JEE Challenger
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MathHardAdv 2014 • Paper 2

Mapping Properties of Piecewise and Composite Functions

Consider functions f1:R→Rf_1: \mathbb{R} \rightarrow \mathbb{R}, f2:[0,∞)→Rf_2: [0,\infty) \rightarrow \mathbb{R}, f3:R→Rf_3: \mathbb{R} \rightarrow \mathbb{R} and f4:R→[0,∞)f_4: \mathbb{R} \rightarrow [0, \infty) given by:

f1(x)={∣x∣if x<0,exif x≥0;f_1(x) = \begin{cases} |x| & \text{if } x < 0, \\ e^x & \text{if } x \ge 0; \end{cases}

f2(x)=x2;f_2(x) = x^2;

f3(x)={sin⁡xif x<0,xif x≥0f_3(x) = \begin{cases} \sin x & \text{if } x < 0, \\ x & \text{if } x \ge 0 \end{cases}

and

f4(x)={f2(f1(x))if x<0,f2(f1(x))−1if x≥0.f_4(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0, \\ f_2(f_1(x)) - 1 & \text{if } x \ge 0. \end{cases}

Match List II with List I:

List IIList IP. f4 is1. onto but not one-oneQ. f3 is2. neither continuous nor one-oneR. f2∘f1 is3. differentiable but not one-oneS. f2 is4. continuous and one-one\begin{array}{ll} \text{List II} & \text{List I} \\ \text{P. } f_4 \text{ is} & \text{1. onto but not one-one} \\ \text{Q. } f_3 \text{ is} & \text{2. neither continuous nor one-one} \\ \text{R. } f_2 \circ f_1 \text{ is} & \text{3. differentiable but not one-one} \\ \text{S. } f_2 \text{ is} & \text{4. continuous and one-one} \end{array}

Options

A

P →\rightarrow 3, Q →\rightarrow 1, R →\rightarrow 4, S →\rightarrow 2

B

P →\rightarrow 1, Q →\rightarrow 3, R →\rightarrow 4, S →\rightarrow 2

C

P →\rightarrow 3, Q →\rightarrow 1, R →\rightarrow 2, S →\rightarrow 4

D

P →\rightarrow 1, Q →\rightarrow 3, R →\rightarrow 2, S →\rightarrow 4

Correct

Topics & Concepts

Relations and Functions (Advanced)FunctionsContinuityDifferentiability

Step-by-Step Solution

To determine the correct matching between List II and List I, we analyze each function step-by-step.

1. Analysis of SS: Function f2(x)f_2(x)

The function f2:[0,∞)→Rf_2: [0,\infty) \rightarrow \mathbb{R} is defined as: f2(x)=x2f_2(x) = x^2

  • Continuity: The polynomial function x2x^2 is continuous on its domain [0,∞)[0, \infty).
  • One-One (Injectivity): For any x1,x2∈[0,∞)x_1, x_2 \in [0, \infty), if f2(x1)=f2(x2)f_2(x_1) = f_2(x_2), then: x12=x22  ⟹  x1=x2(since x1,x2≥0)x_1^2 = x_2^2 \implies x_1 = x_2 \quad (\text{since } x_1, x_2 \ge 0) Thus, f2f_2 is continuous and one-one.

Therefore, S→4\text{S} \rightarrow 4.

2. Analysis of RR: Function (f2∘f1)(x)(f_2 \circ f_1)(x)

First, let us construct the composite function f2(f1(x))f_2(f_1(x)):

  • For x<0x < 0, f1(x)=∣x∣=−x>0f_1(x) = |x| = -x > 0. Since f1(x)∈[0,∞)f_1(x) \in [0, \infty), we can apply f2f_2: f2(f1(x))=(−x)2=x2f_2(f_1(x)) = (-x)^2 = x^2
  • For x≥0x \ge 0, f1(x)=ex≥1>0f_1(x) = e^x \ge 1 > 0. Thus: f2(f1(x))=(ex)2=e2xf_2(f_1(x)) = (e^x)^2 = e^{2x}

Hence,

(f2∘f1)(x)={x2if x<0e2xif x≥0(f_2 \circ f_1)(x) = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} & \text{if } x \ge 0 \end{cases}

  • Continuity at x=0x = 0: lim⁡x→0−(f2∘f1)(x)=lim⁡x→0−x2=0\lim_{x \to 0^-} (f_2 \circ f_1)(x) = \lim_{x \to 0^-} x^2 = 0 lim⁡x→0+(f2∘f1)(x)=lim⁡x→0+e2x=1\lim_{x \to 0^+} (f_2 \circ f_1)(x) = \lim_{x \to 0^+} e^{2x} = 1 Since the left-hand limit ≠\neq right-hand limit, f2∘f1f_2 \circ f_1 is discontinuous at x=0x = 0.

