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Mathematics Test - 15

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Mathematics Test - 15
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  • Question 1
    1 / -0

    Find the cube root of (54 + 30√3) .

    Solution
    Work from the options (3+3)3=27+33+93(3+3)=54+303
  • Question 2
    1 / -0
    sin3xsinxdx=
    Solution
    sin3xsinxdx=3sinx4sin3xsinxdx
    3dx4sin2xdx=3x2(1cos2x)dx+c
    =3x2x+sin2x+c=x+sin2x+c
    Standard Integrals
    (1) Algebric Function
    (i) xndx=xn+1n+1+c,n1
    (ii) (ax+b)ndx=1a(ax+b)n+1n+1+cn1
    (2) Fraction Function
    (i) 1xdx=log|x|+c
    (ii) 1ax+bdx=1a(log|ax+b|)+c
    (3) exdx=ex+c
    (4) axdx=axlogea+c
    (5) sinxdx=cosx+c
    (6) cosxdx=sinx+c
    (7) sec2xdx=tanx+c
    (8) cosec2xdx=cotx+c
    (9) secxtanxdx=secx+c
    (10) cosecxcotxdx=cosecx+c
    (11) tanxdx=log|cosx|+c=log|secx|+c
    (12) cotxdx=log|sinx|+c=log|cosecx|+c
    (13) secxdx=log|secx+tanx|+c=log|tan(π4+x2)|+c
    (14) cosecxdx=log|cosecxcotx|+c=log|tanx2|+c
    (15) dx1x2=sin1x+c=cos1x+c
    (16) dxa2x2=sin1xa+c=cos1xa+c
    (17) dx1+x2=tan1x+c=cot1x+c
    (18) dxa2+x2=1atan1xa+c=1acot1xa+c
    (19) dxxx21=sec1x+c=cosec1x+c
    (20) dxxx2a2=1asec1xa+c=1acosec1xa+c
    In any of the fundamental integration formulae, if x is replaced by ax+b, then the same formulae is applicable but we must divide by coefficient of x or derivative of (ax+b) le., a in general, if f(x)dx=ϕ(x)+c, then f(ax+b)dx=16ϕ(ax+b)+c
    sin(ax+b)dx=16cos(ax+b)+c
    sec(ax+b)dx=1alog|sec(ax+b)+tan(ax+b)|+cetc
    Some more results
    (i) 1x2a2=12alog|xax+a|+c
    (11) 1a2x2dx=12alog|a+xax|+c
    (ii) dxx2d2=log{|x+x2a2|}+c
    (iv) dxx3+87=log{|x+x2+a2|}+c
    (v) a2x2dx=12xa2x2+12a2sin1(xa)+c
    (vi) x2a2dx=12xx2a212a2log|x+x2+a2|+c
    (vii x2+a2dx=12xx2+a2+12a2log|x+x2+a2|+c
    Example 1 : Evaluate the integral 2sin(x)cos(x)dx
    Solution:
    The "extra cosine" next to dx just calls for the sine subsutution
    f2sin(x)cos(x)dx=|y=sin(x)dy=cos(x)dx|=2ydy=y2+C=sin2(x)+C,xIR
  • Question 3
    1 / -0

    If a, b are irrational roots of lx2 + mx + n = 0 (l, m, n Q), then

    Solution

    Irrational roots, always in conjugate pair. So, both roots are conjugate surds.

    Therefore, Option 3 is correct

  • Question 4
    1 / -0

    The line drawn from (4,−1,2)4 to the point (−3, 2,3) meets a plane at right angles at the point (−10,5,4), then the equation of plane is

    Solution
    R. of line is 7 . 3.1
    since given line is perpendicular to the plane
    Required plane is, 7(x+10)3(y5)(z4)=0 or 7x3yz+89=0
  • Question 5
    1 / -0

    Directions: In the question given below, there are two statements labelled as Assertion (A) and Reason (R). Which of the following is correct?

    Assertion (A): The polynomial x2 + 1 has no zero in the real number system.

    Reason (R): A polynomial with coefficients which are complex numbers has all its zeroes in the complex number system.

    Solution

    x2 + 1 = 0 => x2 = - 1 => x = +i, -i which are imaginary!

    Hence, assertion is correct.

