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Mathematics Test - 16

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Mathematics Test - 16
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  • Question 1
    4 / -1

    If no husband and wife play in the same game, then the number of ways in which a mixed double game can be arranged from among nine married couples is-

    Solution

    Out of 9 men, 2 men can be chosen in 9C2 ways.

    Since no husband and wife are to play in the same game, we have to select 2 women from the remaining 7 women.

    This can be done in 7C2 ways.

    If M1, M2 and W1, W2 are chosen, then a team can be constituted in 2 ways, viz. M1 W1 or M1 W2.

    Thus, number of ways of arranging the game = 9C2 x 7C2 x 2 = 1512

  • Question 2
    4 / -1

    The number of ways of dividing 20 persons into 10 couples is

    Solution

    Here, the order of the couples is not important. So, required number of ways is 20!/21010!

    Hence option D is correct.

  • Question 3
    4 / -1

    The solution of the differential equation dy/dx + y/x = x2 is

    Solution
    The given equation \(\frac{d y}{d x}+\frac{y}{x}=x^{2}\) is of the form \(\frac{d y}{d x}+P y=Q\), so \(I F=e^{I \frac{1}{x} d x}=e^{\log x}=x\)
    Hence, required solution \(x y=\int x \cdot x^{2} d x+c\) \(\Rightarrow x y=\frac{x^{4}}{4}+c\)
    \(\Rightarrow 4 x y=x^{4}+4 c\)
  • Question 4
    4 / -1
    The smallest interval [a, b], such that \(\int_{0}^{1} \frac{1}{\sqrt{1+x^{4}}} d x \in[a, b]\) is given by
    Solution
    \(1-\int_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}},\) Here, \(0 \leq x \leq 1 \Rightarrow 1 \leq 1+x^{4} \leq 2\)
    \(\Rightarrow 1 \leq \sqrt{1+x^{4}} \leq \sqrt{2} \Rightarrow \frac{1}{\sqrt{2}} \leq \frac{1}{\sqrt{1+x^{4}}} \leq 1\)
    \(\Rightarrow \frac{1}{\sqrt{2}} \leq \int_{0}^{1} \frac{d x}{\sqrt{1+x^{4}}} \leq 1\)
    Hence, \(|1 / \sqrt{2}, 1|\) is the smallest interval, such that \(\mid \in[1 / \sqrt{2}, 1]\) Note: - If \(\mathrm{m}\) - least value of \((\mathrm{x})=\mathrm{M}\) - greatest value of \(\mathrm{f}(\mathrm{x})\) is \(\mathrm{Ia}\), bl. then \(m(b-a) \leq \int_{\frac{1}{3}}^{b} f(x) d x \leq M(b-a)\)
  • Question 5
    4 / -1

    If A2 - A + I = 0, then inverse of A is

    Solution

    A2 – A + I = 0

    I = A – (A × A)

    IA-1 = AA-1 – A(AA-1)

    A-1 = I - A

  • Question 6
    4 / -1

    What is the number of arrangements that can be made using all the letters of the word 'LAUGH', keeping the vowels adjacent?

    Solution

    Considering AU as one letter, we have 4 letters: L, AU, G, H, which can be permuted in 4! ways. But, AU can be put together in 2! ways. Thus, the required number of arrangements is 4! × 2! = 48.

  • Question 7
    4 / -1

    Assuming that f is continuous everywhere, (1/c) \(\int_{a c}^{b c} f(x / c) d x\)is equal to

    Solution
    We have \(\mid-(1 / c) \int_{0}^{b c} f(x / c) d x\)
    Putting \(x / c=t, d x=c\) dt, we get, \(\mid-(1 / c) \int_{\frac{1}{2}}^{b} f(t) c d t-\int_{\frac{1}{2}}^{b} f(t) d t-\int_{\frac{1}{2}}^{b} f(x) d x\)
  • Question 8
    4 / -1

    The minimum value of √(ex2 - 1) is

    Solution
    Note that \(e^{x^{4}}-1 \geq 0\) for each \(x^{e} R\) and \(e^{0^{4}}-1=0\) \(\sqrt{e^{x^{4}}-1}\) acquires the minimum value 0 at \(x=0\)
  • Question 9
    4 / -1

    If (√8+i)50 = 349 (a + ib), then a2 + b2 is

    Solution
    \((\sqrt{8}+i)^{50}=3^{49}(a+i b)\)
    Taking magnitude and squaring on both sides, we get \((8+1)^{50}=3^{98}\left(a^{2}+b^{2}\right)\)
    \(9^{50}=3^{98}\left(a^{2}+b^{2}\right)\)
    \(3^{100}=3^{98}\left(a^{2}+b^{2}\right)\)
    \(\Rightarrow a^{2}+b^{2}=9\)
  • Question 10
    4 / -1

    Cos-1 (cos7π/6) =

    Solution
    \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\cos \left(\pi+\frac{\pi}{6}\right)\right.\)
    \(=\cos ^{-1}\left(-\cos \frac{\pi}{6}\right)=\pi-\cos ^{-1}\left(\cos \frac{\pi}{6}\right)\)
    \(=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\)
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