Algebra QEE

8/12/2019 Algebra QEE

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QEE Group I

1. The number of real roots of the equation ecosx

- e-cosx

-4 = 0 is(A) 0 (B) 1(C) 2 (D) None of these

1. (A)

ecosx = t t2 –4t – 1 = 0

t = ecosx = 2 5Since ecosx [1/e, e], so number of real roots is 0.

2. The number of real roots of the quadratic equation

n

1k

20kx (n > 1) is

(A) 1 (B) 2(C) n (D) 0

2. (D) Each term of

n

1k

2kx is non-negative, so no real root.

3. The set of values of ‘a’ for which the equation x 3 – 3x + a has three distinct real roots, is

(A) ( -, ) (B) (-2, 2)

(C) ( -1, 1) (D) none of these

3. (B) Let f(x) = x3 – 3x +a

f (x) = 3x2

–3.For three distinct real roots (i) f (x) = 0 should have two distinct real roots and

and (ii) f() f() < 0

Here = 1, = −1.

Now f() f() < 0

(1− 3 + a) (−1+ 3 + a) < 0 (a – 2) (a + 2) < 0

−2 < a < 2.

4. If a1, a2, a3 (a1 > 0) are in G. P. with common ratio r, then the value of r, for which theinequality 9a1 + 5 a3 > 14 a2 holds, can not lie in the interval

(A) [1, ) (B) [1, 9/5](C) [4/5, 1] (D) [5/ 9, 1]

4. (B)Since a1, a2, a3 (a1 > 0) are in G.P.So, a2 = a1 r ; a3 = a1 r

2 Given inequality9a1 + 5 a3 > 14 a2 9 a1+ 5 a1 r

2 > 14 a1r



2

5r 2 – 14 r +9 > 0(r – 1) ( r – 9/5) > 0r > 9/ 5 and r < 1

r [1, 9/ 5].

5. Consider the graph of f(x) = ax2 +bx +c in the adjacent

figure. We can conclude that

(A) c > 0 (B) a< 0(C) b < 0 (D) a+ b+ c < 0

x

f x

5. (A), (B), (C)

Clearly figure represents a downward parabola having its vertex

a4

D,

a2

b in the

second quadrant.

a < 0 , -b/2a < 0 b< 0

also , ax2 + bx + c = 0 has roots of opposite signs c/a < 0 c > 0.

6. Let a, b, c, a1 , b1, c1 R and ax2 +bx + c > 0 x R and a1x2 +b1x +c1 >0 x R.

Then

(A) aa1x2 + bb1 x +cc1 0 x R

(B) aa1x2 + bb1 x + cc1 0 x R

(C) aa1x2 + bb1 x +cc1 = 0 , will have real roots.

(D) Nothing can be said in general about the nature of roots of aa 1 x2 +bb1 x +cc1 =

0,

6. (D)We have, a > 0, c > 0, b2 – 4ac < 0Similarly , a1 > 0, c1 > 0, b1

2 – 4a1c1 < 0From the given information sign of (bb1)

2 – 4aa1 cc1 cannot be checked

7. Consider the equation x2 +x – n = 0, where n is an integer lying between 1 to 100.Total number of different values of ‘n’ so that the equation has integral roots is(A) 6 (B) 4(C) 9 (D) None of these

7. (C)x2 +x –n = 0, discriminent = 1+ 4n = odd number = D(say)Now given equation would have a integral soluton if D is a perfect square .

Let D = ( 2 +1)2 n = +2 = ( +1) = even number

n can be 2, 6, 12, 20, 30, 42, 56, 72, 90.

8. Number of real solutions of the system x + y = 2, xy – z2 = 1 is(A) 0 (B) 1(C) 2 (D) infinite

8. (B)



3

x + y = 2 x = 2 – y

Also, xy – z2 =1 2y – y2 – z2 = 1 z2 + (y –1)2 = 0

z = 0, y = 1, x = 1

9. Let p(x) = 0 be a polynomial equation of least possible degree, with rational coefficients,

having

33

497 as one of its roots. Then the product of all the roots of p(x) = 0 is(A) 7 (B) 49(C) 56 (D) 63

9. (C)

x = 33 497

x3 = 7 + 49 + 3 3333 49749.7

x3 – 21x – 56 = 0

Product of root = 56.

10. Number of positive integers n for which n2 + 96 is a perfect square is

(A) 4 (B) 8(C) 12 (D) infinite

10. (A)Let m be a positive integer for which n2 + 96 = m2

m2 – n2 = 96 (m + n)(m – n) = 96

(m + n) {(m + n) – 2n} = 96

m + n and m – n must be both even

96 = 2 48 or 4 24 or 6 16 or 8 12Number of solution = 4.

11. If the equations ax2 + 2bx + 3c = 0 and 3x2 + 8x +15 = 0 have a common root, where

a, b, c are the lengths of the sides of a ABC then sin2

A + sin2

B + sin2

C is equal to(A) 1 (B) 3/2

(C) 2 (D) 2

11. (D)Discriminant of 3x2 + 8x +15 = 0 is negative. So, the roots are imaginary andtherefore conjugate of each other .So, both roots are common.

15

c3

8

b2

3

a a : b : c = 3 : 4 : 5

ABC is right triangle.

sin

2

A + sin

2

B + sin

2

C = 2.

12. If (y2 – 5y + 3)(x2 +x +1) <2x for all x R, then the interval in which y lies is

(A)

2

55,

2

55 (B) ( -, -2]

(C)

3

2,2 (D) ( 1, 4)



4

12. (A)

( y2 – 5y +3) (x2 + x +1) < 2x x R

y2 – 5y + 3 <1xx

x22

Let

1xx

x22

= p px2 + ( p –2)x +p = 0

Since x is real, ( p –2)2 – 4p2 0

-2 p 3

2

Minimum value of1xx

x22

is -2

So, y2 – 5y +3 < -2 y2 – 5y + 5 < 0

y

2

55,

2

55 .

13. A point (, 2

) lies inside the triangle formed by the coordinate axes and the linex + y = 6. If is a root of f(x) = x2 + ax + b = 0 then which of the following is alwaystrue ?(A) f(0) > 0 (B) f(2) > 0

(C) f() 0 for atleast one (0, 2) (D) –4 < a < 0

13. (C)

14. If p and q are odd integers, then the equation x2 + 2px +2 q = 0(A) has no integral roots (B) has no rational roots(B) has no irrational roots (D) has no imaginary roots.

14. (A), (B)x can not be odd integer for if x is odd, x2 is odd but 2px + 2q is even;

so x2+2 px + 2q 0x can not be even integer for if x is even, x2 +2 px is a multiple of 4 but 2q is not.

So x2 + 2px + 2q 0 Also (x +p)2 = p2 – 2q

If x is fraction then (x +p)2 is also a fraction but p2 –2q is an integer. So, rootscannot be integer or rational numbers.

15. If all the solutions ‘x’ of acosx + a –cosx = 6 (a > 1) are real, then set of values of a is

(A) [3+2 2 , ) (B) (6, 12)

(C) (1, 3 + 2 2 ) (D) none of these.

