InterviewSolution
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InterviewSolution
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Let `a ,b ,c`be real numbers `a!=0.`If `alpha`is a root of `a^2x^2+b x+c=0.beta`is the root of `a^2x^2-b x-c=0a n d0A. `gamma=(alpha+beta)/(2)`B. `gamma=alpha+(beta)/(2)`C. `gamma=alpha`D. `alphaltgammalt beta` |
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Answer» Correct Answer - D Since, `alpha` is a root of `a^(2)x^(2)+bx+c=0` `impliesa^(1)alpha^(2)+balpha+c=0" "...(i)` `and beta"is a root of"a^(2)x^(2)-bx-c=0` `impliesa^(2)beta^(2)-b beta-c=0" "...(ii)` Let `f(x)=a^(2)alpha^(2)+2bx+2c` `therefore f(alpha)a^(2)alpha^(2)+2balpha+2c` `=a^(2)alpha^(2)-2a^(2)alpha^(2)=-a^(2)alpha^(2)" "["from Eq."(i)]` and `f(beta)=alpha^(2)beta^(2)+2b beta+2c` `= a^(2)beta^(2)+2a^(2)beta^(2)=3a^(2)beta^(2)" "["from Eq."(ii).]` `implies f(alpha) f(beta)lt0` f(x) must have a root laying in the open interval `(alpha, beta).` `therefore alpha lt gamma lt beta` |
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