Question

Sum of values of 'x' for which sum of middle terms of expansion $(1+\log_{10}x)^{7+\lim_{t\to 5^+} \frac{[t]-5}{e^{\{t\}}-1}}$ is zero, is (where {.} & [.] denotes fractional part function and greatest integer function respectively)

Answer 1

1.1

Answer 2

1. Evaluate the limit: Let $t = 5+h$ where $h \to 0^+$. Then $[t]=5$ and $\{t\}=h$. The limit is $\lim_{h\to 0^+} \frac{5-5}{e^h-1} = \lim_{h\to 0^+} \frac{0}{e^h-1} = 0$.
2. The exponent of the expansion is $n = 7+0 = 7$.
3. The expansion $(1+\log_{10}x)^7$ has $7+1=8$ terms. The middle terms are the 4th ($T_4$) and 5th ($T_5$) terms.
4. The general term is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For $n=7, a=1, b=\log_{10}x$: $T_4$ (r=3): $\binom{7}{3} (\log_{10}x)^3 = 35 (\log_{10}x)^3$. $T_5$ (r=4): $\binom{7}{4} (\log_{10}x)^4 = 35 (\log_{10}x)^4$.
5. Sum of middle terms: $35 (\log_{10}x)^3 + 35 (\log_{10}x)^4 = 35 (\log_{10}x)^3 (1 + \log_{10}x)$.
6. Set sum to zero: $35 (\log_{10}x)^3 (1 + \log_{10}x) = 0$.
7. This implies $(\log_{10}x)^3 = 0$ or $1 + \log_{10}x = 0$.
8. Solving for $x$: $\log_{10}x = 0 \implies x = 10^0 = 1$. $\log_{10}x = -1 \implies x = 10^{-1} = 0.1$.
9. The sum of values of $x$ is $1 + 0.1 = 1.1$.