JEE Main 2018 — Application Of Derivatives Question with Solution

To determine the absolute maximum (M) and absolute minimum (m) of the function $f(x)=2x^3-9x^2+12x+5$ over the interval $[0,3]$, we need to examine its critical points and endpoints.

First, we find the derivative of the function, $f^{\prime }(x)$, to locate the critical points:

$$f^{\prime }(x)=\frac{d}{dx}(2x^3-9x^2+12x+5)=6x^2-18x+12$$

Next, we set the derivative equal to zero to find the critical points:

$$6x^2-18x+12=0$$

Simplifying this equation by dividing by 6:

$$x^2-3x+2=0$$

We solve this quadratic equation using the factorization method:

$$(x-1)(x-2)=0$$

So, the critical points are:

$$x=1\text{and}x=2$$

We now evaluate the function $f(x)$ at the critical points and at the endpoints of the interval [0, 3]:

1. At $x=0$:

$$f(0)=2(0{)}^3-9(0{)}^2+12(0)+5=5$$

2. At $x=1$:

$$f(1)=2(1{)}^3-9(1{)}^2+12(1)+5=2-9+12+5=10$$

3. At $x=2$:

$$f(2)=2(2{)}^3-9(2{)}^2+12(2)+5=16-36+24+5=9$$

4. At $x=3$:

$$f(3)=2(3{)}^3-9(3{)}^2+12(3)+5=54-81+36+5=14$$

We now identify the absolute maximum (M) and absolute minimum (m) from the above values:

• Maximum value $M=14$ at $x=3$
• Minimum value $m=5$ at $x=0$

Thus, the difference $M-m$ is:

$$M-m=14-5=9$$

Therefore, the correct answer is:

Option B: 9