Given a x2-b x-c - 0- b x2-c x-a - 0- c x2-a x-b - 0 where q-b-c and a- b- c - R- Now- b 2- c 2 cannot take the valuesA- B- -7C- 6-3D- -3-2

The correct options are A 7/9 B 2/7 C 16/3 D 3/2 a gt;0, b gt;0, c gt;0 and b^2 4ac ≥ 0, c^2 4ab ≤ 0, a^2 4bc ≤ 0 ⇒ a^2 + b^2 + c^2 ≤ 4 (ab ...

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Byju's Answer

Standard XII

Mathematics

nth Term of A.P

Given a x2+b ...

Question

Given $a x^{2} + b x + c \geq 0, b x^{2} + c x + a \geq 0, c x^{2} + a x + b \geq 0$ where $a \neq b \neq c$ and $a, b, c ϵ R$ . Now
$\frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}$ cannot take the value(s)

$\frac{7}{9}$

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$\frac{2}{7}$

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$\frac{16}{3}$

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$\frac{- 3}{2}$

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Solution

The correct options are
A
$\frac{7}{9}$

B
$\frac{2}{7}$

C
$\frac{16}{3}$

D
$\frac{- 3}{2}$

a>0, b >0, c >0 and
$b^{2} - 4 a c \geq 0, c^{2} - 4 a b \leq 0, a^{2} - 4 b c \leq 0$
$\Rightarrow a^{2} + b^{2} + c^{2} \leq 4 (a b + b c + c a)$
And $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 0$
$\Rightarrow 1 \leq \frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a} \leq 4$

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Similar questions

Given $a x^{2}$ $+$ $b x$ $+ c$ $\geq 0$ , $b x^{2} + c x + a \geq 0$ , $c x^{2} + a x + b \geq 0$ where $a \neq b \neq c$ and $a, b, c ϵ R$ . Now $\frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}$ cannot take the value(s)

Q. Assertion :

Consider the function f (x) = {log}_{c} (a x^{3} + (a + b) x^{2} + (b + c) x + c)

Domain of the functions is

(- 1, \infty) \sim {- (b / 2 a)},

where

a > 0, b^{2} - 4 a c = 0

Reason:

Consider the function f (x) = {log}_{c} (a x^{3} + (a + b) x^{2} + (b + c) x + c)

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

has equal roots when

b^{2} - 4 a c = 0

Q. Assertion :If

a + b + c > 0

a < 0 < b < c

, then roots of the equation

a (x - b) (x - c) + b (x - c) (x - a) + c (x - a) (x - b) = 0

are real. Reason: Roots of the equation

A x^{2} + B x + K = 0

are real if

B^{2} - 4 A K > 0

Q. If the equation

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

and

b x^{2} + c x + 1 = 0

have a common root, then prove that either

b + c + 1 = 0

b^{2} + c^{2} + 1 = b c + b + c

Q. The quadratic equation,

a (x - b) (x - c) + b (x - c) (x - a) + c (x - a) (x - b) = 0

where

0 < a < b < c,

has:

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Given ax2+bx+c≥0,bx2+cx+a≥0,cx2+ax+b≥0 where a≠b≠c and a,b,cϵR. Now a2+b2+c2ab+bc+ca cannot take the value(s)

The correct options are A 79 B 27 C 163 D −32 a>0, b >0, c >0 and b2−4ac≥0,c2−4ab≤0,a2−4bc≤0 ⇒a2+b2+c2≤4(ab+bc+ca) And (a−b)2+(b−c)2+(c−a)2≥0 ⇒1≤a2+b2+c2ab+bc+ca≤4

Given $a x^{2} + b x + c \geq 0, b x^{2} + c x + a \geq 0, c x^{2} + a x + b \geq 0$ where $a \neq b \neq c$ and $a, b, c ϵ R$ . Now
$\frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}$ cannot take the value(s)