# Unit-04_QEE_Test Solution.doc

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### Unit-04_QEE_Test Solution.doc

• 1. Problem -1: The set of values of p for which the roots of the equation 3x2 +2x +p(p-1) = 0 are of opposite sign is (A) (-, 0 ) (B) (0, 1) (C) (1, ) (D) (0, ) Solution: Since the roots of the given equation are of opposite sign, product of the roots < 0    0 3 1 p p    p(p-1) < 0  p (0, 1). Hence (B) is the correct answer. Problem -2: The number of real roots of (6 –x)4 + (8 –x)4 = 16 is (A) 0 (B) 2 (C) 4 (D) none of these Solution: Let y = 7 – x. Then the given equation becomes (y + 1)4 + (y –1)4 = 16  y4 + 6y2 –7 = 0  (y2 –1) (y2 + 7) = 0  y2 –1 = 0 [ y2 + 7  0]  y =  1  7 –x =  1  x = 6, 8 Hence (B) is the correct answer. Problem -3: Given that tan A and tan B are the roots of x2 –px + q = 0, then the value of sin2 (A + B) is (A)  2 2 2 q 1 p p   (B) 2 2 2 q p p  (C)  2 2 2 q 1 p q   (D)  2 2 q p p  Solution: We have: tan A + tan B = p and tan A tan B = q.  tan (A + B) = q 1 p B tan A tan 1 B tan A tan     Now, sin2 (A + B) = 2 1 [1 –cos 2(A + B)] =               B A tan 1 B A tan 1 1 2 1 2 2 =          2 2 2 2 2 2 2 2 2 q 1 p p q 1 p 1 q 1 p B A tan 1 B A tan           Hence (A) is the correct answer. Problem -4: If p, q  {1, 2, 3, 4}, the number of equations of the form px2 + qx + 1 = 0 having real roots is (A) 15 (B) 9 (C) 7 (D) 8 Solution: For real roots, we must have Disc  0  q2 -4p  0  q2  4p If p = 1, then q2  4p  q2  4  q = 2, 3, 4 If p = 2, then q2  4p  q2  8  q = 3, 4 If p = 3, then q2  4p  q2  12  q = 4 If p = 4, then q2  4p  q2  16  q = 4 Thus, we see that there are 7 cases. Hence (C) is the correct answer.
• 2. Problem -5: The harmonic mean of the roots of the equation (4 + 2 )x2 – (4 + 5 )x+ (8 +2 5 )=0 is (A) 2 (B) 4 (C) 7 (D) 8 Solution: Let ,  be the roots of the given equation. Then,  +  = 2 4 5 4   and  = 2 4 5 2 8   Let H be the H.M. of  and . Then, H = 5 4 5 4 16 2        = 4. Hence (B) is the correct answer. Problem -6: In a triangle PQR, R = 2  . If tan       2 P and tan       2 Q are the roots of the equation ax2 + bx + c (a  0). Then (A) a + b = c (B) b + c = 0 (C) a + c = b (D) b = c Solution: R = 2   P + Q = 2   4 2 Q 2 P     tan        2 Q 2 P = tan 4   2 Q tan 2 P tan 1 2 Q tan 2 P tan   = 1  1 a c 1 a b              a c 2 Q tan 2 P tan and a b 2 Q tan 2 P tan   a b a c 1     a –c = -b  a + b = c. Hence (A) is the correct answer. Problem -7: If the roots of the equation x2 –2ax + a2 + a –3 = 0 are real and less than 3, then (A) a < 2 (B) 2  a  3 (C) 3 < a  4 (D) a > 4 Solution: Let f (x) = x2 –2ax + a2 + a –3 = 0. Since f (x) has real roots both less than 3. Therefore, Disc > 0 and f (3) > 0  a2 –(a2 + a –3) > 0 and a2 –5a + 6 > 0  a < 3 and (a –2) (a –3) > 0  a < 3 and a < 2 or a > 3  a < 2. Hence (A) is the correct answer. Problem -8: If  and  are the roots of the equation, 2x2 –3x –6 =0, then equation whose roots are 2 +2, 2 +2 is (A) 4x2 + 49x +118 = 0 (B) 4x2 - 49x +118 = 0 (C) 4x2 - 49x –118 = 0 (D) x2 - 49x +118 = 0 Solution:  +  = 3/2 ,  = -6/2 = -3 S = 2 +2 +4 = (+)2 -2 +4 = 4 49
• 3. P = 2 2 +2 (2 +2 ) +4 = 2 2 +4 +2 [( +)2 -2 ] = 4 118 . Therefore, the equation is x2 - 0 4 118 x 4 49                4x2 – 49x +118 = 0 Hence (B) is the correct answer. Problem -9: If the roots of the equation x2 –px + q = 0 differ by unity then (A) p2 = 1- 4q (B) p2 = 1+ 4q (C) q2 = 1- 4p (D) q2 = 1+ 4p Solution: Suppose the equation x2 – px + q = 0 has the roots  + 1 and  then  +1+ = p  2 = p –1 . . . . (1) and (+1)  = q  2 +  = q . . . .. (2) Putting the value of  from (1) in (2) , we get   q 2 1 p 4 1 p 2      (p-1)2 + 2(p-1) = 4q  p2 –1 =4q  p2 = 4q +1 . Hence (B) is the correct answer. Problem -10: If p and q are the roots of the equation x2 +px +q = 0 , then (A) p =1, q = -2 (B) p =0 , q = 1 (C) p = –2, q = 0 (D) p = –2, q = 1 Solution: Since p and q are roots of the equation x2 +px + q = 0 , p + q = - p and pq = q pq = q  q = 0 or p = 1 if q = 0, then p = 0 and if p =1, then q = –2 Hence (A) is the correct answer. Problem -11: Let ,  be the roots of the equation (x - a) (x - b) = c , c  0. Then the roots of the equation (x - )(x - ) + c = 0 are (A) a, c (B) b, c (C) a, b (D) a+c , b+c Solution: By given condition (x-a) (x-b) – c  (x-)(x-) or (x-)(x-) +c  (x-a) (x-b). This shows that the roots of ( x- ) ( x-) + c = 0 are a and b. Hence (C) is the correct answer. Problem -12 x4 –4x –1 = 0 has (A) atmost one positive real root (B) atmost one negative real root (C) atmost two real roots (D) none of these . Solution: Let f (x) = x4 –4x –1 + – –  atmost one positive real root. f (–x) = x4 + 4x –1 + + –  atmost one negative real root.  atmost two real root. Hence (C) is the correct answer. Problem -13 If x2 +ax +b is an integer for every integer x then (A) ‘a’ is always an integer but ‘b’ need not be an integer (B) ‘b’ is always an integer but ‘a’ need not be an integer (C) a+b is always an integer (D) a and b are always integers.