15. (A)



5

Let acosx = t

t +t

1 = 6

t2 – 6t + 1 = 0

t =

2

4366 = 3 2 2

acosx = 3 2 2

cosx = loga(3 2 2 )

since a > 1 , for all the roots to be real,

y = loga(3+ 22 )+1

-1

y = logax

y = loga(3 – 22 )

O

we must have loga(3 + 2 2 ) 1 and loga(3 - 2 2 ) -1,

Both are true for a 3 + 2 2 .

16. The equation (1 +a2)x2 +2a2 x + a2 + b2 –1 = 0 has roots of opposite sign if a +ib lies(A) on a straight line x + y =1 .(B) inside a circle of centre (0, 0) and radius 1.

(C) on a parabola of vertex ( 0, 0) and focal length 1.(D) none of these.

16. (B) As roots are of opposite sign, product of roots < 0

a2 + b2 -1 < 0

a2 + b2 < 1

| a+ ib| < 1So, the point a + ib lies inside a circle of centre ( 0, 0) and radius 1.

17. If the equation x2 + nx + n = 0, n I, has integral roots, then n2 – 4n can assume(A) no integral value (B) one integral value

(C) two integral values (D) three integral values

17. (B)The roots of the equation

x =2

n4nn 2 may be integers if n2 – 4n = I2 where I is an integer.

n2 – 4n – I2 = 0 . . . (1)

n = 2 2I4

Now n is an integer 2I4 should be an integer.

4 + I2 = k2 where k is an integer .

k2 – I2 = 4

which is possible only when k = 2 , I = 0 .putting I = 0 in (1) ,

n2 – 4n = 0 n = 0, 4. For both these values,x2 +nx + n = 0 has integral roots.

n = 0, 4 n2 – 4n = 0 .

18. If 4ac > b2 and a + c > b for real numbers a, b and c, then which of the following is true?



6

(A) a > 0 (B) c > 0(C) a + b + c > 0 (D) 4a + c > 2b

18. (A), (B), (C), (D)Let f(x) = ax2 +bx +cIts discriminant D = b2 - 4ac < 0

And f(-1) = a- b +c > 0So, f(x) > 0 for all x R

So, a > 0

f(0) >0 c > 0

f(1) > 0 a +b +c > 0

f( -2) > 0 4a – 2b + c > 0 .

19. The least value of the expression x2 + 4y2 + 3z2 – 2x – 12y – 6z +14 is(A) 0 (B) 1(C) no least value (D) none of these.

19. (B)Let f(x, y, z) = x2 + 4y2+ 3z2 – 2x –12y – 6z + 14= ( x- 1)2 + (2y – 3)2 + 3( z- 1)2 + 1For least value of f ( x, y, z)x-1 = 0 ; 2y – 3 =0 and z – 1 =0

x = 1 ; y = 3/2 ; z = 1Hence least value of f(x, y, z) is f( 1, 3/2, 1) = 1 .

20. If (b2 – 4ac)

2(1 + 4a

2) < 64a

2, a < 0, then maximum value of quadratic expression

ax2 + bx + c is always less than

(A) 0 (B) 2

(C) – 1 (D) – 2

20. B

22

22

a41

4

a16

)ac4b(

Now, max (ax2 + bx + c) = –

a4

ac4b2

2a41

2

< –

a4

ac4b2 <

2a41

2

So, max value always less than 2 (when a 0).

21. The least integral value of k such that (k –2)x2

+ 8x +k + 4 is positive for all real valuesof x is(A) 1 (B) 2(C) 3 (D) 5

21. (D)For ( k-2) x2 + 8x +k +4 > 0(k-2 ) > 0 and (8)2 – 4( k-2)( k +4) < 0i.e. k > 2 . . . . (1)



7

and 24 – k2 – 2k < 0i.e. k2 +2k – 24 > 0i.e. ( k+6) ( k-4) > 0

k < -6 or k > 4 . . . . . (2)From (1) and (2), we have k > 4 .

The least integral value of k = 5 .

22. Let S be the set of values of a for which ( a-4)sec4x+( a –3)sec2x +1 = 0 has realsolutions. Then S is

(A) R (B) (- , 3]

(C) (4, ) (D) [3, 4)

22. (D)

sec2 x=a4

1

, -1

Rejecting the negative value

Sec2x =

a4

1

4 –a > 0 a < 4 . . . . (1)

Also,a4

1

1, 4 – a 1 . . . . (2)

a 3Combining (1) and (2) , we get the solution [3, 4) .

23. If both roots of equation ax2 + x + c – a = 0 are imaginary and c > – 1, then

(A) 3a > 2 + 4c (B) 3a < 2 + 4c

(C) c < a (D) none of these

23. B

Let f (x) = ax

2

+ x + c –

af (1) = c + 1 > 0 ( c > – 1)

Given expression is positive for every x R

So, f

2

1 > 0

2

1

4

a + c – a > 0 4c – 3a + 2 > 0

4c + 2 > 3a.

24. If ax2 + bx + c = 0 and cx2 + bx + a = 0 ( a, b, c R) have a common non-real root,then(A) -2|a| < b < 2 |a| (B) – 2|c| < |b| < 2|c|

(C) a = c (D) a = c

24. (A), (B), (D)D1 = b2 –4ac < 0, D2 = b2 – 4ac < 0, as the root is non-real

Both roots will be common.

1a

c

b

b

c

a a = c

Now, b2 - 4ac < 0 b2 – 4a2 ( or 4c2) < 0

|b| < 2 |a| ( or 2|c|).



8

25. If a, b, c be the sides of ABC and equations ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0

have a common root, then C is(A) 60° (B) 90°(C) 120° (D) 45°

25. (B)since 5x2 + 12x + 13 = 0 has imaginary roots as D = 144 – 4 5 13 < 0So, both roots of ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0 will be common

13

c

12

b

5

a a2 + b2 = c2 C = 90°

26. The equation x2 + nx + m = 0, n, m I, can not have(A) integral roots (B) non-integral rational roots(B) irrational roots (D) complex roots

26. (B)

The given equation can be written as x (x +n) +m = 0If x is a non-integral rational number, then both x and x + n will have the same

denominator ( 1) and x(x + n) will not be an integer .The sum of a non-integer and an integer can never be zero.

the given equation can not have non-integral rational roots.

27. If z, z1 , z2 , z3 are four complex numbers such that2|z – z1| = |z – z2| + |z – z3|

2|z – z2| = |z – z1| + |z – z3|

2|z – z3| = |z – z1| + |z – z2|

Then if the points A, B and C have position vectors represented by complex numbers z1,

z2 and z3 respectively, then the point whose position vector is represented by the

complex number z, is

(A) the circumcentre of ABC (B) the incentre of ABC

(C) the orthocentre of ABC (D) the centroid of ABC

27. (A)Let the given equations be numbered as (1), (2) and (3)Then, subtracting (1) from (2),2( |z –z2 | – |z –z1 | ) = |z –z1| – |z –z2 |

|z –z1 | = |z –z2 |

Similarly solving other pairs of equations and combining the results,|z –z1| = |z –z2| = | z – z3|

z is the circumcentre of ABC.

28. The value of 5724057240 is equal to

(A) 10 (B) -4(C) 4 (D) -10



9

28. (D)

I2 = 2/12 4080)57(22405724057

= 114 – 2 { 2500 + 49 + 700 – 3200 }1/2

= 100 I = –10, as I is negative.

