Discussion Room : Quant

[Benvin Thomas]

SOLVE
(X^4+X^2+1)/(X^2-4X-5) < 0

SOLVE
(X^2+6X-7)/(X^2 + 1)

[saksham thakur]

N = 2³ * 5³ * 7², how many sets of two factors of N are co-prime

[saksham thakur]

@vinaycat2017 45:72:80
197--78800
72-28800

[SwanandK12]

If F(x) = x^4 - 360x^2 + 400 (for any integral value of x) and if F(x) is a prime number, then what is the sum of all possible values of F(x)? Cold anyone help [email protected] sir please help.

[sumit agarwal]

@saksham-thakur said in Discussion Room : Quant:

two

[099vaibhavs]

find the maximum value of n such that
42×57×92×91×52×62×63×64×65×66×67 is perfectly divisible by 42*n

[aneeeshp]

@swanandk12 I think it is infinity. As x increases the value of F(x) will keep on increasing. So possible values of F(x) are infinite. Hence the sum of possible values F(x) is infinity. Thoughts?

[rohitchopra01]

If 76/(4+√7+√11)=p+q√7+r√11+s√77 then p+q+r+s=?

[aneeeshp]

@rohitchopra01 said in Discussion Room : Quant:

If 76/(4+√7+√11)=p+q√7+r√11+s√77 then p+q+r+s=?

Multiply both sides by (4 + 71/2 + 111/2)

76 = (4 + 71/2 + 111/2)p + 71/2(4 + 71/2 + 111/2)q + 111/2(4 + 71/2 + 111/2)r + 771/2(4 + 71/2 + 111/2)s

Expand out:
76 = 4p + 71/2p + 111/2p + 71/24q + 7q + 771/2q + 111/24r + 771/2r + 11r + 771/24s + 111/27s + 71/211s

Now get everything together:
76 = (4p+7q+11r) + 71/2(p+4q+11s) + 111/2(p+4r+7s) + 771/2(q+r+4s)

Rewrite the LHS:
76 + 0 * 71/2 + 0 * 111/2 + 0 * 771/2 = (4p+7q+11r) + 71/2(p+4q+11s) + 111/2(p+4r+7s) + 771/2(q+r+4s)

So we have the following system of equations:
4p + 7q + 11r = 76
p + 4q + 11s = 0
p + 4r + 7s = 0
q + r + 4s = 0

4 equations in 4 unknowns. Solve.

[shyam verma]

Find sum of
√{5+√[11+√【19+√(29..........To infinity

[Adarsh Rathi]

@zabeer x=8 the value turns out to be 108 u should always look for a no which can create max difference between the give nos like (8 and 6) and (11 and 8)