List IList IIP. The number of polynomials f(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $(P)$ Let $f(x) = ax^2 + bx$,where $a, b \in \mathbb{W}$ (since $f(0)=0$).$\int_0^1 (ax^2 + bx) dx = \frac{a}{3} + \frac{b}{2} = 1 \implies 2a + 3

Vedclass · Accepted Answer

$2 \quad 3 \quad 1 \quad 4$

Vedclass · Answer

(D) $(P)$ Let $f(x) = ax^2 + bx$,where $a, b \in \mathbb{W}$ (since $f(0)=0$).$\int_0^1 (ax^2 + bx) dx = \frac{a}{3} + \frac{b}{2} = 1 \implies 2a + 3b = 6$.Possible non-negative integer solutions $(a, b)$ are $(3, 0)$ and $(0, 2)$.Thus,the number of such polynomials is $2$.$(Q)$ $f(x) = \sqrt{2} \sin(x^2 + \frac{\pi}{4})$.$f(x)$ is maximum when $x^2 + \frac{\pi}{4} = 2n\pi + \frac{\pi}{2} \implies x^2 = 2n\pi + \frac{\pi}{4}$.For $n=0$,$x^2 = \frac{\pi}{4} \approx 0.785 \in [0, 13)$.For $n=1$,$x^2 = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4} \approx 7.068 \in [0, 13)$.For $n=2$,$x^2 = 4\pi + \frac{\pi}{4} = \frac{17\pi}{4} \approx 13.35 
otin [0, 13)$.Since $x^2$ can take $2$ values for each $n$,and $x$ can be $\pm \sqrt{x^2}$,we check the interval $(-\sqrt{13}, \sqrt{13})$.For $n=0$,$x = \pm \sqrt{\pi/4} = \pm \sqrt{\pi}/2$ ($2$ points).For $n=1$,$x = \pm \sqrt{9\pi/4} = \pm 3\sqrt{\pi}/2$ ($2$ points).Total points $= 4$. Wait,checking the options,$Q$ matches $3$ ($4$ points).$(R)$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx = \int_0^2 3x^2 (\frac{1}{1+e^x} + \frac{1}{1+e^{-x}}) dx = \int_0^2 3x^2 (\frac{1+e^x}{1+e^x}) dx = \int_0^2 3x^2 dx = [x^3]_0^2 = 8$.$(S)$ The numerator is $\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx$. Since $\cos 2x$ is even and $\log(\frac{1+x}{1-x})$ is odd,the product is odd. Thus,the integral is $0$.

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 $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

Similar Questions

The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to:

If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} [x], & x \leq 2 \\ 0, & x>2 \end{cases}$,where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$,then the value of $(4I-1)$ is

$\int_{0}^{2} ( |2x^2 - 3x| + [x - \frac{1}{2}] ) dx$,where $[ \cdot ]$ is the greatest integer function,is equal to:

Let $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n x \, dx$. Then $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \dots$ are in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module