29. If ,x

asin13x4

3

x2

then a is equal to

(A)2

)1n2( (B)

2)1n4(3

(C) 3 (1 + 4n) (D) None of these

29. (C)

LHS =x

asin1

3

)6x(

3

3)6x(

3

39x12x 222

Since 1x

a

sin , x = 6 16

a

sin a = 3(1+4n)

30. If both the roots of the equation x2 – (p – 4) x + 2 e2lnp – 4 =0 are negative then pbelongs to

(A) 4,2 (B) 4,2

(C) 2,4 (D) 2,

30. (B) Both roots negative sum of roots < 0 and product of roots > 0

p – 4 < 0 and pln2e2 - 4 > 0

p < 4 and p2 > 2

p ( 2 , 4) .

31. If , be the roots of quadratic equation ax2 + bx + c = 0, then value of

ba

ca

ba

ca 22

is

(A)a4

)ac2b(b 2 (B)

a2

ac4b2

(C)ca

)ac2b(b2

2 (D) none of these

31. C

a2 + c = – b and a + b = –

c

Given expression isc

b(2 + 2) =

ca

)ac2b(b2

2 .

32. Product of the real roots of the equation t2x2 + |x| + 9 =0 ( t 0)



10

(A) is always positive (B) is always negative(C) does not exist (D) none of these

32. (C)

33. If the expression kx + x

1

– 2 is non-negative for all positive real x then the minimum

value of k is(A) 1 (B) 2(C) 1/4 (D) 1/2

33. (A)

34. If ‘1’ lies between the roots of equation x2 +(a –1)x + a2 =0 then(A) 1 < a < 2 (B) -2 < a < -1(C) 0 < a< 1 (D) -1 < a< 0

34. (D) If ‘1’ lies between the roots thenf(1) < 0

a2 +a < 0

-1 < a < 0.

35. If the equation 5x2 –10x + log1/5a = 0 has real roots then the minimum value of ais(A) 1/ 55 (B) 1/1010 (C) 1/510 (D) none of these.

35. (A)

36. If the equations ax2 +bx +1 =0 and ax2 +x +b =0 have exactly one common root then(A) a – b =2 (B) a – b = 1(C) a + b = 2 (D) a + b = -1

36. (D) Let be the common root

a2 +b +1 =0 and a2 + +b = 0

= 1Therefore , a +b = -1

37. Consider the equation x2 +x – a = 0, a N. If equation has integral roots then(A) a = 2 (B) a = 6

(C) a = 12 (D) a = 20

37. (A), (B), (C) and (D)

Discriminant, D= a41

1+4a should be a perfect square. As 1+4a is always odd

1+4a = (2+1)2, I+

a= (+1).



11

38. If the equation cx2 +bx – 2a =0 has no real roots and a <2

cb then

(A) a c < 0 (B) a < 0

(C) a

2

ac

(D) a

8

b2c

38. (A), (B), (C) and (D)Let f(x) = cx2 + bx – 2a

a2

cb

b +c –2a > 0 f(1) > 0

Now f(x) =0 has no real root.Therefore f(x) > 0 for all x or f(x) < 0

But f(1) > 0. Therfore f(x) > 0 for all x f(0) >0 a < 0.Clearly c> 0 .

ac < 0 . Also f(-1) > 0 c – b > 2a

similarly, f(1/2) > 0 a8

b2c

.

39. Equation x2 +x +a = 0 will have exactly one root in the interval (0, 1) if

(A) –2 a< 0 (B) –2 < a < -1

(C) 1 a < 4 (D) 0 a < 1

39. (B)f(0).f(1) < 0

a( a+2) < 0

-2 < a < 0.

40.

1x5 24x9xlog 25

(A) x R (B) x ( 0, )(C) x ( -, 0) (D) none of these

40. (A), (B) and (C)

1x5 24x9xlog 25

x2 – 9x +2 4 > x –1

x2 – 10x + 25 > 0 ( x – 5)2 > 0.

Which is true for x R.

41. The integer(s) ‘p’ for which the inequality x2 – 2(p –1) x + ( 2p +1) > 0 is valid for all x(A) 1 (B) 2(C) 3 (D) none of these .

41. (A), (B) and (C)

42. If and are the roots of the equation x2 –2x +2 =0, then the integral solutions to the

equation n = n will be given by

(A) n = 2k, k I (B) n= 4k, k I(C) n = 6k , k I (D) none of these

42. (B) (1+ i) and (1- i) are the roots of the equation x2 – 2x +2 = 0.



12

1i1

i1 n

n = 4k , k I

43. The square of sum of the roots of the equation x3 +2x2 +2x+1 =0 equals to(A) -2 (B) 2

(C) 4 (D) – 4

43. (C) x3 +2x2 +2x +1 = 0 ( x +1)(x2 +x +1) = 0

x = -1 , , 2 Therefore , ( sum of roots)2 = ( - 1+ +2) = ( -1 + ( +2 +1) – 1)2 = ( -2)2 = 4 .

44. Least natural number ‘a’ for which x+ ax-2 > 2 x ( 0, ) is(A) 1 (B) 2(C) 5 (D) none of these

44. (B)

Let f(x) = x + ax-2 f (x) = 1 – 2ax-3 = 0 x = (2a)1/3

f (x) = 6ax-4 > 0 x ( 0 , ) (as ‘a’ is a natural number) so x = ( 2a)1/3 is a point of global minimaThus (2a)1/3 +a(2a)-2/3> 2

a >27

32 a 2 . As ‘a’ is a natural number.

Alternative solutionx + ax-2 > 2

x3 – 2x2 +a > 0Let f(x) = x3 – 2x2 + a

Since f(x) > 0 x ( 0, ), f(0) > 0 and min f(x) > 0 a> 0 and for minimum f(x)

f (x) = 3x2 – 4x

x = 0, 4/3

f(4/3) > 0 a >27

32 .

45. If p(x) = ax2 + bx and q(x) = lx2 + mx + n with p(1) = q(1), p(2) – q(2) = 1, and p(3) – q(3)= 4, then p(4) – q(4) is equal to(A) 7 (B) 16(C) 9 (D) none of these

45. (C)Let f(x) = p(x) – q(x)

f(1) = p(1) – q(1) = 0 f(x) = (x – 1) (t1x + t2)f(2) = 1= 2t1 + t2

f(3) = 4 = 2 (3t1 + t2) t1 = 1, t2 = -1

f(4) = p(4) – q(4) = 3 (4t1 + t2) = 3(4 – 1) = 9



13

46. If the roots of the equation b2x2 + (1+a)x – a2 – b2 + 4 = 0 are of opposite signs, then thepoint represented by the complex number Z = a + ib(A) is located inside a circle centred at origin of radius 2.(B) is located outside a circle centred at origin of radius 2

(C) lies on a straight line

(D) None of these

46. (B)Roots are of opposite sign

a2 + b2 – 4 > 0a2 + b2 > 4

|z| > 2.

47. If x2 + ax + b = 0 and x2 + bx + a = 0, (a b) have a common root, then a + b is equal to

(A) 0 (B) 1

(C) -1 (D) none of these

47. (C)Let be a common root

2 + a + b = 0

and 2 + b + a = 0

ab

1

abba 22

2

= 1

a + b+ 1 = 0 a + b = -1.

