Differentiation (Part-38) | S N De

In previous chapter, we have discussed  Quadratic Equation : Long Answer Type Questions (33-38) of S.N. Dey math exercise .  In this article we will continue our journey for the same.

 39. If $\,x\,$ is real, show that  the expression $\,\,\frac{x^2-ab}{2x-a-b}\,\,$ has no real value between $\,a\,$ and $\,b.$

Sol.  Let $\,y=\frac{x^2-ab}{2x-a-b} \\ \Rightarrow x^2-2xy+ay+by-ab=0 \,\cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore (-2y)^4-4(ay+by-ab) \geq 0 \\ \Rightarrow y^2-(a+b)y+ab \geq 0 \\ \Rightarrow (y-a)(y-b) \geq 0 \\ \therefore \,\,\, -\infty < y \leq a \,\,\text{and}\,\, b \leq y < \infty \\ \text{or,}\,\,\, -\infty <y \leq b\,\,\text{and}\,\,a \leq y <\infty\cdots (2)$

Hence from (2), we can conclude that if $\,x\,$ is real,  the expression $\,\,\frac{x^2-ab}{2x-a-b}\,\,$ has no real value between $\,a\,$ and $\,b.$

40. If $\,x\,$ is real, show that the expression has $\,\frac{3x-5}{x^2-1}\,$ has no value between $\,\frac 12\,$ and $\,\frac 92.$

Sol. Let $\,\,y=\frac{3x-5}{x^2-1} \\ \Rightarrow x^2y-3x-(y-5)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore \,\,(-3)^2-4 \times y \times [-(y-5)] \geq 0 \\ \Rightarrow 9+4y^2-20y \geq 0 \\ \Rightarrow 4y^2-20y+9 \geq 0 \\ \Rightarrow (2y-1)(2y-9) \geq 0 \cdots(2) \\ \therefore \,\,\, -\infty <y \leq \frac 12\,\,\text{and}\,\,\frac 92 \leq y <\infty$

Hence from (2), we can conclude that  if $\,x\,$ is real, the expression has $\,\frac{3x-5}{x^2-1}\,$ has no value between $\,\frac 12\,$ and $\,\frac 92.$

41. (i) If x is real, find the maximum and minimum values of $\,\,\frac{x^2-x+1}{x^2+x+1}\,\,$; also find the corresponding values of $\,x.$ 

Sol. Let $\,\,y=\frac{x^2-x+1}{x^2+x+1} \\ \Rightarrow (y-1)x^2+(y+1)x+(y-1)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore (y+1)^2 -4 \times (y-1) \times (y-1) \geq 0 \\ \Rightarrow (y^2+2y+1)-4(y^2-2y+1) \geq 0 \\ \Rightarrow 3y^2-10y+3 \leq 0 \\ \Rightarrow (3y-1)(y-3) \leq 0 \\ \therefore \,\,\frac 13 \leq y \leq 3.$

Hence, the maximum and minimum values of $\,\,\frac{x^2-x+1}{x^2+x+1}\,\,$ is $\,3\,$ and $\,\frac 13\,$ respectively.

2nd part : Now, from (1) we get , when $\,\,y=\frac 13 \\ -\frac 23 x^2+\frac 43 x-\frac 23=0 \\ \text{or,}\,\,x^2-2x+1=0 \\ \text{or,}\,(x-1)^2=0 \\ \therefore \,\,\, x=1.$

Again , from (1) we get ,

 when $\,\,y=3 \\ 2x^2+4x+2=0 \\ \Rightarrow x^2+2x+1=0 \\ \Rightarrow (x+1)^2=0 \\ \Rightarrow x+1=0 \\ \Rightarrow x=-1. $

Hence, we can say for $\,\,y=3\,$ the corresponding value of $\,\,x=-1\,$ and for $\,\,y=\frac 13\,$ the corresponding value of $\,\,x=1.$

41. (ii) If $\,x\,$ be real, find the greatest value of $\,\,\frac{x+2}{2x^2+3x+6}.$

Sol. Let $\,\,y=\frac{x+2}{2x^2+3x+6} \\ \Rightarrow 2yx^2+(3y-1)x+(6y-2)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore \,\, (3y-1)^2 -4 \times 2y \times (6y-2) \geq 0 \\ \Rightarrow 9y^2-6y+1-48y^2+16y \geq 0 \\ \Rightarrow 39y^2-10y-1 \leq 0 \\ \Rightarrow (13y+1)(3y-1) \leq 0 \\ \therefore \,\,\, -\frac{1}{13} \leq y \leq \frac 13 \cdots (2)$

