Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 - 3x2 - 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q40 with Solution

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Correct answer: 114

We have $f(x)=2x^3-3x^2-12x.$

We need the points of local maximum and local minimum, say at $x=a$ and $x=b$ , and then the total area bounded by:

Differentiate: $f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x-2)(x+1).$

So critical points are:

\quad x=2.$$ Now use the second derivative: $$f''(x)=12x-6.$$ - At $x=-1$: $$f''(-1)=-18<0,$$ so $x=-1$ is a **local maximum**. Hence, $$a=-1.$$ - At $x=2$: $$f''(2)=18>0,$$ so $x=2$ is a **local minimum**. Hence, $$b=2.$$ --- 3. **Check where the curve lies between $x=a$ and $x=b$** We need area between $x=-1$ and $x=2$. Factorize $f(x)$: $$f(x)=2x^3-3x^2-12x=x(2x^2-3x-12).$$ Its zeros are at $x=0$ and the roots of $2x^2-3x-12=0$, i.e. $$x=\frac{3\pm\sqrt{105}}{4}.$$ Among these, only $x=0$ lies in $[-1,2]$. Now test sign: - On $[-1,0]$, take $x=-\tfrac12$: $$f(-\tfrac12)>0.$$ - On $[0,2]$, take $x=1$: $$f(1)=2-3-12=-13<0.$$ So total area is $$A=\int_{-1}^{0} f(x)\,dx-\int_{0}^{2} f(x)\,dx.$$ --- 4. **Integrate** An antiderivative is $$\int f(x)\,dx=\frac{x^4}{2}-x^3-6x^2.$$ Let $$F(x)=\frac{x^4}{2}-x^3-6x^2.$$ Then: ### Area from $-1$ to $0$ $$\int_{-1}^{0} f(x)\,dx=F(0)-F(-1).$$ Compute: $$F(0)=0,$$ $$F(-1)=\frac{1}{2}-(-1)-6=\frac12+1-6=-\frac92.$$ Hence, $$\int_{-1}^{0} f(x)\,dx=0-\left(-\frac92\right)=\frac92.$$ ### Area from $0$ to $2$ $$\int_{0}^{2} f(x)\,dx=F(2)-F(0).$$ Compute: $$F(2)=\frac{16}{2}-8-24=8-8-24=-24.$$ So $$\int_{0}^{2} f(x)\,dx=-24.$$ Therefore the area on $[0,2]$ is $$-\int_{0}^{2} f(x)\,dx=24.$$ Thus total area is $$A=\frac92+24=\frac{57}{2}.$$ --- 5. **Find $4A$** $$4A=4\cdot \frac{57}{2}=114.$$ --- 6. **Comparison with stored answer** Our derived answer is $114$, which matches the stored correct answer.

View written solutionFree

Correct answer: 114

We have $f(x)=2x^3-3x^2-12x.$

We need the points of local maximum and local minimum, say at $x=a$ and $x=b$ , and then the total area bounded by:

Differentiate: $f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x-2)(x+1).$

So critical points are:

\quad x=2.$$ Now use the second derivative: $$f''(x)=12x-6.$$ - At $x=-1$: $$f''(-1)=-18<0,$$ so $x=-1$ is a **local maximum**. Hence, $$a=-1.$$ - At $x=2$: $$f''(2)=18>0,$$ so $x=2$ is a **local minimum**. Hence, $$b=2.$$ --- 3. **Check where the curve lies between $x=a$ and $x=b$** We need area between $x=-1$ and $x=2$. Factorize $f(x)$: $$f(x)=2x^3-3x^2-12x=x(2x^2-3x-12).$$ Its zeros are at $x=0$ and the roots of $2x^2-3x-12=0$, i.e. $$x=\frac{3\pm\sqrt{105}}{4}.$$ Among these, only $x=0$ lies in $[-1,2]$. Now test sign: - On $[-1,0]$, take $x=-\tfrac12$: $$f(-\tfrac12)>0.$$ - On $[0,2]$, take $x=1$: $$f(1)=2-3-12=-13<0.$$ So total area is $$A=\int_{-1}^{0} f(x)\,dx-\int_{0}^{2} f(x)\,dx.$$ --- 4. **Integrate** An antiderivative is $$\int f(x)\,dx=\frac{x^4}{2}-x^3-6x^2.$$ Let $$F(x)=\frac{x^4}{2}-x^3-6x^2.$$ Then: ### Area from $-1$ to $0$ $$\int_{-1}^{0} f(x)\,dx=F(0)-F(-1).$$ Compute: $$F(0)=0,$$ $$F(-1)=\frac{1}{2}-(-1)-6=\frac12+1-6=-\frac92.$$ Hence, $$\int_{-1}^{0} f(x)\,dx=0-\left(-\frac92\right)=\frac92.$$ ### Area from $0$ to $2$ $$\int_{0}^{2} f(x)\,dx=F(2)-F(0).$$ Compute: $$F(2)=\frac{16}{2}-8-24=8-8-24=-24.$$ So $$\int_{0}^{2} f(x)\,dx=-24.$$ Therefore the area on $[0,2]$ is $$-\int_{0}^{2} f(x)\,dx=24.$$ Thus total area is $$A=\frac92+24=\frac{57}{2}.$$ --- 5. **Find $4A$** $$4A=4\cdot \frac{57}{2}=114.$$ --- 6. **Comparison with stored answer** Our derived answer is $114$, which matches the stored correct answer.