We have a function called “f” that has a special property. It says that if you plug in a number “x” into this function, and you also plug in the number “5 + x,” you get the same result as if you plug in the number “5 – x.”

Now, we know that when you put “f(x) = 0,” it has four different real solutions (or “roots”). We want to find the sum of these four solutions.

Here’s how you can approach this problem:

• Start by considering that “f(x) = 0” has four real solutions. These are four different numbers that, when you plug them into the function, make it equal to zero.
• Use the special property of the function: “f(5 + x) = f(5 – x).” This tells us that if “5 + x” is one of the solutions, then “5 – x” is also a solution because they give the same result when plugged into the function.
• So, for every solution “x,” you also have a corresponding solution “5 – x.” Since there are four solutions, you have pairs of solutions like this.
• Now, you need to find the sum of all these solutions. You can pair them up, and each pair adds up to 5 (because when you add “x” and “5 – x,” you get 5).
• Since you have four pairs of solutions, you can add up these four “5”s to find the sum of all the solutions.

So, the sum of the four real roots is 4 * 5 = 20.

That’s the answer: the sum of the roots is 20.

log4^5 = log4^y × log6^√5

log4^5 ÷ log4^y = logy^5 = log6^√5 = k (Let’s assume this as k)

6^k = √5 —> On squaring this {6^k}^2 = 5

log4^5 = log4^y × log6√5

logy^5 = log6√5

logy^5 = k

y^k = 5

{6^k}^2 = y^k = 5

{6^2}k = y^k

36^k = y^k

y = 36.

The question is “If log4^ 5 = (log4 ^y) (log6^ √5), then y equals”

Hence, the answer is, ” 36″

Step 1: We’re given an equation: 2x + 2^(-x) = 2 – (x – 2)^2

Step 2: We want to find out if there are any real values of “x” that make this equation true. To do that, we first analyze two parts of the equation:

The first part, 2x + 2^(-x), can have a minimum value of at least 2.

The second part, 2 – (x – 2)^2, can have a maximum value of 2.

Step 3: Since the equation equates these two parts, the only way they can be equal is if both parts are equal to 2.

Step 4: However, after some calculations, we find that there is no value of “x” that makes both sides of the equation equal to 2.

Step 5: So, there are no real values of “x” that satisfy this equation, which means there are zero real-valued solutions.

In simple terms, the equation can’t be true for any real value of “x,” so the answer is 0 real solutions.

So, in response to the question, “The number of real-valued solutions of the equation 2x + 2^(-x) = 2 – (x – 2)^2 is,” the answer is indeed 0 real solutions.

1. We start with the equation 64^2 = 4096. This equation is telling us that 6 raised to the power of 2 is equal to 4096.
2. Next, we have an expression for “x,” which is x = (4096)^7 + 4√3. This means “x” is equal to 4096 raised to the power of 7 plus 4 times the square root of 3.
3. To simplify the expression further, we rewrite 4096 as 642. So, x = (642)^7 + 4√3.
4. We’re interested in finding the square root of “x,” which we represent as √x.
5. We want to isolate √x on one side of the equation. To do that, we multiply both sides by a certain fraction, which is 1 ÷ (7 + 4√3).
6. We simplify this fraction by multiplying the numerator and denominator by its conjugate (which is 7 – 4√3) to get rid of the square root in the denominator.
7. After simplifying the fraction, we end up with √x times (7 – 4√3) on the left side of the equation.
8. On the right side, 64 is multiplied by the simplified fraction.
9. To continue simplifying, we distribute √x times (7 – 4√3) to get x times (7/2 – 2√3) on the left side.
10. Finally, we have x times (7/2 – 2√3) = 64 on the right side.

