Question

The first two terms in the expansion of e^x in ascending powers of (x - 2) are​

Answer 1

Answer:

x² will be the perfect answer of this question.

Answer 2

Answer:

The first two terms in the expansion of $e^{x}$ in ascending powers of (x - 2) will be  $e^{2}$  and e(x-2)

Step-by-step explanation:

We will use Taylor's Theorem to solve this question.  

According to Taylor's Theorem, for a function f(x), in ascending powers of (x-a)

f(x) = f(a) + f'(a)*(x - a) + f''(a) * $\frac{(x-a)^{2}}{2!}$ + . . . . . . . . . + fⁿ⁺¹(a) * $\frac{(x-a)^{n+1}}{(n+1)!}$

where, f'(a) = First derivative

f''(a) = Second derivative

and so on

Here, in this question, f(x) = $e^{x}$  and a = 2  ;

=>f(a) = $e^{2}$

Therefore, f'(x) = $e^{x}$   f'(1) = e

                 f''(x) = $e^{x}$  f''(1) = e

Since we need the first two terms,

=> $e^{x}$ = $e^{2}$  + e * (x-2)

=> $e^{x}$ = $e^{2}$  + e(x-2)

Therefore, the first two terms in the expansion of $e^{x}$ in ascending powers of (x - 2) will be  $e^{2}$  and e(x-2)