Let f: \mathbbR arrow \mathbbR and g: \mathbbR arrow \mathbbR be functions defined by f(x)= beginarrayll x|x| \sin (\frac1x), & x eq 0, 0, & x=0, \endarray \quad and g(x)= \begincases1-2 x, & 0 \leq x \leq \frac12, 0, & otherwise .\endcases. Let a, b, c, d \in \mathbbR . Define the function h: \mathbbR arrow \mathbbR by h(x)=a f(x)+b(g(x)+g(\frac12-x))+c(x-g(x))+d g(x), x \in \mathbbR . Match each entry in List-I to the correct entry in List-II . table width: 100%; border-collapse: collapse; th background-color: #1E90FF; color: white; th, td border: 1px solid lightgrey; padding: 8px; text-align: left; body[data-theme='dark'] th, body[data-theme='dark'] td border: 1px solid white; List-I List-II (P) If a = 0 , b = 1 , c = 0 , and d = 0 , then (1) h is one-one. (Q) If a = 1 , b = 0 , c = 0 , and d = 0 , then (2) h is onto. (R) If a = 0 , b = 0 , c = 1 , and d = 0 , then (3) h is differentiable on \mathbbR . (S) If a = 0 , b = 0 , c = 0 , and d = 1 , then (4) the range of h is [0, 1] . (5) the range of h is \0, 1\ . The correct option is

(P) arrow(5)(Q) arrow(3)(R) arrow(2)(S) arrow(4)

(P) arrow(4)(Q) arrow(3)(R) arrow(1) (S) arrow (2)

(P) arrow(5)(Q) arrow(2)(R) arrow(4) (S) arrow(3)

(P) arrow(4)(Q) arrow(2)(R) arrow(1)(S) arrow(3)

Limits Continuity and Differentiability PYQ — JEE Advanced 2024 · Shift 1 · Q34 with Solution

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be functions defined by

$f(x)=\left\{\begin{array}{ll} x|x| \sin \left(\frac{1}{x}\right), & x eq 0, \\ 0, & x=0, \end{array} \quad \text { and } g(x)= \begin{cases}1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise } .\end{cases}\right.$

Let $a, b, c, d \in \mathbb{R}$ . Define the function $h: \mathbb{R} \rightarrow \mathbb{R}$ by

$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in \mathbb{R} .$

Match each entry in List-I to the correct entry in List-II.

List-I	List-II
(P) If $a = 0$ , $b = 1$ , $c = 0$ , and $d = 0$ , then	(1) $h$ is one-one.
(Q) If $a = 1$ , $b = 0$ , $c = 0$ , and $d = 0$ , then	(2) $h$ is onto.
(R) If $a = 0$ , $b = 0$ , $c = 1$ , and $d = 0$ , then	(3) $h$ is differentiable on $\mathbb{R}$ .
(S) If $a = 0$ , $b = 0$ , $c = 0$ , and $d = 1$ , then	(4) the range of $h$ is $[0, 1]$ .
	(5) the range of $h$ is $\{0, 1\}$ .

The correct option is

A
$(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(1)$ (S) $\rightarrow$ (2)
B
$(\mathrm{P}) \rightarrow(5)(\mathrm{Q}) \rightarrow(2)(\mathrm{R}) \rightarrow(4)$ (S) $\rightarrow(3)$
C
$(\mathrm{P}) \rightarrow(5)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(2)(\mathrm{S}) \rightarrow(4)$
D
$(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(2)(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(3)$

View written solutionFree

Correct answer: C

We analyze each case separately.

1. Given functions

f(x)= \begin{cases} x|x|\sin\left(\frac1x\right), & x\ne 0,\\ 0,& x=0, \end{cases}

and

g(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Also,

h(x)=af(x)+b\left(g(x)+g\left(\frac12-x\right)\right)+c(x-g(x))+dg(x).

We now evaluate each item in List-I.

2. Case (P): $a=0, b=1, c=0, d=0$

Then

h(x)=g(x)+g\left(\frac12-x\right).

