A) \[\frac{17}{4}\]
B) \[\frac{13}{4}\]
C) \[\frac{19}{4}\]
D) \[\frac{5}{4}\]
Correct Answer: C
Solution :
| The polynomial is an every where differentiable function. Therefore, the points of extremum can only be the roots of the derivative. Furthermore, the derivative of a polynomial is a polynomial. The polynomial of the least degree with root x=1 and x=3. |
| form \[a\,\,(x-1)(x-3).\] |
| Hence, \[P'(x)=a\,\,(x-1)\,\,(x-3).\] |
| Since at x=1, we must have P (1) =6, we have |
| \[P(x)=\int\limits_{1}^{x}{P'(x)\,\,dx+6=a}\int\limits_{1}^{x}{({{x}^{2}}-4x+3)\,\,dx+6}\] |
| \[=a\left( \frac{{{x}^{3}}}{3}-2{{x}^{2}}+3x-\frac{4}{3} \right)+6\] |
| Also \[P(3)=2\,\,so\,\,a=3.\] |
| Hence \[P(x)={{x}^{3}}-6{{x}^{2}}+9x+2.\] |
| Thus, \[\int_{0}^{1}{P(x)\,dx=\frac{1}{4}-2+\frac{9}{2}+2=\frac{19}{4}}\] |
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