Tiêu đề Quadratic Equation(Column).pdf ✅

MATHEMATICS Q.1 Column-I Column-II (A) The set of real value of x (P) [2, 4] for which log0.2 x x + 2  1 is (B) The solution set of the (Q)         − 2 5 1 0, inequality log10(x2–16)  log10 (4x–11) (C) If logcosx sin x  2 and (R) (4, 5) 0  x  3 then sin x lie in the interval (D) The set of real values (S)       − −  2 5 ,  of x satisfying (0, + ) log1/2(x2–6x + 12)  –2 is [A → S, B → Q, C → P, D → R] Q.2 Match the following columns : Column-I Column-II (A) The number of solutions (P) 1 of the equation |cosx| = 2[x] is (B) If the domain of f(x) is [–3,2], (Q) 0 then the integers included in the domain of f([|x|]) + f([2x + 3]) is/are (C) If (R) 3 x [a] Lt → ([2+x] + [1–x]+|[a]–x|) =2, then a can be (D) If x2 –3[sin (x – /6)] =3, (S) –1 then the value of x is/are In above all cases [x] represents greatest integer  x [A → Q, B → S, C → P,Q,R,S D → Q,R] Q.3 If ax2 + bx + c = 0, where a  0 is satisfied by , ,  2 and  2 then match the items of column-I with that of column-II Column-I Column-II (A) The value of  +  can be (P) 3 (B) The number of unordered pairs (Q) 1 (,) is (where   0) (C) The value of  can be (R) 2 (D) The value of  2 +  2 can be (S) –1 Sol. A→ Q, R , B→ P, C→ Q, S, D → Q, R, S Equation is satisfied by    2 and  2   2 &  2 must belong to ( ). Let  2 =   2 =   (, ) is (0, 1) or (1, 1) or (0, 0) or  2 = ;  2 =    4 =     is (1, 1) or (,  2 ) or  2 = ,  2 =    2 =  2   = ±   ( ) is (–1, 1), (–1, 1) , (1, 1)   +  can be equal to 1, 2 , 0 Also number of unordered pairs   such that   0 is (1, 1), (–1, 1) or (,  2 )   could be equal to 1 or –1 The value of  2 +  2 could be 0, 1, 2 or –1 Q.4 Match the following Column I Column II (A) The roots of the equation (P) 6 x 2 + 2(a–3) x + 9 = 0 lie between – 6 and 1 then [a] can be ([] denotes integral part) (B) If cosec ( – ), cosec , (Q) 3 cosec ( + ) are in AP where         −   2 , 2 then number of value of  is (C) If the equation (R) 2
ax2 + b|x| + c = 0 has two distinct real roots then value of a c could be (D) The inequation (S) 0 x + ax–2 > 2 is satisfied for all x  (0, ) and a  W, then a can be [A → P, B → S, C → Q, D→ P,Q,R] Q.5 Match the following : Column-I (A) The roots of the equation ax2 + bx + c = 0 are two consecutive odd positive integers, then (B) The equation x2 + bx + 4a2 = 0 has real roots which exceeds a number c then (C) (a2– b 2 ) x2 + 2(ac – b) x + (c2 –1) = 0 has real roots then (D) The equation ax2 –bx + c = 0 has complex roots & its roots are reciprocal then where a, b, c  R Column-II (P) – b > 2c (Q) |b|  4|a| (R) |b| < 2 |c| (S) it is valid for any condition between a, b and c [A → Q, B → P, C → S, D → R] Q.6 Match the following: Column-I (A) The number of roots of x2 –2 = [sin x] is (Here [x] is greatest integer  x) (B) Let f : D → R and f(x) = n                 − + 16 33 2 x n n ..... n x 2 n times     then the value(s) of n for which f(x) is onto is (are) (C) The number of common points on the graphs of y = |sin |x|| and y = x + |x| for x  [–2, 2], is (D) If f(x) is differentiable function for  x  R and exactly two of its tangents are parallel to x-axis, then number of real values of c for which f(c) = 0 could be Column-II (P) 0 (Q) 1 (R) 2 (S) 3 [A → R, B → R,S, C → S, D → P,Q,R,S] Q.7 Match the following: Column-I (A) The roots of the equation ax2 + bx + c = 0 are two consecutive odd positive integers, then (B) The equation x2 + bx + 4a2 = 0 has real roots which exceeds a number c then (C) (a2– b 2 ) x2 + 2(ac – b) x + (c2 –1) = 0 has real roots then (D) The equation ax2 –bx + c = 0 has complex roots & its roots are reciprocal then where a, b, c  R Column-II (P) – b > 2c (Q) |b|  4|a| (R) |b| < 2 |c| (S) it is valid for any condition between a, b and c [A → Q, B → P, C → S, D → R] Q.8 Match the following : Column I Column II (A) Q1 (x) = x2–mx + 1 is (P) (–3/2, 1/2) negative for values of x in (1, 2), if m lies in the interval (B) Q2 (x) = x2 + 2 (m – 1) x+m+5 (Q) (5/2, ) is positive for all x if m lies in the interval
