A) \[1\]
B) \[-1\]
C) \[2\]
D) \[- 2\]
Correct Answer: B
Solution :
\[f(x)=2{{x}^{3}}+a{{x}^{2}}+bx\] \[let,a=-1,b=1\] Given that f(x) satisfy Roll theorem in interval [?1, 1] f(x) must satisfy two conditions. (1)\[f(a)=f(b)\] (2)\[f'(a)=f(b)\](c should be between a and b) \[f(a)=f(1)=2{{(1)}^{3}}+a{{(1)}^{2}}+b(1)=2+a+b\] \[f(b)=f(-1)=2{{(-1)}^{3}}+a{{(-1)}^{2}}+b(-1)\] \[=-2+a-b\] \[f(a)=f(b)\] \[2+a+b=-2+a-b\] \[2b=-4\] \[b=-2\] (given that\[c=\frac{1}{2}\]) \[f'(x)=6{{x}^{2}}+2ax+b\] at\[x=\frac{1}{2},f'(x)=0\] \[0=6{{\left( \frac{1}{2} \right)}^{2}}+2a\left( \frac{1}{2} \right)+b\] \[\frac{3}{2}+a+b=0\] \[\frac{3}{2}+a-2=0\] \[a=2-\frac{3}{2}=\frac{1}{2}\] \[2a+b=2\times \frac{1}{2}-2=1-2=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
