Reduce the expression y -x2-x-1-x2-x-1 to the form a x2-b x-c and give condition for x to be real-A- y-12-2 - 0B- y-12-4y-12- 0C- y-12-4y-12- 0D- y-12-4y-12- 0

The correct option is B y= x^2 x+1/x^2+x+1 ⇒ x^2(y 1)+x(y + 1)+(y 1) = 0 This is in the form of ax^2 + bx +c =0 Condition for x to be real is D ≥ 0 ⇒ b^2 ...

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Standard XI

Mathematics

Range

Reduce the ex...

Question

Reduce the expression $y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ to the form $a x^{2} + b x + c$ and give condition for x to be real.

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Solution

The correct option is B

$y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ $\Rightarrow$ $x^{2} (y - 1) + x (y + 1) + (y - 1)$ = 0

This is in the form of $a x^{2} + b x + c = 0$

Condition for x to be real is D $\geq$ 0

$\Rightarrow b^{2} - 4 a c \geq 0$

$\Rightarrow (y + 1)^{2} - 4 (y - 1)^{2} \geq 0$

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Reduce the expression y=x2−x+1x2+x+1 to the form ax2+bx+c and give condition for x to be real.

The correct option is B y=x2−x+1x2+x+1 ⇒ x2(y−1)+x(y+1)+(y−1) = 0 This is in the form of ax2+bx+c=0 Condition for x to be real is D ≥ 0 ⇒b2−4ac≥0 ⇒(y+1)2−4(y−1)2≥0

Reduce the expression $y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ to the form $a x^{2} + b x + c$ and give condition for x to be real.

The correct option is B

$y = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ $\Rightarrow$ $x^{2} (y - 1) + x (y + 1) + (y - 1)$ = 0

This is in the form of $a x^{2} + b x + c = 0$

Condition for x to be real is D $\geq$ 0

$\Rightarrow b^{2} - 4 a c \geq 0$

$\Rightarrow (y + 1)^{2} - 4 (y - 1)^{2} \geq 0$