Question

Observe the following lists List I and List II

(A) $\sum _{n=0}^{\infty }\frac{x^n({\log }_{e}a{)}^n}{n!}$ $\left(1\right)\frac{e^x-e^{-x}}{2}$
(B) $\sum _{n=0}^{\infty }\frac{x^{2n}}{\left(2n\right)!}$ $\left(2\right)e^{-ax}$
(C) $\sum _{n=0}^{\infty }\frac{x^{2n+1}}{(2n+1)!}=$ $\left(3\right)a^x$
(D) $\sum _{n=0}^{\infty }\frac{(-1{)}^n.(ax{)}^n}{n!}$ $\left(4\right)\frac{a^x-a^{-x}}{2}$
$\left(5\right)\frac{e^x+e^{-x}}{2}$
The correct match of List I to List II is:

• A. A - 1, B - 4, C - 3, D - 5
• B. A - 4, B - 2, C - 1, D - 3
• C. A - 3, B - 5, C - 1, D - 2
• D. A - 2, B - 3, C - 5, D - 4

Answer 1

Answer: (C)

A - 3, B - 5, C - 1, D - 2

(A) When we expand it, we will get

$1+\frac{x^{}{{\log }_{e}a}^{}}{1}+\frac{x^2({\log }_{e}a{)}^2}{2!}+....$ This is a general expression for the expansion of $a^x$

So, A - 3

(B) Expansion is given by $1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}..$ which is $\frac{e^x+e^{-x}}{2}$

So, B - 5

(C) Expansion is given by $1+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}..$ and this is $\frac{e^x-e^{-x}}{2}$

So, C - 1

(D) Expansion is given by $1-\frac{ax}{1!}+\frac{(ax{)}^2}{2!}-\frac{(ax{)}^3}{3!}....$ which is an expansion of $e^{-ax}$

So, D - 2