48. Let P(x) and Q(x) be two polynomials. If f(x) = P(x4) + xQ(x4) is divisible by x2 +1, then

(A) P(x) is divisible by (x-1) (B) Q(x) is divisible by (x-1)

(C) f(x) is divisible by (x-1) (D) none of these

48. (A), (B), (C)f(x) = P(x4) + xQ(x4) is divisible by (x + i) (x – i)

f(i) = P(1) + iQ(1) = 0f(-i) = P(1) – iQ(1) = 0

P(1) = Q(1) = f(1) = 0

P(x), Q(x) and f(x) are divisible by (x - 1).

49. If the equation ax2 + bx + c = 0 (a < 0) has two roots and such that < -3 and > 3,

then

(A) 9a + 3|b| + c > 0 (B) c > 0

(C) 4a + 2|b| + c > 0 (D) none of these

49. (A), (B), (C)Let f(x) = ax2 + bx + c

–3 and 3 are lying between the roots of f(x) = 0

f(-3) > 0 and f(3) > 0

9a + 3|b| + c > 0 Also, 0, -2, 2 are lying between the rootstherefore c > 0 and 4a + 2|b| + c > 0



14

50. If sin, cos are the the roots of the equation cx2 + bx + a = 0, then a, b,c, areconnected by the relation(A) b2 + 2ac – c2 = 0 (B) c2 – 2ac+ b2 = 0(C) b2 – 2ac – c2 = 0 (D) 2ac – b2 – c2 = 0

50. C

sin + cos = – c

band sin cos =

c

a

sin 2 + cos2 + 2sin cos =2

2

c

b

1+22

2

c

b

c

a

c2+2ac –b2 = 0

51. The greatest integer x for which the inequality 022x9x

3x2

is satisfied, is equal to

(A) – 12 (B) –11(C) 2 (D) 3



15

Group II

1. A

02x11x

3x0

22x2x11x

3x2

Largest integral value of x = –12

2. If a 2 + b2 +c2 + d2 = 1, then the maximum value of ab + bc + cd +da is(A) zero (B) One(C) Two (D) None of these

2. BSince ( a –b)2 + (b –c)2 + (c –d)2 + (d –a)2 = 2(a2 + b2 + c2 + d2) – 2(ab+ bc + cd + da)= 2 – 2(ab + bc+ cd+da)The value of LHS of (1) is minimum when ab + bc +cd +da is maximum and theminimum value of LHS is equal to zero .

Therefore maximum value of ab + bc + cd + da= 1

3. The number of real solutions of the equation

cos5 x+sin3x=1 in the interval [0,2] is(A) 2 (B) 1(C) 3 (D) Infinite

3. C

cos5 x cos2 x

and sin3 x sin2 xso cos5 x + sin3 x = 1 is possible only whencos5 x = cos2x and sin3 x = sin2 x

which is possible only at x = 0, /2 and 2

4. Let f(x) = ax3 + bx2 + x +d has local extrema at x = and such that . < 0, f(), f() >0; Then the equation f(x) = 0

(A) has 3 distinct real roots

(B) has only one real root, which is positive if a f() < 0

(C) has only one real root, which is negative if a f() > 0(D) has 3 equal real roots

4. B, C

5. The set of values of ‘a’ for which 1 lies between the roots of x 2 – ax – a+3 = 0 is

(A) (-, -6) (B) -, -6](C) (-, -6) (2, ) (D) (2, )

5. DLet f(x) = x2 – ax – a+3Since 1 lies between the roots

So f(1) < 0 a > 2 so a ( 2, )



16

6. If sin, sin and cos are in GP, then roots of x2 + 2xcot + 1 = 0 are always(A) equal (B) real(C) imaginary (D) greater than 1

6. B

Sin , sin , cos are in GP sin2 = sin cos Or , cos 2 = 1 – sin 2 0Now discriminate of the given equation is

4cot2 – 4 = 4cos2. cosec2 0Hence roots are always real.

7. Let a, b,c, R such that 2a + 3b + 6c = 0. Then the quadratic equation ax 2 + bx + c = 0has(A) at least one root in (0,1) (B) at least one root in ( -1, 0)(C) both roots in (1,2) (D) imaginary roots

7. A

Let g (x) = .cx

2

bx

3

ax 22

Then g(0) = 0.

Also g(1) = 0c6b3a26

1c

2

b

3

a

Thus , by Rolle’s theorem there exists at least one root of ax2 + bx +c = 0 in (0,1)

8. If ax2 + bx + 1=0 does not have 2 distinct real roots then least value of 2a – b is ____________

8:2

1

since ax2 + bx + 1 = 0 does not have 2 distinct real roots. So either f(x) ax 2 + bx + 1 0 x R or ax2 + bx + 1 0 x R but f(0) = 1 so f(x) 0 x R.

f(-2) 0 2a – b - 1/2.

9. If , , are the roots of the equation, x3 + P0x2 + P1x + P2 = 0, then (1 - 2) (1 - 2)

(1 - 2) is equal to(A) (1 + P1)

2 - (P0 + P2)2 (B) (1 + P1)

2 + (P0 + P2)2

(C) (1 - P1)2 - (P0 - P2)

2 (D) None of these

9. A

x3 + P0x2 + P1x + P2 = (x - ) (x - ) (x - )

Put x = 1 and multiply corresponding sidesWe get the result

10. The set of values of ‘a’ for which the inequality x2 – (a + 2)x-(a + 3) < 0 is satisfied for atleast one positive real x is _________.

10. [ –3, )Let f(x) = x2 – ( a+2) x – (a+3)F(x) < 0 is true for at least one +ve real x if f(x) < 0 have at least one +ve real root for

which a –3



17

11. If roots of quadratic equation ax2 + 2bx +c = 0 are not real, thenax2 + 2bxy + cy2 + dx + ey+ f = 0 represent(A) Ellipse (B) Circle(C) Parabola (D) Hyperbola

11. A

Since roots are not real b2 < ac ax2 + 2bxy+ cy2 + dx +ey + f = 0 can represent only an ellipse,

12. The equation1 1

2

2cos x cos x1

2 a 2 a 02

has only one real root then

(A) 1 a 3 (B) a 1, a –3(C) 1 < a < 3 (D) none of these

12. B

1 1cos x

< 2 1cos x2

< .

Hence 2 should lie between the roots of2 21

t a t a 02

where t =

1cos x2

.

a2 + 2a – 3 0 a ( –, –3] [1, ).

13. If e|x| + 2 =2a a |x|e

has exactly one solution then the values of a are

(A) 1, –2 (B) –1, 2(C) 1, 2 (D) none of these

13. A

|x| + 2 = a2 + a – |x| has only one solution

|x| = 0 a2 + a – 2 = 0

a = 1, –2.

14. If the sum of the squares of the roots of the equation x2 – (cos 2 – 2)x – (1 + cos 2) = 0

is least then tan is equal to(A) –1 (B) 1(C) zero (D) none of these

14. C

, be roots of given quadratic equation then + = cos 2 – 2

= –(1 + cos 2)now 2 + 2 = ( + )2 – 2 = (cos 2 – 2)2 + 2(1 + cos 2)= cos2 2 + 4 – 4 cos 2 + 2 + 2 cos 2 = cos2 2 – 2 cos 2 + 6

= (cos 2 – 1)2 + 5



18

2 + 2 is minimum when cos 2 = 1

= n or tan = 0.

