Qualcomm Equation Quiz-1

mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.

in this case b=0
eqn is ax^2+c = 0 => x=sqrt(-c/a)
for x to be real (-c/a) should +ve => c & a are of opposite sign
so c>a or a>c

for ex x^2-4=0 => x1=-2,x2=2 & modx1=modx2=2 [a>c]
or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 [c>a]

(d) none

x^3-5x+c=0
if x=1, 1-5+c=0 or c=4
if x=-1, -1+5+c=0 or c=-4
if x=2, 8-10+c=0 or c=2
if x=-2, -8+10+c=0 or c=-2
we see that,for c=-2,2,-4,4 we get x=-2,2,-1,1 etc
i.e
we get integral roots of x for infinite values of c.

d)infinite

1/a+1/b+1/c=1/(a+b+c)
or (bc+ac+ab)/(abc)=1/(a+b+c)
or (ab+bc+ac)(a+b+c)=abc
or abc+(a^2)c+(a^2)b+(b^2)c+abc+(b^2)a
(c^2)b+(c^2)a+abc=abc
or 2abc+(a^2)c+(a^2)b+(b^2)c+(b^2)a+(c^2)b+(c^2)a=0
also,
( a+ b ) ( b + c ) ( c + a )=2abc+(a^2)c+(a^2)b+
(b^2)c+(b^2)a+(c^2)b+(c^2)a
or ( a+ b ) ( b + c ) ( c + a )=0

The solution would be 66. Let's first consider the first
and fourth quadrant only

x= 2,5,8,......,50

Now corresponding to these values we have:-

y=32,30,.......,0
and
y=-32,-30, ...... 0

and now values of x in 2nd and 3rd quadrant

x=-2,-5,-8,.......,-50

and

y=32,30,.......,0
and
y=-32,-30, ...... 0 again

So total number of soltions to it are

(17*4) - 2=66

Some of the solutions are
(2,32) , (-2,32) , (-2,-32) , (2,-32)

the correct option is 3 as by putting different values of x at a
constant value of k, we get values of y which first increases
and then decreases with a value of x.

151001 / 11^4 = 10 with remainder = 4591

4591 / 11^3 = 3 with remainder = 598

598 / 11^2 = 4 with remainder = 114

114 / 11^1 = 10 with remainder = 4

Therefore 151001=10(11^4)+3(11^3)+4(11^2)+10(11^1)+4

i.e., (v,w,x,y,z) is (10,3,4,10,4). These values satisfies given
conditions and is one valid option.

Let's analyze if further option available satisfying the given
condition. v,w,x,y,z should be less than 11 and greater
than zero. The solution we got has v as 10. Suppose v=9.
With this, the maximum value possible under the given constraints is
9(11^4)+10(11^3)+10(11^2)+10(11^1)+10=146409

f(x) = 2-x/x-2 = (2-x)(2-x)/(x-2)(2-x) = (2-x)^2 / (-1) (2-x)^2 = -1

So, the right answer is -1.

1/12 = 1/z - 1/y

y = 12 * z / ( 12 - z )

0 < z < 12

Only for z = 3 ,4 ,6 ,8 , 9 ,10 and 11 , y is intiger

Hence seven solutions

(2x+9)^2 / (x+3) = {2(x+3)+3}^2 / (x+3) = 4(x+3) + 12 + 9/(x+3)

for 9/(x+3) to be the integer

x = -12,-6,-4,-2,0,6
6 values

D=b^2- 4 a c >0 for rational roots
= (-5)^2-4*1*c > 0
hence values of c can be either positive or negative

c=0,1,,2,3,4,5,-1,-2,-3,-4,-5...
there is no unique values for c hence ans. will be infinte..
option d is correct