For any real number x , let [x] denote the largest integer less than equal to x . Let f be a real valued function defined on the interval [-10,10] by f(x)= beginarraylx-[x], if [x] is odd 1+[x]-x, if [x] is even .\endarray. Then the value of \frac\pi^210 \int_-10^10 f(x) \cos \pi x \,d x is :

Definite Integration PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q29 with Solution

View written solutionFree

Correct answer: A

Understand the function on each unit interval

For any integer $n$ , on the interval $[n,n+1)$ we have $[x]=n$ . So

if $n$ is odd, then $f(x)=x-n,$
if $n$ is even, then $f(x)=1+n-x=1-(x-n).$

Thus on each unit interval, $f(x)$ is a triangular/sawtooth function.

Let $x=n+t,\qquad t\in[0,1).$ Then

t, & n\text{ odd},\\ 1-t, & n\text{ even}. \end{cases}$$ Also, $$\cos(\pi x)=\cos(\pi(n+t))=\cos(n\pi+\pi t)=(-1)^n\cos(\pi t).$$ So on $[n,n+1)$, $$f(x)\cos(\pi x)=\begin{cases} t(-1)^n\cos(\pi t), & n\text{ odd},\\ (1-t)(-1)^n\cos(\pi t), & n\text{ even}. \end{cases}$$ Since $(-1)^n=-1$ for odd $n$ and $=1$ for even $n$, the contribution from each interval is $$I_n:=\int_n^{n+1} f(x)\cos(\pi x)\,dx$$ with $$I_n=\begin{cases} -\displaystyle\int_0^1 t\cos(\pi t)\,dt, & n\text{ odd},\\ \displaystyle\int_0^1 (1-t)\cos(\pi t)\,dt, & n\text{ even}. \end{cases}$$ 2. **Show both cases give the same value** Let $$A=\int_0^1 t\cos(\pi t)\,dt, \qquad B=\int_0^1 (1-t)\cos(\pi t)\,dt.$$ Now, $$B=\int_0^1 \cos(\pi t)\,dt-\int_0^1 t\cos(\pi t)\,dt.$$ But $$\int_0^1 \cos(\pi t)\,dt=\left[\frac{\sin(\pi t)}{\pi}\right]_0^1=0.$$ Hence $$B=-A.$$ So for even $n$, $I_n=B=-A$, and for odd $n$, $I_n=-A$ as well. Therefore **every unit interval contributes the same amount**: $$I_n=-A.$$ 3. **Compute $A$** Using integration by parts: Take $$u=t,\qquad dv=\cos(\pi t)dt.$$ Then $$du=dt,\qquad v=\frac{\sin(\pi t)}{\pi}.$$ So $$A=\left[\frac{t\sin(\pi t)}{\pi}\right]_0^1-\int_0^1 \frac{\sin(\pi t)}{\pi}\,dt.$$ The boundary term is $0$, hence $$A=-\frac{1}{\pi}\int_0^1 \sin(\pi t)\,dt.$$ Now $$\int_0^1 \sin(\pi t)\,dt=\left[-\frac{\cos(\pi t)}{\pi}\right]_0^1=\frac{2}{\pi}.$$ Thus $$A=-\frac{1}{\pi}\cdot \frac{2}{\pi}=-\frac{2}{\pi^2}.$$ Therefore $$I_n=-A=\frac{2}{\pi^2}.$$ 4. **Sum over all intervals from $-10$ to $10$** The interval $[-10,10]$ consists of $20$ unit intervals: $$[-10,-9),[-9,-8),\dots,[9,10].$$ (Changing values at finitely many endpoints does not affect the integral.) Hence $$\int_{-10}^{10} f(x)\cos(\pi x)\,dx=20\cdot \frac{2}{\pi^2}=\frac{40}{\pi^2}.$$ 5. **Evaluate the required expression** $$\frac{\pi^2}{10}\int_{-10}^{10} f(x)\cos(\pi x)\,dx =\frac{\pi^2}{10}\cdot \frac{40}{\pi^2}=4.$$ 6. **Check options** - A: $4$ ✅ - B: $2$ ❌ - C: $1$ ❌ - D: $0$ ❌ Therefore the correct answer is **A**.

