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    5. Question Bank - Equations - Set 1

    Question Bank - Equations - Set 1

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    • Amar Rajput last edited by @anurag_chauhan

      @anurag_chauhan said in Question Bank - Equations - Set 1:

      1/(a+b) + 1/(b+c) + 1/(c+a)

      4

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      • Amar Rajput last edited by @anurag_chauhan

        @anurag_chauhan
        is it -1 ?

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        • Amar Rajput last edited by @anurag_chauhan

          @anurag_chauhan 2

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          • Amar Rajput last edited by @anurag_chauhan

            @anurag_chauhan 2

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            • Amar Rajput last edited by @anurag_chauhan

              @anurag_chauhan

              -3

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              • Amar Rajput last edited by @anurag_chauhan

                @anurag_chauhan
                6 ?

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                • Amar Rajput last edited by @anurag_chauhan

                  @anurag_chauhan 2

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                  • Amar Rajput last edited by @anurag_chauhan

                    @anurag_chauhan 0

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                    • Amar Rajput last edited by @anurag_chauhan

                      @anurag_chauhan 6

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                      • Amar Rajput last edited by @anurag_chauhan

                        @anurag_chauhan 10

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                        question bank 47 theory of equations 14
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                          @sumit-agarwal In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1 if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76 if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511 if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10 Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2). You can try out with various numbers (may be smaller numbers) so that this can be verified.
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                          We can do it another way too: Put x= y,F(2x) = f(x)^2Put y=2xF(3x)= f(x)^3=>f(nx)=f(x)^n F(-8)=F(-2×4)=f(4)^-2=1/9
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                          @Rowdy-Rathore please give the solution
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                          @Rowdy-Rathore F(x) = x^4-360x^2+400F(x) = (x^2+20)^2 - 400x^2F(x) = (x^2-20x+20)(x^2+20x+20) = Prime NumberSince the value of second factor is always positive and greater than 1 therefore, 1st factor hasto be equals to 1.x^2-20x+20 = 1x^2-20x+19= 0On solving,x = 1, 19.Putting back the values in the original equation we get f(x) = 41 and f(x) = 761. So, the sum will be 41+761 = 802
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                          14 to 28 : 15 students 30, 31, 32 and 1 to 12 : 15 students so total 32
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