Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7$, then $\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}=$
Options:
Solution:
1403 Upvotes
Verified Answer
The correct answer is:
$\frac{53}{3}$
$$
\begin{aligned}
& \mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
& \Rightarrow \mathrm{f}^{\prime}(1)=30-56+30-63+6=-53
\end{aligned}
$$
Now, $\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}$
$$
\begin{aligned}
& =-\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{(1-\alpha)-1} \times \frac{1}{\alpha^2+3} \\
& =-f^{\prime}(1) \times \frac{1}{3}=\frac{53}{3}
\end{aligned}
$$
\begin{aligned}
& \mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
& \Rightarrow \mathrm{f}^{\prime}(1)=30-56+30-63+6=-53
\end{aligned}
$$
Now, $\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}$
$$
\begin{aligned}
& =-\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{(1-\alpha)-1} \times \frac{1}{\alpha^2+3} \\
& =-f^{\prime}(1) \times \frac{1}{3}=\frac{53}{3}
\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.