0 + 1 = 0 Ans.

22. (ABC)Sol. ax2 + (b )x + (a b ) = 0

roots are real D 0

2 + 2(2a b) + b2 4a2 + 4ab 0 R D1 0

a(a b) 00 < a b

b a < 0