New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Definite Integration question

2022 · 29 Jun · Shift 1 · Q33

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Definite Integration
2022 · 29 Jun · Shift 1 · Q33

Definite Integration question

2022 · 29 Jun · Shift 1 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx}

, where [t] denotes greatest integer less than or equal to t, is equal to:

A
$-$ 3
B
$-$ 2
C
2
D
0

View written solutionFree

Correct answer: D

We need to evaluate $I=\int_0^5 \cos\left(\pi\left(x-\left\lfloor \frac{x}{2}\right\rfloor\right)\right)dx.$

Here, the floor term $\left\lfloor \frac{x}{2}\right\rfloor$ is constant on intervals of length $2$ .

Split the interval $[0,5]$ according to the value of $\left\lfloor \frac{x}{2}\right\rfloor.$

For $0\le x<2$ , $\left\lfloor x/2\right\rfloor=0$
For $2\le x<4$ , $\left\lfloor x/2\right\rfloor=1$
For $4\le x\le 5$ , $\left\lfloor x/2\right\rfloor=2$

So, $I=\int_0^2 \cos(\pi x)\,dx+\int_2^4 \cos(\pi(x-1))\,dx+\int_4^5 \cos(\pi(x-2))\,dx.$

Evaluate each part.

First integral: $I_1=\int_0^2 \cos(\pi x)dx=\left[\frac{\sin(\pi x)}{\pi}\right]_0^2=0.$

Second integral: Since $\cos(\pi(x-1))=\cos(\pi x-\pi)=-\cos(\pi x),$ we get

= -\left[\frac{\sin(\pi x)}{\pi}\right]_2^4=0.$$ Third integral: $$\cos(\pi(x-2))=\cos(\pi x-2\pi)=\cos(\pi x),$$ so $$I_3=\int_4^5 \cos(\pi(x-2))dx=\int_4^5 \cos(\pi x)dx =\left[\frac{\sin(\pi x)}{\pi}\right]_4^5=0.$$ Thus, $$I=I_1+I_2+I_3=0+0+0=0.$$ 4. Checking options: - A: $-3$ ❌ - B: $-2$ ❌ - C: $2$ ❌ - D: $0$ ✅ Therefore, the value of the integral is $$\boxed{0}.$$

More from Definite Integration

Let f be a real valued continuous function on [0, 1] and $f (x) = x + 0 \int 1 (x - t) f (t) d t$ . Then, which of the following points (x, y) lies on the curve y = f(x) ?2022 · MCQ
If $0 \int 2 (2 x - 2 x - x^{2}) d x = 0 \int 1 (1 - 1 - y^{2} - \frac{y ^{2}}{2}) d y + 1 \int 2 (2 - \frac{y ^{2}}{2}) d y + I$ , then I equals2022 · MCQ
Let $f (t) = 0 \int t e^{x^{3}} (\frac{x ^{8}}{( x ^{6} + 2 x ^{3} + 2 ) ^{2}}) d x$ . If $f (1) + f^{'} (1) = α e - \frac{1}{6}$ , then the value of 150 $α$ is equal to .2022 · Numerical
Let f : R $\to$ R be a continuous function. Then $x \to \frac{π}{4} lim \frac{\frac{π}{4} 2 \int s e c ^{2} x f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}$ is equal to :2021 · MCQ
Let $J_{n, m} = 0 \int \frac{1}{2} \frac{x ^{n}}{x ^{m} - 1} d x$ , $\forall$ n > m and n, m $\in$ N. Consider a matrix $A = [a_{ij}]_{3 \times 3}$ where $a_{ij} = {j_{6 + i, 3} - j_{i + 3, 3}, 0, i \leq j i > j$ …2021 · MCQ
The function f(x), that satisfies the condition $f (x) = x + 0 \int π /2 sin x . cos y f (y) d y$ , is :2021 · MCQ
Let f : R $\to$ R be a continuous function such that f(x) + f(x + 1) = 2, for all x $\in$ R. If $I_{1} = 0 \int 8 f (x) d x$ and $I_{2} = - 1 \int 3 f (x) d x$ , then the value of I1 + 2I2 is equal to …2021 · Numerical
If the normal to the curve y(x) = $0 \int x (2 t^{2} - 15 t + 10) d t$ at a point (a, b) is parallel to the line x + 3y = $-$ 5, a > 1, then the value of | a + 6b | is equal to .2021 · Numerical