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For real x, the function `(x-a)(x-b)/(x-c)` will assume all real values providedA. `a gt b gt c`B. `a lt b lt c`C. `a gt c lt b`D. `a le c le b` |
Answer» Correct Answer - D Let `y=(x^(2)-(a+b)x+ab)/(x-c)` `impliesyx-cy=x^(2)-(a+b)x+ab` `impliesx^(2)-(a+b+y)x+(ab+cy)=0` For real roots, `D ge 0` `implies(a+b+y)^(2)-4(ab+cy)ge0` `implies(a+b)^(2)+y^(2)+2(a+b)y-4ab-4cyge0` `impliesy^(2)+2(a+b-2x)y+(a-b)^(2)ge0` which is true for all real values of y. `therefore" "D le0` `4(a+b-2a)^(2)-4(a-b)^(2)le0` `implies4(a+b-2c+a-b)(a+b-2c-a+)le0` `implies(2a-2c)(2b-c)le0` `implies(a-c)(b-c)le0` `implies(c-a)(c-b)le0` `implies` c mule lie between a and b `i.e. a le c le b or b le c lea ` |
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