Question

The function $f:[-\frac{1}{2},\frac{1}{2}]\to [-\frac{\pi }{2},\frac{\pi }{2}]$ defined by ${\sin }^{-1}(3x-4x^3)$ is

• A. both one-one onto
• B. onto but not one-one
• C. one-one but not onto
• D. niether one-one nor onto

Answer 1

The correct option is A both one-one onto

$f\left(x\right)={sin}^{-1}(3x-4x^3)=3{sin}^{-1}x$ as $xϵ[-\frac{1}{2},\frac{1}{2}]$
Now $f\left(x\right)={3sin}^{-1}xϵ[-\frac{\pi }{2},\frac{\pi }{2}]\Rightarrow {sin}^{-1}xϵ[-\frac{\pi }{6},\frac{\pi }{6}]xϵ[-sin\frac{\pi }{6},sin\frac{\pi }{6}]\Rightarrow xϵ[-\frac{1}{2},\frac{1}{2}]$
Now let $x_1,x_2ϵ[-\frac{1}{2},\frac{1}{2}]$
Let $f\left(x_1\right)=f\left(x_2\right)\Rightarrow 3x_1-4{x_1}^3=3x_2-4{x_2}^3\Rightarrow x_1-x_2=0\Rightarrow x_1=x_2$
Therefore $f\left(x\right)$ is one one and onto
Hence, option 'A' is correct.