30 Challenging Math Problems and Solutions for Different Levels of Students.

What is the value of 2+2?
Solution: 4

What is the square root of 49?
Solution: 7

Simplify (4x^2 + 3x^2 - 7x + 5) - (5x^2 - 2x + 4).
Solution: 2x^2 - 9x + 1

Solve the equation 3x + 2 = 11.
Solution: x = 3

Simplify (3x^2 + 2x - 5) + (5x^2 - 3x + 1).
Solution: 8x^2 - x - 4

Evaluate the expression 2(3x + 7) - 4 - 2x) when x = 2.
Solution: 12

Factor the expression x^2 + 7x + 10.
Solution: (x + 5)(x + 2)

Solve the equation x^2 - 5x + 6 = 0.
Solution: x = 2 or x = 3

Simplify the expression (x^2 - 4)/(x - 2).
Solution: x + 2

Evaluate the limit as x approaches 3 of (x^2 - 9)/(x - 3).
Solution: The limit is undefined, since the denominator approaches 0 and the numerator does not.

Solve the system of equations:
2x + y = 7
3x - 4y = -10
Solution: x = 2, y = 3

Simplify the expression (2x + 5)(x^2 - 3x + 2).
Solution: 2x^3 - x^2 - 5x + 10

Find the derivative of f(x) = 3x^2 - 5x + 2.
Solution: f'(x) = 6x - 5

Evaluate the definite integral from 0 to 2 of x^2 dx.
Solution: ∫(0 to 2) x^2 dx = 8/3

Solve the inequality 2x + 3 > 5x - 2.
Solution: x < 5/3

Simplify the expression log10(100.
Solution: 3

Find the slope of the line that passes through the points (2, 3) and (5, 7).
Solution: The slope is (7-3)/(5-2) = 4/3.

Simplify the expression sin^2(x) + cos^2(x).
Solution: 1

Solve the equation e^x = 5.
Solution: x = ln(5)

Simplify the expression (2x^3y^2z^4)(3x^2yz^3).
Solution: 6x^5y^3z^7

Find the derivative of f(x) = ln(x^2 + 1).
Solution: f'(x) = 2x/(x^2 + 1)

Evaluate the limit as x approaches infinity of (3x^2 - 2x)/(4x^2 + 7).
Solution: The limit is 3/4.

Factor the expression x^2 - 6x + 9.
Solution: (x - 3)^2

Solve the inequality x^2 - 3




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Solution to problem 24:
x^2 - 3 < 8
x^2 < 11
x < √11 or x > -√11
So the solution set is (-∞, -√11) U (√11, ∞).

Find the value of x that maximizes the function f(x) = 3x^2 - 4x + 1.
Solution: To find the maximum, we take the derivative of f(x) and set it equal to 0:
f'(x) = 6x - 4
6x - 4 = 0
x = 2/3
So the maximum value of the function occurs at x = 2/3, and f(2/3) = 5/3.


Find the indefinite integral of (x^2 + 1)/(x^3 + x) dx.
Solution: We can use partial fractions to express the integrand as A/x + B/(x+1) + C/(x^2 + 1), where A, B, and C are constants. Then, we can integrate each term using logarithms and inverse tangent functions to obtain the final answer:
∫ (x^2 + 1)/(x^3 + x) dx = 1/2 ln|x^2 + 1| - ln|x| + arctan(x) + C

Evaluate the limit as x approaches infinity of (2x^2 - 5x)/(3x^2 + x + 1).
Solution: We can divide the numerator and denominator by x^2 to get:
(2 - 5/x)/(3 + 1/x + 1/x^2)
As x approaches infinity, both 5/x and 1/x^2 go to zero, so we are left with:
2/3
Therefore, the limit is 2/3.

Determine the equation of the normal line to the curve y = x^2 - 4x + 5 at the point (2, 1).
Solution: The slope of the tangent line at (2, 1) is the derivative of y with respect to x, evaluated at x = 2:
y' = 2x - 4
y'(2) = 0
Therefore, the tangent line is horizontal and has slope 0. The normal line is perpendicular to the tangent line, so it must have slope -1/0, which is undefined. Therefore, the equation of the normal line is simply x = 2.

Find the value of x that maximizes the function f(x) = x^3 - 3x^2 + 2x + 1.
Solution: We can find the critical points of f(x) by setting its derivative equal to zero:
f'(x) = 3x^2 - 6x + 2
3x^2 - 6x + 2 = 0
x = (3 ± sqrt(3))/3
We can check that x = (3 - sqrt(3))/3 gives a local maximum, while x = (3 + sqrt(3))/3 gives a local minimum. Therefore, the value of x that maximizes f(x) is x = (3 - sqrt(3))/3, and the maximum value of f(x) is f((3 - sqrt(3))/3) = 4 - 2sqrt(3).

Find the roots of the equation x^4 + 4x^3 + 6x^2 + 4x + 1 = 0.
Solution: We can use the rational root theorem to check for possible rational roots of the equation. The only possible rational roots are ±1, ±1/2. We can see that x = -1 is a root of the equation, so we can divide the equation by (x + 1) to obtain a quadratic equation:
x^3 + 3x^2 + 3x + 1 = 0
We can factor this equation as (x + 1)^3 = 0, so the other three roots are -1, -1, and -1. Therefore, the roots of the original equation are -1, -1, -1, and -1.