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Quadratic Equation and Inequalities question

2017 · Shift 0 · Q27
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  5. /2017 · Shift 0 · Q27

Quadratic Equation and Inequalities question

2017 · Shift 0 · Q27

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If for a positive integer n, the quadratic equation x(x+1)+(x+1)(x+2)+....+(x+n−1‾)(x+n)x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)x(x+1)+(x+1)(x+2)+....+(x+n−1​)(x+n) =10n= 10n=10n has two consecutive integral solutions, then n is equal to :
  1. A
    9
  2. B
    10
  3. C
    11
  4. D
    12
View written solutionFree

Correct answer: C

  1. Form the quadratic equation

We are given

(x)(x+1)+(x+1)(x+2)+⋯+(x+n−1)(x+n)=10n.(x)(x+1) + (x+1)(x+2) + \cdots + (x+n-1)(x+n) = 10n.(x)(x+1)+(x+1)(x+2)+⋯+(x+n−1)(x+n)=10n.

The general term is

(x+k)(x+k+1),k=0,1,2,…,n−1.(x+k)(x+k+1), \quad k=0,1,2,\dots,n-1.(x+k)(x+k+1),k=0,1,2,…,n−1.

So

∑k=0n−1(x+k)(x+k+1)=10n.\sum_{k=0}^{n-1} (x+k)(x+k+1) = 10n.k=0∑n−1​(x+k)(x+k+1)=10n.
  1. Expand the general term
(x+k)(x+k+1)=(x+k)2+(x+k).(x+k)(x+k+1) = (x+k)^2 + (x+k).(x+k)(x+k+1)=(x+k)2+(x+k).

Hence

∑k=0n−1(x+k)(x+k+1)=∑k=0n−1(x+k)2+∑k=0n−1(x+k).\sum_{k=0}^{n-1} (x+k)(x+k+1) = \sum_{k=0}^{n-1}(x+k)^2 + \sum_{k=0}^{n-1}(x+k).k=0∑n−1​(x+k)(x+k+1)=k=0∑n−1​(x+k)2+k=0∑n−1​(x+k).

Now,

∑k=0n−1(x+k)2=∑k=0n−1(x2+2xk+k2)=nx2+2x∑k=0n−1k+∑k=0n−1k2,\sum_{k=0}^{n-1}(x+k)^2 = \sum_{k=0}^{n-1}(x^2+2xk+k^2) = nx^2 + 2x\sum_{k=0}^{n-1}k + \sum_{k=0}^{n-1}k^2,k=0∑n−1​(x+k)2=k=0∑n−1​(x2+2xk+k2)=nx2+2xk=0∑n−1​k+k=0∑n−1​k2,

and

∑k=0n−1(x+k)=nx+∑k=0n−1k.\sum_{k=0}^{n-1}(x+k)=nx+\sum_{k=0}^{n-1}k.k=0∑n−1​(x+k)=nx+k=0∑n−1​k.

Using

∑k=0n−1k=n(n−1)2,∑k=0n−1k2=(n−1)n(2n−1)6,\sum_{k=0}^{n-1}k = \frac{n(n-1)}{2}, \qquad \sum_{k=0}^{n-1}k^2 = \frac{(n-1)n(2n-1)}{6},k=0∑n−1​k=2n(n−1)​,k=0∑n−1​k2=6(n−1)n(2n−1)​,

we get

∑k=0n−1(x+k)(x+k+1)=nx2+xn(n−1)+(n−1)n(2n−1)6+nx+n(n−1)2.\sum_{k=0}^{n-1}(x+k)(x+k+1) = nx^2 + x n(n-1) + \frac{(n-1)n(2n-1)}{6} + nx + \frac{n(n-1)}{2}.k=0∑n−1​(x+k)(x+k+1)=nx2+xn(n−1)+6(n−1)n(2n−1)​+nx+2n(n−1)​.

Combine the xxx-terms:

xn(n−1)+nx=nx[(n−1)+1]=n2x.x n(n-1)+nx = nx[(n-1)+1]=n^2x.xn(n−1)+nx=nx[(n−1)+1]=n2x.

Combine constants:

(n−1)n(2n−1)6+n(n−1)2=n(n−1)(2n−16+36)=n(n−1)2n+26=n(n−1)(n+1)3.\frac{(n-1)n(2n-1)}{6} + \frac{n(n-1)}{2} = n(n-1)\left(\frac{2n-1}{6}+\frac{3}{6}\right) = n(n-1)\frac{2n+2}{6} = \frac{n(n-1)(n+1)}{3}.6(n−1)n(2n−1)​+2n(n−1)​=n(n−1)(62n−1​+63​)=n(n−1)62n+2​=3n(n−1)(n+1)​.

Therefore the equation becomes

nx2+n2x+n(n−1)(n+1)3=10n.nx^2+n^2x+\frac{n(n-1)(n+1)}{3}=10n.nx2+n2x+3n(n−1)(n+1)​=10n.

Since n>0n>0n>0, divide throughout by nnn:

x2+nx+(n−1)(n+1)3=10.x^2+nx+\frac{(n-1)(n+1)}{3}=10.x2+nx+3(n−1)(n+1)​=10.

So

x2+nx+n2−13−10=0,x^2+nx+\frac{n^2-1}{3}-10=0,x2+nx+3n2−1​−10=0,

that is,

x2+nx+n2−313=0.x^2+nx+\frac{n^2-31}{3}=0.x2+nx+3n2−31​=0.
  1. Use the condition of two consecutive integral roots

Let the roots be two consecutive integers, say mmm and m+1m+1m+1.

For the monic quadratic

x2+nx+n2−313=0,x^2+nx+\frac{n^2-31}{3}=0,x2+nx+3n2−31​=0,

we have:

  • sum of roots =−n= -n=−n
  • product of roots =n2−313= \dfrac{n^2-31}{3}=3n2−31​

Since roots are consecutive integers,

m+(m+1)=2m+1=−n.m+(m+1)=2m+1=-n.m+(m+1)=2m+1=−n.

Thus nnn must be odd.

Now test the given odd options: n=9n=9n=9 and n=11n=11n=11.

  1. Check n=9n=9n=9

Equation becomes

x2+9x+81−313=0⇒x2+9x+503=0,x^2+9x+\frac{81-31}{3}=0 \Rightarrow x^2+9x+\frac{50}{3}=0,x2+9x+381−31​=0⇒x2+9x+350​=0,

which does not even have integer constant term, so it cannot have integral roots.

  1. Check n=11n=11n=11

Equation becomes

x2+11x+121−313=0=x2+11x+30=0.x^2+11x+\frac{121-31}{3}=0 = x^2+11x+30=0.x2+11x+3121−31​=0=x2+11x+30=0.

Factorizing,

x2+11x+30=(x+5)(x+6)=0.x^2+11x+30=(x+5)(x+6)=0.x2+11x+30=(x+5)(x+6)=0.

So the roots are

x=−5, −6,x=-5,\,-6,x=−5,−6,

which are consecutive integers.

Hence

n=11.\boxed{n=11}.n=11​.
  1. Option-wise conclusion
  • A: 9 — Incorrect
  • B: 10 — Incorrect
  • C: 11 — Correct
  • D: 12 — Incorrect

Therefore the correct option is

C\boxed{\text{C}}C​
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