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Mathematics Test-7

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Mathematics Test-7
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  • Question 1
    4 / -1

    If the repetition is not allowed, how many numbers lying between \(1000\) and \(10,000\) can be formed with the digits \(1, 2, 3, 4\), and \(5\)?

    Solution

    Each number that lies between \(1000\) and \(10,000\) is a \(4-\) digit number. So, we have to find the permutation of \(5\) digits i.e. \(1, 2, 3, 4\), and \(5\) taken \(4\) at a time.

    So, the number of numbers are:

    \({ }^{5} {P}_{4}=\frac{5 !}{(5-4) !}\)

    \(=\frac{5 !}{1 !}\)

    \(=5 \times 4 \times 3 \times 2\)

    \(=120\)

  • Question 2
    4 / -1

    The exponent of \(x\) occurring in the \(7^{\text {th }}\) term of expansion of \(\left(\frac{3 x}{2}-\frac{8}{7 x}\right)^{9}\) is:

    Solution

    We have the general term of \((x+a)^{n}\) is

    \(T_{r+1}={ }^{n} C_{r} (x)^{n-r} a^{r}\)

    Now, consider \(\left(\frac{3 x}{2}-\frac{8}{7 x}\right)^{9}\)

    Here \( n=9 \) and \( r+1=7\)

    \(\Rightarrow r=6\)

    Also, \(x=\frac{3 x}{2} \) and \( a=-\frac{8}{7 x}\)

    \(\therefore T_{7}=T_{6+1}={ }^{9} C_{6} \left(\frac{3 x}{2}\right)^{3}\left(\frac{-8}{7 x}\right)^{6}\)

    \(={ }^{9} C_{6} \left(\frac{3}{2}\right)^{3}\left(\frac{-8}{7}\right)^{6} x^{-3}\)

    Hence, the exponent of \(x=-3\)

  • Question 3
    4 / -1

    The value of \(\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) is:

    Solution

    Given:

    \(\mathrm{I}=\int_{2}^{8} \frac{\sqrt{10-{x}}}{\sqrt{{x}}+\sqrt{10-{x}}} {dx} \quad \ldots\).(i)

    Properties of definite integral:

    \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(b+a-x) d x\)

    \(\mathrm{I}=\int_{2}^{8} \frac{\sqrt{10-(8+2-{x})}}{\sqrt{8+2-{x}}+\sqrt{10-(8+2-{x})}} {dx}\)

    \(\Rightarrow \mathrm{I}=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \quad \ldots\) (ii)

    Adding (i) and (ii), we get

    \(2\mathrm{I}=\int_{2}^{8} \frac{\sqrt{10-{x}}+\sqrt{{x}}}{\sqrt{{x}}+\sqrt{10-{x}}} {dx}\)

    \(=\int_{2}^{8} {dx}\)

    \(=[{x}]_{2}^{8}\)

    \(=8-2=6\)

    \(\mathrm{I}=3\)

  • Question 4
    4 / -1

    The modulus of \(5+4 i\) is:

    Solution

    Let \(Z=5+4 i\)

    Now, modulus of \(Z\) is calculated as

    \(|Z|=\sqrt{\left(5^{2}+4^{2}\right)}\)

    \(\Rightarrow|Z|=\sqrt{(25+16)}\)

    \(\Rightarrow|Z|=\sqrt{41}\)

    So, the modulus of \(5+4 i\) is \(\sqrt{41}\)

  • Question 5
    4 / -1

    If letters of the work \(KUBER\) are written in all possible orders and arranged as in a dictionary, then the rank of the word \(KUBER\) will be:

    Solution

    Given,

    Total words starting with \(B=4 !=24\)

    Total words starting with \(E=4 !=24\)

    Total words starting with \(K B=3 !=6\)

    Total words starting with \(KE =3 !=6\)

    Total words starting with \(K R=3 !=6\)

    If Starting word will be KUBER, then the rank of KUBER \(=24+24+18+1=67\)

  • Question 6
    4 / -1

    A box of \(600\) bulbs contains \(12\) defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

    Solution

    Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

    also, Probability<1

    \(P(\)non-defective bulb\() =1-P(\)Defective bulb\()\)

    \(=1-(\frac{12}{600})\)

    \(=\frac{(600-12)}{600}\)

    \(=\frac{588}{600}\)

    \(=\frac{147}{150}\)

  • Question 7
    4 / -1

    Consider three sets \(E_{1}=\{1,2,3\}, F_{1}=\{1,3,4\}\) and \(G_{1}=\{2,3,4,5\}\). Two elements are chosen at random, without replacement, from the set \(E_{1}\), and let \(S_{1}\) denote the set of these chosen elements. Let \(E_{2}=E_{1}-S_{1}\) and \(F_{2}=F_{1} \cup S_{1} .\) Now two elements are chosen at random, without replacement, from the set \(F_{2}^{2}\) and let \(S_{2}\) dentoe the set of these chosen elements.

