For the numbers below -5, f(x) is positive.
For the numbers between -5 and -2, the value f(x) is negative.
For the numbers between -2 and 3, the value of f(x) is positive.
For the numbers between 3 and 9, the value of f(x) is negative.
For the numbers above 9, the value of f(x) is positive.
Hence the range should be -5 < x < -2 and 3 < x < 9.

|n – 60| < |n – 100| < |n – 20|

To simplify the inequality, we can break it down into two parts:

1. |n – 60| < |n – 100|
   This inequality holds true when n is between 60 and 100 (exclusive). The distance between n and 60 should be smaller than the distance between n and 100. So, n must be greater than 60 and less than 100.
2. |n – 100| < |n – 20|
3. This inequality holds true when n is greater than 100. The distance between n and 100 should be smaller than the distance between n and 20. So, n must be greater than 100.
4. Combining the two conditions, we find that n must be greater than 100.
5. Therefore, the only possible values for n that satisfy the inequality are integers greater than 100.
6. Hence, the correct answer is 19.

|x² – 4x – 13| = r has exactly three distinct real roots.
Let’s complete the squares of the quadratic equation.
|(x – 2)² – 17| = r
(x – 2)² – 17 = r and (x – 2)² – 17 = -r
(x – 2)² = r + 17 and (x – 2)² = -r + 17
Hence, we get exactly three roots when r = 17.
The answer is 17.

Given that, xn + 2 = 1+xn+1x
x0 = 1
x1 = 2,
x2 = 1+x1x0
 = 1+21
 = 3
x3 = 2,
x4= 1,
x5= 1, x6= 2, x7= 3, x8= 2, x9= 1
The sequence is going on steps of 5.
Hence, x2021 = 2

log₂[3+log₃{4+log₄(x-1)}]- 2= 0
log₂[3 + log₃{4 + log₄(x – 1)}] = 2
3 + log₃{4 + log₄(x – 1)} = 4 ( ∵ log_aN = x ⇒ N = a^x )
log₃{4 + log₄(x – 1)} = 1
4 + log₄(x – 1) = 3
log₄(x – 1) = -1
x – 1 = 1414
4x = 5

x² – xy – x= 22.
y² – xy + y= 34.
If we add these two equations,we get
(x – y)² – x + y = 56 or (y – x)² – x + y = 56
Since we need to find for x > y, we should take (x – y)² – x + y = 56
(x – y)² – (x – y) = 56
(x – y) (x – y – 1) = 56
If we take x – y = n, then we will boil down to n (n – 1) = 56
Therefore, x – y = 8

You’re given that x, y, and z are in an arithmetic progression.
You’re also given the equation xyz = 5(x + y + z).
Recognizing that xyz = 5(3y) because of the arithmetic progression, you simplify it to xz = 15.
You correctly identify the possible combinations for x and z as (3, 5) and (1, 15) based on the equation xz = 15.
You note that z – x > 4 because y – x > 2 (since x, y, z are in an arithmetic progression).
Finally, you conclude that the values of x, y, and z that satisfy these conditions are x = 1, y = 8, and z = 15. This leads to z – x = 15 – 1 = 14.

We have a function f(x) = x2+2x+4/2×2+4x+9

We can complete the squares in the numerator as
f(x) = x^2+2x+1+3/2x^2+4x+9

f(x) = (x+1)2+3/2x^2+4x+9

Now we try to get (x + 1)2in the denominator as well
f(x) = (x+1)2+3/2(x+1)2+7 
From here we can take out (x+1)2 + 3 from both, the numerator and the denominator and put it as ‘k’
Replacing (x+1)2 + 3 with k we get
f(x) = k/2k+1
The minimum value of k = 3, as the square part is always positive and another positive number is added to the equation.
So the fraction will have the minimum value when k = 3
f(x) = 3/2(3)+1= 3/7 
And, the maximum value when k approaches its maximum value. We can observe that the fraction has a 2k + 1 in the denominator and so it will always be one more than twice the numerator. As the value of k increases the value of the addition of one gets less and less of a value to the overall denominator. The fraction approaches 1/2 , but never 1/2 itself.
Hence the range becomes [3/7,1/2).

 2.25 ≤ 2 + 2^n + 2 ≤ 202

So, 2 + 2^ n+2 has to be either lesser than or equal to 202.

2^7 = 128 and so will satisfy the inequality, but 2^8 wont.

Hence, the value of n can be 5 at most.

