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Quadratic Equation and Inequalities question

2008 · Shift 0 · Q52
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  5. /2008 · Shift 0 · Q52

Quadratic Equation and Inequalities question

2008 · Shift 0 · Q52

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
STATEMENT - 1 : For every natural number n≥2,11+12+........+1n>n.n \ge 2,{1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + ........ + {1 \over {\sqrt n }} \gt \sqrt n .n≥2,1​1​+2​1​+........+n​1​>n​. STATEMENT - 2 : For every natural number n≥2,n \ge 2,n≥2,, n(n+1)<n+1.\sqrt {n\left( {n + 1} \right)} \lt n + 1.n(n+1)​<n+1.
  1. A
    Statement - 1 is false, Statement - 2 is true
  2. B
    Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for statement - 1
  3. C
    Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
  4. D
    Statement - 1 is true, Statement - 2 is false
View written solutionFree

Correct answer: B

  1. Statement 1:

We need to check whether

1+12+13+⋯+1n>nfor all n≥2.1+\frac1{\sqrt2}+\frac1{\sqrt3}+\cdots+\frac1{\sqrt n}>\sqrt n \quad \text{for all } n\ge 2.1+2​1​+3​1​+⋯+n​1​>n​for all n≥2.

A useful identity is

1r=rr.\frac1{\sqrt r}=\frac{\sqrt r}{r}.r​1​=rr​​.

Now compare it with

r+1−r=(r+1)−rr+1+r=1r+1+r.\sqrt{r+1}-\sqrt r=\frac{(r+1)-r}{\sqrt{r+1}+\sqrt r}=\frac1{\sqrt{r+1}+\sqrt r}.r+1​−r​=r+1​+r​(r+1)−r​=r+1​+r​1​.

Since

r+1+r>r,\sqrt{r+1}+\sqrt r>\sqrt r,r+1​+r​>r​,

we get

r+1−r<1r.\sqrt{r+1}-\sqrt r<\frac1{\sqrt r}.r+1​−r​<r​1​.

So for each r=1,2,…,nr=1,2,\dots,nr=1,2,…,n,

1r>r+1−r.\frac1{\sqrt r}>\sqrt{r+1}-\sqrt r.r​1​>r+1​−r​.

Summing from r=1r=1r=1 to nnn,

∑r=1n1r>∑r=1n(r+1−r).\sum_{r=1}^n \frac1{\sqrt r}> \sum_{r=1}^n (\sqrt{r+1}-\sqrt r).r=1∑n​r​1​>r=1∑n​(r+1​−r​).

The right-hand side telescopes:

(2−1)+(3−2)+⋯+(n+1−n)=n+1−1.(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{n+1}-\sqrt n)=\sqrt{n+1}-1.(2​−1)+(3​−2​)+⋯+(n+1​−n​)=n+1​−1.

Hence

∑r=1n1r>n+1−1.\sum_{r=1}^n \frac1{\sqrt r}>\sqrt{n+1}-1.r=1∑n​r​1​>n+1​−1.

But to prove Statement 1, we need a stronger inequality: ∑1/r>n\sum 1/\sqrt r>\sqrt n∑1/r​>n​.

So let us use a sharper comparison:

1r>2r+1+r=2(r+1−r)?\frac1{\sqrt r}>\frac{2}{\sqrt{r+1}+\sqrt r}=2(\sqrt{r+1}-\sqrt r)?r​1​>r+1​+r​2​=2(r+1​−r​)?

Check this:

1r>2r+1+r  ⟺  r+1+r>2r  ⟺  r+1>r,\frac1{\sqrt r}>\frac{2}{\sqrt{r+1}+\sqrt r} \iff \sqrt{r+1}+\sqrt r>2\sqrt r \iff \sqrt{r+1}>\sqrt r,r​1​>r+1​+r​2​⟺r+1​+r​>2r​⟺r+1​>r​,

which is true. Thus,

1r>2(r+1−r).\frac1{\sqrt r}>2(\sqrt{r+1}-\sqrt r).r​1​>2(r+1​−r​).

Summing from r=1r=1r=1 to nnn,

∑r=1n1r>2∑r=1n(r+1−r)=2(n+1−1).\sum_{r=1}^n \frac1{\sqrt r}>2\sum_{r=1}^n (\sqrt{r+1}-\sqrt r)=2(\sqrt{n+1}-1).r=1∑n​r​1​>2r=1∑n​(r+1​−r​)=2(n+1​−1).

So it is enough to show

2(n+1−1)>n.2(\sqrt{n+1}-1)>\sqrt n.2(n+1​−1)>n​.

This is equivalent to

2n+1>n+2.2\sqrt{n+1}>\sqrt n+2.2n+1​>n​+2.

Squaring both sides (both sides positive):

4(n+1)>n+4+4n4(n+1)>n+4+4\sqrt n4(n+1)>n+4+4n​ 3n>4n.3n>4\sqrt n.3n>4n​.

For n≥2n\ge 2n≥2, this is true because squaring gives

9n2>16n  ⟺  9n>16,9n^2>16n \iff 9n>16,9n2>16n⟺9n>16,

which holds for n≥2n\ge 2n≥2. Therefore,

∑r=1n1r>n.\sum_{r=1}^n \frac1{\sqrt r}>\sqrt n.r=1∑n​r​1​>n​.

Hence Statement 1 is true.

  1. Statement 2:

We need to check whether

n(n+1)<n+1for all n≥2.\sqrt{n(n+1)}<n+1 \quad \text{for all } n\ge 2.n(n+1)​<n+1for all n≥2.

Since both sides are positive, square both sides:

n(n+1)<(n+1)2.n(n+1)<(n+1)^2.n(n+1)<(n+1)2.

This becomes

n2+n<n2+2n+1n^2+n<n^2+2n+1n2+n<n2+2n+1 0<n+1,0<n+1,0<n+1,

which is true for every natural number nnn. Thus Statement 2 is true.

  1. Does Statement 2 explain Statement 1?

From Statement 2,

r(r+1)<r+1.\sqrt{r(r+1)}<r+1.r(r+1)​<r+1.

Dividing by rr+1\sqrt r\sqrt{r+1}r​r+1​,

1<r+1rr+1=r+1r.1<\frac{r+1}{\sqrt r\sqrt{r+1}}=\frac{\sqrt{r+1}}{\sqrt r}.1<r​r+1​r+1​=r​r+1​​.

More usefully, Statement 2 implies

1r>rr+1.\frac1{\sqrt r}>\frac{\sqrt r}{r+1}.r​1​>r+1r​​.

But the standard telescoping form connected to it is

1r>2r+1+r=2(r+1−r),\frac1{\sqrt r}>\frac{2}{\sqrt{r+1}+\sqrt r}=2(\sqrt{r+1}-\sqrt r),r​1​>r+1​+r​2​=2(r+1​−r​),

which indeed follows from

r(r+1)<r+1.\sqrt{r(r+1)}<r+1.r(r+1)​<r+1.

So Statement 2 gives the key inequality needed in the telescoping proof of Statement 1. Hence Statement 2 is a correct explanation of Statement 1.

  1. Final option

Both statements are true, and Statement 2 correctly explains Statement 1.

Therefore, the correct option is

B\boxed{\text{B}}B​
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