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If `a ,b in R ,a!=0`and the quadratic equation `a x^2-b x+1=0`has imaginary roots, then `(a+b+1)`isa. positiveb. negativec. zero d. Dependent on thesign of `b`A. positiveB. negativeC. zeroD. dependent on the sign of b |
Answer» Correct Answer - 1 `D = b^(2) - 4a lt 0 rArr a gt 0` Therefore the graph is concave upwards. `f(x) gt 0, AA x in R` `rArr f(-1) gt 0` `rArr a + b + 1 gt 0` |
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