Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

If the function f : R → R is defined by f(x) = |x|(x - sinx), then which of the following statements is TRUE ?

solution : f is modulus function so let's break it.

f(x) = -x² + x sinx , x < 0

= x² - x sinx , x ≥ 0

here it is clear that function is continuous for all real value of x.

and f(x) = |x|(x - sinx)

f(-x) = |x|(-x + sinx) = -f(x)

i.e., f is also odd function. so the function is symmetrical about origin.

now $f(-\infty)=\displaystyle\lim_{x\to-\infty}(-x^2)\left(1-\frac{sinx}{x}\right)=-\infty$

$f(\infty)=\displaystyle\lim_{x\to\infty}x^2\left(1-\frac{sinx}{x}\right)=\infty$

it is clear that Range of function is all real value.

i.e., co - domain = range

so f is onto function.

now differentiating f with respect to x,

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.

so the correct option is (C) f is both one and onto