Differentiation (Part-38) | S N De

33. If $\,x\,$ is real then find the greatest and least values of :

$(i)\,\frac{x^2+14x+9}{x^2+2x+3}\,\,\,(ii) \,\frac{x^2-2x+2}{x^2+3x+9}$

Sol. (i) Let $\,y=\frac{x^2+14x+9}{x^2+2x+3} \\ \Rightarrow (y-1)x^2+2(y-7)x+3(y-3)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore [2(y-7)]^2-4 \times 3 \times (y-1)(y-3) \geq 0 \\ \Rightarrow 4[(y-7)^2-4 \times 3 \times (y-1)(y-3)] \geq 0 \\ \Rightarrow (y^2-14y+49)-3(y^2-4y+3) \geq 0 \\ \Rightarrow -2y^2-2y+40 \geq 0 \\ \Rightarrow y^2 +y-20 \leq 0 \\ \Rightarrow (y+5)(y-4) \leq 0 \\ \therefore -5 \leq y \leq 4 \cdots(2)$

Hence from (2), we can conclude that the greatest and least values of  (1) are $\,4\,$ and $\,-5\,$ respectively. 

Sol. (ii)  Let $\,\,y=\frac{x^2-2x+2}{x^2+3x+9} \\ \Rightarrow (y-1)x^2+(3y+2)x +(9y-2)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore (3y+2)^2-4(y-1)(9y-2) \geq 0 \\ \Rightarrow 9y^2+12y+4-4(9y^2-11y+2) \geq 0 \\ \Rightarrow 27y^2-56y+4 \leq 0 \\ \Rightarrow (27y-2)(y-2)\leq 0 \\ \therefore \frac{2}{27} \leq y \leq 2 \cdots(2)$

Hence from (2), we can conclude that the greatest and least values of  (1) are $\,2\,$ and $\, \frac{2}{27} \,$ respectively. 

34. Show that for all real values of $\,x\,$ the value of $\,\frac{x^2-3x+4}{x^2+3x+4}\,$always lies between $\,\frac 17\,$ and $\,7.$

Sol. Let $\,\,y=\frac{x^2-3x+4}{x^2+3x+4} \\ \Rightarrow (y-1)x^2+3(y+1)x+4(y-1)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore [3(y+1)]^2-4 \times 4(y-1) \times (y-1) \geq 0 \\ \Rightarrow 9(y^2+2y+1)-16(y^2-2y+1) \geq 0 \\ \Rightarrow -7y^2+50y-7 \geq 0 \\ \Rightarrow 7y^2-50y+7 \leq 0 \\ \Rightarrow (7y-1)(y-7) \leq 0 \\ \therefore \frac 17 \leq y \leq 7.$

35. If $\,x\,$ is real , show that the value of $\,\,\frac{x}{x^2-5x+9}\,$always lies between $\,1\,$ and $\,\frac{1}{11}.$

Sol. Let $\,y=\frac{x}{x^2-5x+9} \\ \Rightarrow yx^2-(5y+1)x+9y=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore [-(5y+1)]^2-4 \times y \times 9y \geq 0 \\ \Rightarrow 25y^2+10y+1-36y^2 \geq 0 \\ \Rightarrow -11y^2+10y+1 \geq 0 \\ \Rightarrow 11y^2-10y-1 \leq 0 \\ \Rightarrow(11y+1)(y-1)\leq 0 \\ \therefore -\frac{1}{11} \leq y \leq 1\,\,\text{(proved)}$

36. If $\,x\,$ is real , show that the expression $\,\frac{x^2+34x-71}{x^2+2x-7}\,$ has no value between $\,5\,$ and $\,9.$

Sol. Let $\,y=\frac{x^2+34x-71}{x^2+2x-7} \\ \Rightarrow (y-1)x^2+2(y-17)x-(7y-71)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore [2(y-17)]^2-4 \times (y-1) \\ \times [-(7y-71)] \geq 0 \\ \Rightarrow y^2-34y+289+7y^2-78y+71 \geq 0 \\ \Rightarrow 8y^2-112y+360 \geq 0 \\ \Rightarrow y^2-14y+45 \geq 0 \\ \Rightarrow (y-9)(y-5) \geq 0 \\ \therefore \,\,\,\, -\infty <y \leq 5 \,\,\,\text{or,}\,\,\,\, 9 \leq y < \infty. \cdots(2)$

So, from (2) we can say that the expression $\,\frac{x^2+34x-71}{x^2+2x-7}\,$ has no value between $\,5\,$ and $\,9.$

37. If $\,x\,$ is real , show that the expression $\,\frac{(x-1)(x+3)}{(x-2)(x+4)}\,$ has no value between $\,\frac 49\,$ and $\,1.$

Sol. Let $\,y=\frac{(x-1)(x+3)}{(x-2)(x+4)}\\ \Rightarrow y=\frac{x^2+2x-3}{x^2+2x-8} \\ \Rightarrow (y-1)x^2+2(y-1)x-(8y-3)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore [2(y-1)]^2-4 \times (y-1) \times [-(8y-3)] \geq 0 \\ \Rightarrow  4(y^2-2y+1)+4(8y^2-11y+3) \geq 0 \\ \Rightarrow (y^2-2y+1)+ (8y^2-11y+3) \geq 0 \\ \Rightarrow 9y^2-13y+4 \geq 0 \\ \Rightarrow (9y-4)(y-1)\geq 0 \\ \therefore \,\,\,\, -\infty <y \leq \frac 49 \,\,\,\text{or,}\,\,\, 1 \leq y < \infty \cdots(2)$ 

So, from (2) we can say that  the expression $\,\frac{(x-1)(x+3)}{(x-2)(x+4)}\,$ has no value between $\,\frac 49\,$ and $\,1.$

38. If $\,x\,$ is real , show that the expression $\,\frac{x^2+2x-11}{x-3}\,$ assumes those values  which do not lie between $\,a\,$ and $\,b\,$ ; find $\,a\,$ and $\,b.$

Sol. Let $\,y=\frac{x^2+2x-11}{x-3} \\ \Rightarrow -x^2+(y-2)x-(3y-11)=0 \cdots(1)$

If $\,\,x \in \mathbb R,\,\text{discriminant of (1) : }\,\, D \geq 0 \\ \therefore (y-2)^2-4 \times (-1) \times [-(3y-11)] =0  \\ \Rightarrow y^2-4y+4 -12y+44 \geq 0 \\ \Rightarrow y^2-16y+48 \geq 0 \\ \Rightarrow (y-4)(y-12) \geq 0 \\ \therefore \,\,\,-\infty <y \leq 4\,\,\,\text{or,}\,\,\,12 \leq y < \infty \cdots (2)$

So, from (2) we can say that $\,y\,$ can not lie between $\,4\,$ and $\,\,12.$

By question, we can also say $\,\,a=4\,\,\text{and}\,\,\,b=12.$

To get full PDF of S.N. Dey  Math Solutions on Quadratic Equations , click here. 

>>>>>>>>>>>>>>>>>>>>> To continue with the Quadratic Equations (Part-12), click here .