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Mathematics Test - 6
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Mathematics Test - 6

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  • Question 1/10
    4 / -1

    Let f(x) = x3/2, then f′(0) =

    Solutions
    Rf(0)=limh→0f(0+h)−f(0)h=limh→0h3/2−0h=limh→0h1/2=0
    Lf(0)=limh→0f(0−h)−f(0)−h=limh→0(−h)3/2−0(−h)limh→0(−h)1/2
    which is imaginary So. Rf (o) L f(o). Hence, f (o) does not exist.
  • Question 2/10
    4 / -1
    Let f(x) be a function satisfying f(x+y)=f(x)f(y) for all x,y ∈R&f(x)=1+xg(x), where limx→0g(x)=1.f(x) is equal to
    Solutions
    f(x)=limh→0f(x+h)−f(x)h=limh→0f(x)f(h)−f(x)h=f(x)limh→0f(h)−1h=
    f(x)(limh→01+hg(h)−1h)
    limh→0g(h)=f(x)=1+xg(x)
  • Question 3/10
    4 / -1

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

    Solutions
    Assin⁡(A+B+C)=1
    ⇒A+B+C=90∘
    As, tan⁡(A−B)=1/3&sec⁡(A+C)=2
    ⇒A−B=30&A+C=60∘
    From the above statements we can conclude, A=60∘,B=30∘,C=0∘
    OR
    Assin⁡(A+B+C)=1
    ⇒A+B+C=90∘
    INote: we can easily rule out options ( 1 ) \& (3) as A+B+ C>90∘}
    Now checking option (b) sin⁡90∘=1,tan⁡30=1/3,sec⁡60=2, thus satisfy.
  • Question 4/10
    4 / -1

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

    Solutions

    sin 30° + cos 60° + tan 45°

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

    = 2

  • Question 5/10
    4 / -1
    Evaluate ∫t3−e−t−4e−tdt
    Solutions
    Before doing the integral we need to break up the quotient and do some simplification.
    ∫t3−e−t−4e−tdt=∫t3−e−te−t+4e−tdt=∫t3−1+4etdt
    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 aa+" in the process.
    At this point there really isn't too much to do other than to evaluate the integral.
    ∫t3−e−t−4e−tdt=∫t3−1+4etdt=[14t4−t+4et+c
  • Question 6/10
    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

    Solutions
    We have f(x+y)=f(x)f(y), for all x,yl^R Putting x=y=0, we get f(0)=f(0)f(0)
    pf(o)(1−f(o))=0
    pf(0)=1(∵f(0)1 o)
    Now, f(0)=2
    Ph→0t(0+h)−f(0)h=2
    ph→0limθf(0)f(h)−f(0)hh=2
    Pf(0)limh→0f(h)−1h=2
    Pn−0limtt(h)−1h=2[usin⁡gf(0)=1]
    We have f(4)=f(4+h)−f(4)h
    =lim1→0f(4)+(h)−f(4)h
    (limn→∞f(n)−1h)f(4)=2f(4) [from (1)]
    =2×4=8
  • Question 7/10
    4 / -1

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

    sin 2α =

    Solutions
    sinα+β)+(α−β=sin⁡(α+β)cos⁡(α−β)+sin⁡(α−β)cos⁡(α+β)
    sin⁡(α+β)=513andcos⁡(α−β)=45
    sin⁡2α=sin⁡α+β1+α−β
    =sin⁡(α+β)cos⁡(α−β)+sin⁡(α−β)cos⁡(α+β)
    =513×45+1213×35=5665
  • Question 8/10
    4 / -1

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

    Solutions


    Required area =∫−13(x2−x1)dy=∫−13{(y+1)−(y2−12)}dy
    −12∫−13(2y⋅3−y2)dy
    12[y2+3y−y33]−1−3
    =12[9+53]=163
  • Question 9/10
    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

    Solutions


    Required area =∫01(x2−x1)dy
    =∫01(4−y2−3y)dy
    =[12y4−y2+12(4)sin−1⁡y2−3y22]01
    =32+2sin−1⁡(12)−32−2sin−1⁡0
    =32+2(π/6)−32=π3
  • Question 10/10
    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

    Solutions
    Let E= Events of numbers divisible by 2 and 3
    (i.e., divisible by 6) =(6,12,…,96)
    n(E)=16
    ∴ Required probability =16100C3
    =16×15×143×2×1
    100×99×983×2×1
    =41155
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