Let f(x)=x3-3 x2+2 x. If the equation f(x)=k has exactly one positive and one negative solution then the value of k equals

Question

Let f(x)=x3-3 x2+2 x. If the equation f(x)=k has exactly one positive and one negative solution then the value of k equals (A) -(2 √3/9) (B) -(2/9) (C) (2/3 √3) (D) (1/3 √3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/39fbc29bc867791e74ea8c7fbbefe588-.png" /> f(x)=x(x2-3 x+2) f(x)=x(x-2)(x-1) graph of y=f(x) is as shown now f ( x )= k to have exactly one positive and negative solution we have, k = f (1+(1/√3)) (think!) 1-(1/√3) and 1+(1/√3) are the roots of f prime( x )=0

Q. Let $f (x) = x^{3} - 3 x^{2} + 2 x$ . If the equation $f (x) = k$ has exactly one positive and one negative solution then the value of $k$ equals

$- \frac{2 3}{9}$

$- \frac{2}{9}$

$\frac{2}{3 3}$

$\frac{1}{3 3}$

$f (x) = x (x^{2} - 3 x + 2)$
$f (x) = x (x - 2) (x - 1)$
graph of $y = f (x)$ is as shown
now $f (x) = k$ to have exactly one
positive and negative solution
we have, $k = f (1 + \frac{1}{3})$ (think!) $1 - \frac{1}{3}$ and $1 + \frac{1}{3}$ are the roots of $f^{'} (x) = 0$

Q. Let f(x)=x3−3x2+2x. If the equation f(x)=k has exactly one positive and one negative solution then the value of k equals

−923​​

−92​

33​2​

33​1​