A) \[a=4\]
B) \[6=-\,6\]
C) \[c=2\]
D) \[a=3\]
Correct Answer: B
Solution :
| \[f(x)=a{{e}^{2x}}+b{{e}^{x}}+cx\] | ||
| Since, \[f(0)=a+b\] | ||
| \[i.e.,a\,+b=-\,1\] | ??..(i) | |
| \[f'(x)=2a{{e}^{2x}}+b{{e}^{x}}+c\] | ||
| \[\therefore f'\,(log2)=2a{{e}^{2\log 2}}+b{{e}^{\log 2}}+c\] | ||
| \[=8a+2b+c=31\] | ??.(ii) | |
| \[\int\limits_{0}^{\ell n4}{(a{{e}^{2x}}+{{e}^{x}}+cx-cx)dx}\] | ||
| \[\int\limits_{0}^{\ell n4}{(a{{e}^{2x}}+b{{e}^{x}})dx}\] | |
| \[\left( \frac{a{{e}^{2x}}}{2}+b{{e}^{x}} \right)_{0}^{\ell n4}\] | |
| =\[\frac{a{{e}^{2\ell n4}}}{2}+b{{e}^{\ell n4}}-\frac{a}{2}-b\]=\[8a+4b-\frac{a}{2}-b=\frac{15a}{2}+3b\]=\[\frac{39}{2}\] | |
| \[i.e.,15a+6b=39\] | ??.(iii) |
| from equation (i), (ii) & (iii) | |
| 9a = 45 | |
| \[\therefore a=5,\]\[b=-6\] and \[c=3\] | |
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