Question

Let $\lambda \in \{-2,-1,0,1,2\}$ then the chance that the equation $|x^2+3x|+\lambda -\lambda x=0$ possesses exactly three solutions, given that it possesses atleast $2$ solutions is ?

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{2}{3}$
• D. $0$

Answer 1

The correct option is A $\frac{1}{3}$

The given equation : |$x^2$ + $3x$| + $\lambda$ - $\lambda$$x$ = 0 can be rewritten as :

|$x^2$ + $3x$| =$\lambda$ - $\lambda$$x$

$\Rightarrow$ |$x^2$ + $3x$| = $\lambda$($1$ - $x$) $\Rightarrow$ $\lambda$ = $\frac{|x^2+3x|}{1-x}$

Hints for plotting the graph of above equation :

Looking at the equation one can clearly say that $\lambda <0$ only when $x>1$.

Since x is a function of $\lambda$, now :

$Takingx>0$:

We have , $\lambda$ = $\frac{x^2+3x}{1-x}$

So, $\frac{d\lambda }{dx}$ = $\frac{(1-x)(2x+3)-(-1)(x^2+3x)}{(1-x{)}^2}$

On Solving, we get: $\frac{d\lambda }{dx}$ = $\frac{(-x^2+2x+3)}{(1-x{)}^2}$ OR $\frac{-(x^2-2x-3)}{(1-x{)}^2}$ OR $\frac{-(x-3)(x+1)}{(1-x{)}^2}$

For critical points, $\frac{d\lambda }{dx}$ = 0 = $\frac{-(x-3)(x+1)}{(1-x{)}^2}$

$\Rightarrow$ critical points are $x=3$ and $x=-1$( rejected because initially we had taken $x>0$) with $x=1$ as a vertical asymptote.

When $\frac{d\lambda }{dx}$ > 0, $\lambda$ will increase $x>0$ and when $\frac{d\lambda }{dx}$ <0, $\lambda$ will decrease

So $\frac{d\lambda }{dx}$ > 0 at $-1<x<3$ $\Rightarrow$ $\lambda$ will increase

But initially we have taken $x>0$, therefore for $0<x<1$ and $1<x<3$ the function increases.

Also, $\frac{d\lambda }{dx}$ < 0 at $x>3$ $\Rightarrow$ $\lambda$ will decrease

But for $x>3$ only, the function will decrease because initially we have taken $x>0$

$Takingx<0$:

$\lambda$ will be throughout positive and the critical points will be $x=-1$ andn $x=-3$

So, the graph of this function can be easily drawn by above hints.

Now in the question, it is given that the equation posses atleast 2 solutions for $\lambda$ = {$-2,-1,0,1,2$}

So, chance that the equation possess 3 solutions given that it possess rwo solutions is:

Let the event defining the equation possessing 3 solutions = A

and let the event defining the equation already possess atleast two solution = B

So, $P$($\frac{A}{B}$) = $P$(A$\cap$B) / $P$(B) , here P means probability or chance

= $\frac{1}{3}$