JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(f ( x )\) be a polynomial of degree \(6\) in \(x ,\) in which the coefficient of \(x^{6}\) is unity and it has extrema at \(x=-1\) and \(x=1\). If \(\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1,\) then \(5 \cdot f (2)\) is equal to .............
- A \(121\)
- B \(144\)
- C \(169\)
- D \(121\)
Answer & Solution
Correct Answer
(B) \(144\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f\) as \(\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1\) non-zero finite So, \(d=e=f=0\) and \(f(x)=x^{3}\left(x^{3}+a x^{2}+b x+c\right)\) Hence, \(\lim _{x \rightarrow 0} \frac{ f ( x )}{ x ^{3}}= c =1\) Now, as…
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