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Mathematics Test - 11

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Mathematics Test - 11
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  • Question 1
    1 / -0

    State the nature of the given quadratic equation (x + 1) (x + 2) + 2 = 0

    Solution
    let event \(A=\) minimun is 3 let event \(B =\) maximum is 6
    total ways \(=\cdot{ }^{8} C_{3}\) \(P(A \cap B)=\frac{2}{.^{8} C_{3}}=\frac{2}{\frac{8 !}{3 ! 5 !}}\)
    \(=\frac{2 \times 3 !}{8 \times 7 \times 6}=\frac{1}{28}\)
    \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
    \(=\frac{\frac{1}{28}}{\frac{s^{5} C_{2}}{s C_{3}}}\)
    \(=\frac{\frac{2}{s^{8} C_{3}}}{\frac{s C_{2}}{s C_{3}}}\)
    \(=\frac{2}{5^{5} C_{2}}\)
    \(=\frac{2 \times 2}{5 \times 4}=\frac{1}{5}\)
    option 1 is correct
  • Question 2
    1 / -0

    If \(\mathrm{A} \& \mathrm{B}\) are two events then \(P\{(A \cap \bar{B}) \cup(\bar{A} \cap B)\}=\)

    Solution


    \(P(A \cup B)-P(A \cap B)\)
  • Question 3
    1 / -0

    The number of four digit odd numbers that can be formed using 1,2,3,4,5,6,7,8,9 (repetition of digits is allowed ) are

    Solution
    For a number to be odd, the units digit should be either of 1,3,5,7 or 9
    \(\therefore\) Total number of ways of choosing units digit \(=5\) since, it is given that the digit may be repeated. So, we have 9 choices for ten's digit Similarly, for rest two digits we can choose any of the 9 digits
    \(\therefore\) No. of ways of choosing first and second digit is also 9 for each.
    \(\therefore\) Total number of ways to form odd numbers is \(5 \times 9 \times 9 \times 9=5 \times 9^{3}\)
    Hence, the number of 4 digit odd numbers can be formed is \(5 \times 9^{3}\)
  • Question 4
    1 / -0

    The number of nine digit numbers that can be formed with different digits is

    Solution
    The first digit is one of (1,2,3,4,5,6,7,8,9) (i.e., all
    expect 0 ). So, the no. of ways \(=9\) The remaining 8 digits are to be taken from the remaining 9 numbers. So, the no. of ways \(=9 P_{8}\) So, total no. of ways \(=9 \times^{9} P_{8}\) \(=9 \times 9 !\)
  • Question 5
    1 / -0

    What are the nature of  quadratic equation x+ 3x + 4 = 0 ?

    Solution
    Using \(D=(3)^{2}-(4)(1)(4)\)

    \(D=9-16\)

    \(D=-7\)

    \(D<0\)

    So the given quadratic equation have imaginary roots.
  • Question 6
    1 / -0

    If the first term of an AP is 5 and the common difference is −2, then the sum of the first 6 terms is

    Solution
    Sum of an AP is:
    \(\frac{n}{2} \times[2 a+(n-1) d]\)
    \(a=5, d=-2, n=6\)
    Sum \(=\frac{6}{2} \times[(2 \times 5)+(6-1) \times(-2)]\)
    \(=3 \times(10-10)\)
    \(=0\)
  • Question 7
    1 / -0

    The value of \(\frac{\sin 4 \theta}{\cos \theta}\) is

    Solution
    The value of \(\frac{\sin 4 \theta}{\cos \theta}\) is
    \(=\frac{2 \sin 2 \theta \cos 2 \theta}{\cos \theta}\)
    \(=\frac{4 \sin \theta \cos \theta \cos 2 \theta}{\cos \theta}\)
    \(=4 \sin \theta \cos 2 \theta\)
    Hence, answer is option B.
  • Question 8
    1 / -0
    Find the value of \(\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x}\)
    Solution
    The value of \(\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x}\) is defined only when
    \(x, \frac{1}{x} \epsilon[-1,1]\)
    which is possible only when \(x=\pm 1\) For which \(\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+\cos ^{-1} x+\cos ^{-1} \frac{1}{x}=\pi\)
  • Question 9
    1 / -0

    \(\int e^{\sin x} \sin 2 x d x=\)

    Solution
    \(\int e^{\sin x} 2 \sin x \cos x d x\)
    \(=2 \int e^{\sin x} \sin x d(\sin x)\)
    Take \(\sin x=t\)
    \(=2 \int e^{t} t d t=2\left(t e^{t}-e^{t}\right)\)
    \(=2 e^{t}(t-1)\)
    \(=2 e^{\sin x}(\sin x-1)\)
    \(\Rightarrow \int e^{\sin x} \sin 2 x d x=2 e^{\sin x}(\sin x-1)+c\)
  • Question 10
    1 / -0

    \(\int(x+2) \sqrt{x+1} d x=\)

    Solution
    \(\int(x+2) \sqrt{x+1} d x\)
    \(\int[(x+1)+1](\sqrt{x+1}) d x\)
    \(\int(x+1)^{\frac{3}{2}} d x+\int \sqrt{x+1} d x\)
    \(=\frac{2}{5}(x+1)^{5 / 2}+\frac{2}{3}(x+1)^{3 / 2}+c\)
    \(=\frac{2}{15}\left[3(x+1)^{5 / 2}+5(x+1)^{3 / 2}\right]+c\)
    \(=\frac{2}{15}(x+1)^{3 / 2}(3 x+8)+c\)
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