Q52 / 105 · JEE Main 2008 · Mathematics

Statement - 1: For every natural number $n \geq 2, \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$. Statement $-2$ : For every natural number $n \geq 2, \sqrt{n(n+1)} < n+1$.

  • A
    Statement $-1$ is false, Statement $-2$ is true
  • B
    Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$
  • Correct answer
    C
    Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.
  • D
    Statement $-1$ is true, Statement $-2$ is false.

Solution

$$ \begin{aligned} & P(n)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}} \\ & P(2)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2} \end{aligned} $$ Let us assume that $P(k)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k}$ is true $\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$ has to be true. L.H.S. $>\sqrt{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}+1}}=\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$ Since $\sqrt{k(k+1)}>k \quad(\forall k \geq 0)$ $$ \therefore \frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}>\frac{\mathrm{k}+1}{\sqrt{\mathrm{k}+1}}=\sqrt{\mathrm{k}+1} $$ Let $P(n)=\sqrt{n(n+1)} < n+1$ Statement $-1$ is correct. $$ P(2)=\sqrt{2 \times 3} < 3 $$ If $P(k)=\sqrt{k(k+1)} < (k+1)$ is true Now $P(k+1)=\sqrt{(k+1)(k+2)} < k+2$ has to be true Since $(k+1) < k+2$ $$ \therefore \sqrt{(\mathrm{k}+1)(\mathrm{k}+2)} < (\mathrm{k}+2) $$ Hence Statement $-2$ is not a correct explanation of Statement $-1$.

Hence, the correct answer is option: 'C'.

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