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Mathematics Test - 24

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Mathematics Test - 24
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  • Question 1
    4 / -1

    The value of \(81\left(\frac{1}{\log _{5} 3}\right)+27^{\log _{0} 36}+3^{\frac{4}{\log _{7} 9}}\) is equal to

    Solution
    \(\begin{aligned} & 81^{\left(\frac{1}{\log _{5} 3}\right)}+27^{\log _{9} 36}+3^{\frac{4}{\log _{7} 9}} \\=& 3^{\log _{3} 5^{4}+3^{\log _{3} 3} 6^{\frac{3}{2}}}+3^{\log _{3} 7^{\frac{4}{2}}} \\=& 5^{4}+36^{\frac{3}{2}}+7^{2}=890 \end{aligned}\)
  • Question 2
    4 / -1

    \(\lim _{n \rightarrow \infty} n \cdot \cos \left(\frac{\pi}{4 n}\right) \cdot \sin \left(\frac{\pi}{4 n}\right)\) is equal to

    Solution
    \(\quad \lim _{x \rightarrow \infty} n \cdot \cos \left(\frac{\pi}{4 n}\right) \sin \left(\frac{\pi}{4 n}\right)\)
    \(=\lim _{n \rightarrow \infty} \frac{n}{2} \sin \frac{\pi}{2 n}\)
    \(=\frac{\pi}{4}\)
  • Question 3
    4 / -1

    \(\sin ^{-1}|\cos x|-\cos ^{-1}|\sin x|=a\) has at least one solution if \(\mathbf{a} \in\)

    Solution
    \(\sin ^{-1}|\cos x|-\cos ^{-1}|\sin x|=\frac{\pi}{2}-\cos ^{-1}|\cos x|-\frac{\pi}{2}+\sin ^{-1}|\sin x|=a\)
    \(\Rightarrow \sin ^{-1}|\sin x|-\cos ^{-1}|\cos x|=a\)
    \(\Rightarrow a=0 \)
  • Question 4
    4 / -1

    \(\lim _{n \rightarrow \infty} \frac{a^{n}+b^{n}}{a^{n}-b^{n}}\) where \(a>b>1,\) is equal to

    Solution
    \(\quad \lim _{n \rightarrow \infty} \frac{a^{n}+b^{n}}{a^{n}-b^{n}}\)
    \(a>b>1\)
    \(=\operatorname{n}_{n \rightarrow \infty} \frac{1+\left(\frac{b}{a}\right)^{n}}{1-\left(\frac{b}{a}\right)^{n}}\)
    \(=1\)
  • Question 5
    4 / -1

    If \(\int\left(\frac{x-1}{x+1}\right) \frac{d x}{\sqrt{x^{3}+x^{2}+x}}=2 \tan ^{-1} \sqrt{f(x)}+C,\) find \(f(x)\)

    Solution
    \begin{array}{l}
    \begin{aligned}
    I &=\int \frac{(x-1) d x}{(x+1) x \sqrt{x+1+\frac{1}{x}}}=\int \frac{(x-1)(x+1) d x}{(x+1)^{2} x \sqrt{x+1+\frac{1}{x}}} \\
    &=\int \frac{\left(1-\frac{1}{x^{2}}\right) d x}{\left(x+\frac{1}{x}+2\right) \sqrt{x+\frac{1}{x}+1}} \\
    & \text { Put } x+1+\frac{1}{x}=t^{2} \\
    &\left(1-\frac{1}{x^{2}}\right) d x=2 t d t \\
    &=\int \frac{2 t d t}{\left(t^{2}+1\right) t} \\
    &=2 \tan ^{-1} t+c \\
    &=2 \tan ^{-1}(\sqrt{x+\frac{1}{x}+1})+c
    \end{aligned} \\
    f(x)=x+\frac{1}{x}+1
    \end{array}
  • Question 6
    4 / -1
    Let \(n(U)=700, n(A)=200, n(B)=300\) and \(n(A \cap B)=100\),
    Then \(n\left(A^{c} \cap B^{c}\right)=\)
    Solution
    \(n\left(A^{c} \cap B^{c}\right)=n(U)-n(A \cup B)\)
    \(=n(U)-[n(A)+n(B)-n(A \cap B)]\)
    \(=700-[200+300-100]=300\)
  • Question 7
    4 / -1

