Jee advanced (2020) Maths
Answers
Answer
f is both one - one and onto
Step-by-step explanation:
F(x) = |x| (x-sinx)
As we know that for onto function co -domain must be equal to range . and Range of f(x) is R
when x >0
- F(x). = x (x-sinx)
- F(x) = x² - xsinx
When x < 0
- f(x ) = -x (x-sinx )
We have to find f is odd or even
∴ f(-x) = |-x| {-x-(sin(-x) }
f(-x) = - |x|{x-sinx}
f(-x) = - f(x) hence it is odd function i..e symmetric about origine.
Now f (-∞ )
= lim x -> -∞ (-x)² (1-sinx/x) = -∞
f(∞) = limx->∞ (x²) (1- sinx /x ) = ∞
Hence , It is clear that range of the function is real no.
i.e , codomain = range
∴ it is onto function.
f'(x) = {-2x + sinx + x cosx , x < 0
= {2x - sinx - x cosx , x ≥ 0
for (0, ∞)
- f'(x) = 2x - sinx - x cosx
- = (x - sinx) + x(1 - cosx)
we know, x > sinx for all x belongs to R
and 1 - cosx ≥ 0 for all x belongs to R
- so, f'(x) > 0 in (0, ∞)
- at x = 0, f'(x) = 0
- so, f'(x) ≥ 0, in [0, ∞)
also for (-∞, 0)
- f'(x) = -(x - sinx) - x(1 - cosx)
- f'(x) > 0 for all (-∞, 0)
it is clear that f is always increasing. so it can't be many one function. Therefore f is one one function