Tiêu đề 04. Quadratic Equations Easy Ans.pdf ✅

1.)(a) Equation 2 2 a x a x ( 1) ( 1) 0 + − + =  ( 1) 0 2 2 ax − a + x + a =  (ax −1)(x − a) = 0  a x a 1 = , . 2.)(b) Equation 8 9 0 4 2 x − x − =  9 9 0 4 2 2 x − x + x − =  ( 9) 1( 9) 0 2 2 2 x x − + x − =  ( 1)( 9) 0 , 3 2 2 x + x − =  x = i  . 3.)(c) We have 4 4 0 2 ix − x − i =  4 4 0 2 x + ix − =  2 2 4 0 2 x + ix + ix − =  (x + 2i)(x + 2i) = 0  x = −2i, − 2i . 4.)(c) 2 0 2 / 3 1 / 3 x + x − =  ( ) 1( ) 2 0 1 / 3 2 1 / 3 x + x − = Let , 1 / 3 a = x then 2 0 2 a + a − =  a = 1,−2 Hence x = 1,−8 (by 1 / 3 a = x ). 5.)(b) 2 / 3 1 / 3 2 / 3 1 / 3 x = 2 + 2 + 2  x − 2 = 2 + 2 Cubing both sides, we get 8 6 12 6 6( 2) 3 − − + = + −  x x x x  6 6 2 3 2 x − x + x = . 6.)(d) 8 sec 6 sec 1 0 2  −  + =  2 1 sec = or 4 1 sec = , but sec  1 or sec  −1 . Hence the given equation has no solution. 7.)(d) Given equation is 3 1 1 x x + + =  3x + 1 = x − 1 Squaring on both sides, we get 3x + 1 = x + 1 − 2 x  2 x + 2x = 0 (irrational function) Thus x  0 and x  1 , since equation is non-quadratic equation. 8.)(b) Let the required number is x So, x = x + 12  x − 12 = x  25 144 0 2 x − x + =  16 9 144 0 2 x − x − x + =  x = 16 Since x = 9 does not hold the condition. Trick : By inspection, since 16 exceeds its positive square root i.e.,4 by 12. 9.)(b) Given equation is 3 10 .3 9 0 2 − + = x x can be written as (3 ) 10(3 ) 9 0 2 − + = x x Let x a = 3 , then it reduces to the equation 10 9 0 ( 9)( 1) 0 9,1 2 a − a + =  a − a − =  a = Now 3 9 3 3 3 2 2 a =  =  =  x = x x x and x 1 = 3 3 3 0 0  =  x = x . Hence roots are 0, 2. 10.)(d) Given that 7 10 0 2 / 3 1 / 3 x − x + = . Given equation can be written as ( ) 7( ) 10 0 1 / 3 2 1 / 3 x − x + = Let 1 / 3 a = x , then it reduces to the equation 7 10 0 ( 5)( 2) 0 5, 2 2 a − a + =  a − a − =  a = Putting these values, we have 125 3 a = x  x = and 8. 11.)(c) 25 2 2 x + y = and xy = 12
 25 144 25 0 12 4 2 2 2  =  + − =      + x x x x  ( 16)( 9) 0 2 2 x − x − =  16 2 x = and 9 2 x =  x =  4 and x = 3 . 12.)(a) 9 2 log (1 ) = −x x x  log (9) log (1 ) ( log ) 2 x a N N x a x x = x −  =  =  9 (1 ) 1 2 9 0 2 2 = − x  + x − x − =  2 8 0 2 x − x − =  (x + 2)(x − 4) = 0  x = −2,4 . 13.)(a) If one root of a quadratic equation with rational coefficients is irrational and of the form  +  , then the other root must also be irrational and of the form  −  . 14.)(d) Given 2 | | 3| | 2 0 x x − + = Here we consider two cases viz. x  0 and x  0 Case I : x  0 This gives 3 2 0 2 x + x + =  (x + 2)(x + 1) = 0  x = −2,−1 Also x = −1,−2 satisfy x  0, so x = −1 , – 2 is solution in this case. Case II : x  0 . This gives 3 2 0 2 x − x + =  (x − 2)(x −1) = 0  x = 2,1 , so x = 2 , 1 is solution in this case. Hence the number of solutions are four i.e. x = −1,1, 2,− 2 Aliter : | | 3| | 2 0 2 x − x + =  (| x | −1)(| x | −2) = 0  | x |= 1 and | x | = 2  x = 1, x = 2 . 15.)(d) Given equation 4 0 sin sin − − = x − x e e Let e y x = sin , then given equation can be written as 4 1 0 2 y − y − =  y = 2  5 But the value of x y e sin = is always positive, so y = 2 + 5 ( 2  5)  log = log (2 + 5 ) e e y  sin x = loge (2 + 5 )  1 which is impossible, since sin x cannot be greater than 1. Hence we cannot find any real value of x which satisfies the given equation. 16.)(b) Here two cases arise viz. Case I : 4 3 0 2 x + x +  This gives 4 3 2 5 0 2 x + x + + x + =  6 8 0 2 x + x + =  (x + 2)(x + 4) = 0  x = −2,−4 x = −2 is not satisfying the condition 4 3 0 2 x + x +  , so x = −4 is the only solution of the given equation. Case II : 4 3 0 2 x + x +  This gives –( 4 3) 2 5 0 2 x + x + + x + =  2 2 0 2 2 0 2 2 − x − x + =  x + x − =  (x + 1 + 3)(x + 1 − 3) = 0  x = −1 + 3,−1 − 3 Hence x = −(1 + 3) satisfy the given condition 4 3 0 2 x + x +  , while x = −1 + 3 is not satisfying the condition. Thus number of real solutions are two.
