Question

Let $f\left(x\right)=tanx-ta{n}^{3}x+ta{n}^{5}x-ta{n}^{7}x+\cdots \infty$ , where $x\in (0,\frac{\pi }{4})$,

then which of the following is/are correct ?

• A. ${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx=\frac{1}{8}$
• B. $f^{\prime }\left(\frac{\pi }{12}\right)=\frac{1}{2}$
• C. $\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)}{x}=1$
• D. $f\left(x\right)$ is an odd function

Answer 1

Answer: (A), (C), (D)

The correct options are
A ${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx=\frac{1}{8}$
C $\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)}{x}=1$
D $f\left(x\right)$ is an odd function

$f\left(x\right)=tanx-{tan}^{3}x+{tan}^{5}x-{tan}^{7}x+....=\frac{tanx}{1+{tan}^{2}x}=\frac{sin\left(2x\right)}{2}limx\to 0^{+}f\left(x\right)/x=\frac{sin\left(2x\right)}{2x}=1sin\left(x\right)isanoddfunctionsin(-x)=-sin\left(x\right){\int }_0^{\frac{\pi }{6}}\frac{sin\left(2x\right)}{2}=\frac{1}{8}$
Hence, options 'A', 'C' and 'D' are correct.