More from Definite Integration

Let $[t]$ denote the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{- 3}^{101} ([sin (π x)] + e^{[c o s (2 π x)]}) d x$ is equal to2022 · MCQ
Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + 0 \int x (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to …2022 · Numerical
Let $a_{n} = - 1 \int n (1 + \frac{x}{2} + \frac{x ^{2}}{3} + ..... + \frac{x ^{n - 1}}{n}) d x$ for every n $\in$ N. Then the sum of all the elements of the set {n} $\in$ N : an $\in$ (2, 30)} is …2022 · Numerical
The value of $0 \int π \frac{e ^{c o s x} s i n x}{( 1 + c o s ^{2} x ) ( e ^{c o s x} + e ^{- c o s x} )} d x$ is equal to:2022 · MCQ
If $b_{n} = \int_{0}^{\frac{π}{2}} \frac{c o s ^{2} n x}{s i n x} d x, n \in N$ , then2022 · MCQ
The value of b > 3 for which $12 3 \int b \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x = lo g_{e} (\frac{49}{40})$ , is equal to .2022 · Numerical
If $n (2 n + 1) \int_{0}^{1} (1 - x^{n})^{2 n} d x = 1177 \int_{0}^{1} (1 - x^{n})^{2 n + 1} d x$ , then $n \in N$ is equal to …2022 · Numerical
$0 \int 20 π (∣ sin x ∣ + ∣ cos x ∣)^{2} d x is equal to$ 2022 · MCQ

View written solutionFree

Correct answer: A

Understand the function on each unit interval

For any integer $n$ , on the interval $[n,n+1)$ we have $[x]=n$ . So

if $n$ is odd, then $f(x)=x-n,$
if $n$ is even, then $f(x)=1+n-x=1-(x-n).$

Thus on each unit interval, $f(x)$ is a triangular/sawtooth function.

Let $x=n+t,\qquad t\in[0,1).$ Then

t, & n\text{ odd},\\ 1-t, & n\text{ even}. \end{cases}$$ Also, $$\cos(\pi x)=\cos(\pi(n+t))=\cos(n\pi+\pi t)=(-1)^n\cos(\pi t).$$ So on $[n,n+1)$, $$f(x)\cos(\pi x)=\begin{cases} t(-1)^n\cos(\pi t), & n\text{ odd},\\ (1-t)(-1)^n\cos(\pi t), & n\text{ even}. \end{cases}$$ Since $(-1)^n=-1$ for odd $n$ and $=1$ for even $n$, the contribution from each interval is $$I_n:=\int_n^{n+1} f(x)\cos(\pi x)\,dx$$ with $$I_n=\begin{cases} -\displaystyle\int_0^1 t\cos(\pi t)\,dt, & n\text{ odd},\\ \displaystyle\int_0^1 (1-t)\cos(\pi t)\,dt, & n\text{ even}. \end{cases}$$ 2. **Show both cases give the same value** Let $$A=\int_0^1 t\cos(\pi t)\,dt, \qquad B=\int_0^1 (1-t)\cos(\pi t)\,dt.$$ Now, $$B=\int_0^1 \cos(\pi t)\,dt-\int_0^1 t\cos(\pi t)\,dt.$$ But $$\int_0^1 \cos(\pi t)\,dt=\left[\frac{\sin(\pi t)}{\pi}\right]_0^1=0.$$ Hence $$B=-A.$$ So for even $n$, $I_n=B=-A$, and for odd $n$, $I_n=-A$ as well. Therefore **every unit interval contributes the same amount**: $$I_n=-A.$$ 3. **Compute $A$** Using integration by parts: Take $$u=t,\qquad dv=\cos(\pi t)dt.$$ Then $$du=dt,\qquad v=\frac{\sin(\pi t)}{\pi}.$$ So $$A=\left[\frac{t\sin(\pi t)}{\pi}\right]_0^1-\int_0^1 \frac{\sin(\pi t)}{\pi}\,dt.$$ The boundary term is $0$, hence $$A=-\frac{1}{\pi}\int_0^1 \sin(\pi t)\,dt.$$ Now $$\int_0^1 \sin(\pi t)\,dt=\left[-\frac{\cos(\pi t)}{\pi}\right]_0^1=\frac{2}{\pi}.$$ Thus $$A=-\frac{1}{\pi}\cdot \frac{2}{\pi}=-\frac{2}{\pi^2}.$$ Therefore $$I_n=-A=\frac{2}{\pi^2}.$$ 4. **Sum over all intervals from $-10$ to $10$** The interval $[-10,10]$ consists of $20$ unit intervals: $$[-10,-9),[-9,-8),\dots,[9,10].$$ (Changing values at finitely many endpoints does not affect the integral.) Hence $$\int_{-10}^{10} f(x)\cos(\pi x)\,dx=20\cdot \frac{2}{\pi^2}=\frac{40}{\pi^2}.$$ 5. **Evaluate the required expression** $$\frac{\pi^2}{10}\int_{-10}^{10} f(x)\cos(\pi x)\,dx =\frac{\pi^2}{10}\cdot \frac{40}{\pi^2}=4.$$ 6. **Check options** - A: $4$ ✅ - B: $2$ ❌ - C: $1$ ❌ - D: $0$ ❌ Therefore the correct answer is **A**.

Definite Integration question

Definite Integration question

More from Definite Integration

Definite Integration question

Definite Integration question

More from Definite Integration