CAT 2021 Question Paper Slot 3

Now we have :
$$f(g(x))-3x$$
so we get f(x+3)-3x
= $$\left(x+3\right)^2-7\left(x+3\right)-3x$$
=$$x^2-4x-12$$
Now minimum value of expression = $$-\frac{D}{4a}$$ $$ \frac{\left(4ac-b^2\right)}{4a}$$
We get - (16+48)/4
= -16

Now Anil Paints in 12 Days 
Barun paints in 16 Days 
Now together Arun , Barun and Chandu painted in 6 Days 
Now let total work be W 
Now each worked for 6 days 
So Anil's work = 0.5W 
Barun's work = $$\frac{6W}{16}=\frac{3W}{8}$$
Therefore Charu's work = $$\frac{W}{2}-\frac{3W}{8}=\ \frac{W}{8}$$
Therefore proportion of charu =$$\frac{24000}{8}=\ 3,000$$

$$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$$

$$(\sqrt[7]{10})^{1+2+...+n}>999$$

$$10^{\frac{1+2+...+n}{7}}>999$$

For minimum value of n,

$$\frac{1+2+...+n}{7}=3$$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$$\frac{4!}{3!}$$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$$\frac{4!}{2!\cdot2!}=\ 6$$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$$\frac{4!}{3!}$$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$$\frac{4!}{2!}=\ 12$$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively


  We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once 
So the possible cases :

Now we get $$\frac{4!}{2!}\times\ 3$$= 36 ( When one digit is used twice and the remaining two once )
$$\frac{4!}{3!}\times\ 2$$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$$\frac{4!}{2!\times\ 2!}=\ 6$$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

Applying cosine rule in triangle ACD, 

$$100+X^2-2\times\ 10\times\ X\cos30=400$$

$$X^2-10X\sqrt{\ 3}-300=0$$

Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$

Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$

= $$25(\sqrt{3}+\sqrt{15})$$

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with x >= 0, y >= 0,

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test x >= 0, y < 0,

3x - y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 - 2 = 0.