NIMCET 2007 — Mathematics PYQ

To find the number of solutions for the given equation in the interval $x\in [-\pi ,\pi ]$, we can break down the equation by separating the polynomial function and the trigonometric function.

The given equation is:

$$x^3+x^2+3+2\sin x=0$$

Let's rewrite it as:

$$x^3+x^2+3=-2\sin x$$

Now, let's analyze the behavior of both sides of the equation within the given interval $x\in [-\pi ,\pi ]$ by breaking it into two intervals: negative and positive values of $x$.

Case 1: When $x\in [0,\pi ]$

• Left-Hand Side (LHS): $f(x)=x^3+x^2+3$

  Since $x\ge 0$, both $x^3\ge 0$ and $x^2\ge 0$. Therefore:

  $$x^3+x^2+3\ge 3$$

  So, the minimum value of LHS in this interval is $3$.

• Right-Hand Side (RHS): $g(x)=-2\sin x$

  We know that for $x\in [0,\pi ]$, the value of $\sin x$ lies between $0$ and $1$ ($0\le \sin x\le 1$). Multiplying by $-2$ reverses the inequality:

  $$-2\le -2\sin x\le 0$$

  So, the maximum value of RHS in this interval is $0$.

Since the minimum value of LHS is $3$ and the maximum value of RHS is $0$, they can never meet in this interval. Thus, there are no roots in $[0,\pi ]$.

Case 2: When $x\in [-\pi ,0)$

Let's analyze the derivative of the polynomial part, $f(x)=x^3+x^2+3$, to see how it behaves:

$$f^{\prime }(x)=3x^2+2x=x(3x+2)$$

The critical points are $x=0$ and $x=-\frac{2}{3}$.

• For $x\in \left(-\pi ,-\frac{2}{3}\right)$, f'(x) > 0, meaning the function is increasing.

• For $x\in \left(-\frac{2}{3},0\right)$, f'(x) < 0, meaning the function is decreasing.

Let's calculate the values of $f(x)$ at key boundaries to find its range:

• At $x=-\pi \approx -3.14$:

  $$f(-\pi )=(-\pi {)}^3+(-\pi {)}^2+3\approx -31.00+9.86+3=-18.14$$

• At $x=-\frac{2}{3}$:

  $$f\left(-\frac{2}{3}\right)={\left(-\frac{2}{3}\right)}^3+{\left(-\frac{2}{3}\right)}^2+3=-\frac{8}{27}+\frac{4}{9}+3=\frac{-8+12+81}{27}=\frac{85}{27}\approx 3.15$$

• At $x=0$:

  $$f(0)=0+0+3=3$$

Now let's compare this with RHS $=-2\sin x$:

For $x\in [-\pi ,0)$, $\sin x$ is negative, which means $-2\sin x$ is strictly positive and bounded:

0 < -2 \sin x \le 2

Let's look at the intersection behavior:

1. As $x$ goes from $-\pi$ to $-\frac{2}{3}$, $f(x)$ increases continuously from a highly negative value ($-18.14$) up to a positive value ($3.15$). Because it moves through this range while RHS stays between $0$ and $2$, the two curves must intersect exactly once.

2. As $x$ goes from $-\frac{2}{3}$ to $0$, $f(x)$ decreases from $3.15$ down to $3$. Since the lowest value of $f(x)$ here is $3$ and the maximum value of RHS is $2$, the curves cannot cross each other again in this sub-interval.

Therefore, there is exactly $1$ root in the interval $[-\pi ,0)$.

Conclusion

Combining both intervals, the equation has exactly $1$ real root. Since $1$ is not listed in options (a), (b), or (c), the correct choice is "none".

Correct Answer

The correct option is (d) none.