Match the expressions/statements in Column I with the expressions/statements in Column II and indicate your answer by darkening the appropriate bubbles in the matrix given in the ORS. [2007]

Question

Let $f (x) = \frac{x^{2} - 6 x + 5}{x^{2} - 5 x + 6}$

Match the expressions/statements in Column I with the expressions/statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2007]

	Column I		Column II
(A)	If $- 1 < x < 1$ , then $f (x)$ satisfies	(p)	$0 < f (x) < 1$
(B)	If $1 < x < 2$ , then $f (x)$ satisfies	(q)	$f (x) < 0$
(C)	If $3 < x < 5$ , then $f (x)$ satisfies	(r)	$f (x) > 0$
(D)	If $x > 5$ , then $f (x)$ satisfies	(s)	$f (x) < 1$

Smart Achievers · Accepted Answer

(2)

(A) → (r), (s), (p); (B) → (q), (s); (C) → (q), (s); (D) → (r), (s), (p)

$f (x) = \frac{x^{2} - 6 x + 5}{x^{2} - 5 x + 6} = \frac{(x - 5) (x - 1)}{(x - 2) (x - 3)}$

(A) If $- 1 < x < 1$ then $f (x) = \frac{(- ve) (- ve)}{(- ve) (- ve)} = + ve$

$f (x) < 0$ $f (x) > 0 (r)$

Also, $f (x) - 1 = \frac{- x - 1}{x^{2} - 5 x + 6} = \frac{- (x + 1)}{(x - 2) (x - 3)}$

For $- 1 < x < 1,$ $f (x) - 1 = \frac{- (+ ve)}{(- ve) (- ve)} = - ve$

$\Rightarrow f (x) - 1 < 0 \Rightarrow f (x) < 1 (s)$

$∴$ $0 < f (x) < 1 (p)$

(B) If $1 < x < 2,$ then $f (x) = \frac{(- ve) (+ ve)}{(- ve) (- ve)} = - ve$

$∴$ $f (x) < 0 (q)$ and so, $f (x) < 1 (s)$

(C) If $3 < x < 5,$ then $f (x) = \frac{(- ve) (+ ve)}{(+ ve) (+ ve)} = - ve$

$∴$ $f (x) < 0 (q)$ and so, $f (x) < 1 (s)$

(D) For $x > 5,$ $f (x) > 0 (r)$

Also, $f (x) - 1 = \frac{- (x + 1)}{(x - 2) (x - 3)} < 0$

For $x > 5$ , $f (x) < 1 (s)$

$∴$ $0 < f (x) < 1 (p)$

Solution

1 (A) → (q); (B) → (r), (s), (p), (s); (C) → (q), (s); (D) → (r), (s), (p)

2 (A) → (r), (s), (p); (B) → (q), (s); (C) → (q), (s); (D) → (r), (s), (p)

3 (A) → (q); (B) → (q), (s), (s); (C) → (r), (s), (p); (D) → (r), (s), (p)

4 (A) → (q); (B) → (s), (s); (C) → (r), (s); (D) → (r), (s), (p)

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