17.)(c) Given equation is ( ) ( ) ( ) 0 2 p − q x + q − r x + r − p = 2( ) ( ) ( ) 4( )( ) 2 p q r q q r r p p q x − −  − − − − =  ,1 2( ) ( ) ( 2 ) p q r p x p q r q q r p x − −  = − −  + − = 18.)(c) Put x = 4 in 12 0 2 x + px + = , we get p = −7 Now second equation 0 2 x + px + q = have equal roots. Therefore 4 49 4 2 p = q  q = 19.)(d) If x  1 , multiplying each term by (x − 1) , the given equation reduces to x(x − 1) = (x − 1) or ( 1) 0 2 x − = or x = 1 , which is not possible as considering x  1 . Thus given equation has no roots. 20.)(d) 2 0 1 2 1 + =  + − = x x x x ( x  0)  2 1 0 2 x − x + =  ( 1) 0 2 x − =  x = 1,1 . 21.)(a) According to condition   = − + 2 1 2  2 1 0 1, 1 2  +  + =   = − − . 22.)(c) 3 7 30 2 7 5 5 2 2 x − x − + x − x − = x + 3 7 30 ( 5) 2 7 5 2 2 x − x − = x + − x − x − on squaring, 2 7 5 5 2 x − x − = 2 7 30 0 6 2 x − x − =  x = . 23.)(b) Let +  + = + 2 ..... 1 2 1 x 2  x x 1 = 2 + (on simplification)  x = 1  2 But the value of the given expression cannot be negative or less than 2, therefore 1 + 2 is required answer. 24.)(d) 2 .3 9 2 3 /( 1) = x + x x − Taking log, we get log 3 2 log 3 1 3 ( 2)log 2  =      − + + x x x  log 3 0 1 1 ( 2) log 2  =      − + + x x  x = −2 or log 3 log 2 1 1 = − x  log 2 log 3 1 − x =  log 2 log 3 x = 1 − 25.)(d) Given 1 0 2 x + x + =  ( 1 3) 2 1 ( 1 3), 2 1 [ 1 3] 2 1 x = −  i = − + i − − i 2 = , But  =  =  19 19 and . 7 14 2  =  =  Hence the equation will be same. 26.)(c) Here x = 2 and 3 are the critical points. When x  2,| x − 2|= −(x − 2),| x − 3|= −(x − 3)  The given equation reduces to 2 − x + 3 − x = 7  x = −1  2  x = −1 is a solution. When 2  x  3, | x − 2|= x − 2,| x − 3|= −(x − 3)
 The equation reduces to x − 2 + 3 − x = 7  1=7  No solution in this case. When x  3 , the equation reduces to x − 2 + x − 3 = 7  x = 6  3 Hence we get, x = 6 or –1 Trick : By inspection, we have that both the values x = 6,−1 satisfy the given equation. 27.)(d) Since quadratic equation 0 2 ax + bx + c = has three distinct roots so it must be an identity. So a = b = c = 0 . 28.)(d) The equation (| x | −4)(| x | −3) = 0  | x |= 4  x = 4  | x |= 3  x = 3 . 29.)(d) We have 2 log( 2) log 5 log( 1) 2 = − + + x x  2 2 2 2 log{5(x + 1)} = log(x − 2)  5(x + 1) = (x − 2)  2 1 4 4 1 0 2 x + x + =  x = − But for log( 2) 2 1 x = − x − is not meaningful. Hence it has no root. 30.)(a) We have x = 7 + 4 3  7 4 3 . 7 4 3 7 4 3 7 4 3 1 1 + − − = + = x = 7 − 4 3  7 4 3 7 4 3 1 + = + + − x x = ( 3 + 2) + (2 − 3) = 4 . 31.)(d) We have y y x x 2 2 2 2 log 1 log 3 1 3 log 1 log + = + = +  3 1 log2 x = 3,log2 y = ( x  y)  3 x = 2 and 1 / 3 1 / 3 y = 2  x + y = 8 + 2 . 32.)(c) x = 2 + x  2 0 2 x − x − =  (x − 2)(x + 1) = 0  x = 2,−1 But 2 + 2 + .....  −1 , so it is equal to 2. 33.)(b) Equation, 2 2 1 1 2 1 4 3 3 2 − − + − = − x x x x  2 1 2 1 2 2 1 2 2 3 3 + − − + = + x x x x  3 (1 3) 2 1 2 1 2 1 2  = +      + x− x  3 .4 2 3 2 . 2 1 2 − = x x  2 3 2 3 2 3 − − = x x Taking log both sides  (2x − 3)log 2 = (x − 3 / 2)log 3  log 3 2 3 2x log 2 − 3 log 2 = x log 3 −  log 3 2 3 x log 4 − x log 3 = 3 log 2 −