So, after these simplifications, the equation becomes x times (7/2 – 2√3) = 64. This is the final equation after simplifying the original expression.

x + 1/x= y

y^2 – 3y + 2 = 0

y ≥ 2 & y ≤ – 2

(y – 1) (y – 2) = 0

y = 1(Not possible) or y = 2

Now, x + 1/x= 2

This has only one option x = 1. So, only one real root

x₀ = 100 (If all 100 had their birthdays on January)
If x₀ = 3 (Not possible, because all x₁ ,…..x₁₂ together will be only 36)
So x₀ will be lesser when the numbers are as close as possible
1001210012=8.5
So if we add 8 (12 times) = 96
So, need to have 8,8,8,8,8,8,8,8,9,9,9,9 (Adding these 12 we will get 100)
x₀ = 9

x^2-7x+11)(x^{2}-13x+42) = 1
Only possible thing is x^2-13x+42 = 0
Or x^2-7x+11 = 1
Solving these two x^2-13x+42 = 0
(x – 6) (x – 7) = 0
x = 6 or x = 7
x^2-7x+10 = 0
(x – 2) (x – 5) = 0
x = 2 or x = 5
At this moment we have 4 values possible, But there is one more way we can arrive this
(-1)Even = 1
So, x^2-7x+11 = -1
(x – 3) (x – 4) = 0
x = 3 or x = 4
So, 4 + 2 = 6 values

|x| – y = 1

y = |x| – 1

We got a trapezium with h = 1

And parallel sides = 2 and 4

= 1/2

(2+4)×1

= 3 square units 

Taking log on both sides 

y^2log35 log 2 = log23 log 5  

Choosing the base to be 3 

y^2 log35 log3^2 = log23 log35 

y^2 log32 = log23 (Cancelling log35) 

y^2 = log23/log32 

y^2 = (log23)^2 

y = – log23 (Given y is a negative number)  

 y = log2^1/3

x = 2 and y = 19 (Not possible), Given the condition x ≥ y ≥ – 20

x should increase/decrease in the coefficient of y and same for y

x = 7 and y = 17 (Not possible)

x = 12 and y = 15 (Not possible)

x = 17 and y = 13 (Possible)

. y = 11

. y = 9

. .

. .

. .

. y = -19 (Can’t go any further)

By counting totally 17 values are there.

logaa – logab 

 logbb – logba 

Adding 

2 – { loga^b + 1/ logb^a} [loga^b + 1/ logb^a] = [x + 1/x] and value is ≥ 2 and ≤ -2  

From the options, 

x + 1/x ≠ 1 

for 1 < a ≤ b cannot be equal to”

Hence, the answer is, “1”

 x + x + 8 < x + 9 + 5

2x + 8 < x + 14

x < 6

Maximum possible of x = 5, then y = 13

2x + y = 2(5) + 13 = 23.

f(x+1) = f(x+1)^2 + a(x+1) + b

f(x-1) = f(x-1)^2 + a(x-1) + b

g(x) = (x+1)^2 + a(x+1) + b – (x-1)^2 – a(x-1) – b

= 4x + 2a

g(20) = 72 = 4(20) + 2a

a = – 4

x^2 – 4x + b ≥ 0

(x-2)^2 +b – 4 ≥ 0

Value of b is at least 4

We can have three possibilities for the equality of form xn = 1

First n = 0, then x = 1 and then x = -1 and n = even (-1)Even

Using first x = -1, and substituting in given equation

(-1)^2 -5(-1) + 7 = 8 (Acceptable)

Using second condition, x^2

Solving x = 2 or x = 3 (Both are acceptable)

