AP ICET 2nd May 2018 Shift-2

The diagram can be constructed like this:

As we can see, triangle HBB' is a a 30-60-90 triangle with one of the sides known. 
HB is the width of the river. 
HB = BB'/(root 3) = 20*(root 3) 

As $$\left(x-\alpha\ \right)$$ is a factor of $$3x^3 + 4x^2 + 2x + 5$$,

$$3x^3 + 4x^2 + 2x + 5$$ = 0 at $$x=\alpha\ $$
I.e. $$3\alpha\ ^3+4\alpha\ ^2+2\alpha\ +5=0$$

Dividing both sides of equation by $$\alpha^3$$ , we get

$$\frac{5}{\alpha^3}+\frac{2}{\alpha^2}+\frac{4}{\alpha\ }+3=0$$

This result is equivalent to saying that if $$x\ =\ \frac{1}{\alpha\ }$$ , then 
$$5x^3+2x^2+4x+3=0$$

It means that both (x-2) and (x+2) are factors of $$x^4 + ax^3 + bx^2 - 5x + 4$$

This implies that $$x^4 + ax^3 + bx^2 - 5x + 4$$ at x = 2 is 0 
i.e. 4a + 2b = -5

Also, it implies that $$x^4 + ax^3 + bx^2 - 5x + 4$$ at x = -2 is 0
i.e. 4a - 2b = 15

Adding up the two equations and solving them gives us 
a = 5/4, and b = -5

Hence, a - b = 25/4

Applying remainder theorem or performing actual division/factorization is the theoretical approach in such questions. 

However, one can perform a quick addition of the coefficients and see that their sum = 0
This is one of the most basic tests and it means that (x-1) is a factor of the given polynomial. 

If (x-2) is a factor of $$p\left(x^2+4\right)$$, it means that at x = 2, $$p\left(x^2+4\right)$$ = 0 
That is, $$p\left(2^2+4\right)=0$$ i.e. p(8) = 0

This is equivalent to saying that whatever polynomial function p(x) is, (x-8) is a factor of p(x) as we can see that p(8) = 0

Let the two numbers be 5x and 6x. When 3 is added to both, ration becomes 7:8
$$\frac{5x+3}{6x+3}=\frac{7}{8}$$

Cross multiplying, we get: x = 3/2
Hence, sum of the two numbers = 11x = 33/2

Multiplying the matrices on the LHS and equating with the one on RHS, we get:
3x - 2y = 11
-x + 2y = -5

Adding the 2 equations, we get x = 3, y = -1

Hence, 3x - 7y = 9-7 = 2

$$S_1=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$$

$$S_2=n\left[2a+\left(2n-1\right)d\right]$$

$$S_3=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]$$

The desired operation results in: 
$$\frac{3\left[2a+\left(3n-1\right)d\right]}{2\left[2a+\left(2n-1\right)d\right]-2a-\left(n-1\right)d}$$
=$$\frac{\left[6a+9nd-3d\right]}{2a+3nd-d}=3$$

Let the five consecutive even integers in ascending order be x, x+2, x+4, x+6, x+8

$$\Rightarrow$$ x + x + 2 + x + 4 + x + 6 + x + 8 = S

$$\Rightarrow$$ 5x + 20 = S

$$\Rightarrow$$  x = $$\frac{S-20}{5}$$

$$\therefore\ $$Least number among them = x = $$\frac{S-20}{5}$$

Consider the most basic case of (x-1)*(x-2) 
The co-efficient of $$x^{2-1}=x$$ is -3

We can reject option A and C outright. Option D gives value -1, hence, incorrect. 
Only option B gives correct answer.