Hence, by (2) we can conclude that the greatest value of $\,\,\frac{x+2}{2x^2+3x+6}\,\,$ is $\,\,\frac 13.$

42. If $\,x\,$ is real, show that each of the following expressions is capable of assuming all real values:

$(i)\,\,\frac{2x^2+4x+1}{x^2+4x+2}\,\,(ii)\,\,\frac{2x^2+5x+2}{x^2+6x+7} \\ (iii)\,\,\frac{p^2}{1-x}-\frac{q^2}{1+x}$

Sol. (i) Let $\,\,y=\frac{2x^2+4x+1}{x^2+4x+2} \\ \Rightarrow (y-2)x^2+4(y-1)x+(2y-1)=0 \cdots(1)$

Now, the discriminant (D) of (1) is : $=[4(y-1)]^2-4 \times (y-2)\times (2y-1)\\=4[4(y^2-2y+1)-(2y^2-5y+2)]\\=4[2y^2-3y+2]\\=8[y^2-\frac 32 y+1]\\=8[y^2-2.y.\frac 34+\frac{9}{16}-\frac{9}{16}+1]\\=8[(y-\frac 34)^2+\frac{7}{16}]>0\,\,\forall y \in \mathbb R \\ \therefore D >0 ,\,\,\forall y \in \mathbb R$

Hence, if $\,x\,$ is real,   the aforesaid expression given by (1) is capable of assuming all real values.

Sol. (ii) Let $\,\,y=\frac{2x^2+5x+2}{x^2+6x+7} \\ \Rightarrow (y-2)x^2+(6y-5)x+(7y-2)=0 \cdots(1)$

Now, the discriminant (D) of (1) is : $=(6y-5)^2-4 \times (y-2)\times (7y-2)\\=(36y^2-60y+25)-4(7y^2-16y+4)\\=8y^2+4y+9\\=8\left[y^2+\frac 12 y+\frac 98\right]\\=8\left[y^2+2 \times y \times \frac 14 +\frac{1}{16}+\frac 98-\frac{1}{16}\right]\\=8\left[(y+\frac 14)^2+\frac{17}{16}\right] >0,\,\,\forall y \in \mathbb R \\ \therefore \,\,D >0,\,\,\forall y \in \mathbb R$

Clearly, for all real values of $\,y,\,\,\,x\,\,$ is real. 

So, the given expression can assume any real values.

Sol. (iii) Let $\,\,y= \frac{p^2}{1-x}-\frac{q^2}{1+x} \\ \Rightarrow y=\frac{(p^2-q^2)+(p^2+q^2)x}{1-x^2}\\ \Rightarrow yx^2+(p^2+q^2)x+(p^2-q^2-y)=0\cdots(1)$

Now, the discriminant (D) of (1) is : $=(p^2+q^2)^2-4.y.(p^2-q^2-y)\\=4y^2-4(p^2-q^2)y+(p^2+q^2)^2\\=4y^2-4(p^2-q^2)y+(p^2-q^2)^2+4p^2q^2\\=(2y-p^2+q^2)^2+4p^2q^2 >0,\,\,\forall y\in \mathbb R.$

Clearly, for all real values of $\,y,\,\,\,x\,\,$ is real. 

So, the given expression can assume any real values.

43. If $\,3x^2+2(p+q+r)x+ pq+qr+rp\,\,$ be a perfect square, prove that $\,p=q=r.$

Sol. We have ,$\,\,\,\,3x^2+2(p+q+r)x+ pq+qr+rp \\=3 \left[x^2+\frac 23 (p+q+r)x+\frac 13(pq+qr+rp)\right]\\=3[x+2.x.(\frac{p+q+r}{3})+(\frac{p+q+r}{3})^2 \\+\frac 13(pq+qr+rp)-\frac 19(p+q+r)^2]\\=3\left[\left(x+\frac{p+q+r}{3}\right)^2-\frac 19(p^2+q^2+r^2\\-pq-qr-rp)\right]\\=3[\left(x+\frac{p+q+r}{3}\right)^2-\frac{1}{18}(2p^2+2q^2+2r^2\\-2pq-2qr-2rp)]\\=3[\left(x+\frac{p+q+r}{3}\right)^2-\frac{1}{18}\{(p-q)^2+(q-r)^2\\+(r-p)^2\}] \cdots(1)$

Now, for (1) to be a perfect square , $\,\,(p-q)^2+(q-r)^2+(r-p)^2=0 \\ \Rightarrow \,\,p=q=r\,\,\,\text{(proved.)}$

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