| List I | | List II | | --- | --- | --- | | P. | The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$, satisfying $f(0)=0$ and $\int_{0}^{1} f(x) dx=1$, is | 1. 8 | | Q. | The number of points in the interval $[-\sqrt{13}, \sqrt{13}]$ at which $f(x)=\sin \left(x^{2}\right)+\cos \left(x^{2}\right)$ attains its maximum value, is | 2. 2 | | R. | $\int_{-2}^{2} \frac{3 x^{2}}{\left(1+e^{x}\right)} dx$ equals | 3. 4 | | S. | $\frac{\left(\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x\right)}{\left(\int_{0}^{\frac{1}{2}} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x\right)}$ equals | 4. 0 | P Q R S

Definite Integrals — JEE Advanced · 2014 · 25 May 2014 · Paper 2 Question with Solution

List I		List II
P.	The number of polynomials $f (x)$ with non-negative integer coefficients of degree $\leq 2$ , satisfying $f (0) = 0$ and $\int_{0}^{1} f (x) d x = 1$ , is	1. 8
Q.	The number of points in the interval $[- 13, 13]$ at which $f (x) = sin (x^{2}) + cos (x^{2})$ attains its maximum value, is	2. 2
R.	$\int_{- 2}^{2} \frac{3 x ^{2}}{( 1 + e ^{x} )} d x$ equals	3. 4
S.	$\frac{( \int _{- \frac{1}{2}}^{\frac{1}{2}} c o s 2 x l o g ( \frac{1 + x}{1 - x} ) d x )}{( \int _{0}^{\frac{1}{2}} c o s 2 x l o g ( \frac{1 + x}{1 - x} ) d x )}$ equals	4. 0

P Q R S

Answer: (D)

P → 2: Let

f (x) = a x^{2} + b x + c

. From

f (0) = 0

, we get

c = 0

. From

\int_{0}^{1} f (x) d x = 1

\frac{a}{3} + \frac{b}{2} = 1

, giving

2 a + 3 b = 6

. Non-negative integer solutions: (a,b) = (0,2), (3,0), giving 2 polynomials.

Q → 3:

f (x) = 2 sin (x^{2} + \frac{π}{4})

attains maximum when

x^{2} + \frac{π}{4} = \frac{π}{2} + 2 nπ

, i.e.,

x^{2} = \frac{π}{4} + 2 nπ

. In

[- 13, 13]

x^{2} \leq 13

. Values:

\frac{π}{4}, \frac{9 π}{4}

(both

< 13

), giving 4 points (±√(π/4), ±√(9π/4)).

R → 1 (equals 8): Using

\int_{- a}^{a} \frac{f ( x )}{1 + e ^{x}} d x = \int_{- a}^{a} f (x) d x

when

f

is even:

\int_{- 2}^{2} \frac{3 x ^{2}}{1 + e ^{x}} d x = \int_{- 2}^{2} \frac{3 x ^{2}}{2} d x = \frac{3}{2} \cdot \frac{2 x ^{3}}{3} ∣_{- 2}^{2} = x^{3} ∣_{- 2}^{2} = 8

.

S → 4 (equals 0): The integrand

cos 2 x lo g (\frac{1 + x}{1 - x})

is odd (product of even and odd functions), so

\int_{- 1/2}^{1/2} = 0

, making the ratio 0.

Answer: P→2, Q→3, R→1, S→4, which is 2314.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option