Question

please solve this question ​

Answer 1

Answer:

⇒a∈[0,1]

Step-by-step explanation:

Let

x

2

+4x+3a

x

2

+2x+a

=k

⇒x

2

+2x+a=k(x

2

+4x+3a)

⇒x

2

+2x+a−k(x

2

+4x+3a)=0

⇒(1−k)x

2

+(2−4k)x+(a−3ka)=0

Since x∈R,D≥0

⇒(2−4k)

2

−4×(1−k)(a−3ka)≥0

⇒4+16k

2

−16k−4(a−3ka−ak+3k

2

a)≥0

⇒4+16k

2

−16k−4a+12ka+4ak−12k

2

a≥0

⇒1+4k

2

−4k−a+3ka+ak−3k

2

a≥0

⇒k

2

(4−3a)+k(4a−4)+1−a≥0

⇒4−3a>0

⇒3a<4

∴a<

3

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

⇒(1−a)(−a)≤0

⇒(a−1)a≤0

⇒a∈[0,1]