Maths Rocks Quiz ( Mrq)

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Quizzes Created: 1 | Total Attempts: 101
| Attempts: 101
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  • 1/15 Questions

    HARD QUESTION 6 Sigma(k=1 to k=n) arctan(2k/2+k^2+k^4)

    • Pi/4
    • (n-2)*pi/8
    • Arc tan(n^2 + n +1) - pi/4
    • 2pie
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About This Quiz


Maths rocks is facebook page for promoting maths
We had organised a quiz on the facebook page. Those who cant give the quiz or want to give quiz again can solve the questions here:1. There will be 5 easy questions carrying 1 point each
There will be 4 average questions carrying 2 points each
There will be 6 tough ones carrying 3 points each.
2. Time limit
easy 8 mins
average 9min
hard 10 min
Maths rocks
www. Facebook. Com/mathsrocks Some Important Instructions1. NC0 , nC1 , nC2 ,…………. . NCn are represented by C0 , C1 , C2 , …………. Cn.
2. X^y means ‘x’ raise to the power ‘y’.
3. Arc sinx denotes the sin inverse of x.
4. Provide the answer of probability questions in rational form.
5. In any triangle ABC ‘a’,’b’ and’c’ denotes the side lengths opposite to the angle A,B and C respectively.

Maths Rocks Quiz ( Mrq) - Quiz

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  • 2. 

      EASY QUESTION 2 Find the no. of non negative integral solutions of 3x + y + z =24

    • 438

    • 351

    • 117

    • 108

    Correct Answer
    A. 117
    Explanation
    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.

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  • 3. 

    HARD QUESTION 2 Five ordinary dice are rolled at random and the sum of the numbers shown on them is 16. What is the probability that the numbers shown on each is anyone from 2,3,4 and 5?

    • 5/81

    • 1/8

    • 137/735

    • 5/648

    Correct Answer
    A. 137/735
    Explanation
    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.

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  • 4. 

    EASY QUESTION 1  Find sum of series C0 + C1/2 + C2/3 +………Cn/(n+1).

    • 2^(n+1) - 1/(n+1)

    • 4(5+n)+3(2-n)

    • 2^(n+1)/(n+1)

    • 4(5-n)+3(2+n)

    Correct Answer
    A. 2^(n+1) - 1/(n+1)
    Explanation
    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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  • 5. 

    HARD QUESTION 4 How many 4 letter words can be formed by taking by taking letters from ALLAHABAD

    • 63

    • 25840

    • 287

    • 2994

    Correct Answer
    A. 287
  • 6. 

    AVERAGE QUESTION 4 Find the area bounded interiorly by graph of |y|=e^(-|x|) – 1/2 .

    • 1

    • 6 + ln2

    • Ln4 - 2

    • Log 2^-3

    Correct Answer
    A. Ln4 - 2
    Explanation
    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.

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  • 7. 

    AVERAGE QUESTION 2 Integrate {(arctan(ax) – arctan(x))/x} limit from (0 to infinity).

    • Pi/2 ln a

    • 1

    • -1

    • Pi/4

    Correct Answer
    A. Pi/2 ln a
    Explanation
    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.

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  • 8. 

    HARD QUESTION 5 if (1+x)^n = C0+C1x+C2x^2……..Cnx^n. then find the sum of C1^2+2(C2)^2+3(C3)^2………….n(Cn)^2

    • [(2n-1)!]/[(n-1)!]^2

    • N(1+x)^(n-1)

    • (2n+1)C(n+1)/(n+2)

    • 2n + 1

    Correct Answer
    A. [(2n-1)!]/[(n-1)!]^2
    Explanation
    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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  • 9. 

    Easy question 5  Find the rank of the word CRICKET in the dictionary which prints only 7 letter words made from CRICKET only.

    • 117

    • 554

    • 531

    • 553

    Correct Answer
    A. 531
  • 10. 

    average question 1 let (xr,yr) (r is subscript) : r={1,2,3,4} be the points of intersection of the parabola y^2=4ax and the circle x^2+y^2+2gx+2fy+c=0. Find y1+y2+y3+y4.

    • 0

    • 12.6

    • -1

    • 1

    Correct Answer
    A. 0
  • 11. 

    HARD QUESTION 1 If x^3 + 3px^2 + 3qx +r=0 and x^2 + 2px+q=0 have a common root, then value of 4(p^2-q)(q^2-pr) = ?

    • (pq-r)^2

    • Pq-r

    • 0

    • 1

    Correct Answer
    A. (pq-r)^2
    Explanation
    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.

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  • 12. 

    EASY QUESTION 3 Angle between pair of tangents drawn to the ellipse 3(x)^2 + 2(y)^2 = 5 from the point (1,2) is ?

    • 90

    • 180

    • Arctan(6)- arctan(1)

    • Arctan(12/root5)

    Correct Answer
    A. Arctan(12/root5)
    Explanation
    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).

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  • 13. 

    EASY QUESTION 4 No. of real roots of the equation e^(sinx) +e^(-sinx) -2 = 0 in the interval [-pi,+pi]

    • 3

    • 4

    • 2

    • 8

    Correct Answer
    A. 3
    Explanation
    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].

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  • 14. 

    AVERAGE QUESTION 3 Sigma(sum)(from n=0 to n=infinty) {1/(3n+1) – 1/(3n+2)}.

    • Ln 2/3

    • Ln 2

    • 2e/9 root 5

    • pi/3*(root3)

    Correct Answer
    A. pi/3*(root3)
    Explanation
    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).

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  • 15. 

    HARD QUESTION 3 If the area bounded by the curve f(x)=x^3/3 –x^2+a and lines x=0,x=a and y=0 is minimum then find value of ‘a’?

    • 2/3

    • Sqrt 3

    • 0

    • 1/3

    Correct Answer
    A. 2/3
    Explanation
    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.

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  • Dec 08, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Feb 25, 2012
    Quiz Created by
    Keshushivang
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