    Let \(G_{2}=G_{1} \cup S_{2}\). Finally, two elements are chosen at random, without replacement from the set \(G_{2}\) and let \(S_{3}\) denote the set of these chosen elements.

    Let \(E_{3}=E_{2} \cup S_{3}\). Given that \(E_{1}=E_{3}\), let \(p\) be the conditional probability of the event \(S_{1}=\{1,2\}\). Then the value of \(p\) is:

    Solution

    Given:

    Three sets \(E_{1}=\{1,2,3\}, F_{1}=\{1,3,4\}\) and \(G_{1}=\{2,3,4,5\}\)

    \(E_{2}=E_{1}-S_{1}\) and \(F_{2}=F_{1} \cup S_{1} \)

    \(G_{2}=G_{1} \cup S_{2}\)

    \(E_{3}=E_{2} \cup S_{3}\)

    \(E_{1}=E_{3}\), let \(p\)

    \(S_{1}=\{1,2\}\)

    We will follow the tree diagram,

    \(P\left(E_{1}=E_{3}\right)\)

    \(=\frac{1}{3}\left[\frac{1}{2} \times \frac{1}{10}+\frac{1}{2} \times 0+\frac{1}{2} \times \frac{1}{10}+\frac{1}{2} \times \frac{1}{6}+\frac{2}{3} \times \frac{1}{10}+\frac{1}{3} \times 0\right]\)

    \(=\frac{1}{3}\left[\frac{1}{4}\right]\)

    Required probability \(=\frac{\frac{1}{3}\left[\frac{1}{2} \times \frac{1}{10}\right]}{\frac{1}{3} \times \frac{1}{4}}=\frac{1}{5}\)

  • Question 8
    4 / -1
    The modulus-amplitude form of \(\sqrt{3}+i,\) where \(i=\sqrt{-1}\) is:
    Solution
    Modulus amplitude form of \((\sqrt{3}+i)\) is:
    \(z=r(\cos \theta+i \sin \theta)\)
    \(\Rightarrow\)\(r \cos \theta=\sqrt{3}, r \sin \theta=1\)
    \(\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=4\)
    \(\Rightarrow r^{2}=4\)
    \(\Rightarrow r=2\)
    \(2 \cos \theta=\sqrt{3}\) and \(2 \sin \theta=1\)
    \(\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\) and \(\sin \theta=\frac{1}{2}\)
    \(\Rightarrow \theta=\frac{\pi}{6}\) and \(\theta=\frac{\pi}{6}\)
    \(\therefore z=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\)
  • Question 9
    4 / -1

    Evaluate the expression \((\mathrm{y}+1)^{4}-(\mathrm{y}-1)^{4}\).

    Solution

    By using Binomial theorem,

    The expression \((y+1)^{4}-(y-1)^{4}\) can be expanded as

    \(=(y+1)^{4}={ }^{4} C_{0} y^{4}+\)

    \({ }^{4} C_{1} y^{3}+{ }^{4} C_{2} y^{2}+{ }^{4} C_{3} y^{1}+{ }^{4} C_{4} y^{0}\)

    and,

    \((y-1)^{4}={ }^{4} C_{0} y^{4}-{ }^{4} C_{1} y^{3}+\)

    \({ }^{4} C_{2} y^{2}-{ }^{4} C_{3} y^{1}+{ }^{4} C_{4} y^{0}\)

    Now,

    \((y+1)^{4}-(y-1)^{4}\)

    \(=\left({ }^{4} C_{0} y^{4}+{ }^{4} C_{1} y^{3}+{ }^{4} C_{2} y^{2}+{ }^{4} C_{3} y^{1}+{ }^{4} C_{4} y^{0}\right)-\left({ }^{4} C_{0} y^{4}-{ }^{4} C_{1} y^{3}+{ }^{4} C_{2} y^{2}-{ }^{4} C_{3} y^{1}+{ }^{4} C_{4} y^{0}\right)\)

    \(=2\left({ }^{4} C_{1} y^{3}+{ }^{4} C_{3} y^{1}\right)\)

    \(=8\left(y^{3}+y^{1}\right)\)

     

  • Question 10
    4 / -1

    In the figure given below, if \(\mathrm{O}\) is the centre of the circle, \(\angle \mathrm{AOB}=\angle \mathrm{EOF}=\angle \mathrm{COD}=60^{\circ}\) and \(A B=4\) units, then find \(E F+C D\).

    Solution

    Given, \(\angle A O B=\angle E O F=\angle C O D=60^{\circ}\), AB = 4 units

    Since the central angle of all the triangles are same and they are formed from the common center with radius as their sides, the length of the chord would also be the same. So, the chords will become equal as well \(A B=4\) units

    Therefore, \(A B=E F=C D=4\) units

    \(\mathrm{EF}+\mathrm{CD}=4+4=8\) units

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