15. If sin x + cosec x + tan y + cot y = 4, where x and y 0,

2

, theny

tan

2

is root of the

equation

(A) 2 + 2 + 1 = 0 (B) 2 + 2 1 = 0

(C) 22 2 1 = 0 (D) none of these

15. BGiven that sin x + cosec x + tan y + cot y = 4

x =2

and y =

4

tan y = 1

2

2

2tany / 2 y y1 tan 2 tan 1 0

2 21 tan y / 2

.

16. If both the distinct roots of the equation |sin x|2 + |sin x| + b = 0 in [0, ] are real, thenvalues of ‘b’ are

(A) [ 2, 0] (B) ( 2, 0)

(C) [ 2, 0) (D) ( 2, 0]

16. BGiven that |sin x|2 + |sin x| + b = 0

|sin x| =1 1 4b

2

0

1 1 4b

2

< 1

2 < b < 0.

17. The set of values of ‘a’ for which all the solutions of the equation 4sin4 x + asin2x + 3 = 0are real and distinct is(A) (2, 6) (B) (2, 4)

(C) (0, 1) (D) [-7, -4 3 )

17. DPut sin2 x = t and f (t) = 4t2 + at + 3For required condition both the roots of the equation f (t) = 0 should be distinct andshould lie in the interval [0, 1]

for this, D > 0 a2 –48 > 0 a < -4 3 , a > 4 3

f (0) 0, f (1) 0 a -7

Since 0 < sum of roots < 2 0 < -4

a < 2 -8 < a < 0

so required ‘a’ will be –7 a < -4 3 .

18. If f (x) = |a2x2 + b|x| + c| is non-differentiable at two points then the sum of the roots of

the equation a2x2 + bx + c = 0 is always



19

(A) positive (B) negative

(C) zero (D) can’t say

18. C

Given that f (x) = |a2x2 + b|x| + c|

f (x) =2 2

2

2 2

a x b | x | c bx2a x

| x || a x b | x | c |

|a2x2 + b|x| + c| will have two roots.

Since |a2x2 + b|x| + c| is an even function, so the roots will be numerically equal sum

of the roots = 0.

19. If the equation sin2 x – a sin x + b = 0 has only one solution in (0, ) then which of thefollowing statements are correct.

I: a (- , 1] [2 , )II: b (- , 0] [1, )III: a = 1 + b

(A) I and III (B) II and III(C) I and II (D) I, II and III

19. D

sin2 x a sin x + b = 0 has only one solution in (0, ) sin x = 1 gives one solution and sin x = gives other solution such that > 1 or 0 (sin x 1) (sin x ) is the same equation as sin2 x a sin x + b = 0

1 + = a and = b

1 + b = a and b > 1 or b 0 b ( , 0] [1, )and a ( , 1] [2, ).

20. The complete set of non-zero values of k such that the equation |x2 10x + 9| = kx issatisfied by atleast one and atmost three values of x, is

(A) [4, ) ( , 16] (B) [16, ), ( , 16]

(C) [4, ) ( , 4] (D) [16, ), ( , 4]

20. AFor the given condition to be satisfied the vale of k must liebetween the slopes of line OA and OBfor slope of line OA

(x2 10x + 9) = kx must have equal roots

(k 10)2 36 = 0 k = 4, 16 k = 4.

For slope of OB x2 10x + 9 = k must have equal roots

(k + 10)2 36 = 0 k = 4, 16

k = 16

k should lie between [4, ) ( , 16].

y

x

A

B

O 1 9

21. For x2 – (a + 3)|x| + 4 = 0 to have real solutions, the range of a is

(A) ( –, –7] [1, ) (B) ( –3, )(C) ( –, –7] (D) [1, )



20

21. D

a =2x 4

3| x |

22 x 4

3| x |

(A.M G.M)

a 1.

22. If the roots of the equation ax2 – bx + c = 0 are , then the roots of the equationb2cx2 – ab2 x + a3 = 0 are

(A)2

1

,

2

1

(B)

3

1

,

3

1

(C)4

1

,

4

1

(D) none of these

22. A

Multiplying the given equation by3

c

a, we get

2 2 22

3 2

b c b cx x c 0

a a

2

2 2

bc bca x b x c 0

a a

– 2

bcx ,

a

( + ) x = ,

x = 1 1,( ) ( )

.

23. If exactly one of the roots of the equation x2 + (a + 3)x + a = 0 lies in [1, 3] then the

minimum value of21 a

a

is

(A)3

2 (B)

77

18

(C) – 9

2 (D) – 2

23. A

(1 + (a + 3) + a) ( 9 + (a + 3)3 + a) 0

9

a 22

Now, as21 a 1

aa a

is decreasing in a, its minimum value =

1 3( 2)

( 2) 2

.



21

24. In the quadratic equation 4x2 – 2 (a + c – 1)x + ac – b = 0 (a > b > c)(A) both roots are greater than a(B) both roots are less than c

(C) both roots lie betweenc a

and2 2

(D) exactly one of the roots lies betweenc a

and2 2

24. DHere f (x) = (2x – a) (2x – c) + (2x – b)

So fa

2

= a – b, fc

2

= c – b.

As fa

2

fc

2

= (a – b)(c – b) < 0 (a > b > c)

exactly one of the roots lies betweenc a

and2 2

.

25. If all roots of equation z3 + az2+ bz + c = 0 (a, b, c R) are of unit modulus, then

(A) |a| 3 (B) |b| < 3

(C) |c| 3 (D) none of these

25. Az1 + z2 + z3 = –a

| z1 + z2 + z3| = |a| |z1| + |z2| + |z3|

1 + 1 + 1

|a| 3.

26. If the equation |x2

+ bx + c| = k has four real roots, then

(A) b2 – 4c > 0 and 0 < k <24c b

4

(B) b2 – 4c < 0 and 0 < k <

24c b

4

(C) b2 – 4c > 0 and k >24c b

4

(D) none of these

26. AFor the equation to have four real roots the line y = k must intersect |x 2 + bx + c| at four

points D > 0 and k D

0,4a

.

27. f(x) =3 2 2

x x x and > 0. If f(x) > 0 for x 2 then lies in the interval

(A)1

0,2

(B) (0, 2)

(C) (2, ) (D)1

,2

27. D



22

f(x) = 2x x 1 0

> 0 x – 1 > 0

1

x

1 1

x 2

max

1x

12 .

28. The number of integral solutions of1 1

1 1 2b c

is

(A) 1 (B) 2(C) 4 (D) 0

28. B

2b 1

c 1

b is integer when

2

c 1 is integer i.e. c = ( –1, 0, 2, 3).

–1, 0 get rejected as these do not satisfy the original equation i.e. solutions are (2, 3), (3,2).

29. If f(-1) < 1 , f(1) > -1, f(3) < -4 are known to be satisfied for a certain function

f(x) = ax2 + bx + c, a 0, then

(A) a >1

8

(B) a <

1

8

(C) a R – { –1/8} (D) None of these

29. Bf(x) = ax2 +bx +cgiven f(-1) = a –b +c < 1

f(1) = a + b + c > -1, f(3) = 9a +3b +c < -4 .Now a + b + c > -1 i.e. – a-b –c < 1- 3a – 3b – 3c < 39a + 3b + c < -46a – 2c < -1 . . . . (1)

Also 3a – 3b +3c < 39a + 3b + c < -412a + 4c < -1 . . . . . (2)

From equation (1) and (2)12a – 4c < -212a + 4c < -1

24 a < – 3 a < -1/8 < 0 .

