The point of inflection of the function y=₀^{x}(t²-3t+2)dt is

Correct Answer is : (32,34)≤ft(3/2, 3/4)(23,43)

Correct Answer is : (32,34)≤ft(3/2, 3/4)(23,43)

Test Index

UPSEE 2017 Solved Paper

Question : 134 of 150

Marks: +1, -0

y=\int\limits_{0}^{x}(t^{2}-3t+2)dt

Solution:

Given,

y=\int\limits_{0}^{x}(t^{2}-3t+2)dt

......(i)

On differentiating w.r.t.

x'

, we get

\frac{dy}{dx}=x^{2}-3x+2

........(ii)

Again, on differentiating w.r.t. '

x

', we get

\frac{d^{2}y}{dx^{2}}=2x-3

.......(iii)

We know that, at point of inflection

\frac{d^{2}y}{dx^{2}}=0

\therefore

From Eq. (iii), we get

2x-3=0 \Rightarrow x=\frac{3}{2}

Now, we have to check behaviour of

\frac{d^{2}y}{dx^{2}}

at point

x=\frac{3}{2}

x=\frac{3}{2}

Clearly, at

x=\frac{3}{2}

sign at

\frac{d^{2}y}{dx^{2}}

changes

\therefore \left(\frac{3}{2}, \frac{3}{4}\right)

is point of inflection.

Go to Question:

Prev Question

Next Question