Question

Find number of solutions of the equation $\sin x=x^2+x+1$ .

Answer 1

Hint: To solve the question given above, we will first find out the range of the function $f\left(x\right)=x^2+x+1$ . To find the range, we will find out its maximum value and minimum value as infinite and for an equation $a{x}^2+bx+c=0as-\left({\displaystyle \frac{b^2-4ac}{4a}}\right)$ . Then, we will find out at what x, $f\left(x\right)$ us minimum using $x={\displaystyle \frac{-b}{2a}}$ . Then, at that value of x, we will check whether the $\sin x$ is more than the minimum value of $f\left(x\right)$ or not. On this basis, we will determine how many solutions are there for the above equations.

Complete step-by-step answer:
To start with, we will first find out the range of the function $f\left(x\right)=x^2+x+1$ . Now, to find the range of $f\left(x\right)$ we will find its maximum value and minimum value. We can clearly see that the maximum value of $f\left(x\right)$ will be infinite. Now, we know that the minimum value of the equation $a{x}^2+bx+c=0as-\left({\displaystyle \frac{b^2-4ac}{4a}}\right)$ . In our case, a = 1, b = 1 and c = 1. Thus, minimum value of $f\left(x\right)=-\left({\displaystyle \frac{{\left(1\right)}^2-4\left(1\right)\left(1\right)}{4}}\right)$ . Thus, the minimum value of $f\left(x\right)={\displaystyle \frac{-3}{4}}$ . Now, the minimum value of $f\left(x\right)$ occurs at $x={\displaystyle \frac{-b}{2a}}$ . Thus, $x={\displaystyle \frac{-1}{2}}$ . Thus, the rough graph of $y=x^2+x+1$ is:

Now, we have to determine the value of $\sin x$ at $x={\displaystyle \frac{-1}{2}}$ . We know that in $[{\displaystyle \frac{-\pi }{2}},0)$ , $\sin x$ is negative i.e. it is not possible for $\sin x$ to interest $f\left(x\right)$ in that interval. Also, the maximum value of $\sin x$ is 1. Thus, the combined graph of $y=\sin xandy=x^2+x+1$ is shown below:

Thus, we can see that both the graphs do not intersect at all. Thus, there will be no solution of $\sin x=x^2+x+1$ .
Hence, there are 0 solutions of the equation $\sin x=x^2+x+1$ .

Note: he minima of $f\left(x\right)=x^2+x+1$ can also be find out by an alternate method as shown below:
We know that if $f\left(x\right)$ is a quadratic function of the form $a{x}^2+bx+c$ , where $a>0$ then its minimum value will occur at $x=\alpha$ where $\alpha$ is solution of $f^{\prime }\left(x\right)=0$ . Thus, we have:
 $f\left(x\right)=x^2+x+1$
On differentiating both sides, we have:
 $\Rightarrow f^{\prime }\left(x\right)=2x+1$
Now, $f^{\prime }\left(x\right)=0$ . So, we have
 $\begin{array}{rl}& \Rightarrow 2x+1=0\\ & \Rightarrow x={\displaystyle \frac{-1}{2}}\end{array}$
At $x={\displaystyle \frac{-1}{2}}$ , there will be minima. The value of this minima will be $=f\left({\displaystyle \frac{-1}{2}}\right)$ . Thus,
 $\begin{array}{rl}& f\left({\displaystyle \frac{-1}{2}}\right)={\left({\displaystyle \frac{-1}{2}}\right)}^2+\left({\displaystyle \frac{1}{2}}\right)+1\\ & f\left({\displaystyle \frac{-1}{2}}\right)={\displaystyle \frac{-3}{4}}+{\displaystyle \frac{3}{2}}={\displaystyle \frac{3}{4}}\end{array}$