KOUSHIK SOM
Forum Replies Created
-
Posts
-
$\mathbf{A}=$ area of given rectangle, $\mathbf{I}=$ length $(\mathbf{5 6}), \mathbf{w}=$ width
$A=I x w(2016)$
[
\begin{array}{l}
2016=56 w \
w=2016 / 56 \
w=36
\end{array}
]
Hence, the dimension of the given rectangle is $56 \times 36$
Next we determine the size of the small squares to cut out. Its side is the greatest common multiple (gcf) of 56 and 36
The gef of 36 and 56 can be obtained like this:
. The factors of 36 are 36,18,12,9,6,4,3,2,1
. The factors of 56 are 56,28,14,8,7,4,2,1
. The common factors of 36 and 56 are 4,2,1 , intersecting the two sets above.
. In the intersection factors of 36 n factors of 56 the greatest element is 4Therefore, the greatest common factor of 36 and 56 is 4 . That makes $4 \times 4$ the
size of small square cut outs.Let $y=\frac{x^{2}+34 x-71}{x^{2}+2 x+7}$
$\Rightarrow \quad x^{2} y+2 x y-7 y=x^{2}+34 x-71$
$\Rightarrow \quad(x)+2 x y-7 y-x^{2}-34 x+71=0$
$(y-1) x^{2}+(2 y-34) x+(71-7 y)=0$
It is in the form of $ax^2+bx+c=0$Where a = $a=(y-1) \quad b=2 y-34, c=71-7 y$
Since $x$ is real $ \Rightarrow \Delta \geq 0 \Rightarrow b^{2}-4 a c \geq 0$
$\Rightarrow(2 y-34)^{2}-4(y-1)(7-7 y) \geq 0$
$\Rightarrow \quad 4 y^{2}+1156-136 y-4\left(7 y-7 y^{2}-71+7 y\right) \geq 0$
$\Rightarrow 4 y^{2}+1156-136 y-4\left(78 y-7 y^{2}-71\right) \geq 0$
$\Rightarrow 4 y^{2}+1156-136 y-312 y+28 y^{2}+284 \geq 0$
$\Rightarrow \quad 32 y^{2}-448 y+1440 \geq 0$
$\Rightarrow \quad y^{2}-14 y+45 \geq 0$
$\Rightarrow \quad y^{2}-9 y-5 y+45 \geq 0$
$\Rightarrow y(y-9)-5(y-9)>0$
$\Rightarrow(y-9)(y-5) \geq 0 \Rightarrow y \in(-\infty, 5] \cup[9, \infty)$
$\Rightarrow$ y doest not lie tetween 5 and 9
$\Rightarrow$ Given Expresion doesnot lie between 5 and 9
