A) -1
B) -2
C) -3
D) -4
Correct Answer: C
Solution :
| [c] We first find the derivatives of \[f\left( x \right)\]at \[x=-1\] and at \[x=0.\] |
| We have, |
| \[f'(-1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(-1+h)-f(-1)}{h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{[2{{(-1+h)}^{2}}+3(-1+h)-5]-[2{{(-1)}^{2}}+3(-1)-5]}{h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{h}^{2}}-h}{h}=\underset{h\to 0}{\mathop{\lim }}\,(2h-1)=2(0)-1=-1\] |
| and \[f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{[2{{(0+h)}^{2}}+3(0+h)-5]-[2{{(0)}^{2}}+3(0)-5]}{h}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{h}^{2}}+3h}{h}=\underset{h\to 0}{\mathop{\lim }}\,(2h+3)=2(0)+3=3\] |
| Clearly, \[f'(0)=-3\,f'(-1)\] |
You need to login to perform this action.
You will be redirected in
3 sec
