Let $f_{1}: \mathbb{R} \rightarrow \mathbb{R}, f_{2}:[0, \infty) \rightarrow \mathbb{R}, f_{3}: \mathbb{R} \rightarrow \mathbb{R}$ and $f_{4}: \mathbb{R} \rightarrow[0, \infty)$ be defined by \[ \begin{aligned} & f_{1}(x)=\begin{cases}|x| & \text { if } x<0, \\ e^{x} & \text { if } x \geq 0 ;\end{cases} \\ & f_{2}(x)=x^{2} ; \\ & f_{3}(x)=\begin{cases}\sin x & \text { if } x<0, \\ x & \text { if } x \geq 0\end{cases} \end{aligned} \] and \[ f_{4}(x)= \begin{cases}f_{2}\left(f_{1}(x)\right) & \text { if } x<0, \\ f_{2}\left(f_{1}(x)\right)-1 & \text { if } x \geq 0 .\end{cases} \] | List I | | List II | | --- | --- | --- | | P. | $f_{4}$ is | 1. onto but not one-one | | Q. | $f_{3}$ is | 2. neither continuous nor one-one | | R. | $f_{2} \circ f_{1}$ is | 3. differentiable but not one-one | | S. | $f_{2}$ is | 4. continuous and one-one | P Q R S

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Continuity and Differentiability

Let

f_{1} : R \to R, f_{2} : [0, \infty) \to R, f_{3} : R \to R

and

f_{4} : R \to [0, \infty)

be defined by

f_{1} (x) = {∣ x ∣ e^{x} if x < 0, if x \geq 0; f_{2} (x) = x^{2}; f_{3} (x) = {sin x x if x < 0, if x \geq 0

and

f_{4} (x) = {f_{2} (f_{1} (x)) f_{2} (f_{1} (x)) - 1 if x < 0, if x \geq 0.

List I		List II
P.	$f_{4}$ is	1. onto but not one-one
Q.	$f_{3}$ is	2. neither continuous nor one-one
R.	$f_{2} \circ f_{1}$ is	3. differentiable but not one-one
S.	$f_{2}$ is	4. continuous and one-one

P Q R S

Answer: (D)

P → 1 (onto but not one-one):

f_{4} (x) = {x^{2} e^{2 x} - 1 x < 0 x \geq 0

. For

x < 0

, range is

(0, \infty)

; for

x \geq 0

, range is

[0, \infty)

. Combined range is

[0, \infty)

, so onto. Not one-one since

f_{4} (- 1) = f_{4} (1) = e^{2} - 1

(approximately).

Q → 3 (differentiable but not one-one):

f_{3} (x) = {sin x x x < 0 x \geq 0

. Continuous and differentiable everywhere. Not one-one since

sin x

oscillates for

x < 0

.

R → 2 (neither continuous nor one-one):

f_{2} \circ f_{1} (x) = {x^{2} e^{2 x} x < 0 x \geq 0

. At

x = 0

: left limit = 0, right limit = 1, so not continuous. Not one-one.

S → 4 (continuous and one-one): On domain

[0, \infty)

f_{2} (x) = x^{2}

is continuous and strictly increasing, hence one-one.

Answer: P→1, Q→3, R→2, S→4, which is 1324.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option