A) f is an odd function
B) x =1 is a point of minima and \[x=1\] is a point of maxima of f.
C) \[f\left( 1 \right)4f\left( 1 \right)=4\]
D) x =1 is a point of maxima and \[x=1\] is a point of minimum of f.
Correct Answer: B
Solution :
| [b] \[\because \] f(x) is a five degree polynomial such that |
| \[\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{f(x)}{{{x}^{3}}} \right)=4\] |
| let \[f(x)=a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}\] |
| \[\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{a\,{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}}{{{x}^{3}}} \right)=4\] |
| \[\Rightarrow \,\,2+c=4\Rightarrow \,c=2\] |
| Now, \[f'(x)=5a{{x}^{4}}+4b{{x}^{3}}+3c{{x}^{2}}\] |
| \[={{x}^{2}}(5a{{x}^{2}}+4bx+3c)\] |
| \[\because \,\,f'(1)\,=0\Rightarrow \,5a+4b+6=0\] |
| and \[f'(-1)=0\Rightarrow \,5a-4b+6=0\] |
| \[\therefore \,\,b=0,\,a=-\frac{6}{5}\] |
| \[\therefore \,f(x)=-\frac{6}{5}{{x}^{5}}+2{{x}^{3}}\] |
| \[f'(x)=-6{{x}^{4}}+6{{x}^{2}}=-\,6{{x}^{2}}(x+1)(x-1)\] |
| It is clear that maxima at x = 1 and minima at\[x=1\]. |
| and \[f\left( 1 \right)4f\left( 1 \right)=4\] |
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