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1 Multiple Select

Consider the following pseudocode function, which takes three non-negative integers

p_{\text{init}},\ q_{\text{init}},\ n

as input. ```pseudocode function calculate(p_init, q_init, n) { p = p_init q = q_init for i from 1 to n { if (p > q) { p = p - q q = q + i } else { q = q - p p = p + i } } return p + q } ```` Which of the following statements is/are correct?

A

The function call `calculate(10, 5, 3)` returns

10.

B

For any non-negative integers

p_{\text{init}},\ q_{\text{init}},\ n,

the function calls `calculate(p_init, q_init, n)` and `calculate(q_init, p_init, n)` return the same value.

C

The function call `calculate(7, 7, 2)` returns

12.

D

For any non-negative integers

p_{\text{init}}

and

q_{\text{init}},

the function call `calculate(p_init, q_init, 0)` returns

0.

View Solution

Let us check each statement. For `calculate(10, 5, 3)`: * After

i=1,

(p,q)=(10,5)\to(5,6).

* After

i=2,

(p,q)=(5,6)\to(7,1).

* After

i=3,

(p,q)=(7,1)\to(6,4).

Hence the return value is

p+q=6+4=10,

so statement 1 is true. Now consider one iteration starting from values

a

and

b

at step

i.

* If

a>b,

the next pair is

(a-b,\ b+i).

* If

a\le b,

the next pair is

(a+i,\ b-a).

In either case, the unordered pair of next values is

{|a-b|,\ \min(a,b)+i},

which depends only on the unordered pair

{a,b},

not on their order. Therefore, if we start with

(p_{\text{init}},q_{\text{init}})

or with

(q_{\text{init}},p_{\text{init}}),

then after every iteration we get the same two values, possibly in different order. Hence the final sum

p+q

is the same in both cases. So statement 2 is true. For `calculate(7, 7, 2)`: * After

i=1,

(p,q)=(7,7)\to(8,0).

* After

i=2,

(p,q)=(8,0)\to(8,2).

Thus the return value is

p+q=8+2=10,

not

12.

So statement 3 is false. If

n=0,

the loop does not execute at all, so the function returns

p_{\text{init}}+q_{\text{init}},

which is not always

0.

Therefore statement 4 is false. Hence the correct statements are:

1 \text{ and } 2.

2 Multiple Select

Consider the function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=x|x|.

Which of the following statements is/are true?

A

f(x)

is continuous at

x=0.

B

f(x)

is differentiable at

x=0.

C

The derivative

f'(x)

is continuous at

x=0.

D

The second derivative

f''(0)

exists.

View Solution

Write

f(x)= \begin{cases}x^2, & x\ge 0,\\ -x^2, & x<0.\end{cases}

First, at

x=0,

\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}x^2=0,

and

\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(-x^2)=0.

Also,

f(0)=0.

Hence

f

is continuous at

x=0,

so statement 1 is true. Now,

f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}.

For

h>0,

\frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h\to 0,

and for

h<0,

\frac{f(h)-f(0)}{h}=\frac{-h^2}{h}=-h\to 0.

So

f'(0)=0,

and statement 2 is true. For

x>0,

f'(x)=2x,

while for

x<0,

f'(x)=-2x.

Thus

f'(x)= \begin{cases}2x, & x>0,\\ 0, & x=0,\\ -2x, & x<0,\end{cases} =2|x|.

Since

\lim_{x\to 0}f'(x)=\lim_{x\to 0}2|x|=0=f'(0),

the derivative

f'(x)

is continuous at

x=0.

Hence statement 3 is true. Finally, to check

f''(0),

compute

f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h}.

For

h>0,

\frac{f'(h)-f'(0)}{h}=\frac{2h-0}{h}=2,

while for

h<0,

\frac{f'(h)-f'(0)}{h}=\frac{-2h-0}{h}=-2.

The two one-sided limits are different, so

f''(0)

does not exist. Therefore statement 4 is false. Hence the correct statements are:

1,\ 2,\ 3.

3 Multiple Select

Let

A

be an

n\times n

matrix with real entries, where

n\ge 3

and

\det(A)=5.

Which of the following statements is/are true?

A

Let

B

be the matrix obtained from

A

by multiplying its first row by

2

and its second column by

3

. Then

\det(B)=30.

B

Let

C

be the matrix obtained from

A

by swapping its first and third rows, and then replacing its second column by

C_2\to C_2-2C_1.

Then

\det(C)=-5.

C

Let

D

be the matrix obtained from

A

by multiplying every entry of

A

by

-1

. If

n

is even, then

\det(D)=5.

D

Let

E

be the matrix obtained from

A

by sequentially applying the row operations

R_1\to R_1+R_2,\qquad R_2\to R_2-R_1.

Then

\det(E)=5.

View Solution

We use standard determinant properties. For statement 1, multiplying one row by

2

multiplies the determinant by

2

, and multiplying one column by

3

multiplies the determinant by

3

. Therefore

\det(B)=2\cdot 3\cdot \det(A)=6\cdot 5=30.

So statement 1 is true. For statement 2, swapping the first and third rows changes the sign of the determinant, so it becomes

-\,\det(A)=-5.

Then the column operation

C_2\to C_2-2C_1

does not change the determinant. Hence

\det(C)=-5.

So statement 2 is true. For statement 3, multiplying every entry of

A

by

-1

gives

D=-A.

Hence

\det(D)=\det(-A)=(-1)^n\det(A).

If

n

is even, then

(-1)^n=1

, so

\det(D)=\det(A)=5.

