Tiêu đề 04. QUADRATIC EQUATIONS MEDIUM Ans.pdf ✅

3 19 3 5(5) 6 = − S = ∵ 1 0 3 2 19 x − x + =  3 19 3 0 2 x − x + = 10.)(a) Since  and  are roots of equation, 0 2 x − ax + b = , therefore  +  = a ,  = b Now, ( )( ) ( ) 1 1 1 1 1 + + − − + = + = + + − + n n n n n n Vn           1 1 . . Vn+ = a Vn −b Vn− 11.)(d) Let  and 2  be the roots then + = −p 2   , = q 2 . Now ( ) 3 ( ) 2 3 3 6 3 2  + =  + +   +  p q q 3 pq 3 2 − = + −  (1 3 ) 0 3 2 p + q + q − p = 12.)(d) Roots of 1 0 2 x + x + = are 2 , 2 1 3 , 2 1 1 4 =   −  = −  − = i x Take 2  =,  =  19 19 7 2 7 14 2  = w = w,  = (w ) = w = w  Required equation is 1 0 2 x + x + = 13.)(b) Given root 2 5 1 2 5 2 5 1 = − + − − = + = ,  other root = −2 − 5 Again, sum of roots = – 4 and product of roots = – 1. The required equation is 4 1 0 2 x + x − = 14.)(b) If  is the coincident root, then 0 2  + a + b = and 0 2  + b + a =  a b b a b − a = − = − 1 2 2 2   ( ) 2  = − a + b ,  = 1  −(a + b) = 1  (a + b) = −1 15.)(a) Since roots are less than a real number, roots must be real  4a2 – 4(a2 + a – 3)  0  a  3.) ... (1) Let f(x) = x2 – 2ax + a2 + a – 3. Since 3 lie outside the roots, f(3) > 0  a < 2 or a > 3. . . . .(2) From (1), (2) we have a < 2 16.)(b) Since the roots of the given equation are of opposite sign, product of the roots < 0  ( ) 0 3 p p 1  −  p(p – 1) < 0  0 < p < 1. For real roots 2 2 1 p 1 3 3 −   +  0 < p < 1 17.)(d) We have x2 + 2x + 2 = (x + 1)2 + 1 > 0,  x  R. Therefore, 5 x 2x 2 mx 3x 4 2 2  + + + +  mx2 + 3x + 4 < 5 (x2 + 2x + 2)  (m – 5)x2 – 7x – 6 < 0,  x  R. This is possible if D = b2 – 4ac = 49 + 24(m – 5) < 0 and m – 5 < 0  m < 24 71
18.)(b) 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 19.)(d) 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 20.)(a) 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) 21.)(d) By wavy curve method. ( )( ) 0 x 2 x 3 x 1 2  − − −  x  (–, –1)  (1, 2)  (3, ). Therefore largest negative integer is –2 22.)(c) x = log37 = log 3 1 7 23.)(c) log5 loga × log a log x = 2  log5x = 2  x = 52 24.)(c) x = log(3/4) 7 = log(3/ 4) log 7 = log 3 – log 4 log 7 25.)(a) loge       + 2 a b = 2 1 log (ab) 2 a + b = ab a + b – 2 ab = 0 ( a – b ) 2 = 0  a = b  a = b 26.)(a) Let / = m/n  b 2 /ac = (m+n)2 /mn i.e., (3a–1)2 /(a2 –5a +3)2 = (2+1)2 /2.1 (As m : n = 2 : 1) 9a2 – 6a + 1 = 9a2 – 45a + 27  39a =26  a = 2/3 27.)(a) Let x= 1+ x or x2 = 1 + x; i.e., x2 –x– 1= 0 x =1± 5 /2 so x = 1+ 5 /2 (As x  1– 5 /2) 28.)(b) Here +=a–2,=–(a–1), Now, 2+ 2=(+) 2 – 2 = (a– 2)2 + 2 (a +1) = a2 – 2a + 6 = (a–1)2 + 5 So  2 +  2 will be maximum at a = 1. 29.)(b) Here  –  =  – (As , ,,  are in A.P.) (–) 2 = (–)   D1/a1 2 = D2/a2 2 = D1/D2 = a1 2 /a2 2=a2 /p2 30.)(a) LHS being the sum of a number and its reciprocal, is greater than or equal to 2, whereas RHS is less than 2, so, no solution.(LHS 31.)(c) Let  be common root, then
2(c – b) c – b 1/ 2 a 2b c 0 a 2c b 0 2 2  =   =  +  + =  +  + = putting  = 1/2 in a 2 + 2c + b = 0, we get a + 4b + 4c = 0 32.)(c) As ax2 +bx+c = 0 has roots of opp. signs when a > 0 , c < 0 so, k2 – 3k + 2 < 0  (k – 1) (k – 2) < 0 or K (1, 2) 33.)(a) Here sin  + cos  = –q/p, sin  cos  = r/p (sin  + cos ) 2 = q2 /p2  sin2 + cos2 + 2 sin  cos  + q2 /p2  1 + 2r/p = q2 /p2  p 2 – q 2 + 2pr = 0 34.)(a) Here b = ac (As a, b, c → GP) So, ax2 + 2 ac + c = 0  (x a + c ) 2 = 0 or x = – c / a , Now, substituting in eqn (2); dc/a – 2e c / a + f = 0  d/a – 2e/ ac + f/c = 0  d/a –2e/b + f/c = 0 35.)(c) Let y = x2+2x + 1/x 2 + 2x + 7  x 2 (1– y) + 2x (1–y)+1–7y = 0, As x is real, so D  0; i.e. 4(1 – y)2 – 4(1 – y) (1 – 7y)  0; 1 + y2–2y–7y2 + 8y – 1  0  – 6y2 + 6y  0 = y2 – y  0  y  [0, 1) (As y  1) 36.)(c) (x+3)2–4 (x+3)+3 = 0 or (x+3)2 + 4 (x+3) + 3 = 0 (x  – 3) (x < – 3)  x 2 + 2x = 0  x 2 + 10 x + 24 = 0 x = 0, – 2 (both possible)  x=–4, – 6 (both possible) so, sum of roots = 0 – 2 – 4 – 6 = – 12 37.)(d) Let ,  2 be roots, then  3 = 1   = 1 , ,  2 Now,  +  2 = –p/3 ; 1 + 1 = –p/3or  +  2 = –p/3 p = – 6 (not possible) or p = 3 ( 1 +  +  2 = 0) 38.)(d)  is a common root, 1 2 – 3 – 3 2 0 2 2  =  +  =    + = so, f() = f(1) = 4(1)2 + 3(1) – 7 = 0 39.)(a) ax2 + bx + c=0, (x  0) or ax2 – bx + c = 0, (x < 0) Here a, b, c > 0 Here a >0, b< 0, c >0 so, both roots are –ve so, both roots are +ve but x  0 so, no soln .)but x < 0 so, no soln . so, Number of soln .s are 0. 40.)(c) x 2 – 3x + 2 > 0 x 2 – 3x – 4  0 (x – 1) (x – 2) > 0 (x – 4) (x + 1)  0 x  (– , 1)  (2, ) x  [–1, 4]