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Functions question

2020 · Shift 1 · Q20
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Functions question

2020 · Shift 1 · Q20

JEE AdvancedMathematicsFunctionsMCQ+3 / −1
If the function f : R →\to→ R is defined by f(x) = |x| (x −-− sin x), then which of the following statements is TRUE?
  1. A
    f is one-one, but NOT onto
  2. B
    f is onto, but NOT one-one
  3. C
    f is BOTH one-one and onto
  4. D
    f is NEITHER one-one NOR onto
View written solutionFree

Correct answer: C

The user wants me to analyze the function f: R -> R defined by f(x) = |x|(x - sin x) and determine if it is one-one (injective) and/or onto (surjective).

1. Analyze Injectivity (One-one)

A function is one-one if it is strictly monotonic. We can determine this by examining its derivative, f'(x).

First, let's write the function in a piecewise form: f(x) = { x(x - sin x), if x >= 0; -x(x - sin x), if x < 0 } f(x)=x2−xsinx,ifx>=0;−x2+xsinx,ifx<0f(x) = { x^2 - x sin x, if x >= 0; -x^2 + x sin x, if x < 0 }f(x)=x2−xsinx,ifx>=0;−x2+xsinx,ifx<0

Now, let's find the derivative f'(x).

Case 1: x > 0 f(x)=x2−xsinxf(x) = x^2 - x sin xf(x)=x2−xsinx f′(x)=d/dx(x2−xsinx)=2x−(1∗sinx+x∗cosx)=2x−sinx−xcosxf'(x) = d/dx(x^2 - x sin x) = 2x - (1 * sin x + x * cos x) = 2x - sin x - x cos xf′(x)=d/dx(x2−xsinx)=2x−(1∗sinx+x∗cosx)=2x−sinx−xcosx To determine the sign of f'(x), we can regroup the terms: f'(x) = (x - sin x) + (x - x cos x) f'(x) = (x - sin x) + x(1 - cos x) For x > 0, we know that:

  • x > sin x, so (x - sin x) > 0.
  • x > 0 and cos x <= 1, so (1 - cos x) >= 0. Thus, x(1 - cos x) >= 0. The term x(1 - cos x) is zero only when x = 2kπ for k being a positive integer. Since f'(x) is the sum of a strictly positive term (x - sin x) and a non-negative term x(1 - cos x), f'(x) is strictly positive for x > 0. So, f'(x) > 0 for x > 0.

Case 2: x < 0 f(x)=−x2+xsinxf(x) = -x^2 + x sin xf(x)=−x2+xsinx f′(x)=d/dx(−x2+xsinx)=−2x+(1∗sinx+x∗cosx)=−2x+sinx+xcosxf'(x) = d/dx(-x^2 + x sin x) = -2x + (1 * sin x + x * cos x) = -2x + sin x + x cos xf′(x)=d/dx(−x2+xsinx)=−2x+(1∗sinx+x∗cosx)=−2x+sinx+xcosx To analyze the sign, let x = -y where y > 0. f'(-y) = -2(-y) + sin(-y) + (-y)cos(-y) f'(-y) = 2y - sin y - y cos y This is the same expression as f'(y) for y > 0. Since we have already shown that f'(y) > 0 for y > 0, it follows that f'(x) > 0 for x < 0.

Case 3: x = 0 We check for the derivative at x = 0 using the definition. f(0) = 0 * (0 - sin 0) = 0. Right-Hand Derivative (RHD): limh−>0+(f(h)−f(0))/h=limh−>0+(h2−hsinh)/h=limh−>0+(h−sinh)=0lim_{h->0+} (f(h) - f(0))/h = lim_{h->0+} (h^2 - h sin h)/h = lim_{h->0+} (h - sin h) = 0limh−>0+​(f(h)−f(0))/h=limh−>0+​(h2−hsinh)/h=limh−>0+​(h−sinh)=0 Left-Hand Derivative (LHD): limh−>0−(f(h)−f(0))/h=limh−>0−(−h2+hsinh)/h=limh−>0−(−h+sinh)=0lim_{h->0-} (f(h) - f(0))/h = lim_{h->0-} (-h^2 + h sin h)/h = lim_{h->0-} (-h + sin h) = 0limh−>0−​(f(h)−f(0))/h=limh−>0−​(−h2+hsinh)/h=limh−>0−​(−h+sinh)=0 Since RHD = LHD = 0, f'(0) = 0.

Conclusion for Injectivity: We have f'(x) > 0 for x != 0 and f'(0) = 0. Since f'(x) >= 0 for all x in R and f'(x) = 0 only at a single point, the function f(x) is strictly increasing on R. A strictly increasing function is always one-one. Thus, f is one-one.

2. Analyze Surjectivity (Onto)

A function f: R -> R is onto if its range is equal to its codomain, R.

First, we note that f(x) is continuous for all x in R, as it is formed by the product and composition of continuous functions (|x|, x, sin x).

Next, we find the limits of f(x) as x approaches +infinity and -infinity.

As x -> +infinity: f(x)=x(x−sinx)=x2−xsinx=x2(1−sin(x)/x)f(x) = x(x - sin x) = x^2 - x sin x = x^2(1 - sin(x)/x)f(x)=x(x−sinx)=x2−xsinx=x2(1−sin(x)/x) limx−>+infinityf(x)=limx−>+infinityx2(1−sin(x)/x)lim_{x->+infinity} f(x) = lim_{x->+infinity} x^2(1 - sin(x)/x)limx−>+infinity​f(x)=limx−>+infinity​x2(1−sin(x)/x) Since -1 <= sin x <= 1, limx−>+infinitysin(x)/x=0lim_{x->+infinity} sin(x)/x = 0limx−>+infinity​sin(x)/x=0. So, limx−>+infinityf(x)=(+infinity)∗(1−0)=+infinitylim_{x->+infinity} f(x) = (+infinity) * (1 - 0) = +infinitylimx−>+infinity​f(x)=(+infinity)∗(1−0)=+infinity.

As x -> -infinity: f(x)=−x(x−sinx)=−x2+xsinx=−x2(1−sin(x)/x)f(x) = -x(x - sin x) = -x^2 + x sin x = -x^2(1 - sin(x)/x)f(x)=−x(x−sinx)=−x2+xsinx=−x2(1−sin(x)/x) limx−>−infinityf(x)=limx−>−infinity−x2(1−sin(x)/x)lim_{x->-infinity} f(x) = lim_{x->-infinity} -x^2(1 - sin(x)/x)limx−>−infinity​f(x)=limx−>−infinity​−x2(1−sin(x)/x) Again, limx−>−infinitysin(x)/x=0lim_{x->-infinity} sin(x)/x = 0limx−>−infinity​sin(x)/x=0. So, limx−>−infinityf(x)=−(infinity)∗(1−0)=−infinitylim_{x->-infinity} f(x) = - (infinity) * (1 - 0) = -infinitylimx−>−infinity​f(x)=−(infinity)∗(1−0)=−infinity.

Conclusion for Surjectivity: Since f(x) is a continuous function on R, and limx−>−infinityf(x)=−infinitylim_{x->-infinity} f(x) = -infinitylimx−>−infinity​f(x)=−infinity and limx−>+infinityf(x)=+infinitylim_{x->+infinity} f(x) = +infinitylimx−>+infinity​f(x)=+infinity, by the Intermediate Value Theorem, the function f(x) must take on every real value. Therefore, the range of f(x) is (-infinity, +infinity), which is R. Since the range R is equal to the codomain R, the function f is onto.

Final Conclusion: The function f(x) is both one-one and onto. This corresponds to option C.

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