Question

Jee advanced 2014 maths ques....
help...
multiple correct​

Answer 1

Step-by-step explanation: Answer (2) &(3):

Given function is f(x)=(log(sec(x)+tan(x)))³

Now,

We know, if f(-x)= -f(x) then the function is odd functuin & if f(-x)= f(x) then the function is even

so,

$f( - x) = (log( \sec( - x) + \tan( - x) ) )^{3} = > f( - x) =( log( - \sec(x) - \tan(x) ) ^{3}$

$= > f( - x) = ( log - ( \sec(x) + \tan(x) ) )^{3}$

So this implies that f(x) is neither even nor odd.

We know that, a function is called onto if its range= co-domain

Now, range of f(x) will be

$- \frac{ \pi}{2} \leqslant x \leqslant \frac{\pi}{2}$

$- \frac{\pi}{4} \leqslant \frac{x}{2} \leqslant \frac{\pi}{4}$

$- \frac{ \pi}{4} + \frac{\pi}{4} \leqslant \frac{x}{2} + \frac{\pi}{4} \leqslant \frac{\pi}{4} + \frac{\pi}{4}$

$0 \leqslant \frac{x}{2} + \frac{\pi}{4} \leqslant \frac{\pi}{2}$

Taking tan in both sides,

$\tan(0) \leqslant \tan( \frac{\pi}{4} + \frac{x}{2} ) \leqslant \tan( \frac{\pi}{2} )$

$0 \leqslant \tan( \frac{\pi}{4} + \frac{x}{2} ) \leqslant \infty$

Taking log both sides,

$log(0) \leqslant log( \tan( \frac{\pi}{4} + \frac{x}{2} ) ) \leqslant log( \infty )$

We know that, log(0)= -infinity and log(infinity)=infinity

So,

$- \infty \leqslant log( \tan( \frac{\pi}{4} + \frac{x}{2} ) ) \leqslant \infty$

$( - \infty )^{3} \leqslant ( log( \tan( \frac{\pi}{4} + \frac{x}{2} )) )^{3} \leqslant { \infty }^{3}$

We know that, tan(π/4+x/2)=sec(x)+ tan(x)

So,

$- \infty \leqslant ( log( \sec(x) + \tan(x) ))^{3} \leqslant \infty$

So f(x) is onto function

Also,

derivative of f(x) is always positive, i.e, f(x) is strictly increasing function.

Hence both option are correct