JEE Mains · Maths · STD 11 - 12. limits
Let \(f(x)=x^{6}+2 x^{4}+x^{3}+2 x+3, x \in R\). Then the natural number \(\mathrm{n}\) for which \(\lim _{x \rightarrow 1} \frac{\mathrm{x}^{\mathrm{n}} \mathrm{f}(1)-\mathrm{f}(\mathrm{x})}{\mathrm{x}-1}=44\) is ...... .
- A \(6\)
- B \(7\)
- C \(8\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(f(n)=x^{6}+2 x^{4}+x^{3}+2 x+3\) \(\lim _{x \rightarrow 1} \frac{x^{n} f(1)-f(x)}{x-1}=44\) \(\lim _{x \rightarrow 1} \frac{9 x^{n}-\left(x^{6}+2 x^{4}+x^{3}+2 x+3\right)}{x-1}=44\) \(\lim _{x \rightarrow 1} \frac{9 n x^{n-1}-\left(6 x^{5}+8 x^{3}+3 x^{2}+2\right)}{1}=44\)…
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