Given ax 2- bx - c - 0- bx 2- cx - a - 0- cx 2- ax - b - 0 where - b - c and a - b - c - R- Now a 2- b 2- c 2- ab - bc - ca cannot take the valuesA- 7 - 9B- 2 - 7C- 16 - 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 ndash; 4ac ≤ 0, c^2 ndash; 4ab ≤ 0 , a^2 ndash; 4bc ≤ 0 ⇒ a^ ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Given ax 2+ 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)

7/9

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2/7

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16/3

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-3/2

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Solution

The correct options are
A
7/9

B
2/7

C
16/3

D
-3/2

$a > 0, b > 0, c > 0 a n d$

$b^{2} - 4 a c \leq 0$ , $c^{2} - 4 a b \leq 0$ , $a^{2} - 4 b c \leq 0$

$\Rightarrow$ $a^{2} + b^{2} + c^{2}$ $\leq 4 (a b + b c + c a)$

$A n d (a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 0$

$\Rightarrow 1 \leq$ $\frac{a^{2} + b^{2} + c^{2}}{a b + b c + c a}$ $\leq 4$

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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. Suppose

a, b, c ϵ R

, a+b+c> 0,

A = b c - a^{2}

B = c a - b^{2}

C = a b - c^{2}

and

∣ ∣ ∣ ∣ \begin{matrix} A & B & C B & C & A C & A & B \end{matrix} ∣ ∣ ∣ ∣ = 49

then

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣

equals :

Q. If the quadratic equation

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

, where

a, b, c

are distinct real numbers, has imaginary roots, then

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. If

α

is common root of both these equations

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

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

,then

\frac{α^{2}}{a b - c^{2}} = \frac{α}{c b - a^{2}} = \frac{1}{a c - b^{2}}

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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 7/9 B 2/7 C 16/3 D -3/2 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)