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A curve with equation , y=ax4+bx3+cx2+d , has zero gradient at point (0,1) and also touches the x-axis at point (-1,0).Then value of x for which the curve has a negative gradient are given by; ans is for x plzz explain


A curve with equation , y=ax4+bx3+cx2+d , has zero gradient at point (0,1) and also touches the x-axis at point (-1,0).Then value of x for which the curve has a negative gradient are given by;



ans is for x<-1


plzz explain


Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
y = ax^{4} + bx^{3} + cx^{2} + d
y' = 4ax^{3} + 3bx^{2} + 2cx
y' = x(4ax^{2} + 3bx + 2c)
There are three roots of y’ = 0
Two roots are given: x =0, -1
y = ax^{4} + bx^{3} + cx^{2} + d
1 = a.0^{4} + b.0^{3} + c.0^{2} + d
d = 1
0 = a.-1^{4} + b.-1^{3} + c.-1^{2} + d
0 = a - b + c + 1
b-c-a = 1
Let x1be the 3rdroot of y’ = 0
y' = x(4ax^{2} + 3bx + 2c) < 0
x(4ax^{2} + 3bx + 2c) < 0
a'x(x+1)(x-x_{1}) < 0
a’ > 0
By wavy curve method, we have
x1> 0
\Rightarrow x < -1, x > x_{1}

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