Last condition x^2 -5x + 7 = -1

x^2 -5x + 8 = 0

No integer values,

So 3 values satisfy the equality

|x|^2 – 2|x| + |a – 2| = 0

|x|^2 – 2|x| + 1 = 0 is the square of a quadratic number

In the above equation the value of constant cannot be more than 1

So |a – 2| = 0 or = 1

|x|^2 – 2|x| = 0

|x|^2 = 2|x|

x = 0 or 2 or -2

For all these possibilities value of a = 2

|x|^2 – 2|x| + 1 = 0

(|x| – 1)^2 = 0

|x| = 1

So, x = 1 or x = -1

Then |a – 2| = 1, a = 3 or a = 1

Four combinations of (x,a) are possible already we have 3

Totally 7 pairs

1. You have an equation: xm = xm + 1 + (m + 1).This equation tells us that when you raise x to the power of m, it’s equal to x to the power of (m + 1) plus (m + 1).
2. You introduced a new variable k, which is k = m + 1.This just makes the equation a bit easier to work with.
3. Now, the equation becomes: xk – 1 = xk + k.This is the same equation, just written using k instead of m.
4. You want to isolate xk on one side of the equation, so you rearrange it: xk = xk – 1 – k.This means that xk is equal to xk – 1 minus k.
5. Rewriting it again: xk = – k + xk – 1.Now, it’s clear that xk is equal to negative k plus xk – 1.
6. To visualize this, let’s calculate some values for x1, x2, x3, and so on:
   • x1 = -1
   • x2 = -2 – 1
   • x3 = -3 – 2 – 1
   • x4 = -4 – 3 – 2 – 1
7. Now, you want to find x100:
   • x100 = -100 – 99 – 98 … – 3 – 2 – 1
8. You notice a pattern where you’re subtracting consecutive numbers from -1 to -100.
9. You can simplify this sum as follows:
   • x100 = -(100 + 99 + 98 + … + 3 + 2 + 1)
10. This is the sum of the first 100 positive integers, which is known as a triangular number.
11. You can calculate it like this:
    • x100 = -(100 * (100 + 1) / 2)
12. Simplify further:
    • x100 = -(100 * 101 / 2)
13. Finally, calculate:
    • x100 = -5050

So, x100 equals -5050.

1. You have two ranges:
   • For x: x can take any value between 2 and 10 but not including 2 and 10 itself. This means x can be 3, 4, 5, 6, 7, 8, or 9.
   • For y: y can take any value between 14 and 23 but not including 14 and 23 itself. This means y can be 15, 16, 17, 18, 19, 20, 21, or 22.
2. You want to find the highest possible value for N, where N is the sum of x and y.
3. The highest value for x is 9, and the highest value for y is 22 (at their respective limits).
4. So, the highest value for N is 9 + 22, which equals 31 (when x = 9 and y = 22).
5. Next, you consider the other possible values for N:
   • N = 9 + 21 = 30 (when x = 9 and y = 21)
   • N = 9 + 20 = 29 (when x = 9 and y = 20)
   • N = 9 + 19 = 28 (when x = 9 and y = 19)
   • N = 9 + 18 = 27 (when x = 9 and y = 18)
   • N = 9 + 17 = 26 (when x = 9 and y = 17)
   • N = 9 + 16 = 25 (when x = 9 and y = 16)
6. You notice that the desired sum, x + y = 25, doesn’t appear in this list.
7. Therefore, the different values that x + y (or N) can take are {31, 30, 29, 28, 27, 26}.
8. In total, there are 6 distinct values for x + y (or N).

So, you can get 6 different values for N based on the different combinations of x and y within their respective ranges.

Let loga30 = A be equation 1. 

And  loga5/3 = -B be equation 2. 

We are told that log2a = ⅓  

Log2a = ⅓   implies loga2= 3 

Dividing eqn(1) and eq(2) by loga2 = 3   

We get; log230 = A/3 and  

log25/3= -B/3 

log230= log2(5 × 3 × 2)  

A/3= log25 + log23 + log22 

A/3=  log25 + log23 + 1, let’s call this equation 3 

log25/3= log25 – log23 

-B/3=  log25 – log23, let’s call this equation 4  

Solving equations 3 and 4 we get, log23 = A+B−3/6 

Dividing this by  log2a = 1/3 we get,  

Log2a= A+B−3/2 

log2a=A+B−3/2 implies  loga2=  2/A+B−3.

 y = |x| is a symmetric function, symmetric on the y axis…

y = |x – 2| + 4 is nothing but a translated graph of y = |x| from (0 , 0) to (2 , 4)

We are asked to find the area, in sq. units, enclosed by the lines x = 2, y = |x – 2| + 4, the X-axis and the Y-axis. Now the enclosed area is the sum of areas of the rectangle and the triangle.

Enclosed Area = 2 × 4 + 1/2

 × 2 × 2 = 8 + 2 = 10.

