Question

Let f : R $/to$ R be given by f(x) = (x - 1)(x - 2)(x - 5). Define $F\left(x\right) = \int\limits^{x}_{0} f\left(t\right) dt, x > 0.$ Then which of the following options is/are correct?

• A. F has a local minimum at x = 1
• B. F has a local maximum at x = 2
• C. F has two local maxima and one local minimum in (0, $\infty$)
• D. \(F\left(x\right) \ne\) 0 for all \(x \in \left(0,5\right)\)

Answer 1

The Correct Option is D

The correct option is(D): \(F\left(x\right) \ne\) 0 for all \(x \in \left(0,5\right)\).

\(f(x) = (x - 1) (x - 2) (x - 5)\)
\(f(x) =\) \(\int\limits^{{x}}_{{0}}\) \(f(t) dt,x > 0\)
\(F'\left(x\right)=f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-5\right), x>0\) 
clearly F(x) has local minimum at x = 1,5 
\(F(x)\) has local maximum at \(x = 2\)
\(f\left(x\right)=x^{3}-8x^{2}+17x-10\)
\(\Rightarrow F\left(x\right)=\)\(\int\limits^{{x}}_{{0}}\)\(\left(t^{3}-8t^{2}+17t-10\right)dt\)
\(F\left(x\right)=\frac{x^{4}}{4}-\frac{8x^{3}}{3}+\frac{17x^{2}}{2}-10x\) 
from the graph of \(y = F(x)\), clearly \(F\left(x\right)\ne0 \forall x\,\in \left(0.5\right)\)