    The value of \(\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots \ldots \ldots\)

    \(\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}\) is equal to

    Solution
    Given expression is
    \[
    \sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots \ldots 20^{\circ}
    \]
    We know that \(\sin 90^{\circ}=1\) or \(\sin ^{2} 90^{\circ}=1\)
    Similarly, \(\sin 45^{\circ}=\frac{1}{\sqrt{2}}\) or \(\sin ^{2} 45^{\circ}=\frac{1}{2}\) and the angles are in A.P. of 18 terms.
    We also know that
    \[
    \sin ^{2} 85^{\circ}=\left[\sin \left(90^{\circ}-5^{\circ}\right)\right]^{2}=\cos ^{2} 5^{\circ}
    \]
    Therefore from the complementary rule, we find
    \[
    \sin ^{2} 5^{\circ}+\sin ^{2} 85^{\circ}=\sin ^{2} 5^{\circ}+\cos ^{2} 5^{\circ}=1
    \]
    Therefore,
    \[
    \begin{array}{c}
    \sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots \ldots+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ} \\
    =(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9 \frac{1}{2}
    \end{array}
    \]
  • Question 8
    4 / -1

    If \(\int \frac{d x}{x \sqrt{1-x^{3}}}=\operatorname{aln}\left(\frac{\sqrt{1-x^{3}}+b}{-\sqrt{1-x^{3}}+1}\right)+k,\) then

    Solution
    \(I=\int \frac{d x}{x \sqrt{1-x^{3}}}\)
    Put \(1-x^{3}=t^{2}\)
    3 \(x^{2} d x=-2 t d t\)
    \(I=\frac{-2}{3} \int \frac{d t}{\left(1-t^{2}\right)}\)
    \(=\frac{-1}{3} \int\left(\frac{1}{1-t}+\frac{1}{1+t}\right) d t\)
    \(=\frac{-1}{3} \log \frac{1+t}{1-t}+c\)
    \(=\frac{-1}{3} \log \frac{1+\sqrt{1-x^{3}}}{1-\sqrt{1-x^{3}}}+c\)
  • Question 9
    4 / -1

    \(\int \frac{(2+\sec x) \sec x}{(1+2 \sec x)^{2}} d x\)

    Solution
    \(\int \frac{(2 \cos x+1)}{(2+\cos x)^{2}} d x=\int \frac{(2+\cos x) \cos x+\sin ^{2} x}{(2+\cos x)^{2}} d x\)
    \(=\int \frac{\cos x}{2+\cos x} d x-\int \frac{-\sin ^{2} x}{(2+\cos x)^{2}} d x=\frac{\sin x}{2+\cos x}+c\)
  • Question 10
    4 / -1
    \(\int \frac{e^{\tan ^{-1} x}}{\left(1+x^{2}\right)}\left[\left(\sec ^{-1} \sqrt{1+x^{2}}\right)^{2}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] d x(x>0)\)
    Solution
    note that \(s e c^{-1} \sqrt{1+x^{2}}=\tan ^{-1} x ; \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x\)
    for \(x>0\)
    \(\left.I=\int \frac{e^{\tan ^{-1 x}}}{1+x^{2}}\left(\tan ^{-1} x\right)^{2}+2 \tan ^{-1} x\right) d x\)
    Put \(\tan ^{-1} x=t\)
    \(=\int e^{t}\left(t^{2}+2 t\right) d t=e^{t} . t^{2}=e^{\tan ^{-1} x}\left(\left(\tan ^{-1} x\right)^{2}\right)+C\)
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