A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\frac{\tan (x-2)\{{{x}^{2}}+(k-2)x-2k\}}{{{x}^{2}}-4x+4}=5\] \[\Rightarrow \]\[\underset{x\to 2}{\mathop{\lim }}\,\frac{\tan (x-2)\{{{x}^{2}}+kx-2x-2k\}}{{{(x-2)}^{2}}}=5\] \[\Rightarrow \]\[\underset{x\to 2}{\mathop{\lim }}\,\frac{\tan (x-2)\{x(x-2)+k(x-2)\}}{(x-2)\times (x-2)}=5\] \[\Rightarrow \]\[\underset{x\to 2}{\mathop{\lim }}\,\left( \frac{\tan (x-2)}{(x-2)} \right)\times \underset{x\to 2}{\mathop{\lim }}\,\left( \frac{(k+2)\cancel{(x-2)}}{\cancel{(x-2)}} \right)=5\] \[\Rightarrow \]\[1\times \underset{x\to 2}{\mathop{\lim }}\,(k+x)=5\] \[\left\{ \because \underset{h\to 0}{\mathop{\lim }}\,\frac{\tanh }{h}=1 \right\}\]or\[k+2=5\]\[\Rightarrow \]\[\]
You need to login to perform this action.
You will be redirected in
3 sec
