Can you please check, there must be a brackets in $B$ option.

@Arjun Sir

Like $(\frac{\lambda + 2}{\lambda})$

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If $a,b$ are positive real variables whose sum is a constant $\lambda$, then the minimum value of $\sqrt{(1+1/a)(1+1/b)}$ is

- $\lambda \: – 1/\lambda$
- $\lambda + 2/\lambda$
- $\lambda+1/\lambda$
- None of the above

0

Can you please check, there must be a brackets in $B$ option.

@Arjun Sir

Like $(\frac{\lambda + 2}{\lambda})$

0

Are the first three options like below?

$\mathrm{A.}~\frac{\lambda-1}{\lambda}$

$\mathrm{B.}~\frac{\lambda+2}{\lambda}$

$\mathrm{C.}~\frac{\lambda+1}{\lambda}$

$\mathrm{A.}~\frac{\lambda-1}{\lambda}$

$\mathrm{B.}~\frac{\lambda+2}{\lambda}$

$\mathrm{C.}~\frac{\lambda+1}{\lambda}$

0

no mistake in editing. brackets are not there...

https://www.isical.ac.in/~admission/IsiAdmission2017/PreviousQuestion/MMA-2015.pdf

+2 votes

__ Answer: __$\mathbf B$

**Explanation:**

Let $a + b = \lambda$

$\therefore b = \lambda - a$

Let $f = \sqrt{({1 + \frac{1}{a}})(1 + \frac{1}{b})}$

Let $g = f^2 =({1 + \frac{1}{a}})(1 + \frac{1}{b})$

Taking derivative of $g$ with respect to $a$ and making it equal to $0$, we get:

$g' = \frac{-1}{a^2}(1+\frac{1}{\lambda-a}) + (1+\frac{1}{a})(\frac{1}{(\lambda-a)^2}) = 0$

Simplifying this we get:

$(\lambda-2a)(\lambda+1) = 0, \;or \; a = \frac{\lambda}{2}$

Again on differentiating $g$ with respect to a, we get:

$g^{''} = (\frac{2}{a^3})(1+\frac{1}{\lambda-a}) - \frac{1}{a^2}\frac{1}{(\lambda-a)^2} + (1+\frac{1}{a})(\frac{2}{(\lambda-a)^3})-\frac{1}{a^2}(\frac{1}{(\lambda-a)^2})$

At $a = \frac{\lambda}{2}$ the value of $g^{''}>0$

$\Rightarrow $ It's a point of minima.

$\therefore$ Minimum value $= \frac{\lambda+2}{\lambda}$

$\therefore \mathbf B$ is the correct option.

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