Thus statement 3 is true. For statement 4, each operation of the form

R_i\to R_i+kR_j

does not change the determinant. First,

R_1\to R_1+R_2

does not change the determinant. Then,

R_2\to R_2-R_1

again does not change the determinant, since this is also a row replacement operation. Therefore

\det(E)=\det(A)=5.

So statement 4 is true. Hence the correct statements are:

1,\ 2,\ 3,\ 4.

4 Multiple Select

Let

f:A\to B \qquad \text{and} \qquad g:B\to C

be two functions. Let

g\circ f:A\to C

be their composition, defined by

(g\circ f)(x)=g(f(x)) \qquad \text{for all } x\in A.

Which of the following statements is/are necessarily true?

A

If

g\circ f

is injective, then

f

is injective.

B

If

g\circ f

is surjective, then

f

is surjective.

C

If

g\circ f

is surjective, then

g

is surjective.

D

If

g\circ f

is injective, then

g

is injective.

View Solution

We check each statement. For statement 1, suppose

g\circ f

is injective. If

f(x_1)=f(x_2),

then applying

g

to both sides gives

g(f(x_1))=g(f(x_2)),

that is,

(g\circ f)(x_1)=(g\circ f)(x_2).

Since

g\circ f

is injective, we get

x_1=x_2.

Hence

f

is injective. So statement 1 is true. For statement 2, this is not necessarily true. Take

A=\{1\}, \qquad B=\{a,b\}, \qquad C=\{c\}.

Define

f(1)=a,

and define

g:B\to C

by

g(a)=c,\qquad g(b)=c.

Then

g\circ f:A\to C

is surjective, since its image is

\{c\}=C.

But

f

is not surjective onto

B,

because

b

is not in the image of

f.

So statement 2 is false. For statement 3, suppose

g\circ f

is surjective. Let

c\in C.

Since

g\circ f

is surjective, there exists some

a\in A

such that

(g\circ f)(a)=c.

That means

g(f(a))=c.

Now

f(a)\in B,

so we have found an element of

B

that maps to

c

under

g.

Hence

g

is surjective. So statement 3 is true. For statement 4, this is not necessarily true. Take

A=\{1\}, \qquad B=\{a,b\}, \qquad C=\{c,d\}.

Define

f(1)=a,

and define

g:B\to C

by

g(a)=c,\qquad g(b)=c.

Then

g

is not injective, because

g(a)=g(b) \quad \text{but} \quad a\ne b.

However,

g\circ f

is injective because its domain

A

has only one element. So statement 4 is false. Hence the correct statements are:

1,\ 3.

5 Multiple Select

Let

f(x)=x|x| \qquad \text{and} \qquad g(x)=|x^2-4|

be two real-valued functions defined on

\mathbb{R}.

Which of the following statements is/are true?

A

The function

f(x)

is differentiable at

x=0.

B

The function

g(x)

is continuous but not differentiable at

x=2.

C

The product function

h(x)=f(x)g(x)

is not differentiable at

x=2.

D

The limit

\lim_{x\to 2}\frac{g(x)}{x-2}

exists.

View Solution

We check each statement. First,

f(x)=x|x|= \begin{cases}x^2, & x\ge 0,\\ -x^2, & x<0.\end{cases}

At

x=0,

\lim_{h\to 0^+}\frac{f(h)-f(0)}{h} =\lim_{h\to 0^+}\frac{h^2}{h} =\lim_{h\to 0^+}h=0,

and

\lim_{h\to 0^-}\frac{f(h)-f(0)}{h} =\lim_{h\to 0^-}\frac{-h^2}{h} =\lim_{h\to 0^-}(-h)=0.

Hence

f

is differentiable at

x=0.

So statement 1 is true. Now consider

g(x)=|x^2-4|.

Since absolute value of a continuous function is continuous,

g

is continuous everywhere, in particular at

x=2.

To check differentiability at

x=2,

note that near

x=2,

g(x)= \begin{cases}x^2-4, & x>2,\\ 4-x^2, & x<2.\end{cases}

So

g'_+(2)=\left.\frac{d}{dx}(x^2-4)\right|_{x=2}=4,

while

g'_-(2)=\left.\frac{d}{dx}(4-x^2)\right|_{x=2}=-4.

These are not equal, so

g

is not differentiable at

x=2.

Therefore statement 2 is true. Now let

h(x)=f(x)g(x).

Since

2>0,

we have

f(x)=x^2

for all

x

near

2.

Hence near

x=2,

h(x)=x^2|x^2-4|= \begin{cases}x^2(x^2-4), & x>2,\\ x^2(4-x^2), & x<2.\end{cases}

Thus

h'_+(2)=\left.\frac{d}{dx}(x^4-4x^2)\right|_{x=2} =\left.(4x^3-8x)\right|_{x=2} =16,

and

h'_-(2)=\left.\frac{d}{dx}(4x^2-x^4)\right|_{x=2} =\left.(8x-4x^3)\right|_{x=2} =-16.

Since these are unequal,

h

is not differentiable at

x=2.

So statement 3 is true. Finally,

\frac{g(x)}{x-2} =\frac{|x^2-4|}{x-2}.

For

x>2,

\frac{|x^2-4|}{x-2} =\frac{(x-2)(x+2)}{x-2}=x+2 \to 4.

For

x<2,

\frac{|x^2-4|}{x-2} =\frac{-(x-2)(x+2)}{x-2}=-(x+2) \to -4.

The one-sided limits are different, so the limit does not exist. Therefore statement 4 is false. Hence the correct statements are:

1,\ 2,\ 3.

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