Question

If
$\frac{ {x}^{3} + 3x}{3 {x}^{2} + 1 } = \frac{341}{91}$
then value of x is:
a)11
b)12
c)13
d)7

Answer 1

Answer:

x=11

Step-by-step explanation:

if (x^3 + 3x)/(3x^2 +1) = 341/91

then x^3 + 3x = 341K where k is constant

and 3x^2 + 1 = 91K

such that above equation is varified

then solving 3x^2 + 1 = 91K

for k=0

x=(-1/3)^(1/2) which is not possible

for k=1

x= (90/3)^(1/2)

=(30)^(1/2)

which is not in option

for k=2

x= (182/3)^(1/2) which is not in option

now k= 3

x=(272/3)^(1/2) which is not in option

for k = 4

x=(363/3)^(1/2)

= (121)^(1/2) = 11

Hence option (a) is correct.

Answer 2

If

$\frac{ {x}^{3} + 3x}{3 {x}^{2} + 1 } = \frac{341}{91}$

then value of x is:

✔️✔️✔️a)11

b)12

c)13

d)7