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Mathematics Test - 6

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Mathematics Test - 6
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  • Question 1
    4 / -1

    Let f(x) = x3/2, then f′(0) =

    Solution
    \(R f(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{3 / 2}-0}{h}=\lim _{h \rightarrow 0} h^{1 / 2}=0\)
    \(L f(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{(-h)^{3 / 2}-0}{(-h)} \lim _{h \rightarrow 0}(-h)^{1 / 2}\)
    which is imaginary So. \(\mathrm{R} \mathrm{f}\) (o) \(\mathrm{L}\) f(o). Hence, \(\mathrm{f}\) (o) does not exist.
  • Question 2
    4 / -1
    Let \(f(x)\) be a function satisfying \(f(x+y)=f(x) f(y)\) for all \(x, y\) \(\in \mathrm{R} \& f(x)=1+x g(x),\) where \(\lim _{x \rightarrow 0} g(x)=1 . f(x)\) is equal to
    Solution
    \(f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h}=f(x) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=\)
    \(f(x)\left(\lim _{h \rightarrow 0} \frac{1+h g(h)-1}{h}\right)\)
    \(\lim _{h \rightarrow 0} g(h)=f(x)=1+x g(x)\)
  • Question 3
    4 / -1

    If sin (A + B + C) = 1, tan (A - B) = 1/√3, sec(A + C) = 2,

    Solution
    \(A s \sin (A+B+C)=1\)
    \(\Rightarrow A+B+C=90^{\circ}\)
    As, \(\tan (A-B)=1 / \sqrt{3} \quad \& \sec (A+C)=2\)
    \(\Rightarrow A-B=30 \& A+C=60^{\circ}\)
    From the above statements we can conclude, \(A=60^{\circ}, B=30^{\circ}, C=0^{\circ}\)
    \(O R\)
    \(A s \sin (A+B+C)=1\)
    \(\Rightarrow A+B+C=90^{\circ}\)
    INote: we can easily rule out options ( 1 ) \& (3) as \(A+B+\) \(\left.C>90^{\circ}\right\}\)
    Now checking option (b) \(\sin 90^{\circ}=1, \tan 30=1 / \sqrt{3}, \sec 60=2,\) thus satisfy.
  • Question 4
    4 / -1

    sin 30° + cos 60° + tan 45° is equal to

    Solution

    sin 30° + cos 60° + tan 45°

    = (1/2) + (1/2) + 1

    = 2

  • Question 5
    4 / -1
    Evaluate \(\int t^{3}-\frac{\mathrm{e}^{-t}-4}{\mathrm{e}^{-t}} d t\)
    Solution
    Before doing the integral we need to break up the quotient and do some simplification.
    \[
    \int t^{3}-\frac{\mathbf{e}^{-t}-4}{\mathbf{e}^{-t}} d t=\int t^{3}-\frac{\mathbf{e}^{-t}}{\mathbf{e}^{-t}}+\frac{4}{\mathbf{e}^{-t}} d t=\int t^{3}-1+4 \mathbf{e}^{t} d t
    \]
    Make sure that you correctly distribute the minus sign when breaking up the second term and don't forget to move the exponential in the denominator of the third term (after splitting up the integrand) to the numerator and changing the sign on the \(t\) to \(a^{a}+"\) in the process.
    At this point there really isn't too much to do other than to evaluate the integral.
    \[
    \int t^{3}-\frac{\mathbf{e}^{-t}-4}{\mathbf{e}^{-t}} d t=\int t^{3}-1+4 \mathbf{e}^{t} d t=\left[\frac{1}{4} t^{4}-t+4 \mathbf{e}^{t}+c\right.
    \]
  • Question 6
    4 / -1

    Let f(x + y) = f(x)f(y), for all x, y ∈ R. If f'(0) = 2 and f(4) = 4, then f'(4) is equal to

