Consider the polynomial \[ f(x)=1+2 x+3 x^{2}+4 x^{3} . \] Let s be the sum of all distinct real roots of $\mathrm{f}(\mathrm{x})$ and let $\mathrm{t}=|\mathrm{s}|$. The area bounded by the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ and the lines $\mathrm{x}=0, \mathrm{y}=0$ and $\mathrm{x}=\mathrm{t}$, lies in the interval

(A) $\left(\frac{3}{4}, 3\right)$

(B) $\left(\frac{21}{64}, \frac{11}{16}\right)$

(D) $\left(0, \frac{21}{64}\right)$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Area Under Curves

Consider the polynomial

f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} .

Let s be the sum of all distinct real roots of

f (x)

and let

t = ∣ s ∣

.

The area bounded by the curve

y = f (x)

and the lines

x = 0, y = 0

and

x = t

, lies in the interval

Answer: (A)

From the previous question,

s \in (- \frac{3}{4}, - \frac{1}{2})

, so

t = ∣ s ∣ \in (\frac{1}{2}, \frac{3}{4})

.

The area is:

A = \int_{0}^{t} f (x) d x = \int_{0}^{t} (1 + 2 x + 3 x^{2} + 4 x^{3}) d x = [x + x^{2} + x^{3} + x^{4}]_{0}^{t} = t + t^{2} + t^{3} + t^{4}

For

t = \frac{1}{2}

A = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}

For

t = \frac{3}{4}

A = \frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} = \frac{192 + 144 + 108 + 81}{256} = \frac{525}{256} \approx 2.05

So the area lies in

(\frac{15}{16}, \frac{525}{256}) \approx (0.94, 2.05)

, which is contained in

(\frac{3}{4}, 3)

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option