Question

consider y=2x/1+x^2, where x is real then the range of the expression y^2+y-2 is [a,b]. find b-4a

Answer 1

(OPEN BRAINLY WEBSITE FOR VIEWING THE ANSWER)$\text{Notice,}\\\begin{cases}(x-1)^2>0\\(x+1)^2>0\end{cases}\\\begin{cases}x^2+1-2x>0\\x^2+1+2x>0\end{cases}\\\Rightarrow\begin{cases}2x<1+x^2\\2x>-1-x^2\end{cases}\\\text{As }1+x^2}>0, \text{I may divide both sides by }1+x^2\\\text{and the sign will not change}\\\Rightarrow -1<\frac{2x}{1+x^2}<1\\\Rightarrow -1<y<1\\\text{Now, complete the square in } y^2-y+2\text{ as}\\y^2-y+2\\=y^2-2\cdot\frac{1}{2}\cdot y+(\frac{1}{2})^2+\frac{7}{4}\\=(y-\frac{1}{2})^2+\frac{7}{4}\\\text{The quadratic expression has minimum value at }y=\frac{1}{2}\\\text{which is in the domain:}\Rightarrow \text{ Minimum value}=\frac{7}{4}\\\text{For maximum value observe the domain} -1<y<1\\\text{and use symmetry arguments about }y=\frac{1}{2}\\\text{As y reaches -1 from }\frac{1}{2}\text{ the increase in value of expression}\\\text{should be more than that when} \text{ y reaches 1 from } \frac{1}{2}\\\text{,as the change in value in y is more in case of former.}\\\text{So, Maximum value }=(-1)^2-(-1)+2=4\\\text{Therefore, Range of expression}=[\frac{7}{4},4]=[a,b]\\\text{Therefore, }b-4a=4-4\cdot\frac{7}{4}=-3$

Answer 2

Answer:

Hence, the value of b-4a=8

Step-by-step explanation:

We are given a variable y in terms of x as:

$y=\dfrac{2x}{1+x^2}$

Now we know that:

$(1+x)^2=1+x^2+2x\geq 0$

Also,

$(x-1)^2=x^2+1-2x\geq 0$

Hence,

$x^2+1\geq2x\\\\\dfrac{2x}{1+x^2}\leq1$

Also,

$x^2+1\geq-2x\\\\\dfrac{-2x}{1+x^2}\leq1\\\\\dfrac{2x}{1+x^2}\geq-1$

Hence, we get that:

$-1\leq\dfrac{2x}{1+x^2}\leq1$

i.e.

$-1\leq y\leq1$

Hence,

$y^2\leq1$

Hence,

$y^2+y-2=y^2+2\cdot y\cdot \dfrac{1}{2}+(\dfrac{1}{2})^2-(\dfrac{1}{2})^2-2\\\\=(y+\dfrac{1}{2})^2-\dfrac{1}{4}-2\\\\=(y+\dfrac{1}{2})^2-\dfrac{9}{4}\\\\=(y+\dfrac{1}{2})^2-(\dfrac{3}{2})^2$

Now,

$-1\leq y\leq1\\\\-1+\dfrac{1}{2}\leq y+\dfrac{1}{2}\leq 1+\dfrac{1}{2}\\\\\dfrac{-1}{2}\leq y+\dfrac{1}{2}\leq\dfrac{3}{2}\\\\\dfrac{1}{4}\leq (y+\dfrac{1}{2})^2\leq (\dfrac{3}{2})^2\\\\\dfrac{1}{4}-(\dfrac{3}{2})^2\leq (y+\dfrac{1}{2})^2-(\dfrac{3}{2})^2\leq 0\\\\-2\leq (y+\dfrac{1}{2})^2-(\dfrac{3}{2})^2\leq 0{$

Hence, the range of the function is:

$[-2,0]$

i.e.

a= -2,b=0

Hence, b-4a=0-4×(-2)

b-4a=8

Hence, the value of b-4a=8