BFSK Modulation Output
BFSK uses a pair of discrete frequencies to transmit binary (0s and 1s) information. \$s_1(t)\$ and \$s_2(t)\$ are the discrete frequency signals here. \$s_1(t)\$ has a frequency of \$1/2T\$ and \$s_2(t)\$ has a frequency of \$1/T\$. But your question does not say which signal is sent for input \$0\$. 

So if \$s_1(t)\$ is sent for input \$0\$ and \$s_2(t)\$ is sent for input \$1\$, then the output is given by 

$$y(t) =  \overline{x(t)}\ s_1(t)+x(t)\ s_2(t)$$

where, \$x(t) \in \{0,1\}\$ is the input and \$\overline{x(t)}=1-x(t)\$. 

EDIT:
In a FSK demodulator, the received signal, \$r(t)\$ is correlated with signal corresponding to each symbol, \$S_i(t)\$ to get \$Z_i(t)\$. 

$$Z_i(t) = \int r(t)\ s_i(t)\ dt\tag1$$
Then the demodulator makes a decision based on the value of \$Z_i(t)\$. The decision is made in favor of i\$^{th}\$ symbol producing maximum correlation product (\$Z_i(t)\$)

So in BFSK, the received signal \$r(t)\$ is correlated with both \$s_1(t)\$ and \$s_2(t)\$. So if 1 be the value of modulating signal, then \$s_2(t)\$ is the received signal and its correlation product with s1(t) and s2(t) are calculated using equation (1). Correlating \$s_2(t)\$ with itself will produce the maximum value and hence the demodulator makes a decision in favor of \$s_2(t)\$.