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  • The graph of   is shown below. Let <img alt="F(x)" src="https://entrancecorner.oncodecogs.com/gif.latex?F

The graph of y=f(x)  is shown below. Let F(x)  be an antiderivative of f(x). Then, F(x) equals
 

Option: 1

point of inflection at x=0, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3} and 2 \pi, a local maximum at  x=\frac{\pi}{2} and a local maximum at x=\frac{3 \pi}{2}


Option: 2

point of inflection at x=0, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3} and 2 \pi, a local minimum at x=\frac{\pi}{2} and a local maximum at x=\frac{3 \pi}{2}


Option: 3

point of inflection at x=\pi, a local maximum at x=\frac{\pi}{2} and a local minimum at x=\frac{3 \pi}{2}


Option: 4

 point of inflection at x=\pi, a local minimum at x=\frac{\pi}{2}and a local maximum at x=\frac{3 \pi}{2}


Answers (1)

best_answer

Since,F(x) is the antiderivative of f(x).


\begin{array} {c c c} \text{Hence, } & \quad \int f(x) d x=F(x) \\ \Rightarrow & \quad F^{\prime}(x)=f(x) and F^{\prime \prime}(x)=f^{\prime}(x)\\ \text{Now,} & \quad F^{\prime}(x)=0 \Rightarrow f(x)=0 \end{array}
i.e., at x=\frac{\pi}{2}, \pi, \frac{3 \pi}{2}
at x=\frac{\pi}{2} F^{\prime}(x) changes sign from +ve to -ve
Hence, y=F(x) has maxima at x=\frac{\pi}{2}
at x=\pi F^{\prime}(x) does not change sign
\Rightarrow  Point of inflection
at x=\frac{3 \pi}{2} \quad F(x)  changes sign from -ve to +ve
\Rightarrow minima at \frac{3 \pi}{2}.

Posted by

Ritika Harsh

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