30. The solution ofy

1|y|

y

y1 2

where y is a real number is

(A) [ –1, 1] (B) ( –1, 1)

(C)

1,

2

1

2

1,1 (D) none of these

30. D



23

0y.y

y1 2

1 y2

– 1 y 1, y 0.

31. The value of ‘a’ for which 2 1 1x (a 1)x 4 sin (sin12) cos (cos12) > 0 x R is

(A) ( –5, 3) (B) ( –, 4)

(C) ( –3, 5) (D) (4, )

31. A1sin (sin12) 12 4 and 1cos (cos12) 4 12

2x (a 1)x 4 12 4 4 12

2x (a 1)x 4 0

f (x) = 2x (a 1)x 4 is always positive for x R

D < 0

2(a 1) 16 0

|a + 1| < 4

a ( –5, 3).

32. The value of k such that the equation 2x 5 x 6 k x has six solution is

(A) 5 2 6 (B) 5 2 6

(C) 5 6 (D) None of these

32. B

The equation 2x 5 x 6 k x has six solutions.

If y = 2x 5x 6 touches kx for x (2, 3).

Descriminant of 2x (k 5)x 6 0 is zero

k 5 2 6 .

33. If , are the roots of the equation x2 – x + q = 0 and S r = r + r , (where r N {0})

then Sk+1 is equal to(A) 1 – q (Sk –1 + Sk –2 + …… + S0) (B) 1 + q (Sk –1 + Sk –2 + …… + S0)(C) (Sk –1 + Sk –2 + …… + S0) (D) none of these

33. A

, are roots of x2 – x + q = 0

+ = 1; = qNow Sk+1 = Sk – qSk –1 Sk = Sk –1 – q Sk –2 ::S2 = S1 – qS0

Adding Sk+1 = S1 – q (Sk –1 + Sk –2 + ……… S0)= 1 – q (Sk –1 + Sk –2 + ……… + S0).



24

34. If the line ax + by + c = 0 lies in exactly two quadrants then of the set of quadraticequations,x2 + ax – ab = 0; x2 + bx – bc = 0; x2 + cx – ac = 0;(A) At least one has identical roots (B) At most two have identical roots(C) Both (A) and (B) (D) exactly one has identical roots

34. DLine ax + by + c = 0 lies in two quadrants

abc = 0; but a and b can’t be zero simultaneously also a and c or b and c cannot bezero simultaneously

a = 0 or b = 0 or c = 0

D1 = 0 or D2 = 0 or D3 = 0

exactly one of the equations has identical roots.

35. The value of a for which the equation 3x 2ax 2 0 and 4 2x 2ax 1 0 have a

common root is

(A) 178

(B) 178

(C)15

8 (D)

15

8

35. B

Let be a common root then 3 4 22a 2 0 and 2a 1 0

i.e. 4 22a 2 0 ….(i) 4 22a 1 0 …(ii)

From (i) and (ii)1

2 1 0

2

So,

31 1

2a 2 02 2

1a 2 0

8

1 17a 2

8 8 .

36. If c < a < b < d, then roots of the equation bx2 + [1- b[c + d]]x + bcd –a = 0(A) real and distinct in which one lie between c and a(B) real and distinct in which one lie between a and b(C) real and distinct in which one lie between c and d(D) roots are not real

36. CEquation can be rewritten as

f (x) = b(x –c) (x –d) + (x –a) = 0f (c) = c –a < 0, f (d) = d –a > 0So one root lies between c and d.

37. If p and q are odd prime numbers, then which of the following statement about the rootsof the equation x2 + 2px –4q = 0 is right(A) always rational but never integer (B) always integer(C) always irrational (D) nothing can be said about the root



25

37. C

D = 4p2 + 16q > 0 as p, q NLet D is a perfect square

p2 + 4q = d2 4q = d2 –p2

Which is not possible as d2 –p2 = 8k (k I).

38. Number of solution of the equation x2 –2 –2[x] = 0 ([.] denotes the greatest integerfunction) is(A) one (B) two(C) zero (D) infinity

38. A

If [x] = 1x2 –2 = –2 x = 0, not possibleif [x] = 0

x2 –2 = 0 x = 2 , not possible

if [x] = 1

x2

–2 = 2 x = 2, not possibleif [x] = 2

x2 –2 = 4 x = 6 , only possible solution is 6

–1 0 1 2 3

39. Least value of the expression xsinbxbx2

1222

, x [-1, 0], b [2, 3] is

(A)4

1 (B) -

4

1

(C)1sin8

12

(D) none of these

39. BExpression will be least if2bx –[x2 + b2 + sin2 x] is maximum

x2 + b2 + sin2 x –2bx is minimum

(x –b)2 + sin2 x is minimumnow |x –b| and | sin x| both are minimum if x = 0, b = 2

so least value is -4

1

40. If p N and the equation x3 –px + 1 = 0 has a rational root, the root must be(A) 1 (B) –1

(C)21 (D) none of these

40. A

Letb

a where a, b I and g.c.d (a, b) = 1 be a root of x3 –px + 1 = 0

So, 01b

a.p

b

a3

3

a3 –pab2 + b3 = 0 a(pb2 –a2) = b3



26

a divides b3 a divides b g.c.d (a, b) = a = 1

so, 1 –pb2 + b3 = 0 b2 (b –p) = –1 b2 divides 1 b = 1so, the root is 1 or –1

But if the root is –1 then p = 0 the root is 1.

41. The roots of x3 + px2 + qx –9 = 0 are each one more than the roots of x 3 –ax2 + bx –c =

0, where a, b, c, p, q R. Then a + b + c is equal to(A) 9 (B) 8(C) 10 (D) none of these

41. A

Let roots of x3 –ax2 + bx –c = 0 be , , a = + + , b = + + , c = Now, ( + 1), ( + 1), ( + 1) are the roots of x3 + px2 + qx –9 = 0

( + 1)( + 1) ( + 1) = 9

+ + + + + + + 1= 9 a + b + c = 8

42. Let x2 + x + 1 is divisible by 3. If x is divided by 3 leaves remainder(A) 0 (B) 2(C) 1 (D) none of these

42. CLet x = 3I, where I is integer then

(3I)2 + 3I + 1 = 9I2 + 3I+ 1 is not divisible by 3, so x 3I.Let x = 3I + 1, then (3I + 1)2 + 3I + 1 = 9I2 + 9I + 3 is divisible by 3Hence when x is divided by 3 leaves 1 as remainder.