Further, subtracting 2 from the first two parts of the inequality,

0.25 ≤ 2n+2 

Or, 2^-2 ≤ 2^n+2

Hence, n ≥ -4

So from the inequality we can say that the integral value of n can be [-4, 5].

For 3 + 3^n+1 to have integral values, n but be greater than or equal to -1 and go up till 5.

Hence the number of values of n can be -1, 0, 1, 2, 3, 4, 5.

7 values

 Let’s assume y to be 3x,

Now, the equation becomes | y – 20 | + | y – 40 | = 20

Let’s comprehend the meaning of | y – 20 |,

| y – 20 | simply means, the distance between y and 20

Let’s jot the three terms, y, 20 and 40 on a number line.

Number line for inequality question  

  Now, y could be anywhere on the number line.

Observe that if y is on the left of 20, the distance between y and ‘40’ will be greater than 20 and the condition | y – 20 | + | y – 40 | = 20 fails.

If y is on the right of 40, the distance between y and ‘20’ will be greater than 20 and the condition | y – 20 | + | y – 40 | = 20 fails.

Hence the point y needs to be between 20 and 40.  

Will any point between 20 and 40 qualify to be ‘y’ ?

Yes, Since the distance between 20 and 40 is exactly 20, for any ‘y’ between 20 and 40, the distance between 20 and ‘y’ plus the distance between ‘y’ and 40 is exactly 20.

So, 20 < y < 40

20 < 3x < 40

20/3 < x < 40/3 

Option B alone holds good.

Alternate Solution:

We know that the minimum value of the modulus function can be zero.

Thus value occurs in the two modulus functions at 20/3 and 40/3. 

So, for numbers less than 20/3 the value of both the functions increases and similarly, the value of the sum of the functions increases when x is greater than 40/3.

Between 20/3 and 40/3, the value of the functions remains the same as one of the inequality increases by the same amount as the decrease in the other.

So, the value of the sum of the modulus remains at 20 for the range of values [20/3 , 40/3].

Which can be rewritten as [6.66,13.33]

The option which satisfies this condition is 7 < x < 12.

For a quadratic equation, a x^2 + b x + c = 0.

the roots are given by x = −b±b^2−4ac√2a

Since the roots, a, b, c are all rational and one of the roots, 2 + √ 3 is irrational, the other root of the equation will also be irrational and a conjugate of 2 + √ 3.

That is, the other root is, 2 – √ 3.

Sum of roots is given by −ba

 = (2 + √ 3) + (2 – √ 3) = 4.

And Product of roots is given by ca

 = (2 + √ 3) (2 – √ 3) = 22 – (√ 3)2 = 4 – 3 = 1

But here the quadratic equation is a x2 – b x + c = 0.

Therefore,

b/a = 4  and c/a = 1

we are given a generic formula in ‘n’ for the sum of first n terms, with alternating signs.

Sn = x1 – x2 + x3 – x4 + ….. + (-1)^n+1 xn = n^2 + 2n = n(n + 2)

(Observe that the even terms have a negative sign, while the odd terms have a positive sign)

S50 = 50 (50 + 2) = 50 (52) = 50 (2) (26) = 2600

S49 = 49 (49 + 2) = 49 (51) = 2499

S48 = 48 (48 + 2) = 48 (50) = (24) (2) (50) = 2400

S50 = S49 – x50

x50 = S49 – S50 = 2499 – 2600 = -101

S49 = S48 + x49

x49 = S49 – S48 = 2499 – 2400 = 99

x49 + x50 = 99 – 101 = -2

Alternate Solution:

Observe that n2 + 2n is an increasing function for all natural numbers.

That is, if x > y; x2 + 2x > y2 + 2y.

Sn = n2 + 2n = n(n + 2) = n2 + 2n

Sn – 1 = (n – 1)(n +1) = n2 – 1

Sn – Sn-1 = n2 + 2n – (n2 – 1) = 2n + 1

So adding the nth term to the sequence increases the sequence by 2n + 1

The even terms of the sequence are subtracted, but then the sum of the terms in the sequence still increases, this means that the nth term is negative if n is even and nth term is positive if n is odd, but the magnitude of the nth term is is always (2n +1)

So, xn = (-1)n+1 (2n + 1)

x50 = (-1)50+1 (2(50) + 1) = -101

x49 = (-1)49+1 (2(49) + 1) = 99

x49 + x50 = 99 – 101 = -2

 g(x) = x + 3

f(x) = x^2 – 7x

f(g(x)) – 3x

= f(x + 3) – 3x

= (x + 3)^2 – 7(x + 3) – 3x

= x^2 + 9 + 6x – 7x – 21 – 3x

= x^2 – 4x – 12

= x^2 – 4x + 4 – 4 – 12

= x^2 – 4x + 4 – 16

= (x – 2)^2 – 16

f(g(x)) – 3x is minimum when (x – 2)^2 – 16 is minimum.

