Welcome (I am proud to be with you..)

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I have started this blog to share some Mathematics e.g. solutions to WBSSC Math. papers, C.U. Math. Hons. & Pass papers,AIEEE Math papers, WBJEE Math. papers, WBCHSE math. papers etc. Also I share some links related to syllebus & result of these Exams. So follow this blog for regular updates.

I look forward to my colleagues and students for their constructive suggestions by either commenting on this blog or E-mailing me.

69 responses to “Welcome (I am proud to be with you..)

  1. You r doing such a nice job Sir. But plzzzzz kindly upload class Xll Philosophy question pattern for H.S-2015. Its my heartest request to you.

    • Let ,at time t radius & area of the ink drop are r & A respectively.
      So, A = 22/7 r^2 Thus dA/st = 2. 22/7. r.dr/dt.
      By the problem dA/dt =2
      So, 2. 22/7. r. dr/dt = 2 Thus 22/7. r. dr/dt = 1
      Integrating we get, t= 1/2. 22/7. r^2 ( with the condition r = 0 at time t= 0)
      At t=28/11, r = 14/11.
      Hence at r = 14/11, dr/dt = 7/22. 11/14 = 1/4 =0.25 cm/sec.

  2. Suppose A₁,A₂ ,A₃ ,……………A₃₀,each of these 30 sets contains 5 terms,
    B₁,B₂,B₃ ,…………….Bᵣ, each of these r sets contains 3 terms,
    A₁ U A₂ U……….. U A₃₀= B₁ U B₂ U B₃ U…………….U Bᵣ=S
    And consider each term of set S is contained in just ten A and nine B sets.
    Find the value of r.

  3. 1)If x=a+a/r+a/r²+…….∞ ,-1<1/r<1
    y=b-b/r+b/r²-………∞ & z=c+c/r+c/r⁴+…….∞
    then show that xy/z=ab/c.
    2)If p-th term be q &q-th term be p in a harmonic progression, then prove that it’s (p+q)-th term will be pq/(p+q).
    3)if A.M & H.M of two numbers are 15&48/5,then find the numbers

    • 1)
      x=a+a/r+a/r²+…….∞ = a/(1-1/r) = ar/(r-1)
      y=b-b/r+b/r²-………∞ = b/(1+1/r) = br/(r+1)
      & z=c+c/r²+c/r⁴+…….∞ = c/(1-1/r²) = cr²/(r²-1)
      Now xy/z = ar/(r-1) . br/(r+1) . (r²-1)/cr² = ab/c

    • 2)
      Let first term & common difference of the corresponding A.P are a & d respectively.
      Then a+(p-1)d = 1/q & a+(q-1)d = 1/p
      So (p-q)d = 1/q-1/p i.e, d = 1/(pq)
      Again a = 1/q-(p-1)/(pq)= 1/(pq)
      So a+(p+q-1)d = 1/(pq)+(p+q-1)/(pq) = 1/(pq)+(p+q)/(pq)-1/(pq) = (p+q)/(pq)
      Thus (p+q)th term is pq/(p+q)

    • 3)
      Let the reqd. nos. are a & b.
      So (a+b)/2 = 15 i.e, a+b = 30
      And 2/(1/a+1/b) = 48/5 or, 2ab/(a+b) = 48/5
      or, ab = 144 [since a+b = 30]
      Now (a-b)² = 900-576 = 324
      So a-b = 18 , -18
      Then a = 24 , 6 & b = 6 , 24
      Thus the reqd, nos, are 6 & 24

  4. If α, β & γ are the coefficient of linear, surface & volume expansion of solids respectively i.e. α=(dl/dθ)1/l β=(ds/dθ)1/s γ=(dv/dθ)1/v
    1)When a sphere of radii ‘r’ is heated, it’s volume increases. In that case prove that α=β/2=γ/3.
    2)When a cylinder of radii ‘r’ & height ‘l’ is heated, it’s volume increases.in that case prove that α=β/2=γ/3.
    (NB.l=length,s=area of surface,v=volume, θ=temperature)

    • 1)α=(1/r) (dr/dθ)
      β=(1/s)( ds/dθ)=(1/4πr^2)8πr(dr/dθ)=2(1/r)( dr/dθ)=2 α
      γ=(1/v)( dv/dθ)=(3/4πr^3)4πr^2 (dr/dθ)
      =3(1/r) (dr/dθ)=3 α
      ∴ α=β/2=γ/3

      2)(1/l) (dl/dθ)= α=(1/r) (dr/dθ)
      β=(1/s)( ds/dθ)=(1/2πrl)[2πr(dl/dθ)+ 2πl(dr/dθ)]
      = (1/l)( dl/dθ)+ (1/r)( dr/dθ)=2 α
      γ=(1/v)( dv/dθ)=(1/πr^2l)[πr^2 (dl/dθ) +2πrl(dr/dθ)]
      = (1/l)( dl/dθ)+ 2(1/r) (dr/dθ)=3 α
      ∴ α=β/2=γ/3

    • Mathematics has no generally accepted definition. Benjamin Peirce defined mathematics as:
      “Mathematics is the science that draws necessary conclusions”

      • I think MATHEMATICS is the spontaneous knowledge from the depth of the good thinking of Human beings…
        We can’t live without Mathematics. Not just we, also the Universe follows the rule of Math I think.
        Ultimately Mathematics was born at the time when the universe learn to love…
        Because I think WITHOUT LOVE WE CAN DO NOTHING WITH MATHEMATICS.
        It’s also true that when we think of true Mathematics then our mind stays in a absolute natural piety phase.🙂
        What do you think?
        plz reply.

