150 Follower Problem!

For the second one what I did was

F(x)=(x-1)(x-2)

P(x)=F(x) +x(x+7) -1

Then let the variable be b

b=F(b)+b(b+7)-1

From the above equation we get. b= -1 ,-1/2

Since we only need the integral value of b

Therefore b= -1

Note for the problem poster: You need to change your question a bit because, for Q-(b), $P(b)=b$ has two solutions $b=(-1),\left(-\frac{1}{2}\right)$ . You should add the fact that we need only integral $b$ .

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For (1):

Using mid-point formula and a diagram, we get that the required square has the vertices $W(150,0)~,~X(0,-150)~,~Y(-150,0)~,~Z(0,150)$ Now, it is easy to get that the side of the square obtained is $150\sqrt{2}$ using distance formula. As such, $a=(150\sqrt{2})^2$

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For (2):

Using method of finite differences, construct a difference table for quadratic $P(x)$ . Denote by $D_i(x)$ the value of the $i^{\textrm{th}}$ difference column for $x$ . We'll get that $D_1(1)=10~,~D_2(1)=4$ . Hence, using the reconstruction method,

$P(x)=P(1)+\sum_{i=1}^2\left(\frac{D_i(1)}{i!}\cdot \prod_{j=1}^i(x-j)\right)=2x^2+4x+1$

$P(b)=b\implies 2b^2+3b+1=0\implies (2b+1)(b+1)=0\implies b=(-1),\frac{-1}{2}$

But we're looking for integral value of $b$ , so $b=(-1)$ .

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For (3):

Note that $P(x)$ can be written compactly using summation notation:

$P(x)+c=\sum_{i=1}^{150}(-1)^iix^i\implies P(-1)+c=\sum_{i=1}^{150}1^ii\\ \implies 0+c=\sum_{i=1}^{150}i\implies c=\frac{150\times 151}{2}$

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Now, we perform a very high level operation on these values, called addition.

$a+b+c=2\times 150^2-1+\frac{150\times 151}{2}=\boxed{56324}$

Once Again, I will give away the answers, I encourage you to try it again on your own.

a) 45000 Sq. Units

b) -1

c) 11325.

Adding, We get 56324. Love the answer xD

Cheers!