A) \[2a\,\sin \,a\]
B) \[{{a}^{2}}\cos \,a\]
C) \[{{a}^{2}}\cos a+2a\,\,\sin \,a\]
D) none of these
Correct Answer: C
Solution :
Here, \[\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{2}}\{\sin (a+h)-\sin \,a\}}{h}\] \[+\,\frac{h\{2a\,\sin (a+h)+h\sin (a+h)\}}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{2}}.2\cos \left[ a+\frac{h}{2} \right].\sin \frac{h}{2}}{2.\frac{h}{2}}\] \[+\,(2a+h)\sin (a+h)\] \[={{a}^{2}}\cos \,a+2a\sin a.\]
You need to login to perform this action.
You will be redirected in
3 sec
