MHT CET · Maths · Limits
If \(\mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7\), then \(\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}=\)
- A \(\frac{53}{3}\)
- B \(\frac{-53}{3}\)
- C \(\frac{52}{3}\)
- D \(\frac{-52}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{53}{3}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
& \Rightarrow \mathrm{f}^{\prime}(1)=30-56+30-63+6=-53
\end{aligned}
\)
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, \(\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}\)
\(
\begin{aligned}
& =-\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{(1-\alpha)-1} \times \frac{1}{\alpha^2+3} \\
& =-f^{\prime}(1) \times \frac{1}{3}=\frac{53}{3}
\end{aligned}
\)
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