Let us compute this carefully.

2.1 Compute $g\left(\frac12-x\right)$

By definition,

g\left(\frac12-x\right)=1-2\left(\frac12-x\right)=2x

whenever

0\le \frac12-x\le \frac12.

This gives

0\le x\le \frac12.

So,

g\left(\frac12-x\right)= \begin{cases} 2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Also,

g(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Hence for $0\le x\le \frac12$ ,

h(x)=(1-2x)+2x=1.

For all other $x$ , both terms are $0$ , so

h(x)=0.

Therefore,

h(x)= \begin{cases} 1, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

So the range is

\{0,1\}.

Thus,

(P)\to (5).

3. Case (Q): $a=1, b=0, c=0, d=0$

Then

h(x)=f(x)= \begin{cases} x|x|\sin\left(\frac1x\right), & x\ne 0,\\ 0,& x=0. \end{cases}

We need to identify the correct property.

3.1 Differentiability of $f$ at $x=0$

For $x\ne 0$ , $f$ is clearly differentiable.

At $x=0$ ,

f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}= \lim_{x\to 0}\frac{x|x|\sin(1/x)}{x} =\lim_{x\to 0}|x|\sin(1/x).

Since

\big||x|\sin(1/x)\big|\le |x|\to 0,

we get

f'(0)=0.

So $f$ is differentiable on all of $\mathbb R$ .

Thus,

(Q)\to (3).

4. Case (R): $a=0, b=0, c=1, d=0$

Then

h(x)=x-g(x).

Now compute piecewise.

For $0\le x\le \frac12$ ,

h(x)=x-(1-2x)=3x-1.

Otherwise, $g(x)=0$ , so

h(x)=x.

Hence

h(x)= \begin{cases} 3x-1, & 0\le x\le \frac12,\\ x, & \text{otherwise}. \end{cases}

We check onto / one-one / range.

4.1 Range on different intervals

For $x<0$ , $h(x)=x$ , so range is $(-\infty,0)$ .
For $0\le x\le \frac12$ , $h(x)=3x-1$ , so range is $[-1,\frac12]$ .
For $x>\frac12$ , $h(x)=x$ , so range is $(\frac12,\infty)$ .

Union gives

(-\infty,0)\cup [-1,\tfrac12]\cup (\tfrac12,\infty)=\mathbb R.

So $h$ is onto.

It is not one-one because, for example,

h\left(\frac13\right)=3\cdot\frac13-1=0,

and

h(0)= -1, \quad h(1)=1,

more directly take

h\left(\frac14\right)=\frac34-1=-\frac14,

and

h\left(-\frac14\right)=-\frac14.

Different inputs give same output, so not one-one.

Thus,

(R)\to (2).

5. Case (S): $a=0, b=0, c=0, d=1$

Then

h(x)=g(x).

h(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

For $x\in[0,\frac12]$ , $1-2x$ takes all values from $1$ to $0$ . Outside this interval, value is $0$ . Hence overall range is

[0,1].

Thus,

(S)\to (4).

6. Final matching

We have obtained:

(P)\to(5),\qquad (Q)\to(3),\qquad (R)\to(2),\qquad (S)\to(4).

This matches Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the stored answer is correct.

More from Limits Continuity and Differentiability

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be functions defined by

Let $a, b, c, d \in \mathbb{R}$ . Define the function $h: \mathbb{R} \rightarrow \mathbb{R}$ by

$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in \mathbb{R} .$

Match each entry in List-I to the correct entry in List-II.

List-I	List-II
(P) If $a = 0$ , $b = 1$ , $c = 0$ , and $d = 0$ , then	(1) $h$ is one-one.
(Q) If $a = 1$ , $b = 0$ , $c = 0$ , and $d = 0$ , then	(2) $h$ is onto.
(R) If $a = 0$ , $b = 0$ , $c = 1$ , and $d = 0$ , then	(3) $h$ is differentiable on $\mathbb{R}$ .
(S) If $a = 0$ , $b = 0$ , $c = 0$ , and $d = 1$ , then	(4) the range of $h$ is $[0, 1]$ .
	(5) the range of $h$ is $\{0, 1\}$ .

