Complex Numbers And Quadratic Equations question 17
Question: If $ x $ is real, the function $ \frac{(x-a)(x-b)}{(x-c)} $ will assume all real values, provided [IIT 1984; Karnataka CET 2002]
Options:
A) $ a>b>c $
B) $ a<b<c $
C) $ a>c<b $
D) $ a<c<b $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y=\frac{(x-a)(x-b)}{(x-c)} $ or $ y(x-c)=x^{2}-(a+b)x+ab $ or $ x^{2}-(a+b+y)x+ab+cy=0 $ $ \Delta ={{(a+b+y)}^{2}}-4(ab+cy) $ $ =y^{2}+2y(a+b-2c)+{{(a-b)}^{2}} $ Since x is real and y assumes all real values, we must have $ \Delta \ge 0 $ for all real values of y.The sign of a quadratic in y is same as of first term provided its discriminant $ B^{2}-4AC<0 $ This will be so if $ 4{{(a+b-2c)}^{2}}-4{{(a-b)}^{2}}<0 $ or $ 4(a+b-2c+a-b)(a+b-2c-a+b)<0 $ or $ 16(a-c)(b-c)<0 $ or $ 16(c-a)(c-b)=- $ ve \ c lies between a and b i.e., $ a<c<b $ …..(i) Where $ a<b $ , but if $ b<a $ then the above condition will be $ b<c<a $ or $ a>c>b $ …..(ii) Hence from (i) and (ii) we observe that (d) is correct answer.
BETA