(x – 2)^2 – 16 is minimum when (x – 2)^2 is minimum.

Since (x – 2)^2 is non-negative, the minimum value it can take is 0.

Hence the minimum value of (x – 2)^2 – 16 = 0 – 16 = -16

Therefore, the minimum value of f(g(x)) – 3x is -16

loga15+loga32/(loga15×loga32)

 = 4

1loga32+1loga15

= 4

log32a+log15a

= 4

log(32×15)a

 = 4

a^4 = 32 × 15 = 480

We know that, 4^4 = 256 and 5^4 = 625

256 < 480 < 625

4^4 < a^4 < 5^4  

4 < a < 5.

 We are given two equations: 3x + 2|y| + y = 7 and x + |x| + 3y = 1

|x| or the modulus of x, is a function of x, that gives the magnitude of x.

|x| = -(x); if x is negative, and

|x| = x; if x is non-negative.

Therefore, depending on whether ‘x’ and ‘y’ are positive or negative, we assume the following cases for the two equations given.

Case (i): ‘x’ and ‘y’ are both positive.

|x| = x and |y| = y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x + 2y + y = 7

x + x + 3y = 1

3x + 3y = 7

2x + 3y = 1

Solving the two equations, we get x = 6 and y = -11/3.

Since this is contradictory to the assumption that y is positive, we discard this case.

Case (ii): ‘x’ and ‘y’ are both negative.

|x| = -x and |y| = -y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x – 2y + y = 7

x – x + 3y = 1

3x – y = 7

3y = 1

Solving the two equations, we get x = 22/9 and y = 1/3.

Since this is contradictory to the assumption that both x and y are both negative, we discard this case.

Case (iii): ‘x’ is negative and ‘y’ is positive.

|x| = -x and |y| = y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x + 2y + y = 7

x – x + 3y = 1

3x + 3y = 7

3y = 1

Solving the two equations, we get x = 2 and y = 1/3.

Since this is contradictory to the assumption that both x is negative, we discard this case.

Case (iv): ‘x’ is positive and ‘y’ is negative.

|x| = x and |y| = -y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x – 2y + y = 7

x + x + 3y = 1

3x – y = 7

2x + 3y = 1

Solving the two equations, we get x = 2 and y = -1.

This satisfies the assumption that x is positive and y is negative.

Hence x = 2 and y = -1

Therefore, x + 2y = 2 + 2(-1) = 0

Hence, x + 2y = 0.

 xn+1 = xn + n – 1

x1 = -1

x2 = x1 + (1 – 1)

x3 = x2 + (2 – 1)

x3 = x1 + (1 – 1) + (2 – 1)

x4 = x3 + (3 – 1)

x4 = x1 + (1 – 1) + (2 – 1) + (3 – 1)

Similarly we see that,

x100 = x1 + (1 – 1) + (2 – 1) + (3 – 1) + … + (99 – 1)

x100 = x1 + (1 + 2 + 3 + 4 + … + 98 + 99) – 99 (1)

x100 = x1 + (1 + 2 + 3 + 4 + … + 98)

x100 = (-1) + (98×992

)

x100 = (-1) + 4851

x100 = 4850

  |1 + mn| < |m + n| < 5

For two numbers ‘a’ and ‘b’,

|a| < |b| is equivalent to a^2 < b^2

So, we can say that:

(1 + mn)^2 < (m + n)^2

1 + 2mn +m^2n^2 < m^2 + n^2 + 2mn

1 – n^2 – m^2 + m^2n^2 < 0

(1 – n^2) – m^2(1 – n^2) < 0

(1 – m^2)(1 – n^2) < 0

For the product to be negative, either one of the two terms has to be negative.

But they cannot simultaneously be 0.

The only possibility for either of the two terms to be positive is when

n = 0 and |m| > 1, or |n| > 1 and m = 0

Now for the case when m = 0 and |n| > 1

|m + n| < 5

|0 + n| < 5

So n can be 

±2, ±3, ±4

Which are 6 cases

Similarly for the case when n = 1 and |m| > 1

|m + n| < 5

|0 + m| < 5

So m can be 

±2, ±3, ±4

Again we have 6 cases.

Hence the answer is 12.