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A curve with equation of the form `y=a x^4+b x^3+c x+d`has zero gradient at the point `(0,1)`and also touches the `x-`axis at the point `(-1,0)`then the value of `x`for which the curve has a negative gradient are:`xgeq-1`b. `x |
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Answer» Correct Answer - C We have ,`(dy)/(dx) =0 =at (0,1) and (-1,0)` ` therefore c= 0 and - 4a + 3 b + c =0` `implies a= (3b)/(4) and c=0` Also , the passes through (0,1) and (-1,0) `therefore d=1 and 0=a - b - c +d` ` a-b-c+1 =0` ….(ii) From (i) and (ii) we get a=3, b=4 c=4, c=0 and d=1 `therefore y=3x^(4)+4x^(3) +1,implies (dy)/(dx)=12x^(3)+12x^(2)` `Now , (dy)/(dx)lt 0 = implies 12x ^(3)+12x^(2)lt 0` `implies x+1lt 0impliesxlt - 1` |
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