Let \f(x)= \begin{cases} \lg x, & x > 0 \\ x+ \int_{0}^{a}3t^{2}dt, & x\leqslant 0 \end{cases}\, if \f(f(1))=1\, then the constant term in the expansion of \(4^{x}-2^{-x})^{a+5}\ is \_\_\_\_\_\_.
Solution
Solution: When \x\leqslant 0\, \f(x)=x+ \int_{0}^{a}3t^{2}dt=x+a^{3}\,
When \x > 0\, \f(x)=\lg x\,
Since \f(f(1))=1\,
Therefore \f(0)=1\, which means \a^{3}=1\,
Thus \a=1\,
Therefore, the expansion of \(4^{x}-2^{-x})^{a+5}\ is equivalent to the expansion of \(4^{x}-2^{-x})^{6}\. The general term formula of the expansion is:
\ C_{6}^{r}(4^{x})^{6-r}\cdot(-2^{-x})^{r}= C_{6}^{r}\cdot(-1)^{r}\cdot2^{12x-3xr}\,
Let \12x-3xr=0\, then \r=4\,
Hence, the constant term in the expansion is \ C_{6}^{4}\cdot(-1)^{4}=15\.
Therefore, the answer is:
By using the knowledge of definite integrals, we find the expression of \f(x)\ when \x\leqslant 0\, and then by \f(f(1))=1\ we get \a=1\. The constant term can be obtained from the general term of the binomial expansion.
This problem examines the application of piecewise functions, the operation of definite integrals, and the use of the binomial theorem, mainly focusing on finding a specific term, and is considered a basic question.
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