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Mathematics Test - 30

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Mathematics Test - 30
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  • Question 1
    1 / -0

    Consider the following statements

    I. \(\lim _{x \rightarrow 0} \sin \frac{1}{x}\) does not exist.
    II. \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}\) exists.
    Which of the above statement(s) is/are correct?
    Solution
    I. \(\lim _{x \rightarrow 0} \sin \frac{1}{x}\)
    \(\mathrm{LHL}=\mathrm{f}(0-0)=\lim _{h \rightarrow 0} f(0-h)\)
    \(=\lim _{h \rightarrow 0} \sin \frac{1}{(-h)}=\lim _{h \rightarrow 0}-\sin \frac{1}{h}=-\sin (\infty)\)
    \(\mathrm{RHL}=\mathrm{f}(0+0)=\lim _{h \rightarrow 0} f(0+h)\)
    \(=\lim _{h \rightarrow 0} \sin \frac{1}{h}=\sin (\infty)\)
    \(\because \mathrm{LHL} \neq \mathrm{RHL}\)
    Hence \(\lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)\) does not exist.
    II. \(\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)\)
    \(\mathrm{LHL}=\mathrm{f}(0-0)=\lim _{h \rightarrow 0} f(0-h)\)
     
    \(=\lim _{h \rightarrow 0}(-h) \sin \left(\frac{-1}{h}\right)\)
    \(=\lim _{h \rightarrow 0} h \sin \frac{1}{h}=0\)
    \(\operatorname{RHL}=f(0+0)=\lim _{h \rightarrow 0} f(0+h)\)
    \(=\lim _{h \rightarrow 0}(h) \sin \left(\frac{1}{h}\right)\)
    \(=0 \times \sin \infty=0\)
    \(\because \mathrm{LHL}=\mathrm{RHL}\)
    Hence, \(\lim _{h \rightarrow 0} x \sin \left(\frac{1}{x}\right)\) exists.
  • Question 2
    1 / -0
    Find the area of the region bounded by \(y^{2}=x, x=1, x=4\) and the \(x-\) axis.
    Solution

  • Question 3
    1 / -0

    Prithvi spent Rs. 89,745 on his college fees, Rs. 51,291 on personality development classes and the remaining 27% of the total amount he had as cash with him .what was the total amount?

    Solution
    \(73 \%=89745+51291=141036\)
    Total amount \(100 \%=\frac{141036}{73} \times 100=\mathrm{Rs} .193200\)
  • Question 4
    1 / -0
    Let \(n(U)=700, n(A)=200, n(B)=300, n(A \cap B)=100,\) then \(n\left(A^{\prime} \cap B^{\prime}\right)\) is equal to
    Solution
    \(\mathrm{n}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{n}(\mathrm{A} \cup \mathrm{B})^{\prime}\)
    \(=\mathrm{n}(\cup)-n(A \cup B)\)
    \(=n(\cup)-[n(A)+n(B)-n(A \cap B)]\)
    \(=700-\{200+300-100\}=300\)
  • Question 5
    1 / -0

    Find the equation of the sphere having the centre (–2, 2, 3) and passing through the point (3, 4, –1).

    Solution
    The equation of the sphere with centre (-2,2,3) is
    \[
    (x+2)^{2}+(y-2)^{2}+(z-3)^{2}=r^{2}
    \]
    Radius, \(r=\sqrt{(3+2)^{2}+(4-2)^{2}+(-1-3)^{2}}=\sqrt{45}\)
    \(\therefore\) Required equation of the sphere is
    \[
    \begin{array}{l}
    (x+2)^{2}+(y-2)^{2}+(z-3)^{2}=(\sqrt{45})^{2} \\
    \Rightarrow x^{2}+y^{2}+z^{2}+4 x-4 y-6 z-28=0
    \end{array}
    \]
  • Question 6
    1 / -0
    What is \(\int e^{x}\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right) d x\) is equal to?
    Solution
    \(\int e^{x}\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right) d x\)
    \(\int e^{x} \sqrt{x} d x+\frac{1}{2} \int \frac{e^{x}}{\sqrt{x}} d x\)
    \(e^{x} \sqrt{x}-\int \frac{1}{2 \sqrt{x}} e^{x} d x+\frac{1}{2} \int \frac{e^{x}}{\sqrt{x}} d x+c\)
    \(=e^{x} \sqrt{x}+c\)
  • Question 7
    1 / -0
    \(\mathrm{A} \mathrm{no} \cdot \frac{129}{2^{2} \mathrm{s}^{7} 7^{5}}\) is
    Solution
    As the prime factorization of the denominator is not of the form \(2^{\mathrm{n}} 5^{\mathrm{m}},\) where \(\mathrm{n}, \mathrm{m}\) are non negative integers. Hence \(\frac{129}{2^{2} 5^{7} 7^{5}}\) is a non terminating repeating decimal expansion.
  • Question 8
    1 / -0

    Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60° and 30° at a point on the road between the pillars. Find the height of each pillar.

    Solution

    Let \(A B \& C D\) be two pillars, each of height h meters Let \(P\) be a point on the road such that \(A P=x\) meters Then \(\mathrm{CP}=(100-\mathrm{x})\) meters. It is given that \(\angle \mathrm{APB}=60^{\circ}\) and \(\angle \mathrm{CPD}=30^{\circ}\)
    In \(\Delta P A B,\) we have \(\tan 60^{\circ}=\frac{A B}{A P}\)
    \(\Rightarrow \sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3 x}\)
    In \(\Delta P C D,\) we have
     
    \(\tan 30^{\circ}=\frac{c D}{\rho c}\)
    \(\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{100-x}\)
    \(\Rightarrow \mathrm{h} \sqrt{3}=100-x\)
    Eliminating h between equation (1)\(\&(2)\) we get \(3 x=100-x \Rightarrow 4 x=100\) \(\Rightarrow x=25\)
    Substituting \(x=25\) in equation (i), we get \(h=25 \sqrt{3}\)
  • Question 9
    1 / -0
    \(\frac{\cot ^{2} 15^{\circ}-1}{\cot ^{2} 15^{\circ}+1}\) is equal to
    Solution
    \(\frac{\cot ^{2} 15^{\circ}-1}{\cot ^{2} 15^{\circ}+1}=\frac{\frac{\cos ^{2} 15^{\circ}}{\sin ^{2} 15^{\circ}}-1}{\frac{\cos ^{2} 15^{\circ}}{\sin ^{2} 15^{\circ}}+1}=\frac{\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}}{\cos ^{2} 15^{\circ}+\sin ^{2} 15^{\circ}}\)
     
    \(=\cos \left(2 \times 15^{\circ}\right)=\cos 30^{\circ}\)
    \(=\frac{\sqrt{3}}{2}\)
  • Question 10
    1 / -0

    The two ends of latus rectum of a parabola are the points (3, 6) and (–5, 6), then the focus is

    Solution
    Focus is the mid-point of latus rectum.
    Let the two end points of latus rectum be
    \(\mathrm{L}_{1}(3,6) \& L_{2}(-5,6)\)
    \(\therefore\) Focus, \(x=\frac{3+(-5)}{2}=-1\)
    \(y=\frac{6+6}{2}=6\)
    Focus \(=(x, y)=(-1,6)\)
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