Let f : R → R be a polynomial function of degree four having extreme values at x = 4 and x = 5.

Question

Let \(f: \mathbf{R} \rightarrow \mathbf{R}\) be a polynomial function of degree four having extreme values at x = 4 and x = 5. If \(\lim\limits _{x \rightarrow 0} \frac{f(x)}{x^{2}}=5,\) then f(2) is equal to :

(1) 12

(2) 10

(3) 8

(4) 14
 

Answer 1

Correct option is: (2) 10 

\(\lim\limits _{x \rightarrow 0} \frac{f(x)}{x^{2}}=5\)

\(\lim\limits _{x \rightarrow 0} \frac{\left.a x^{4}+b x^{3}+c x^{2}+d x+e\right)}{x^{2}}=5\)

\(\mathrm{c}=5\) and \(\mathrm{d}=\mathrm{e}=0\)

\(f(x)=a x^{4}+b x^{3}+5 x^{2}\)

\(f^{\prime}(x)=4 a x^{3}+3 b x^{2}+10 x\)

\(=x\left(4 a x^{2}+3 b x+10\right)\)

has extremes at 4 and so \(\mathrm{f}^{\prime}(4)=0\ \&\ \mathrm{f}^{\prime}(5)=0\)

so \(\mathrm{a}=\frac{1}{8}\ \&\ \mathrm{~b}=\frac{-3}{2}\)

so \(f(2)=\frac{1}{8} \times 2^{4}-\frac{3}{2} \times 2^{3}+5 \times 2^{2}\)

= 2 - 12 + 20 = 10