The correct option is

A
$(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(1)$ (S) $\rightarrow$ (2)
B
$(\mathrm{P}) \rightarrow(5)(\mathrm{Q}) \rightarrow(2)(\mathrm{R}) \rightarrow(4)$ (S) $\rightarrow(3)$
C
$(\mathrm{P}) \rightarrow(5)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(2)(\mathrm{S}) \rightarrow(4)$
D
$(\mathrm{P}) \rightarrow(4)(\mathrm{Q}) \rightarrow(2)(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(3)$

View written solutionFree

Correct answer: C

We analyze each case separately.

1. Given functions

f(x)= \begin{cases} x|x|\sin\left(\frac1x\right), & x\ne 0,\\ 0,& x=0, \end{cases}

and

g(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Also,

h(x)=af(x)+b\left(g(x)+g\left(\frac12-x\right)\right)+c(x-g(x))+dg(x).

We now evaluate each item in List-I.

2. Case (P): $a=0, b=1, c=0, d=0$

Then

h(x)=g(x)+g\left(\frac12-x\right).

Let us compute this carefully.

2.1 Compute $g\left(\frac12-x\right)$

By definition,

g\left(\frac12-x\right)=1-2\left(\frac12-x\right)=2x

whenever

0\le \frac12-x\le \frac12.

This gives

0\le x\le \frac12.

So,

g\left(\frac12-x\right)= \begin{cases} 2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Also,

g(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

Hence for $0\le x\le \frac12$ ,

h(x)=(1-2x)+2x=1.

For all other $x$ , both terms are $0$ , so

h(x)=0.

Therefore,

h(x)= \begin{cases} 1, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

So the range is

\{0,1\}.

Thus,

(P)\to (5).

3. Case (Q): $a=1, b=0, c=0, d=0$

Then

h(x)=f(x)= \begin{cases} x|x|\sin\left(\frac1x\right), & x\ne 0,\\ 0,& x=0. \end{cases}

We need to identify the correct property.

3.1 Differentiability of $f$ at $x=0$

For $x\ne 0$ , $f$ is clearly differentiable.

At $x=0$ ,

f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}= \lim_{x\to 0}\frac{x|x|\sin(1/x)}{x} =\lim_{x\to 0}|x|\sin(1/x).

Since

\big||x|\sin(1/x)\big|\le |x|\to 0,

we get

f'(0)=0.

So $f$ is differentiable on all of $\mathbb R$ .

Thus,

(Q)\to (3).

4. Case (R): $a=0, b=0, c=1, d=0$

Then

h(x)=x-g(x).

Now compute piecewise.

For $0\le x\le \frac12$ ,

h(x)=x-(1-2x)=3x-1.

Otherwise, $g(x)=0$ , so

h(x)=x.

Hence

h(x)= \begin{cases} 3x-1, & 0\le x\le \frac12,\\ x, & \text{otherwise}. \end{cases}

We check onto / one-one / range.

4.1 Range on different intervals

For $x<0$ , $h(x)=x$ , so range is $(-\infty,0)$ .
For $0\le x\le \frac12$ , $h(x)=3x-1$ , so range is $[-1,\frac12]$ .
For $x>\frac12$ , $h(x)=x$ , so range is $(\frac12,\infty)$ .

Union gives

(-\infty,0)\cup [-1,\tfrac12]\cup (\tfrac12,\infty)=\mathbb R.

So $h$ is onto.

It is not one-one because, for example,

h\left(\frac13\right)=3\cdot\frac13-1=0,

and

h(0)= -1, \quad h(1)=1,

more directly take

h\left(\frac14\right)=\frac34-1=-\frac14,

and

h\left(-\frac14\right)=-\frac14.

Different inputs give same output, so not one-one.

