JEE Advanced 2014 Paper 2Question 1
$$\begin{array}{ll}\textbf{List I} & \textbf{List II}\\ \text{P. The number of polynomials } f(x) \text{ with non-negative integer coefficients of degree} \le 2 \text{, satisfying } f(0)=0 \text{ and } \int_0^1 f(x)dx=1 \text{, is} & 1.\ 8\\ \text{Q. The number of points in the interval } [-\sqrt{13},\sqrt{13}] \text{ at which } f(x)=\sin(x^2)+\cos(x^2) \text{ attains its maximum value, is} & 2.\ 2\\ \text{R. } \int_{-2}^{2}\frac{3x^2}{(1+e^x)}dx \text{ equals} & 3.\ 4\\ \text{S. } \dfrac{\left(\int_{-1/2}^{1/2}\cos 2x\,\log\left(\frac{1+x}{1-x}\right)dx\right)}{\left(\int_{0}^{1/2}\cos 2x\,\log\left(\frac{1+x}{1-x}\right)dx\right)} \text{ equals} & 4.\ 0\end{array}$$
Codes (P Q R S):
Definite Integration
Solution
Concept. Do each row separately: P counts solutions of a linear equation, Q finds where $\sin+\cos$ is largest, R uses $x\to-x$, S uses an odd integrand. Formulas used. (1) $\sin\theta+\cos\theta=\sqrt2\sin\big(\theta+\frac\pi4\big)$(2) $\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=1$(3) $\int_{-a}^{a}(\text{odd})\,dx=0$Step 1 - P. $f(0)=0$ gives $f=ax+bx^2$ with $a,b\ge0$ integers. $\int_0^1f=\frac a2+\frac b3=1$, i.e. $3a+2b=6$: $(a,b)=(2,0)$ or $(0,3)$. So $2$ polynomials: P-2. Step 2 - Q. By (1), the maximum $\sqrt2$ occurs when $x^2+\frac\pi4=\frac\pi2+2k\pi$, i.e. $x^2=\frac\pi4+2k\pi$. With $x^2\le13$: $x^2=\frac\pi4$ or $\frac{9\pi}4\approx7.07$ (next is $\frac{17\pi}4\approx13.35>13$). Each value of $x^2$ gives two values of $x$ ($\pm$), so $4$ points: Q-3. Step 3 - R. Put $x\to-x$ and add; by (2), $$2I=\int_{-2}^{2}3x^2\,dx=16,\qquad I=8.$$ R-1. Step 4 - S. $\log\frac{1+x}{1-x}$ is odd and $\cos2x$ is even, so the product is odd. By (3) the numerator is $0$ (the denominator is positive), so the ratio is $0$: S-4. Step 5 - Code. P-2, Q-3, R-1, S-4. Answer: (D)