Given that a, b and c are positive real numbers, such that a+b+c= x, then which of the following is true?

(x-a)(x-b)(x-c)≥8abc

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

1/a + 1/b + 1/c ≥ 9/x

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

(x-a)(x-b)(x-c)≥8/27a³

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

a,b and c

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

Open in App

Solution

The correct option is D a,b and c
Assumption of values is the easiest way to solve. a=b=c=1 and x=3.

Take a=1, b=2 and c=3. x=6. all 3 still hold good. Answer option d.

Note: with answer options like “All of these” and “None of these” you need to be a bit careful. Do not mark the answer based on only one set of values. Take two cases and see which options fit in BOTH the cases.

Suggest Corrections

Similar questions

Q. If

a

b

c

be non-zero real numbers such that

\int_{0}^{1} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x = 0

. Then the equation

a x^{2} + b x + c = 0

will have

Q. Let

a, b, c

be non – zero real numbers such that

\int_{0}^{1} (1 + c o s^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + c o s^{8} x) (a x^{2} + b x + c) d x .

Then the quadratic equation

a x^{2} + b x + c = 0

has

Let a, b, c be non-zero real numbers such that
$\int_{0}^{1} (1 + c o s^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + c o s^{8} x) (a x^{2} + b x + c) d x$
Then the quadratic equation $a x^{2} + b x + c = 0$ has [IIT 1981; CEE 1993]

Q. If

y = \frac{{a x}^{2}}{(x - a) (x - b) (x - c)} + \frac{b x}{(x - b) (x - c)} + \frac{c}{x - c} + 1

then prove that

\frac{d y}{d x} = \frac{y}{x} [\frac{a}{a - x} + \frac{b}{b - x} + \frac{c}{c - x}]

Q. If

y = \frac{a x^{2}}{(x - a) (x - b) (x - c)} + \frac{b x}{(x - b) (x - c)} + \frac{c}{x - c} + 1,

then

\frac{y^{'}}{y}

is equal to

(Here, y^{'} = \frac{d y}{d x})

Join BYJU'S Learning Program