Question

$f\left(x\right)=si{n}^{-1}{\left[4-(x-7{)}^3\right]}^{\frac{1}{5}}$ then $f^{-1}\left(x\right)=$ ...

• A. $(7-si{n}^{5}x{)}^{⅓}$
• B. $7+(4-si{n}^{5}x{)}^{⅓}$
• C. $7+(4+si{n}^{5}x{)}^{⅓}$
• D. $(7+si{n}^{5}x{)}^{⅓}$

Answer 1

The correct option is B

$7+(4-si{n}^{5}x{)}^{⅓}$

Explanation for the correct options:

Find the value of $f^{-1}\left(x\right)$;

Let $f\left(x\right)=y$

$\Rightarrow$ $x=f^{-1}\left(y\right)$

Now,

$si{n}^{-1}{\left[4-(x-7{)}^3\right]}^{\frac{1}{5}}=y$

$\Rightarrow$ ${\left[4-(x-7{)}^3\right]}^{\frac{1}{5}}=\sin y$

$\Rightarrow$ $4-(x-7{)}^3={\sin }^{5}y$

$\Rightarrow$ $4–{\sin }^{5}y={(x–7)}^3$

$\Rightarrow$ ${\left(4–{\sin }^{5}y\right)}^{\frac{1}{3}}=(x–7)$

$\Rightarrow$ $x=7+{\left(4–{\sin }^{5}y\right)}^{\frac{1}{3}}$

$\therefore x=f^{-1}y==\mathbf{7}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}{\mathbf{sin}}^{\mathbf{5}}\mathbf{y}\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{3}}}$

Hence, Option ‘B’ is Correct.