Question

answer with proper explanation...
right answer will be marked brainliest​

Answer 1

$\large\underline{\text{Step 1. Quadratic equation.}}$

Let $y=\dfrac{x^{2}+2x+a}{x^{2}+4x+3a}$. (Attachment 1)

$\iff y=\dfrac{x^{2}+2x+a}{x^{2}+4x+3a}$

$\iff y(x^{2}+4x+3a)=x^{2}+2x+a$

$\iff yx^{2}-x^{2}+4yx-2x+3ay-a=0$

$\iff(y-1)x^{2}+2(2y-1)x+3ay-a=0$

$\large\underline{\text{Step 2. Real solutions.}}$

The equation above shows the rational function graph. Solving this equation for $x$, as solutions need to be real, we get the range of $y\in\mathbb{R}$ by discriminant.

$\iff \dfrac{D}{4}=(2y-1)^{2}-(y-1)(3ay-a)\geq0$

$\iff \dfrac{D}{4}={-3ay^{2}+4ay-a+4y^2-4y+1}\geq0$

$\iff \dfrac{D}{4}=(4-3a)y^{2}+(4a-4)y-a+1\geq0$

Now, this equation should be true for all $y$.

$\large\underline{\text{Step 3. Graph for }y\text{.}}$

Let's draw the graph above. (Attachment 2)

For the graph for $y$;

• to open up, the leading coefficient should be positive.
• to intersect once or never, the discriminant should be non-positive.

Then the graph will have a non-positive discriminant,

$\iff \dfrac{D}{4}=(2a-2)^{2}-(4-3a)(-a+1)\leq0$

$\iff \dfrac{D}{4}=4a^{4}-8a+4+4a-4-3a^{2}+3a\leq0$

$\iff \dfrac{D}{4}=a^{2}-a\leq0$

$\iff\dfrac{D}{4}=a(a-1)\leq0$

$\iff 0\leq a\leq1$

and the leading coefficient will be positive.

$\iff4-3a>0$

$\iff a<\dfrac{4}{3}$

The union is $a\in[0,1]$, which is the option (4).

Answer 2

Answer:

Step-by-step explanation:

• ⇒ x² + 2x + a = k (x² + 4x + 3a)

• ⇒ x² + 2x + a-k (x² + 4x + 3a) = 0

• ⇒ (1 − k) x² + (2 - 4k) x + (a -3ka) = 0

• Since x € R, D≥0

• (2-4k)² - 4x (1-k) (a -3ka) ≥ 0

• ⇒ 4+16k² - 16 k. - 4 (a - 3ka - ak + 3k²a) ≥ 0

• ⇒ 4+16k² - 16k4a + 12ka + 4ak –

• 12k²a > 0

• 1 + 4k² - 4k -a + 3ka +ak – 3k²a > 0

• ⇒ k² (4 - 3a) + k (4a − 4) + 1 - a ≥ 0

• ⇒ 4 - 3a > 0

• → 3a < 4

• D <0

• ⇒ (4a − 4)² – 4 (4 – 3a) (1 − a) ≤ 0

• ⇒ 16(a − 1)² - 4 (4 - 3a) (1 - a) ≤ 0

• ⇒ (1-a) (4 (1-a) - 4+3a) ≤ 0

• a < 4/3

• (4a - 4)² - 4 (4 – 3a) (1 − a) ≤ 0

• ⇒ 16(a − 1)² – 4 (4 – 3a) (1 − a) ≤ 0

• ⇒ (1-a) (4 (1-a) −4+3a) ≤ 0

• ⇒ (1-a) (-a) ≤ 0

• ⇒ (a-1) a ≤0

• ⇒ a € [0, 1]