Maths Rocks Quiz ( Mrq)

The given expression is the sum of arctan(2k/2+k^2+k^4) from k=1 to k=n. The arctan function returns the angle whose tangent is equal to its argument. In this case, the argument is 2k/2+k^2+k^4. The sum of these angles can be simplified to arctan(n^2 + n + 1) - pi/4. This is the correct answer.

The number of non-negative integral solutions of the equation 3x + y + z = 24 can be found using the concept of stars and bars. In this case, we have 24 stars (representing the sum of x, y, and z) and 2 bars (dividing the stars into 3 groups). The number of ways to arrange the stars and bars is given by (24+2-1) choose (2) which is equal to 25 choose 2. Therefore, the number of non-negative integral solutions is 117.

The probability that the numbers shown on each die is anyone from 2, 3, 4, and 5 can be calculated by considering the number of favorable outcomes divided by the total number of possible outcomes. To find the number of favorable outcomes, we can use the concept of combinations. We need to find the number of ways to distribute 16 among 5 dice, where each dice can show a number from 2, 3, 4, and 5. This can be calculated as the sum of the number of ways to distribute 16 - 2*5 = 6 among 5 dice, the number of ways to distribute 16 - 3*5 = 1 among 5 dice, the number of ways to distribute 16 - 4*5 = -4 among 5 dice, and the number of ways to distribute 16 - 5*5 = -9 among 5 dice. The total number of possible outcomes is 4^5 (since each die can show 4 different numbers). Therefore, the probability is (6+1)/4^5 = 137/735.

The given series is a sum of terms where each term is the coefficient of the binomial expansion divided by its corresponding index plus one. The correct answer, 2^(n+1) - 1/(n+1), represents the sum of the series.

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The given integral can be solved using the substitution method. Let u = arctan(ax) - arctan(x), then du = (a - 1)/(1 + a^2x^2) dx. Rearranging the equation, we have dx = (1 + a^2x^2)/(a - 1) du. Substituting these values into the integral, we get the integral of (1 + a^2x^2)/(x(a - 1)) du. Simplifying further, we have the integral of (1 + a^2x^2)/(x(a - 1)) du = (1/(a - 1)) integral of (1/x) du + (a^2/(a - 1)) integral of x du. Evaluating the integrals, we get (1/(a - 1)) ln|x| + (a^2/(2(a - 1))) x^2 + C. Taking the limit as x approaches infinity, the ln|x| term goes to infinity, so the overall result is infinity. Therefore, the answer is pi/2 ln a.

The correct answer is ln4 - 2. To find the area bounded interiorly by the graph of |y|=e^(-|x|) - 1/2, we need to find the integral of the function e^(-|x|) - 1/2. The antiderivative of e^(-|x|) is -e^(-|x|), and the antiderivative of 1/2 is 1/2x. Evaluating the integral from -∞ to ∞ gives us ln4 - 2.

The given expression represents the sum of the squares of the coefficients in the expansion of (1+x)^n. The sum is equal to [(2n-1)!]/[(n-1)!]^2.

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When two equations have a common root, it means that the value of x satisfies both equations. In this case, if x satisfies both equations x^3 + 3px^2 + 3qx + r = 0 and x^2 + 2px + q = 0, then we can substitute the value of x from the second equation into the first equation.

By substituting, we get (x^2 + 2px + q)^2 + 3p(x^2 + 2px + q) + 3qx + r = 0. Simplifying this equation, we get (pq - r)^2 = 0.

Therefore, the value of 4(p^2 - q)(q^2 - pr) is equal to (pq - r)^2.

The angle between a pair of tangents drawn to an ellipse from an external point is given by the formula arctan(2ab/(a^2-b^2)), where a and b are the semi-major and semi-minor axes of the ellipse, respectively. In this case, the equation of the ellipse is 3(x)^2 + 2(y)^2 = 5, which can be rearranged to (x^2)/(5/3) + (y^2)/(5/2) = 1. Comparing this equation with the standard form of an ellipse, we can see that a^2 = 5/3 and b^2 = 5/2. Plugging these values into the formula, we get arctan(2(√(5/3))(√(5/2))/((5/3)-(5/2))) = arctan(12/√5).

The given equation can be rewritten as e^(sinx) + e^(-sinx) = 2. Let's consider the function f(x) = e^(sinx) + e^(-sinx). This function is even, which means f(x) = f(-x). Therefore, we only need to consider the interval [0, pi]. In this interval, the function is increasing and takes the value 2 at x = 0. As the function is continuous, it must cross the x-axis at least once. Therefore, there are at least 3 real roots in the interval [-pi, +pi].

To find the value of 'a' that minimizes the area bounded by the curve and the given lines, we can use calculus.
First, we need to find the derivative of the curve f(x) = x^3/3 - x^2 + a. Taking the derivative, we get f'(x) = x^2 - 2x.
Next, we set f'(x) = 0 to find the critical points. Solving x^2 - 2x = 0, we get x = 0 and x = 2.
We evaluate the area using the definite integral from x = 0 to x = 2: ∫[0,2] (x^3/3 - x^2 + a) dx.
Simplifying the integral, we get [x^4/12 - x^3/3 + ax] evaluated from x = 0 to x = 2.
Plugging in the values, we get (2^4/12 - 2^3/3 + 2a) - (0^4/12 - 0^3/3 + 0a) = 16/12 - 8/3 + 2a.
Simplifying further, we get 4/3 - 8/3 + 2a = -4/3 + 2a.
To minimize the area, we need to find the value of 'a' that makes the area equal to zero.
Setting -4/3 + 2a = 0, we solve for 'a' and get a = 2/3. Therefore, the value of 'a' that minimizes the area is 2/3.

The given expression is a series that involves the difference between two fractions. The numerator of the first fraction is always 1, and the denominator increases by 3 for each term. The numerator of the second fraction is always 1, and the denominator increases by 3 for each term.

By simplifying the expression, we can rewrite it as 1/(3n+1) - 1/(3n+2).

To find the sum of this series, we can use the telescoping series method. By pairing up consecutive terms and simplifying, we can see that most terms cancel out.

After simplification, we are left with the expression (1/1 - 1/2) + (1/4 - 1/5) + (1/7 - 1/8) + ...

This is a telescoping series where most terms cancel out, leaving only the first and last terms.

The sum of this series is equal to 1/1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ... = ln(2/3).

Therefore, the correct answer is ln(2/3).