Questions & Answers

Question

Answers

A. One real root

B. Two real roots

C. Three real roots

D. Four real roots

Answer
Verified

Let us denote the function given to us by $f(x)$.

So, $f(x) = {e^{x - 1}} + x - 2$.

Now, we will find its derivative.

So, we will get:-

$f'(x) = \dfrac{d}{{dx}}({e^{x - 1}}) \times \dfrac{d}{{dx}}(x - 1) + \dfrac{d}{{dx}}(x)$ ……(1)

We know that $\dfrac{d}{{dx}}({e^x}) = {e^x}$ and $\dfrac{d}{{dx}}({x^n}) = {x^{n - 1}}$.

Using these formulas in (1), we will have with us:-

$f'(x) = {e^{x - 1}} \times 1 + 1$

$\therefore f'(x) = {e^{x - 1}} + 1$ ……..(1)

Now, let us draw the graph of ${e^{x - 1}}$ to know about the range of values it can possess.

We clearly see from the graph that it never takes negative value.

So, ${e^{x - 1}} > 0$.

Hence, by putting this in (1), we have:

$\therefore f'(x) = {e^{x - 1}} + 1 > 0 + 1 = 1$.

Hence, $f(x)$ is strictly increasing.

We must carefully observe now that if a graph just keeps on increasing for increasing values of x, we can never have more than one point where the graph touches the x- axis.

The point where the graph touches the x- axis is called the real root of the graph.

Hence, we can have one real root.

The students also might get confused that if it does follow the strictly increasing condition, then it definitely has one real root, but that is not true. This is just a necessary condition not a sufficient one. Example to prove this is take the graph of $y = x$, where $x > 1$. We will have:-

Here, we have all the conditions as per we had in the above solved question. Our derivative is strictly increasing.

But, this does not have any real root because $y \ne 0$ for any value of $x$.