Title: Quadratic Equation Chapter: Quadratic Equation Approximate Time to Complete (to be filled by student): ________________________ Total Marks Scored: Worksheet Compiled By: Prashant Jain (PJ Sir) • • • • • Attempt the worksheet in one go. See answers in one go at the end. All questions carry 4 marks for positive and -1 if you leave and -2 if you attempt incorrectly. All proving or show questions (if done correctly) are of 4 marks. There is no negative marking. If you cheat in worksheet then you are only cheating and such sinners do not get selection in JEE so for your own sake refrain from cheating. Maintain the solutions of this worksheet and share the link of solution pdf in the tracker if you want to be monitored. Exercise – I ONLY ONE OPTION CORRECT TYPE 1. Let a > 0, b > 0 & c > 0. Then both the roots of the equation ax2 + bx + c = 0 (A) are real & negative (B) have negative real parts (C) are rational numbers (D) have positive real parts 2. If the roots of the equation x2 + 2ax + b = 0 are real and distinct and they differ by atmost 2m, then b lies in the interval [16JM110151] (A) (a2 – m2, a2) (B) [a2 – m2, a2) (C) (a2, a2 + m2) (D) none of these 3. The set of possible values of for which x2 – (2 – 5 + 5)x + (22 – 3 – 4) = 0 has roots, whose sum and product are both less than 1, is 5 5 5 (A) −1 , (B) (1, 4) (C) 1 , (D) 1 , 2 2 2 4. If p, q, r, s R, then equaton (x2 + px + 3q) (–x2 + rx + q) (–x2 + sx – 2q) = 0 has (A) 6 real roots (B) atleast two real roots (C) 2 real and 4 imaginary roots (D) 4 real and 2 imaginary roots 5. Find the set of all real values of such that the root of the equation x2 + 2(a + b + c)x + 3 (ab + bc + ca) = 0 are always real for any choice of a, b, c (where a, b, c represents sides of scalene triangle). 4 4 1 5 4 5 (A) −, (B) , (C) , (D) , 3 3 3 3 3 3 6. If coefficients of biquadratic equation are all distinct and belong to the set {–9, – 5, 3, 4, 7}, then equation has (A) atleast two real roots (B) four real roots, two are conjugate surds and other two are also conjugate surds (C) four imaginary roots (D) None of these 7. Let p, q, r, s R, x2 + px + q = 0, x2 + rx + s = 0 such that 2 (q + s) = pr then (A) atleast one of the equation have real roots. (B) either both equations have imaginary roots or both equations have real roots. (C) one of equations have real roots and other equation have imaginary roots (D) atleast one of the equations have imaginary roots. 8. The equation, x = − 2x2 + 6x − 9 has: (A) no solution (B) one solution 9. (C) two solutions If (2 + – 2)x2 + ( + 2) x < 1 for all x R, then belongs to the interval [16JM110152] [16JM110153] (D) infinite solutions (A) (–2, 1) 2 (B) −2, 5 2 (C) , 5 1 (D) none of these 10. Let conditions C1 and C2 be defined as follows : C1 : b2 – 4ac 0, C2 : a, –b, c are of same sign. The roots of ax2 + bx + c = 0 are real and positive, if (A) both C1 and C2 are satisfied (B) only C2 is satisfied (C) only C1 is satisfied (D) none of these 11. If 'x' is real, then (A) c [0, 6] x2 − x + c can take all real values if : x2 + x + 2c (B) c [− 6, 0] (C) c (− − 6) (0, ) [16JM110154] (D) c (− 6, 0) 12. If both roots of the quadratic equation (2 − x) (x + 1) = p are distinct & positive, then p must lie in the interval: (A) (2, ) (B) (2, 9/4) (C) (– , – 2) (D) (– , ) 13. If two roots of the equation (a – 1) (x2 + x + 1)2 – (a + 1) (x4 + x2 + 1) = 0 are real and distinct, then 'a' lies in the interval [16JM110155] (A) (–2, 2) (B) (– , –2) (2, ) (C) (2, ) (D) (– –2) 14. The equations x3 + 5x2 + px + q = 0 and x3 + 7x2 + px + r = 0 have two roots in common. If the third root of each equation is represented by x1 and x2 respectively, then the ordered pair (x1, x2) is: (A) (− 5, − 7) (B) (1, − 1) (C) (− 1, 1) (D) (5, 7) 15. If a, b, c are real and a2 + b2 + c2 = 1, then ab + bc + ca lies in the interval: 1 1 (A) , 2 (B) [0, 2] (C) − , 1 2 2 Answer Key [16JM110156] 1 (D) −1, 2 ONLY ONE OPTION CORRECT TYPE 1. 6. 