We are told that f(x+y) = f(x) × f(y)
f(x) = f(x + 0) = f(x) × f(0)
f(x) = f(x) × f(0)
Assuming that f(x) is non-zero;
This implies: f(0) = 1

f(5) = 4 (Given)
f(10) = f(5+5) = f(5) × f(5) = 4 × 4 = 16
f(10) = 16

we know that f(0) = 1
f(0) = f(10 – 10) = f(10 + (-10)) = f(10) × f(-10) = 1
16 × f(-10) = 1
f(-10) = 116116 = 0.0625

f(10) – f(-10) = 16 – 0.0625 = 15.9375.

Note: You need not know or calculate the exact value of 116116
Whatever 116116 is, we know that it is less than 1212
So, the value 16 – 116116, will be greater than 16 – 1212 or 15.5

 Two generic equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, will have a unique solution if and only if, a1/a2 ≠ b1/b2  

We have the equations kx + y = 3 and 4x + ky = 4,

For them to have unique solutions, they must satisfy the condition 

k/4 ≠ 1/k  

k^2 ≠ 4 

 k ≠ ±2

|k| ≠ 2

1. When you have a quadratic equation in the form ax^2 + bx + c = 0, the solutions for ‘x’ are given by the formula: x = (-b ± √(b^2 – 4ac)) / (2a).
2. For these solutions to be real numbers (not imaginary), the part inside the square root, which is ‘b^2 – 4ac’, needs to be non-negative. This part is called the “Determinant,” and we’ll call it ‘D’ for short.
3. In simple terms, if D is greater than or equal to 0, it means the equation has real solutions. Mathematically, this can be written as ‘D ≥ 0,’ which is the same as ‘b^2 ≥ 4ac.’
4. Now, you’re given two quadratic equations: x^2 + mx + 2n = 0 and x^2 + 2nx + m = 0. Both of these equations have real solutions.
5. This means that for the first equation, m^2 ≥ 4(2n), and for the second equation, (2n)^2 ≥ 4m.
6. Instead of solving these equations, let’s plug in some numbers and see what happens:
   • If we set n = 1, then m^2 should be greater than or equal to 8 (4 times 2). So, m should be greater than or equal to 3.
   • However, when n = 1, the second inequality (n^2 ≥ m) doesn’t hold because 1^2 is not greater than or equal to 3.
   • If we set n = 2, then m^2 should be greater than or equal to 16 (4 times 4). So, m should be greater than or equal to 4.
   • When n = 2 and m = 4, both inequalities hold because 2^2 is indeed greater than or equal to 4.
7. Therefore, the smallest pair of values for m and n that satisfy these conditions is m = 4 and n = 2.
8. So, the minimum sum of m and n is 4 + 2, which equals 6.

In simpler terms, you’ve found that the smallest values for m and n that make both equations have real solutions are m = 4 and n = 2. The minimum sum of m and n is 6.

1. You have a line with the equation x + 9y + c = 0. This line forms one of the diagonals of a parallelogram.
2. The other diagonal of the parallelogram connects the points (2, 1) and (-3, -4).
3. In a parallelogram, the diagonals bisect each other, which means they intersect at their midpoints.
4. To find the midpoint of the line joining (2, 1) and (-3, -4), you average their x-coordinates and y-coordinates separately.
5. For the x-coordinate: (2 + (-3)) / 2 = (-1) / 2 = -0.5. For the y-coordinate: (1 + (-4)) / 2 = (-3) / 2 = -1.5.
6. So, the midpoint is (-0.5, -1.5).
7. Since the line x + 9y + c = 0 passes through the midpoint, it means that when you plug in these coordinates into the equation, it should be true.
8. So, you substitute the values into the equation: (-0.5) + 9(-1.5) + c = 0.
9. Now, simplify the equation step by step:
   • (-0.5) + 9(-1.5) + c = 0
   • (-0.5) – 13.5 + c = 0
   • (-14) + c = 0
10. To isolate ‘c’, add 14 to both sides of the equation:
    • (-14) + c + 14 = 0 + 14
    • c = 14.

So, the value of ‘c’ that makes the line x + 9y + c = 0 pass through the midpoint of the diagonal is 14.