43. If , are roots of the equation x2 + ax + b = 0 then maximum value of expression

– x2

+ ax + b +

4

2 will be

(A)4

b4a2 (B)

4

a4b2

(C) 0 (D) none of these

43. CLet us consider – x2 + ax + b

Maximum value of – x2 + ax + b =

4

ab4

4

b4a 22

=

4

2

So maximum value of – x2 + ax + b +

4

2 = 0

44. Let a, b, c be positive real parameters and Rxcx

bax

2

2 . Then

(A) 4ab c2 (B) 4ac b2

(C) 4bc c2 (D) 4ac < b2



27

44. (A) Let f (x) =2

2

x

bax 0

x

b2ax2xf

3

a

bx 2

Clearlya

bx f has local minima, so global minima

f 2cab4cbab

abac

ab

Alternate: For all x R+,2

2

x

b,ax are real positive numbers

So ab2x

bax

x

bax

2

x

bax

2

2

2

22

2

Rx

ab2 is least value of f (x) =2

2

x

bax , which occurs at x =

a

b

ab2 c 4ab c2

45. The variable quadratic equation (2x –a) (2x –c) + (x –2b) (x –2d) = 0, (where 0 < 4a <4b < c < 4d) has

(A) a root between b and d for all (B) a root between b and d for all –ve (C) a root between b and d for all +ve (D) none of these

45. (D) Let f (x) = (2x –a) (2x –c) + (x –2b) (x –3d)

f (2b) = (4b –a) (4b –c) < 0 (as 4b > 4a 4b > a and a > 0)f (2d) = (4d –a) (4d –c) > 0so f (x) = 0 has a real root between 2b and 2d which may not be necessarily between band d

46. If , , be the roots of x3 + a3 = 0 (a R), then number of equation(s) which roots are22

and

is

(A) 1 (B) 2(C) 3 (D) 6

46. (A) x3 + a3 = 0 x = -a, -a, -a2 where is non real cube root of unity

if = -a, then

1

,1

,2

22

sum is –1, product = 1

if = -a, then

,, 2

22

sum is –1 product = 1



28

if = -a2 , then 2

22

,,

sum is –1 product = 1so we have only one equation x2 + x + 1 = 0

47. If and are non real, then condition for x2 + x + = 0 to have a real roots is(A) ( - ) ( - ) = ( - )2 (B) ( -) ( - ) = ( - )2

(C) ( - ) ( - ) = ( -)2 (D) none of these

47. (B) Let be a real root of x2 + x + = 0

2 + + = 0 ….(1)

take conjugate 2 + + = 0 ….(2) ( = , since is real)

12

2

( - )2 = ( -) ( - )

48. Number of solutions of x2 –2[x] + 3 = 0 is; (where [.] denotes the greatest integerfunction).(A) 1 (B) 2(C) 3 (D) none of these

48. (D) graphs of x2 + 3 and 2[x] would never meet.

49. Total number of solutions of the equation 7|x| ( |5 - |x| | ) =1, is equal to,(A) 12 (B) 2

(C) 3 (D) 4

49. (D)

(5, 0)(-5, 0)

(5, 0)

(0, 1)

y = |5 -|x| |

y = 7- x

| 5 -|x| | = 7-|x|

From the graph, it is clear that there would be 4 solutions.

50. If roots of ax2+bx+c = 0, a 0, (a, b, c are real numbers), are imaginary and a +c < b,then(A) 4a+c = 2b (B) 4a+c>2b(C) 4a+c<2b (D) 4a+c<2b, if a < 0 and 4a+c>2b if a > 0

50. (C)Let f(x) =ax2+bx+c, then f(-1) = 1-b+c<0 (given)



29

Thus f(-2) < 0 (as roots of f(x) = 0 are imaginary) 4a + c < 2b



30

GroupIII

1. If 13x2x

6xx

2

2

, then x belongs to the set

(A) (2,3) (B)

23,1

(C) R (D) R-

2

3,1

1. (B)

-x2+2x-3<0, x, so the given inequality reduces to x2 + x-6<-x2+2x-3

2x2-x-3<0 (2x-3) (x+1) < 0

x

2

3,1

2. Let the equation Ax4 + Bx3 + x2 + x = 0 have a positive root . Then the equation 4Ax3 +

3Bx2 + 2x +1 = 0 has a positive root where

(A) < (B) = (C) > (D) = 2

2. (A)

Since 0 and are the roots of the above equation. Applying Rolle’s theorem, we have

() = 0 where 0 < < .

3. The interval(s) which satisfyx10 – x7 + x4 - x+1 > 0 is (are)

(A) -1 x 0 (B) 0 < x < 1(C) x 1 (D) - < x <

3. (A), (B), (C)

For x 0, All terms are +ve

For 0 < x 1, x10 + x4 (1-x3 ) + 1 –x > 0 Expression is positive

x (-,1], x10 + x4 (1 –x3) + 1 –x > 0

For x > 1, x7 (x3-1)+x(x3-1)+1>0 Expression is +ve x R

4. If x and y are integers, then the number of solutions of the equation (x-8) (x-10)=2y is(A) 1 (B) 2(C) 3 (D) More than 3

4. (B)

x = 9 12y y = 3 and x = 6, 12

5. Suppose f(x) is a quadratic expression which is negative for all real x. If g(x) = f(x) + f (x)

+ f (x), then for any real x(A) g(x) < 0 (B) g(x) > 0



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(C) g(x) = 0 (D) None of these

5. (A)Let f(x) = ax2 + bx+cwhere a < 0 and b2 – 4ac < 0g(x) = ax2 + bx + c+ (2ax+b) + 2a

= ax2 + x (2a+b)+2a+b+cConsider, (2a+b)2 –4a(2a+b+c)

= b2 – 4ac – 4a2 < 0 g(x) < 0 x R

6. If x2 + 2ax + b c, x R, then

(A) b-ca2 (B) c -a b2

(C) a – b c2 (D) None of these

6. (A)Let f(x) = x2 + 2ax + b = (x+a)2 + b-a2 So minimum value of f(x) is b-a2

Since f(x) c, x R, hence b –a2 c i.e., b- c a2

7. If the product of the roots of the equation 2x2 + ax+4 sina = 0 is 1, then roots will beimaginary if

(A) a R (B) a

6

,6

7

(C) a

6

5,

6 (D) None of these

7. (B), (C)

Since product of roots is 2 sin a = 1 or a = n + (-1)n/6, n I. Again since roots are

imaginary, a

2

< 16 i.e., -4 < a < 4. Thereforea =

6

5,

6,

6

7 . Hence choices (B) and (C) are correct.

8. The least value of |a| for which tan and cot are the roots of the equation x2+ax+b = 0is(A) 2 (B) 1

(C)2

1 (D) 0

8. (A)

Since product of roots is one, b =1. Again since roots are real, a2 – 4 0. Thus least

value of |a| is 2.

9. If (a-1)x2 + (a2-3a+2) x+a2-1 = 0 has more than two real roots then a is equal to(A) 2 (B) 1(C) 0 (D) None of these

9. (B)If will happen when simultaneously



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Coefficient of x2 = Coefficient of x = constant term = 0

10. The set of values of ‘a’ for which x2 – ax + sin-1(sin 4) > 0 x R is

(A) R (B) (2, 2)

(C) (D) none of these

11. (C)

Given inequality can be re-written as x2 ax + 4 > 0, x R which is true only when

a2 4( 4) < 0, which is not possible for any ‘a’. Hence the required set is

12. Let a, b, c be three distinct positive real numbers, then number of real roots of ax2 +2b|x| + c = 0 is(A) 0 (B) 1(C) 2 (D) 4

12. (A)

ax2 + 2b|x| + c > 0, x R. Hence the given equation has no real root.