(C) If 2x 3x x 2x 1 3 2 + + − is positive, (R) (1/2, 5/2) then x lies in the interval (D) The interval of x for which (S) (–, –3/2) x 12 – x 9 + x4 – x + 1 > 0 [A → R, B → Q,R, C → R,S, D → P,Q,R,S] Q.9 If ,  are the roots of the equation x 2 – 4x + 1 = 0, then Column-I Column–II (A)  2 +  2 (P) 52 (B)  3 +  3 (Q) 4 (C) | – | (R) 14 (D)  1 +  1 (S) 2 3 [A → R, B → P, C → S, D → Q] Q.10 Match the following equations with the number of real roots Column-I Column-II (A) x 7 + x5 + x3 – 1 = 0 (P) none (B) 4(x4 + x2 + 1) = 5(x2 + x + 1) (Q) one (C) x 3 – x + 1 = 0 (R) two (D) x 2 + 5|x| + 6 = 0 (S) three Sol. (A) → Q ; (B) → R ; (C) → Q ; (D) → P (A) P(x) = x 7 + x5 + x3 – 1 = 0 P(x) = 7x6 + 5x4 + 3x2 = x2 (7x4 + 5x2 + 3) P (x) = 0 at x = 0 but does not change sign in neighbourhood of x = 0 hence there are no points of extreme.  P(x) = 0 will have only one real root. (B) 4(x4 + x2 + 1) = 5 (x2 + x + 1)  4(x2 + x + 1) (x2 –x + 1) – 5(x2 + x + 1) = 0  (x2 + x + 1) (4x2 –4x –1) = 0  x 2 + x + 1) (4x2 –4x –1) = 0  x 2 + x + 1 = 0  no real roots. 4x2 – 4x – 1 = 0  two real roots (C) P(x) = x3 –x + 1, P(x) = 3x2 –1 P(x) is zero for two values of x i.e.  = – 3 1 ,  = 3 1 P(). P() = ( 3 –  + 1) ( 3 –  + 1) but 3 2 –1 = 0 and 3 2 – 1 = 0  P() . P() = 9 1 (4 –3) (4 –3) = 9 16+ 9−12( +) But,  +  = 0 and  = – 3 1  P(). P() = 9 16 − 3 > 0 Hence, P(x) = 0 will have only one real root (D) x 2 + 5|x| + 6 =    + +  − −  (x 2)(x 3) x 0 (x 2)(x 3) x 0 Hence no real roots. Q.11 Consider the equation x2 + 2(a –1)x + a + 5 = 0, where 'a' is a parameter. Match of the real values of 'a' so that the given equation has Column I Column II (A) imaginary roots (P)       −  − 7 8 , (B) one root smaller than 3 (Q) (–1, 4) and other root greater than 3 (C) exactly one root in the (R)       − − 7 8 , 3 4 interval (1, 3) & 1 and 3 are not the root of the equation (D) one root smaller than 1 (S)       −  − 3 4 , and other root greater than 3 Sol. A → Q ; B → P; C → R ; D → S Q.12 Column I Column II (A) Q1 (x) = x2– mx + 1 is (P) (–3/2, 1/2) negative for values of x in (1, 2), if m lies in the interval (B) Q2 (x) = x2 + 2 (m – 1) (Q) (5/2, ) x + m + 5 is positive for
all x if m lies in the interval (C) If 2x 3x x 2x 1 3 2 + + − is (R) (1/2, 5/2) positive, then x lies in the interval (D) The interval of x for (S) (–, –3/2) which x 12 – x 9 + x4 – x + 1 > 0 Sol. A → Q ; B → R; C → Q,R,S ; D → P,Q, R, S Q.13 Let f(x) = x 5x 6 x 6x 5 2 2 − + − + [IIT-2007] Match the expressions/statements in Column- I with expression/statements in Column- II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS. Column- I (A) If –1 < x < 1, then f(x) satisfies (B) If 1 < x < 2, then f(x) satisfies (C) If 3 < x < 5, then f(x) satisfies (D) If x > 5, then f(x) satisfies Column- II (P) 0 < f(x) < 1 (Q) f(x) < 0 (R) f(x) > 0 (S) f(x) < 1 Sol. A → P,R,S ; B → Q, S; C → Q, S ; D → P, R, S Q.14 Match the column : If ,  are the roots of the equation x 2 – 4x + 1= 0, then Column - I Column -II (A)  2 +  2 (P) 52 (B)  3 +  3 (Q) 4 (C) | – | (R) 14 (D)  +  1 1 (S) 2 3 Sol. (A) → (R), (B) → (P), (C) → (S), (D) → (Q) Q.15 Match the column : Column-I Column-II (A) If x = 3 + i 6 , (P) 3 then the value of 4x4 – 24x3 + 57x2 +18x – 40 is (B) One root of the equation (Q) 2 3x2 + px + 3 = 0 (p > 0) is square of the other if p = (C) Characteristic of log of (R) 5 0.0234 is (D) If the roots of the equation (S) –2 x 2 – 3x + 2loge  2e + 1 = 0 are real and equal, then  = Sol. (A) → (R), (B) → (P), (C) → (S), (D) → (Q) Q.16 Match the following : Column – I Column–II (A) If x2 + x – a = 0 has integral roots and aN, then a can be equal to (P) 2 (B) If the equation ax2 + 2bx + 4c = 16 has no real roots and a + c > b + 4 then integral value of c can be equal to (Q) 6 (C) If the equation x 2 + 2bx + 9b – 14 = 0 has only negative roots then integral values of b can be (R) 12 (D) The largest positive term of the H.P. whose first two terms are 5 2 and 23 12 is (S) 20 Sol. A → P, Q, R, S B → Q, R, S C → P, R, S D → Q (A) Discriminant, D = 1+ 4a  1 + 4a should be a perfect square. As 1 + 4a is always odd  1 + 4a = (2 + 1)2 ,   I +  a =  ( + 1) (B) Let f(x) = ax2 + 2bx + 4c – 16 Clearly f(–2) = 4a – 4b + 4c – 16 = 4 (a – b + c – 4) > 0 = f(x) > 0,  x  R  f(0) > 0  4c – 16 > 0  c > 4 (C) Both the roots of the equation f(x) = x2 + 2bx + 9b – 14 = 0 will be negative if D  0  4b2 – 4 (9b – 14)  0 ...(i) 9b – 14 > 0 ...(ii) From (i) b – 9b + 14  0  (b – 2) (b – 7)  0