Differentiation (Part-38) | S N De

 In the previous article , we have discussed few Short Answer Type Questions (From Qst. 31-40) of S.N. Dey Math Exercise. In this article,    we are going to start new chapter : Quadratic Equations (Part-5) as a part of our S.N. Dey Math solution series.  So, let's start.

41. Show that if $\,x\,$ is real and $\,x^2+5 <6x,\,$ then $\,x\,$ must lie between $\,1\,$ and $\,5.$

Sol.  $\,x^2+5 <6x \\ \Rightarrow x^2-5x+5 <0 \\ \Rightarrow (x-5)(x-1) <0 \\ \therefore 1<x<5.$

Hence, if $\,x\,$ is real and $\,x^2+5 <6x,\,$ then $\,x\,$ must lie between $\,1\,$ and $\,5.$

42. If $\,x\,$ is real, find the real values of $\,p\,$ which make $\,\,4x^2+px+1\,\,$ always positive.

Sol. $\,\,4x^2+px+1 \\=(2x)^2+2.2x.\frac p4+\left(\frac p4\right)^2+\left(1-\frac{p^2}{16}\right)\\=\left(2x+\frac p4\right)^2+\left(1-\frac{p^2}{16}\right)$

Since $\,\,x \in \mathbb R,\,\, 4x^2+px+1\,\,$ is always positive, so $\,\,1-\frac{p^2}{16} \geq 0 \\ \Rightarrow 16-p^2 \geq 0 \\ \Rightarrow p^2-16 \leq 0 \\ \Rightarrow (p+4)(p-4) \leq  0 \\ \therefore -4\leq p \leq 4.$

43. If $\,x\,$ is real, find the real values of $\,m\,$ which make the expression $\,\,mx-x^2-1\,\,$ always positive. 

Sol. $\,\,mx-x^2-1 \\=-(x^2-mx+1)\\=-[x^2-2.x.\frac m2+(\frac m2)^2+(1-\frac{m^2}{4})]$

Since $\,\,x ,\,m\,\,$ are real and $\,\,\left(x-\frac m2\right)^2 \geq 0.$

So, if $mx-x^2-1 <0,\,\\ 1-\frac{m^2}{4} \geq 0 \\ \Rightarrow  \frac{m^2}{4}-1 \leq 0 \\ \Rightarrow m^2-4 \leq 0 \\ \Rightarrow (m+2)(m-2) \leq 0 \\ \Rightarrow -2 \leq m \leq 2.$

44. If $\,\,\frac{(x-5)(x^2-2x+1)}{(x-7)(x^2+2x+3)}\,\,$ is positive for all real values of $\,x\,$,  show that $\,x\,$ has no real value between $\,5\,$ and $\,7.$

Sol. Let $\,\,y=\frac{(x-5)(x^2-2x+1)}{(x-7)(x^2+2x+3)} \\ \Rightarrow y=\frac{(x-5)(x-1)^2}{(x-7)[(x+1)^2+2]} \cdots (1)$

From (1), we notice if $\,\,x=7,\,y\,\,$ is undefined. $\,\therefore y \neq 7.$ 

Now, Since $\,\, y \geq 0,\,\frac{(x-5)(x-1)^2}{(x-7)[(x+1)^2+2]} \geq 0. $

Since $\,\,x \in \mathbb R \,\text{and}\,\,(x-1)^2\geq 0 \\ \text{and}\,\,(x+1)^2+2 >0.$

Hence for non-negative values of $\,\,y,\, \,(x-5)(x-7) \geq 0.$

Now, since $\,\,(x-5)(x-7)\geq 0 \\ \Rightarrow \text{either}\,\,x-5 \geq 0,\,x-7 \geq 0 \cdots(2) \\ \text{or,}\,\,x-5 \leq 0 ,\,\,x-7 \leq 0 \cdots(3)$

Hence, from (2) we get, $\,\,7 \leq x < \infty \\ \text{which means}\,\,\, x \in [7,\infty) \\ \therefore x \in (7,\infty)\,\,[\text{Since,}\,\, y \neq 7]$ 

or  from (3), we get $\, -\infty <x \leq 5 \\ \therefore x \in (-\infty,5].$

Hence, $\,x\,$ has no real value between $\,5\,$ and $\,7.$ 

45.  If the equation $\,\,y^2+x^2-10x+21=0\,\,$ is satisfied by real values of $\,\,x\,$ and $\,y\,$, prove that $\,x\,$ lies between $\,3\,$ and $\,7\,$ and $\,y\,$ lies between $\,\,(-2)\,$ and $\,2.$

Sol. $\,\,y^2+x^2-10x+21=0 \cdots(1).$

For all real values of $\,x\,$, the discriminant of $\,(1)\, \,\,\geq 0 \\ (-10)^2-4.1.(y^2+21)\geq 0 \\ \Rightarrow 4[25-y^2-21] \geq 0 \\ \Rightarrow y^2-4 \leq 0 \\ \Rightarrow (y-2)(y+2) \leq 0 \\ \Rightarrow  -2 \leq y \leq2.$

Again, for real values of $\,y\,$, the discriminant of $\,(1)\,\, \geq 0 \\ \Rightarrow 0^2-4 \times 1 \times (x^2-10x+21) \geq 0 \\ \Rightarrow x^2-10x+21 \leq 0 \\ \Rightarrow (x-7)(x-3) \leq 0 \\ \Rightarrow 3 \leq x \leq 7.$ 

Hence,  if the equation $\,\,y^2+x^2-10x+21=0\,\,$ is satisfied by real values of $\,\,x\,$ and $\,y\, ,\,\,\,\,x\,$ lies between $\,3\,$ and $\,7\,$ and $\,y\,$ lies between $\,\,(-2)\,$ and $\,2.$

46. If $\,x\,$ is real, show that $\,\,(x-2)(x-3)(x-4)(x-5)+2\,\,$ is always positive. 

Sol. $\,\,(x-2)(x-3)(x-4)(x-5)+2\\ =[(x-2)(x-5)][(x-3)(x-4)]+2\\=(x^2-7x+10)(x^2-7x+12)+2\\=(x^2-7x)^2+22(x^2-7x)+120+2\\=(x^2-7x)^2+2 \times (x^2-7x) \times 11+11^2+1\\=(x^2-7x+11)^2+1>0\,\,[\,\, x \in \mathbb R]$

   Now, to get the solution of Long Answer Type Questions [Q. 1-7] , click here .

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