13. Roots of the quadratic equation (x2 4x + 3) + (x2 6x + 8) = 0, R will be

(A) always real (B) real only when is positive

(C) real only when is negative (D) always imaginary

13. (A)

Let f(x) = (x2 4x + 3) + (x2 6x + 8)

= (x 1) (x 3) + (x 2) (x 4)Now f(2).f(4) < 0, so f(x) = 0 has a root in (2, 4). Thus roots are always real.

14. The constant term of the quadratic expression

n

1k

k

1x

1k

1x as n is

(A) 1 (B) 0(C) 1 (D) none of these

14. (C)

Constant term =1n

11

1k

1

k

1

k

1.)1k (

1 n

1k

n

1k

1 as n

15. The roots of the equation (x – b)(x – c) + (x – a)(x – c) + (x – a)(x – b) = 0 are always(A) positive (B) negative(C) real (D) imaginary

15. CThe equation is 3x2 – 2x (a + b + c) + ab + bc + ca = 0. The discriminant is4(a + b + c)2 – 12(ab + bc + ca) = 4(a2 + b2 + c2 – ab – bc – ca)= 2[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca] = 2[(a – b)2 + (b – c)2 + (c – a)2] > 0.

16. The values of k, for which the equation kx2 + 1 = kx + 3x – 11x2 has equal roots, are(A) {-11, -3} (B) {5, 7}



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(C) {5, -7} (D) {11, 3}

16. CWe have (k + 11)x2 – (k + 3)x + 1 = 0.The roots are equal

(k + 3)2 – 4(k + 11) = 0

(k + 1)2 = 36 k + 1 = 6 k = – 7, 5

17. If , are the roots of the equation x2 + px + q = 0, then – 1/, – 1/ are the roots of theequation(A) qx2 – px + 1 = 0 (B) qx2 + px + 1 = 0(C) x2 + px + q = 0 (D) x2 – px + q = 0

17. A

Replace x by ( – 1/x). We get 0qx

p

x

12

qx2 – px + 1 = 0

18. If x2 + 3x – 2 =x3x

82 , then x is equal to

(A) {4, 4, – 2, – 2} (B) { – 4, – 1, – 4, – 1}(C) {1, – 1, – 4, – 2} (D) {1, – 1, 4, – 2}

18. CLet x2 + 3x = t

t – 2 = 8/t t2 – 2t – 8 = 0

(t + 2) (t – 4) t = – 2, 4

x2 + 3x + 2 = 0 (x + 1) (x + 2) = 0 x = – 1, – 2

and x2 + 3x – 4 = 0 (x – 1) (x + 4) = 0 x = 1, – 4

19. If p and q are the roots of the quadratic equation x2 + mx + m2 + a = 0, then the value ofp2 + q2 + pq is(A) a (B) –a(C) ap (D) aq

19. Bp2 + q2 + pq = (p + q)2 – pq = m2 – (m2 + a) = – a.

20. If a, b, c are positive real numbers, then the roots of the equation ax2 + bx + c = 0(A) are real and positive (B) are real and negative(C) have negative real part (D) have positive real part.

20. C

x = – a2

ac4bb 2 . a, b, c are positive –

a2

b is negative

If the roots are complex, the real part is negative

If the roots are real, thena2

ac4b2 < –

a2

b.

Hence the root is negative.



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21. This set of value of k, for which roots of the equation (x – a) (x – c) + k(x – b) (x – d) = 0are real given by is (a > b > c > d).

(A) ( – , d) (B) (a, )(C) ( – , ) (D) None of these

21. CLet f(x) = (x – a) (x – c) + k(x – b) (x – d)

f(a) = k(a – b) (a – d) > 0 if k > 0 ; f(a) < 0 if k < 0f(c) = k(c – b)(c – d) < 0 if k > 0 ; f(c) > 0 if k < 0

There is real root of the given equation between a and c.Hence the given equation has real roots.

22. If c + 1 < b, then roots of x2 – bx + c = 0(A) are real and equal(B) are real and distinct(C) are Imaginary(D) nothing can be said about the nature of roots as given condition is not sufficient.

22. BLet f(x) = x2 – bx + c

f(1) = 1 – b + c < 0

roots are real and distinct.

23. The number of real roots of the equation x2 + a|x| + c = 0, a and c being positive realnumbers, is(A) 4 (B) 2(C) 0 (D) none of these

23. C

24. If one root of x2 + 2(a 1)x + a + 5 = 0 is greater than 3 and the other root is smallerthan 3, then 'a' belongs of

(A)

4

3, (B)

4

1,

(C)

7

8, (D) (, )

24. C

25. The number of real roots of the quadratic equation

n

1k

2k x = 0, n > 1; is

(A) 0 (B) 1(C) 2 (D) infinitely many

25. A

26. If one root of equation x2 + px + 12 = 0 is 4 while the equation x2 + px + q = 0 has equalroots, the value of q is equal to



36

Subtracting one equation from other,

(a – b)x = (a – b) x = 1

31. If x2 + x + 1 = 0, then the value of x3n is

(A) –1 if n is odd, 1 if n is even (B) 1

(C) –1 (D) None of these

31. B

x2 + x + 1 = 0 has two roots and 2 3n = (2)3n = 1.

32. (ax + 2y + 1)100 is a polynomial in x and y. If the sum of the coefficients vanishes for

some real , then possible values of a are(A) –2 (B) 2(C) 1 (D) none of these

32. (A), (B)Sum of the coefficients is (a( + (2 + 1)100 which is zero for some ( if quadraticequation (2 + a( + 1 = 0 (in () has real roots i.e. a2 – 4 ( 0 i.e. a ( -2 or a ( 2 .

33. If a, b, c are positive real numbers and the least value of ab2 + ac2 + a2b + bc2 + ca2 +cb2 is k abc, then k equals

(A) 1 (B) 2

(C) 4 (D) 6

34. If 4 + c < 2b, and f(x) = x2 + bx + c, then(A) f(x) = 0 has non real complex roots(B) f(x) = 0 has distinct real roots

(C) f(x) = 0 has equal real roots(D) Nothing can be said about the nature of roots of f(x) = 0

35. If the quadratic equation

x2 + x + a2 + b2 + c2 - ab - bc - ca = 0

has imaginary roots, and > 0 then

(A) 2 ( - ) + (a - b)2 + (b - c)2 + (c - a)2 > 0

(B) 2 ( - ) + (a - b)2 + (b - c)2 + (c - a)2 < 0

(C) 2 ( - ) + (a - b)2 + (b - c)2 + (c - a)2 = 0(D) none of these .

36. The number of solutions of the equation

5 sec - 4 tan = 3 in [0, 2] is

(A) 1 (B) 2

(C) 4 (D) 0

37. If all the real solutions of the equation

4x - (a - 3)2x + (a - 4) = 0



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are non-positive, then

(A) 4 < a 5 (B) 0 < a < 4

(C) a > 4 (D) a < 3

38. Consider the equation x3

– nx + 1 =0, n N , n 3 . Then(A) Equation has atleast one rational root .(B) Equation has exactly one rational root.(C) Equation has atleast one root belonging to (0, 1).(D) Equation has no rational root.

39. The real values of x which satisfy x2 – 3x + 2 0 and x2 – 3x – 4 0 are given by

(A) -1 x 1 (B) 1 x 2(C) 2 x 4 (D) none of these

Algebra QEE

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