      • You are absolutely right. A game whose tasks are determined by arbitrarily stipulated rules. Mathematics is not these type of games. A conceptual system that possess internal necessity and can only be so. Mathematics is like these type of systems.

    • “Mathematics is the language of science and technology”, It may be other defination of Mathematics but for me this is the concept.
      Hom many of you go with my concept?
      Please comment
      Mujhe bhi kuch seekhne milega……

  5. 1)Prithibi theke surjer durotto,prithibi theke chandrer durotter 360 gun ebong chandro o surjer bas porjobekhkhaker chokhe jathakrome 30′ o 32′(second) kon utponno kore.
    surjo o chandrer basardho nirnay koro.
    2)Duti britto emon je , ekti oportir kendrabindugami. prothom britter L doirgher chap tar kendre θ ebong ditiyo britter M doirgher chap tar kendre Ø sommukh kon utponno kore. dekhao je LØ=Mθ.

    • 1)Let the distance from the Earth to Sun and Moon are 360x and x units respectively.
      Let O be the position of the men and ACB is the diameter of the Sun.
      By the problem, Angle AOB = 32 minute = 32/60 degree = (32/60).(π/180) radian.
      Obviously, Angle AOB is very small.
      So, Diameter of the Sun ACB = Arc length AB of the circle whose center is at O and radius is OA or OB = 360x.(32/60).(π/180) units.
      Similarly diameter of the Moon = x.(30/60).(π/180) units.
      So the ratio of the radius of Sun and Moon is 360x.(32/60).(π/180) : x.(30/60).(π/180)
      = 384 : 1
      2) By the problem radius of two circles are same.
      Let r be the radius of the circle.
      So, l = rθ and m = r∅.
      Then, l/θ = m/∅.
      Hence, l∅ = mθ

  6. A police started to chase a thief by a van at a uniform velocity of ‘V’. When the van is ‘d’ distance away from the thief, he started to flee (from rest condition) at a uniform acceleration of ‘F’. Now show that the police will be able to catch the thief if V≥√(2Fd).

    • Let P & T are the position of the police & thief respectively.
      Then PT=d.
      If possible let, after time t at M police catch the thief.
      So, TM=(1/2) Ft^2 & PM=Vt
      Now, PT+TM=PM
      ⇒ d+ (1/2) Ft^2=Vt
      ⇒ Ft^2 – 2Vt+2d=0
      ⇒t=(2V±√(4V^2 – 8Fd))/2F
      = (V±√(V^2 – 2Fd))/F
      Police catch the thief if t is real
      i.e, if V^2 – 2Fd≥0 i.e, if V≥√(2Fd)
      (Note: If V^2 – 2Fd≤0 then time t is imaginary so in this case police unable to catch the thief)

    • 1) √i+√(-i)
      =√( cos π/2 + i sin π/2 ) + √( cos π/2 – i sin π/2 )
      One of the value is cos π/4 + i sin π/4 + cos π/4 – i sin π/4
      =2 cos π/4
      = 2/√2
      = √2

      2) Let z = 2/(1+cos θ + i sin θ)
      =2(1+cos θ – i sin θ)/ 2(1+cos θ)
      =1 – i tan θ/2
      Then |z|=sec θ/2

  7. (x-a)(x-b)=c(c not=o) somikoroner bij duti “alfa” o “bita” hole prove that (x-alfa)(x-bita)+c=0 somikoroner bij duti “a” and “b”

      • IF x+y VARIES AS TO z WHEN y IS CONSTANT AND z+X VARIES AS TO y WHEN z IS CONSTANT THEN SHOW THAT x+y+z VARIES AS TO yz WHEN y AND z ARE VARIABLES.

      • Since (x + y) ∝ z when y is constant

        Therefore, x + y = kz [where, k = constant of variation] when y is constant

        or, x + y + z = kz + z = (k + 1) z when y is constant.

        Therefore x + y + z ∝ z when y is constant [since (k + 1) = constant] ……. (1)

        Again, (x + z ) ∝ y when z is constant.

        Therefore x + z = my [where, m = constant of variation] when z is constant.

        or, x + y + z = my + y = (m + 1) y when z is constant.

        Therefore x + y + z ∝ y when z is constant [since, (m + 1) = constant]….. (2)

        From (1) and (2), using the theorem of joint variation, we get, x + y + z ∝ yz when both y and z vary.

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