The number of solutions of the equation 3 tan x+x3=2 in (0, (π/4)) is

Question

The number of solutions of the equation 3 tan x+x3=2 in (0, (π/4)) is (A) 1 (B) 2 (C) 3 (D) infinite

Tardigrade Admin · Accepted Answer

Let f(x)=3 tan x+x3-2. Then f'(x)=3 sec 2 x+3 x2 > 0 . Hence, f(x) increases. Also, f(0)=-2 and f((π/4)) > 0. So, by intermediate value theorem, f(c)=2 for some c ∈(0, (π/4)) . Hence, f(x)=0 has only one root.

Q. The number of solutions of the equation $3 tan x + x^{3} = 2$ in $(0, \frac{π}{4})$ is

1

2

3

infinite

Let $f (x) = 3 tan x + x^{3} - 2$ .
Then $f^{'} (x) = 3 sec^{2} x + 3 x^{2} > 0.$
Hence, $f (x)$ increases.
Also, $f (0) = - 2$ and $f (\frac{π}{4}) > 0$ .
So, by intermediate value theorem,
$f (c) = 2$ for some
$c \in (0, \frac{π}{4}) .$
Hence, $f (x) = 0$ has only one root.

Q. The number of solutions of the equation 3tanx+x3=2 in (0,4π​) is

1

2

3

infinite

Let f(x)=3tanx+x3−2. Then f′(x)=3sec2x+3x2>0. Hence, f(x) increases. Also, f(0)=−2 and f(4π​)>0. So, by intermediate value theorem, f(c)=2 for some c∈(0,4π​). Hence, f(x)=0 has only one root.

Q. The number of solutions of the equation $3 tan x + x^{3} = 2$ in $(0, \frac{π}{4})$ is

Let $f (x) = 3 tan x + x^{3} - 2$ .
Then $f^{'} (x) = 3 sec^{2} x + 3 x^{2} > 0.$
Hence, $f (x)$ increases.
Also, $f (0) = - 2$ and $f (\frac{π}{4}) > 0$ .
So, by intermediate value theorem,
$f (c) = 2$ for some
$c \in (0, \frac{π}{4}) .$
Hence, $f (x) = 0$ has only one root.