Question

Let $\left\{ \begin{matrix} 2x\tan x–\frac{\pi}{\cos x}; & x \neq \frac{\pi}{2} \\ k & þ \end{matrix} = \frac{\pi}{2} \right.\$ be a function. Define $\frac{\pi}{2}$ by

$\frac{1}{2}$ for all x. Then g is

• A. Onto if f is onto
• B. One-one if f is one-one
• C. Continuous if f is continuous
• D. Differentiable if f is differentiable

Answer 1

Answer: (C)

Continuous if f is continuous

Answer 2

$g ( x ) = | f ( x ) | \geq 0$. So $g ( x )$ cannot be onto. If $f ( x )$ is one-one and $f \left( x _ { 1 } \right) = - f \left( x _ { 2 } \right)$ then $g \left( x _ { 1 } \right) = g \left( x _ { 2 } \right)$. So, ‘$f ( x )$ is one-one’ does not ensure that $g ( x )$ is one-one.

If $f ( x )$ is continuous for $x \in R , | f ( x ) |$ is also continuous for $x \in R$. This is obvious from the following graphical consideration.

So the answer (3) is correct. The fourth answer (4) is not correct from the above graphs $y = f ( x )$ is differentiable at P while $y = | f ( x ) |$ has two tangents at P, i.e. not differentiable at P.