MCQ
જો $f(x) = 4$ અને $f'(x) = 4$, તો $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}} = . . . $
- A$2$
- B$-2$
- ✓$-4$
- D$3$
==> $y = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(2) + 2f(2) - 2f(x)}}{{x - 2}}$
==> $y = \mathop {\lim }\limits_{x \to 2} \frac{{ - 2f(x) + 2f(2) + xf(2) - 2f(2)}}{{(x - 2)}}$
==> $y = \mathop {\lim }\limits_{x \to 2} - 2\frac{{[f(x) - f(2)]}}{{x - 2}} + \mathop {\lim }\limits_{x \to 2} \frac{{f(2).(x - 2)}}{{(x - 2)}}$
==> $y = - 2\mathop {\lim }\limits_{x \to 2} \frac{{f(x) - f(2)}}{{x - 2}} + f(2)$
==> $y = - 2\,\,\mathop {\lim }\limits_{x \to 2} f'(x) + f(2) = - \,8 + 4 = - \,4$.
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