Number of solution(s) satisfying the equation, 3 x2-2 x3= log 2(x2+1)- log 2 x is -

Question

Number of solution(s) satisfying the equation, 3 x2-2 x3= log 2(x2+1)- log 2 x is - (A) 1 (B) 2 (C) 3 (D) none

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7454e92c24f5122c1f606b609dea6e35-.png" /> 3 x 2-2 x 3= log 2( x +(1/ x )), x >0 f ( x )=3 x 2-2 x 3 f prime( x )=6 x -6 x 2 =6 x (1- x ) f ( x ) ≤ f (1) f ( x ) ≤ 1 ⇒ text LHS ≤ 1 text RHS ≥ 1 text LHS = text RHS =1 text for x =1 ∴ equation has exactly one solution

Q. Number of solution(s) satisfying the equation, $3 x^{2} - 2 x^{3} = lo g_{2} (x^{2} + 1) - lo g_{2} x$ is -

1

2

3

none

$3 x^{2} - 2 x^{3} = lo g_{2} (x + \frac{1}{x}), x > 0$
$f (x) = 3 x^{2} - 2 x^{3}$
$f^{'} (x) = 6 x - 6 x^{2}$
$= 6 x (1 - x)$
$f (x) \leq f (1)$
$f (x) \leq 1$
$\Rightarrow LHS \leq 1 & RHS \geq 1$
$LHS = RHS = 1 for x = 1$
$∴$ equation has exactly one solution

Q. Number of solution(s) satisfying the equation, 3x2−2x3=log2​(x2+1)−log2​x is -

1

2

3

none

3x2−2x3=log2​(x+x1​),x>0 f(x)=3x2−2x3 f′(x)=6x−6x2 =6x(1−x) f(x)≤f(1) f(x)≤1 ⇒ LHS ≤1 & RHS ≥1 LHS = RHS =1 for x=1 ∴ equation has exactly one solution

Q. Number of solution(s) satisfying the equation, $3 x^{2} - 2 x^{3} = lo g_{2} (x^{2} + 1) - lo g_{2} x$ is -

$3 x^{2} - 2 x^{3} = lo g_{2} (x + \frac{1}{x}), x > 0$
$f (x) = 3 x^{2} - 2 x^{3}$
$f^{'} (x) = 6 x - 6 x^{2}$
$= 6 x (1 - x)$
$f (x) \leq f (1)$
$f (x) \leq 1$
$\Rightarrow LHS \leq 1 & RHS \geq 1$
$LHS = RHS = 1 for x = 1$
$∴$ equation has exactly one solution