Thus,

(R)\to (2).

5. Case (S): $a=0, b=0, c=0, d=1$

Then

h(x)=g(x).

h(x)= \begin{cases} 1-2x, & 0\le x\le \frac12,\\ 0, & \text{otherwise}. \end{cases}

For $x\in[0,\frac12]$ , $1-2x$ takes all values from $1$ to $0$ . Outside this interval, value is $0$ . Hence overall range is

[0,1].

Thus,

(S)\to (4).

6. Final matching

We have obtained:

(P)\to(5),\qquad (Q)\to(3),\qquad (R)\to(2),\qquad (S)\to(4).

This matches Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the stored answer is correct.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Given functions

2. Case (P): $a=0, b=1, c=0, d=0$

2.1 Compute $g\left(\frac12-x\right)$

3. Case (Q): $a=1, b=0, c=0, d=0$

3.1 Differentiability of $f$ at $x=0$

4. Case (R): $a=0, b=0, c=1, d=0$

4.1 Range on different intervals

5. Case (S): $a=0, b=0, c=0, d=1$

6. Final matching

7. Comparison with stored answer

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Given functions

2. Case (P): $a=0, b=1, c=0, d=0$

2.1 Compute $g\left(\frac12-x\right)$

3. Case (Q): $a=1, b=0, c=0, d=0$

3.1 Differentiability of $f$ at $x=0$

4. Case (R): $a=0, b=0, c=1, d=0$

4.1 Range on different intervals

5. Case (S): $a=0, b=0, c=0, d=1$

6. Final matching

7. Comparison with stored answer

More from Limits Continuity and Differentiability

1. Given functions

2. Case (P): a=0,b=1,c=0,d=0a=0, b=1, c=0, d=0a=0,b=1,c=0,d=0

2.1 Compute g(12−x)g\left(\frac12-x\right)g(21​−x)

3. Case (Q): a=1,b=0,c=0,d=0a=1, b=0, c=0, d=0a=1,b=0,c=0,d=0

3.1 Differentiability of fff at x=0x=0x=0

4. Case (R): a=0,b=0,c=1,d=0a=0, b=0, c=1, d=0a=0,b=0,c=1,d=0

4.1 Range on different intervals

5. Case (S): a=0,b=0,c=0,d=1a=0, b=0, c=0, d=1a=0,b=0,c=0,d=1

6. Final matching

7. Comparison with stored answer

More from Limits Continuity and Differentiability

1. Given functions

2. Case (P): a=0,b=1,c=0,d=0a=0, b=1, c=0, d=0a=0,b=1,c=0,d=0

2.1 Compute g(12−x)g\left(\frac12-x\right)g(21​−x)

3. Case (Q): a=1,b=0,c=0,d=0a=1, b=0, c=0, d=0a=1,b=0,c=0,d=0

3.1 Differentiability of fff at x=0x=0x=0

4. Case (R): a=0,b=0,c=1,d=0a=0, b=0, c=1, d=0a=0,b=0,c=1,d=0

4.1 Range on different intervals

5. Case (S): a=0,b=0,c=0,d=1a=0, b=0, c=0, d=1a=0,b=0,c=0,d=1

6. Final matching

7. Comparison with stored answer

More from Limits Continuity and Differentiability

2. Case (P): $a=0, b=1, c=0, d=0$

2.1 Compute $g\left(\frac12-x\right)$

3. Case (Q): $a=1, b=0, c=0, d=0$

3.1 Differentiability of $f$ at $x=0$

4. Case (R): $a=0, b=0, c=1, d=0$

5. Case (S): $a=0, b=0, c=0, d=1$

2. Case (P): $a=0, b=1, c=0, d=0$

2.1 Compute $g\left(\frac12-x\right)$

3. Case (Q): $a=1, b=0, c=0, d=0$

3.1 Differentiability of $f$ at $x=0$

4. Case (R): $a=0, b=0, c=1, d=0$

5. Case (S): $a=0, b=0, c=0, d=1$