MCQ
If $|x| > 1$, then ${(1 + x)^{ - 2}}$ =
- A$1 - 2x + 3{x^2} - ....$
- B$1 + 2x + 3{x^2} + $....
- C$1 - \frac{2}{x} + \frac{3}{{{x^2}}} - ....$
- ✓$\frac{1}{{{x^2}}} - \frac{2}{{{x^3}}} + \frac{3}{{{x^4}}} - $...
So given expression can be written as
${x^{ - 2}}{\left( {1 + \frac{1}{x}} \right)^{ - 2}} = {x^{ - 2}}\left[ {1 - \frac{2}{x} + \frac{3}{{{x^2}}} - \frac{4}{{{x^3}}} + ....} \right]$
$ = \left[ {\frac{1}{{{x^2}}} - \frac{2}{{{x^3}}} + \frac{3}{{{x^4}}} - \frac{4}{{{x^5}}} + ....} \right]$
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$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4) = 0$
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$
$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.
$(v)$ the signs of $f '(x)$ is given below
On the possible graph of $y = f (x)$ we have 