#### To determine

**To sketch:**

the graph of the function y=ln(tan2x)

#### Answer

See the graph given below.

#### Explanation

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**Steps in Curve tracing**

**A. Domain:** {x|x≠kπ,where k is an integer}

Since “ln 0” is not defined, tan2x≠0

If tan2x=0⇒tanx=0⇒x=kπ

Therefore, the domain is {x|x≠kπ,where k is an integer}

**B. Intercepts:**

x Intercept is obtained by taking y=0

Therefore

ln(tan2x)=0⇒tan2x=1⇒tanx=±1⇒x=π4(2k+1) where k is an integer

x Intercept is x=π4(2k+1)

No y intercept is possible as x can’t be zero

**C. Symmetry:**

Since f(−x)=f(x), the curve is symmetrical about y axis

**D. Asymptotes:**

There are no horizontal asymptotes and vertical asymptotes are given by x=kπ

**E. Interval of increase and decrease**

y=ln(tan2x)=2ln(tanx)y′=2sec2xtanxy′=2(1+tan2x)tanx=2(cotx+tanx)

Since tanx is periodic with period π, we sketch the graph in the region −π≤x≤π

The function is increasing when tanx is positive, therefore f(x) is increasing in (−π,−π2)∪ (0,π2) and it is decreasing in (−π2,0)∪ (π2,π)

**F. Local maxima and minima**

Maxima and minima are obtained by the first derivative test,

i.e., y′=2sec2xtanx=0⇒secx=0⇒x=kπ

But x=kπ is not included in the domain, Hence there are no local maxima and minima

**G. Concavity and inflexion points**

y′=2sec2xtanx=2(1+tan2x)tanx=2(cotx+tanx)y″=2(−csc2x+sec2x)y″=0⇒2(−csc2x+sec2x)=0⇒secx=±cscx⇒cosx=±sinx⇒x=π4(2k+1)

x=π4(2k+1) are the inflexion points

H. Using this information we sketch the curve

**Final statement:**

We could sketch the graph of the given function.