Indian Institute of Foreign Trade 2012 Solved Paper

Let us start by putting the value of x.
When $\mathit{x}=1,$
$7^{\mathit{x}-1}+{11}^{\mathit{x}-1}=1+1=2$
When $\mathit{x}=2$,
$7^{\mathit{x}-1}+{11}^{\mathit{x}-1}=7+11=17$
When $\mathit{x}=3,$
$7^{\mathit{x}-1}+{11}^{\mathit{x}-1}=49+121=170$
When $\mathit{x}=4,$
$7^{\mathit{x}-1}+{11}^{\mathit{x}-1}=243+1331=1574$
As we can see $7^{\mathit{x}-1}+{11}^{\mathit{x}-1}$ is an increasing function and the only value that satisfies $7^{\mathit{x}-1}+{11}^{\mathit{x}-1}=170$,is $\mathit{x}=3$
Thus, the given equation has only 1 solution.
Hence, option B is the correct answer.