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If solution set of Inequality `(x^(2)+x-2)(x^(2)+x-16)ge -40` is `x epsilon(-oo, -4]uu[a,b]uu[c,oo)` then `a+b-c` isA. `-2`B. `-3`C. `-4`D. `-6` |
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Answer» Correct Answer - C If solution set of inequality …………………. Let `x^(2)+x -2 = lambda" then "lambda(lambda -14)+40 ge 0` `(lambda - 4) (lambda -10) ge 0` `lambda le 4` , `lambda ge 10` `x^(2)+x-2le4` , `x^(2)+x-12ge0` `(x+3)(x-2)le0` `x in(-oo, -4]uu[3, oo)` `x in(-oo, -4]uu[-3,2]uu[3,oo)` `a+b-c= -3+2-3= -4` |
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