• 4. Solution: Let f(x) = x2 +ax +b Clearly, f(0) = b  b is an integer . Now f(1) = 1+ a+ b  a is an integer. Hence (C) and (D) are the correct answers. Problem -14 Sum of the real roots of the equation x2 +5|x| +6 = 0 (A) equals to 5 (B) equals to 10 (C) equals to –5 (D) does not exit. Solution: Since x2 , 5|x| and 6 are positive so x2 +5|x| +6 = 0 does not have any real root . Therefore sum does not exist. Hence (D) is the correct answer. Problem -15 If c > 0 and 4a +c < 2b then ax2 –bx +c = 0 has a root in the interval (A) (0, 2) (B) (2, 4) (C) (0, 1) (D) (-2, 0) Solution: Let f(x) = ax2 –bx +c f(0) = c > 0 and f(2) = 4a –2b +c < 0 so that f(x) =0 has a root in the interval (0, 2). Hence (A) is the correct answer. Problem -16 The largest negative integer which satisfies    0 3 x 2 x 1 x 2     is (A) - 4 (B) –3 (C) –1 (D) –2 Solution: By wavy curve method .    0 3 x 2 x 1 x2      x  (-, -1)  (1, 2)  ( 3,  ) Therefore largest negative integer is -2 . Hence (D) is the correct answer. Problem -17 If x2 –4x +log1/2a = 0 does not have two distinct real roots, then maximum value of a is (A) 1/4 (B) 1/ 16 (C) –1/4 (D) none of these Solution: Since x2 –4x +log1/2a = 0 does not have two distinct real roots, discriminant  0  16 – 4 log1/2 a  0  log1/2 a  4  a  1/16 Hence (B) is the correct answer. 1. If |x-2|+|x-9|=7, then the set values of x is (A) {2, 9} (B) (2, 7) (C) {2} (D) [2, 9] 1. D 2. If (m2 -3) x2 + 3mx + 3m + 1= 0 has roots which are reciprocals of each other, then the value of m equals to (A) 4 (B) –3 (C) 2 (D) None of these 2. A 3. If ax2 +bx + 6 =0 does not have two distinct real roots, then the least value of 3a+b is (A) 2 (B) –2 (C) 1 (D) -1 3. B
• 5. 4. If the quadratic equation x2 +x+a2 +b2 +c2 – ab – bc – ca = 0 has imaginary roots, 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 . 4. A 5. The roots  and  of the quadratic equation ax2 +bx +c = 0 are real and of opposite sign. Then the roots of the equation (x-)2 + (x-)2 =0 are (A) positive (B) negative (C) Real and of opposite sign (D) imaginary 5. C 6. If p, q, r are real and p q, then the roots of the equation (p -q)x2 +5(p +q)x -2(p -q)=0 are (A) real and equal (B) complex (C) real and unequal (D) none of these. 6. C 8. The real roots of the equation |x|3 – 3x2 +3|x| - 2 =0 are (A) 0, 2 (B)  1 (C)  2 (D) 1,2 8. C 9. If the equations x2 +ax+b=0 and x2 +bx+a=0 have exactly one common root, then the numerical value of a + b is (A) 1 (B) –1 (C) 0 (D) none of these 9. B 10. If r be the ratio of the roots of the equation ax2 + bx + c = 0, then r ) 1 r ( 2  is equal to (A) bc a2 (B) ac b2 (C) ab c2 (D) none of these 10. B 11. The set of values of a for which the inequality x2 +ax+a2 +6a<0 is satisfied for all x  (1, 2) lies in the interval (A) (1, 2) (B) [1, 2] (C) [-7, 4] (D) None of these 11. D 12. If the product of the roots of the equation x2 -3kx+2e2lnk -1=0 is 7, then for real roots the value of k is equal to (A) 1 (B) 2 (C) 3 (D) 4 12. B 13. If 2 + i3 is a root of x2 + px + q = 0, where p, q are real, then (p, q) is equal to (A) (-4, 7) (B) (4, - 7) (C) (-7 4) (D) (4, 7) 13. A 14. The equations ax2 + bx + a = 0, x3 – 2x2 + 2x – 1 = 0 have two roots in common. Then a + b must be equal to (A) 1 (B) –1 (C) 0 (D) none of these 15. C