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← IIT-JEE 2007 · Paper 2
MathematicsMatrix match+6Question 64 of 66

IIT-JEE 2007 · Paper 2, Mathematics question 64

Let f(x)=x2−6x+5x2−5x+6f(x)=\dfrac{x^2-6x+5}{x^2-5x+6}.

Match the expressions/statements in Column I with expressions/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<1-1<x<1, then f(x)f(x) satisfies
  • B
    If 1<x<21<x<2, then f(x)f(x) satisfies
  • C
    If 3<x<53<x<5, then f(x)f(x) satisfies
  • D
    If x>5x>5, then f(x)f(x) satisfies

Column II

  • P
    0<f(x)<10<f(x)<1
  • Q
    f(x)<0f(x)<0
  • R
    f(x)>0f(x)>0
  • S
    f(x)<1f(x)<1
Mark every match for each entry
PQRS
A
B
C
D
Answer and worked solution

Official answer: A → P, R, S; B → Q, S; C → Q, S; D → P, R, S

Factor. f(x)=(x−1)(x−5)(x−2)(x−3)f(x)=\dfrac{(x-1)(x-5)}{(x-2)(x-3)} and
f(x)−1=−(x+1)(x−2)(x−3).f(x)-1=\frac{-(x+1)}{(x-2)(x-3)}.

  • A, −1<x<1-1<x<1: numerator and denominator of ff both positive, so f>0f>0; f−1<0f-1<0 since x+1>0x+1>0 and the denominator is positive. → P, R, S.
  • B, 1<x<21<x<2: (x−1)(x−5)<0(x-1)(x-5)<0, denominator >0>0: f<0f<0 (so also f<1f<1). → Q, S.
  • C, 3<x<53<x<5: same signs: f<0f<0. → Q, S.
  • D, x>5x>5: f>0f>0 and f−1<0f-1<0. → P, R, S.

Answer: A → P, R, S; B → Q, S; C → Q, S; D → P, R, S