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Let `f:R->[1,oo)` be defined as `f(x)=log_10(sqrt(3x^2-4x+k+1)+10)` If f(x) is surjective then k =A. `k=(1)/(3)`B. `klt(1)/(3)`C. `lgt(1)/(3)`D. `k=1` |
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Answer» Correct Answer - A If f(x) is surjective then range of f(x) must be `[1,oo)`. `therefore" Range of "sqrt(3x^(2)-4x+k+1)+10 in [0,oo)` `rArr" Range of "3x^(2)-4x+k+1 in [0,oo)` `rArr" "D=0` `rArr" "16-12 (k+1)=0` `rArr" "4-3k-3=0` `rArr" "k=(1)/(3)` |
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