11. (B) (A) (D) 2. 7. 12. (B) (A) (B) 3. 8. 13. (D) (A) (B) 4. 9. 14. (B) (B) (A) 5. 10. 15. Solution ONLY ONE OPTION CORRECT TYPE 1. a > 0, b > 0 and c > 0 + = – b/a = – ve, = ax2 + bx + c = 0 c = + ve a –ve real part 2. x2 + 2ax + b = 0 0 < 4a2 – 4b 4m2 3. 0 < | – | 2m 0< a2 – m2 b < a2 Sum of roots < 1 2 – 5 + 5 < 1 Product of roots < 1 ( – 1)( – 4) < 0 (2 – 5) ( + 1) < 0 22 – 3 – 5 < 0 (1) & (2) 1<< ( + )2 – 4 2m b [a2 – m2, a2) –1 < < 1<<4 5 2 ...(2) 5 . 2 4. Dis. of x2 + px + 3q is p2 – 12q D1 2 Dis. of –x + rx + q is r2 + 4q D2 Dis. of –x2 + sx – 2q is s2 – 8q D3 Case 1 : If q < 0, then D1 > 0, D3 > 0 and D2 may or may not be positive Case 2 : If q > 0, then D2 > 0 and D1, D3 may or may not be positive Case 3 : If q = 0, then D1 0, D2 0 and D3 0 from Case 1, Case 2 and Case 3 we can say that the given equation has atleast two real roots. 5. We, know that a + b > c, b + c > a and c + a > b c – a < b, a – b < c, b – c < a squaring on both sides and adding (c – a)2 + (a – b)2 + (b – c)2 < a2 + b2 + c2 a2 + b2 + c2 – 2(ab + bc + ca) < 0 (a + b + c)2 – 4(ab + bc + ca) < 0 (a + b + c)2 <4 ....(i) ab + bc + ca Now roots of equation x2 + 2(a + b + c) x + 3 (ab + bc + ca) = 0 are real, then D 0 (a + b + c)2 4 (a + b + c)2 – 4. 3 (ab + bc + ca) 0 3 ab + bc + ca So 6. 3 (a + b + c)2 <4 ab + bc + ca < 4 3 Let biqhadratic is ax4 + bx3 + cx2 + dx + e = 0 a + b + c + d + e = 0 as a, b, c, d, e {–9, – 5, 3, 4, 7} Hence x = 1 is a root. So real root will be atleast two. ...(1) (A) (A) (C) 7. x2 + px + q = 0 x2 + rx + s = 0 D1 = p2 – 4q D2 = r2 – 4s D1 + D2 = p2 + r2 – 4 (q + s) .....(1) .....(2) [ pr = 2(q + s)] = (p – r)2 > 0 Since D1 + D2 is +ve, so atleast one of the equation has real roots. 8. x = – 2x2 + 6x – 9 D = 36 – 4(–2) (–9) = 36 – 72 < 0 & a < 0 2 So quadratic expression – 2x + 6x – 9 is always negative whereas x is always +ve Equation will not hold for any x. x So x = – 2x2 + 6x – 9 has no solution. 9. ( + 2) ( – 1)x2 + ( + 2)x – 1 < 0 x R ( + 2) ( – 1) < 0 –2 < < 1 ...(1) (a < 0) and ( + 2)2 + 4( + 2) ( – 1) < 0 (D < 0) ( + 2) ( + 2 + 4 – 4) < 0 ( + 2) (5 – 2) < 0 2 –2 < < ...(2) 5 2 (1) & (2) −2, Also = –2 0 < 1 which is true 5 2 Required interval is −2, 5 10. C1 : b2 – 4a c 0 ; C2 : a, – b, c are of same sign ax2 + bx + c = 0 has real roots then D 0 i.e. C1 must be satisfied b b (i) Let a, – b, c > 0 then – >0 (ii) Let a, – b, c < 0 then – >0 2a 2a Hence, for roots to be + ve, C2 must be satisfied. Thus both C1, C2 are satisfied 11. Let y = 12. (2 – x) (x + 1) = p x2 – x + (p – 2) = 0 (1) has both roots distinct & positive x2 − x + c ; x R and y R (y – 1) x2 + (y + 1)x + 2y c – c = 0 x2 + x + 2c xR D0 (y + 1)2 – 4 c(y – 1) (2y – 1) 0 2 2 y + 1 + 2y – 4c [2y – 3y + 1] 0 (1 – 8c)y2 + (2 + 12c) y + 1 – 4c 0 ....... (1) 1 Now for all y R (1) will be true if 1 – 8c > 0 c < and D 0 8 4 (1 + 6c)2 – 4 (1 – 8c) (1 – 4c) 0 1 + 36c2 + 12c – 1 – 32c2 + 12c 0 2 4c + 24c 0 –6c0 But c = –6 and c = 0 will not satisfy given condition c (–6, 0) (i) D > 0 (i) (iii) 13. D>0 (ii) f(0) > 0 p< −b 1 = >0 2a 2 9 4 (ii) (iii) ...(1) −b >0 2a f(0) > 0 p > 2 (always true) 9 (i) (ii) (iii) p 2, . 4 (a – 1) (x2 + x + 1)2 – (a + 1) (x4 + x2 + 1) = 0 ........(1) x4 + x2 + 1 = (x2 + x + 1) (x2 – x + 1) (1) becomes (x2 + x + 1) [(x2 + x + 1) (a – 1) – (a + 1) (x2 – x + 1)] = 0 (x2 + x + 1) (x2 – ax + 1) = 0 Here two roots are imaginary and for other two roots to be real D > 0 a2 – 4 > 0 a (–, –2) (2, ) 14. x1 x2 + + x1 = – 5, + x1 + x1 = p ...(1) + + x2 = – 7, + x2 + x2 = p ...(2) + x1 + x1 = p + x 2 + x 2 = p (x1 – x 2 ) + (x1 – x 2 ) = 0 (x1 – x2) ( – ) = 0 [x1 x2] +=0 x1 = – 5 a2 + b2 + c2 = 1 (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca) 0 1 + 2 (ab + bc + ca) 0 (ab + bc + ca) – a2 + b2 + c2 – (ab + bc + ca) 0 (ab + bc + ca) 1 From (1) and (2) we can say that x3 + 5x2 + px + q = 0 x3 + 7x2 + px + r = 0 Subtracting (2) from (1) 15. x2 = – 7 1 2 1 (ab + bc + ca) − , 1 2 ........(1) ........(2)