  • One-One (Injectivity): Take x1=−2<0x_1 = -2 < 0, then (f2∘f1)(−2)=(−2)2=4(f_2 \circ f_1)(-2) = (-2)^2 = 4. Take x2=ln⁡2>0x_2 = \ln 2 > 0, then (f2∘f1)(ln⁡2)=e2ln⁡2=4(f_2 \circ f_1)(\ln 2) = e^{2\ln 2} = 4. Since (f2∘f1)(−2)=(f2∘f1)(ln⁡2)=4(f_2 \circ f_1)(-2) = (f_2 \circ f_1)(\ln 2) = 4 for −2≠ln⁡2-2 \neq \ln 2, the function is not one-one.

Thus, f2∘f1f_2 \circ f_1 is neither continuous nor one-one.

Therefore, R→2\text{R} \rightarrow 2.

3. Analysis of QQ: Function f3(x)f_3(x)

The function f3:R→Rf_3: \mathbb{R} \rightarrow \mathbb{R} is defined as:

f3(x)={sin⁡xif x<0xif x≥0f_3(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x & \text{if } x \ge 0 \end{cases}

  • Differentiability:

    • For x<0x < 0, f3(x)=sin⁡xf_3(x) = \sin x is differentiable.
    • For x>0x > 0, f3(x)=xf_3(x) = x is differentiable.
    • At x=0x = 0: LHD=lim⁡h→0−sin⁡h−0h=1\text{LHD} = \lim_{h \to 0^-} \frac{\sin h - 0}{h} = 1 RHD=lim⁡h→0+h−0h=1\text{RHD} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1 Since LHD=RHD=1\text{LHD} = \text{RHD} = 1, f3(x)f_3(x) is differentiable at x=0x = 0. Hence, f3(x)f_3(x) is differentiable on all of R\mathbb{R}.

  • One-One (Injectivity): f3(−π)=sin⁡(−π)=0andf3(0)=0f_3(-\pi) = \sin(-\pi) = 0 \quad \text{and} \quad f_3(0) = 0 Since f3(−π)=f3(0)=0f_3(-\pi) = f_3(0) = 0 for −π≠0-\pi \neq 0, f3f_3 is not one-one.

Thus, f3f_3 is differentiable but not one-one.

Therefore, Q→3\text{Q} \rightarrow 3.

4. Analysis of PP: Function f4(x)f_4(x)

The function f4:R→[0,∞)f_4: \mathbb{R} \rightarrow [0, \infty) is defined as:

f4(x)={f2(f1(x))if x<0f2(f1(x))−1if x≥0={x2if x<0e2x−1if x≥0f_4(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0 \\ f_2(f_1(x)) - 1 & \text{if } x \ge 0 \end{cases} = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} - 1 & \text{if } x \ge 0 \end{cases}

  • Onto (Surjectivity):

    • For x≥0x \ge 0, as xx increases continuously from 00 to ∞\infty, e2x−1e^{2x} - 1 increases continuously from 00 to ∞\infty.
    • Therefore, the range of f4f_4 is [0,∞)[0, \infty), which is equal to the given codomain [0,∞)[0, \infty). Thus, f4f_4 is onto.

  • One-One (Injectivity): Take x1=−1<0  ⟹  f4(−1)=(−1)2=1x_1 = -1 < 0 \implies f_4(-1) = (-1)^2 = 1. Take x2=ln⁡22>0  ⟹  f4(ln⁡22)=eln⁡2−1=2−1=1x_2 = \frac{\ln 2}{2} > 0 \implies f_4\left(\frac{\ln 2}{2}\right) = e^{\ln 2} - 1 = 2 - 1 = 1. Since f4(−1)=f4(ln⁡22)=1f_4(-1) = f_4\left(\frac{\ln 2}{2}\right) = 1 with −1≠ln⁡22-1 \neq \frac{\ln 2}{2}, f4f_4 is not one-one.

Thus, f4f_4 is onto but not one-one.

Therefore, P→1\text{P} \rightarrow 1.

Conclusion:

The correct mapping is: P→1,Q→3,R→2,S→4\text{P} \rightarrow 1, \quad \text{Q} \rightarrow 3, \quad \text{R} \rightarrow 2, \quad \text{S} \rightarrow 4

This corresponds to Option D.

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