CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$x = n \pi- tan^{-1} 3$$ is a solution of the equation $$12 tan  2x + \displaystyle \frac{\sqrt{10}}{cos  x} + 1 = 0$$ then 


A
n is any integer
loader
B
n is an even integer
loader
C
n is a positive integer
loader
D
n is an odd integer
loader

Solution

The correct option is C n is an odd integer
Let n is even
Then for $$x=n\pi -tan^{ -1 }3$$

$$12tan2x+\cfrac { \sqrt { 10 }  }{ cosx } +1$$ 

$$ =12tan\left( 2n\pi -2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( n\pi -tan^{ -1 }3 \right) +1$$

$$ =-12tan\left( 2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( tan^{ -1 }3 \right) +1$$ 

$$ =-12tan\left( \pi +tan^{ -1 }\cfrac { 2\times 3 }{ 1-{ 3 }^{ 2 } }  \right) +\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 1+{ 3 }^{ 2 } }  \right)  \right) +1$$ 

$$ =-12tan\left( \pi +tan^{ -1 }\left( -\cfrac { 3 }{ 4 }  \right)  \right) +\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 10 }  \right)  \right) +1$$ 

$$ =-12tantan^{ -1 }\left( -\cfrac { 3 }{ 4 }  \right) +10+1=12\times \cfrac { 3 }{ 4 } +10+1=20$$

And for odd n

$$12tan2x+\cfrac { \sqrt { 10 }  }{ cosx } +1$$

$$ =12tan\left( 2n\pi -2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( n\pi -tan^{ -1 }3 \right) +1$$

$$ =-12tantan^{ -1 }\left( -\cfrac { 3 }{ 4 }  \right) -\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 10 }  \right)  \right) +1\\ =9-10+1=0$$

Hence n is odd

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image