Search any question & find its solution
Question:
Answered & Verified by Expert
If the set of all values of $x$ for which the expansion of $(7-5 x)^{-\frac{2}{3}}$ valid is equal to $(-a, a)$, then $5 a+7$ is equal to
Options:
Solution:
2765 Upvotes
Verified Answer
The correct answer is:
14
$(7-5 x)^{-\frac{2}{3}}=(7)^{-\frac{2}{3}}\left(1-\frac{5 x}{7}\right)^{-\frac{2}{3}}$
For validation, $\left|\frac{-5 x}{7}\right| < 1 \Rightarrow-1 < \frac{5 x}{7} < 1$
$$
\Rightarrow-7 < 5 x < 7 \Rightarrow-\frac{7}{5} < x < \frac{7}{5}
$$
So, $(-a, a)=\left(-\frac{7}{5}, \frac{7}{5}\right)$
$$
\begin{aligned}
\therefore & a=\frac{7}{5} \\
\therefore & 5 a+7=5 \times \frac{7}{5}+7=14
\end{aligned}
$$
For validation, $\left|\frac{-5 x}{7}\right| < 1 \Rightarrow-1 < \frac{5 x}{7} < 1$
$$
\Rightarrow-7 < 5 x < 7 \Rightarrow-\frac{7}{5} < x < \frac{7}{5}
$$
So, $(-a, a)=\left(-\frac{7}{5}, \frac{7}{5}\right)$
$$
\begin{aligned}
\therefore & a=\frac{7}{5} \\
\therefore & 5 a+7=5 \times \frac{7}{5}+7=14
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.