    Solution
    We have \(f(x+y)=f(x) f(y),\) for all \(x, y \hat{l} R\) Putting \(x=y=0\), we get \(f(0)=f(0) f(0)\)
    \(\mathrm{p} \mathrm{f}(\mathrm{o})(1-\mathrm{f}(\mathrm{o}))=0\)
    \(p f(0)=1\left(\because f(0)^{1}\right.\) o)
    Now, \(f(0)=2\)
    \(P_{h \rightarrow 0} \frac{t(0+h)-f(0)}{h}=2\)
    \(p_{h \rightarrow 0} \frac{\lim _{\theta} \frac{f(0) f(h)-f(0)}{h}}{h}=2\)
    \(P f(0) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2\)
    \(P_{n-0}^{\lim _{t} \frac{t(h)-1}{h}}=2[u \sin g f(0)=1]\)
    We have \(f(4)=\frac{f(4+h)-f(4)}{h}\)
    \(=\lim _{1 \rightarrow 0} \frac{f(4)+(h)-f(4)}{h}\)
    \(\left(\lim _{n \rightarrow \infty} \frac{f(n)-1}{h}\right) f(4)=2 f(4)\) [from \(\left.(1)\right]\)
    \(=2 \times 4=8\)
  • Question 7
    4 / -1

    α and β lie between 0 and π/4, cos(α + β) = 12/13 and sin(α - β) = 3/5.

    sin 2α =

    Solution
    \(\left.\sin _{\alpha}+\beta\right)+(\alpha-\beta=\sin (\alpha+\beta) \cos (\alpha-\beta)+\sin (\alpha-\beta) \cos (\alpha+\beta)\)
    \(\sin (\alpha+\beta)=\frac{5}{13} \operatorname{and} \cos (\alpha-\beta)=\frac{4}{5}\)
    \(\sin 2 \alpha=\sin \alpha+\beta_{1}+\alpha-\beta\)
    \(=\sin (\alpha+\beta) \cos (\alpha-\beta)+\sin (\alpha-\beta) \cos (\alpha+\beta)\)
    \(=\frac{5}{13} \times \frac{4}{5}+\frac{12}{13} \times \frac{3}{5}=\frac{56}{65}\)
  • Question 8
    4 / -1

    What is the area of the figure bounded by the curves y2 = 2x + 1 and x - y = 1?

    Solution


    Required area \(=\int_{-1}^{3}\left(x_{2}-x_{1}\right) d y=\int_{-1}^{3}\left\{(y+1)-\left(\frac{y^{2}-1}{2}\right)\right\} d y\)
    \(-\frac{1}{2} \int_{-1}^{3}\left(2 y \cdot 3-y^{2}\right) d y\)
    \(\frac{1}{2}\left[y^{2}+3 y-\frac{y^{3}}{3}\right]_{-1}^{-3}\)
    \(=\frac{1}{2}\left[9+\frac{5}{3}\right]=\frac{16}{3}\)
  • Question 9
    4 / -1

    The area lying in the first quadrant and bounded by the circle x2 + y2 = 4, the line x =√3 y and the x-axis is

    Solution


    Required area \(=\int_{0}^{1}\left(x_{2}-x_{1}\right) d y\)
    \(=\int_{0}^{1}(\sqrt{4-y^{2}}-\sqrt{3} y) d y\)
    \(=\left[\frac{1}{2} y \sqrt{4-y^{2}}+\frac{1}{2}(4) \sin ^{-1} \frac{y}{2}-\frac{\sqrt{3} y^{2}}{2}\right]_{0}^{1}\)
    \(=\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1} 0\)
    \(=\frac{\sqrt{3}}{2}+2(\pi / 6)-\frac{\sqrt{3}}{2}=\frac{\pi}{3}\)
  • Question 10
    4 / -1

    If three natural numbers between 1 and 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is

    Solution
    Let \(\mathrm{E}=\) Events of numbers divisible by 2 and 3
    (i.e., divisible by 6) \(=(6,12, \ldots, 96)\)
    \(n(E)=16\)
    \(\therefore\) Required probability \(=\frac{16}{100} \mathrm{C}_{3}\)
    \(=\frac{16 \times 15 \times 14}{3 \times 2 \times 1}\)
    \(\frac{100 \times 99 \times 98}{3 \times 2 \times 1}\)
    \(=\frac{4}{1155}\)
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