If a>0, b>0 a...

Question

If a > 0, b > 0 and ab = 1, then the least value of the expression (1 + a) (1 + b) is :

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $4$
If ab is constant, then (a+b) takes minimum value
when a = b
$∴$ a = b = 1
$∴$ (1 + a) (1 + b) = (1 + 1)(1 + 1) = 4

Suggest Corrections

Similar questions

a, b \in R

a, b > 0

a b \leq \frac{1}{4} {(a + b)}^{2}

a + b = 1

(1 + \frac{1}{a}) (1 + \frac{1}{b}) .

If A + B = $\frac{π}{4}$ where A, B $\in R^{+}$ , then the minimum value of (1+tanA) (1+tanB) is always equal to

a, b ϵ R, a, b > 0

a b \leq \frac{1}{4} (a + b)^{2}

a, b > 0,

a + b = 1

(1 + \frac{1}{a}) (1 + \frac{1}{b})

a^{2} b^{4} c^{6} - 1

x = 1 + a + a^{2} + . . . . \infty

y = 1 + b + b^{2} + . . . . \infty,

1 + a b + + a^{2} b^{2} + . . . . \infty

(0 < a < 1, 0 < b < 1)