$R a n g e o f f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)} i s w h e r e [x] d e n o t e s t h e g r e a t e s t i n t e g e r f u n c t i o n)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

$f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)} = {0} b e c a u s e o f [x^{2} - x] i s i n t e g e r .$

Suggest Corrections

Similar questions

f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)}

[x]

f (x) = \frac{tan π [x]}{1 + sin π [x] + [x^{2}]}

f (x) = {\begin{matrix} \frac{tan [x] - [x] tan 1}{x} & ; x \neq 0 0 & ; x = 0 \end{matrix}

f^{'} (0^{-})

[

[x]

x]

[{cot}^{- 1} x] + 2 [{tan}^{- 1} x] = 0

[.]

f (x) = \frac{tan (π [x^{2} - x])}{1 + sin (cos x)}

Join BYJU'S Learning Program