    The reason is correct.

  • Question 6
    1 / -0
    secxsecx+tanxdx
    Solution
    Let I=secxsecx+tanxdx
    =secx(secxtanx)sec2xtan2xdx
    =(sec2xsecxtanx)dx
    =tanxsecx+c
    Trigonometric integrals dxa+bcosx and dxa+bsinx
    To evaluate such form of integrais, proceed as follows:
    (A) Put cosx=1tan2x21+tan2x2 and sinx=2tanx21+tan2x2
    (B) Replace 1+tan2x2 in the numerator bysec 2x2
    (C) Put tanx2=tsothat12sec2x2dx=dt
    (D) Now, evaluate the integral obtained which will be of the form dtat2+bt+c by the method discussed earlier.
    Integrais of the form dxa+bcosx+csinx,dxasinx+bcosx
    (1) Integral of the form dxa+bcosx+csinx To evaluate such integrais, we put b=rcosα and c=rsinα so that, r2=b2+c2 and α=tan1cb
    I=dxa+r(cosacosx+sinαsinx)=dxa+cos(xα)
    Again, put xα=tdx=dt, we have I=dxa+rcost
    Which can be evaluated by the method discussed earier
    (ii) Integral of the form dxasinx+bcosx:: To evaluate this type of integrais we substitute
    a=rcosθ,b=rsinθ and sor=a2+b2,θ=tan1ba
    So, dxasinx+bcosx=1rdxsin(x+θ)=1rcosec(x+θ)dx
    =1rlog|tan(x2+θ2)|=1a2+b2log|tan(x2+12tan1ba)|+C
    Note: The integral of the above from can be evaluated by using cosx=1tan2x31/tan223 and
    sinx=2tanx21+tan2x3
  • Question 7
    1 / -0

    A die is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is

    Solution
    p=1/2,q=1/2
    n=5
    Variance =npq
    =5×1/2×1/2
    =5/4
    Classical Definition of Probability
    If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which m are favourable to the occurrence of an event A, then the probability of occurrence of A is given by
    It is obvious that 0mn. If an event A is certain to happen, then m=n, thus P(A)=1
    If A is impossible to happen, them m=0 and soP(A)=0
    Hence we conclude that 0P(A)1
    Further, if A denotes negative of A i.e., event that A doesn't happen, then for above cases m,n, we shall have P(A)=nmn=1mn=1P(A)
    P(A)+P(A)=1
  • Question 8
    1 / -0

    If the arithmetic mean of two unequal real numbers a and b (a > b) is three times their geometric mean, then a : b is

    Solution
    AM=a+b2
    GM=ab
    Given, (AM=a+b2
    GM=ab
    Given, AM=3GM
    a+b2=3abAM=3GMV
    a+b2=3ab
    Squaring both sides, we get
    a2+b2+2ab36ab
    a2+b234ab0
    (ab)234(ab)+1=0
    (ab)=34+11522=17±288
    From option 3+8:38 is equal to 3+838×3+83+8=17+288
    Hence, option (2) is correct
  • Question 9
    1 / -0
    The slope of the tangent to the curve y=0xdt1+t3 at the point where x=1 is
    Solution
    Given y=0xdt1+t3
    dydx=11+x3
    m=(dydx)x=1=11+1=12
    Newton Leibniz Theorem
    (1) If f(x) is continuous and u(x),v(x). are differentiable functions in the interval |a,b ]. then
    ddx(u(x)v(x)f(t)dt)=f{v(x)}ddx{v(x)}f{u(x)}ddx{u(x)}
    (2) If the function ϕ(x) and ψ(x) are defined on [a,b] and differentiable at a point xϵ(a,b), and f(x,t) is continuous, then
    ddx[ϕ(x)ψ(x)f(x,t)dt]=ϕ(x)ψ(x)ddxf(x,t)dt+{dψ(x)dx}f(x,ψ(x)){dϕ(x)dx}f(x,ϕ(x))
  • Question 10
    1 / -0

    Which of the following is logically equivalent to ~(~p⇒q)

    Solution
    As we know that
    pq(pq)
    Непсе, pq(pq)
    Непсе, (pq)≡∼(pq)
    Непсе, (pq)≡∼pq
    Hence, pq is the answer
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