Let the function f : [1, ∞) → R be defined by f(t) = (-1)^(n+1) 2 / (2n+1-t) if t = 2n-1, n ∈ N; -f(

Q: Wait, why do we say the integral is zero at $x=3, 5, 7$? Do I have to integrate the whole thing?

Great question! You absolutely do not need to perform brute-force integration for every segment. Look at the geometry. The function $f(t)$ is periodic and symmetric. The area from $1$ to $2$ is a positive triangle, and the area from $2$ to $3$ is an identical negative triangle. They cancel out perfectly. Since this pattern repeats, the net area (which is $g(x)$) returns to zero at every odd integer. Symmetry is your best friend in JEE Advanced!

Q: Why do we use L'Hopital's Rule for $\beta$? Can't we just solve the integral?

You could solve the integral, but L'Hopital's Rule is the elegant shortcut here. The expression $\lim_{x \to 1^+} \frac{g(x)}{x-1}$ is a classic $0/0$ form because $g(1) = \int_1^1 f(t) dt = 0$. By differentiating the numerator and denominator, you get $g'(x)/1$. By the Fundamental Theorem of Calculus, $g'(x) = f(x)$. So, the limit is just $f(1^+)$. It saves you time and reduces the chance of calculation errors.

Q: The function definition looks terrifying with the $(-1)^{n+1}$ term. How do I handle that?

It looks intimidating, but it's just a mathematical way of saying 'the sign flips every time we move to the next interval.' Don't let the notation paralyze you. Whenever you see a complex piecewise definition, test the first two values of $n$ (like $n=1$ and $n=2$). You will quickly see the pattern emerge, and the 'scary' notation will reveal itself as a simple oscillating sequence.

Question

Let the function $f : [1, \infty) 	o \mathbb{R}$ be defined by
$f(t) = \begin{cases} \frac{(-1)^{n+1} 2}{(2n+1-t)} & ; 	ext{if } t = 2n-1, n \in \mathbb{N} \ -f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1) & ; 	ext{if } 2n-1 < t < 2n+1, n \in \mathbb{N} \end{cases}$
Define $g(x) = \int_1^x f(t)dt, x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \lim_{x 	o 1^+} \frac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a calculus problem; we are peeling back the layers of a function that behaves like a heartbeat. When you first look at the definition of $f(t)$, it is easy to feel overwhelmed by piecewise definitions, $(-1)^{n+1}$ terms, and limits.

In JEE Advanced, the most complex-looking problems often hide the most elegant, simple geometric truths. Let us embark on this journey together.

### Visualizing the Rhythm of $f(t)$

Let us start by decoding $f(t)$. The problem defines it in pieces. For $1 < t < 3$, we are looking at a linear function passing through points $(1, 2)$ and $(3, -2)$.

Using the point-slope form, we find the equation:

$$f(t) = 4 - 2t$$

Imagine standing on the $t$-axis. As $t$ moves from $1$ to $2$, the function is positive, creating a triangle of area $1$. As $t$ moves from $2$ to $3$, the function dips below the axis, creating a triangle of area $-1$.

This is the 'heartbeat' of our function. It rises and falls, creating equal positive and negative areas. This symmetry is the key to the entire problem.

### The Integral as a Net Signed Area

Now, we define $g(x) = \int_1^x f(t) dt$. In calculus, the integral is the accumulation of area, representing the 'net signed area' under the curve from $1$ to $x$.

For the interval $1 < x < 3$, we calculate:

$$g(x) = \int_1^x (4 - 2t) dt$$

Performing the integration, we obtain:

$$g(x) = [4t - t^2]_1^x = (4x - x^2) - (4 - 1) = -x^2 + 4x - 3$$

This is a downward-opening parabola. If we set $g(x) = 0$, we find the roots at $x=1$ and $x=3$. This confirms our geometric intuition: the area starts at $0$ (at $x=1$), grows to a maximum, and then shrinks back to $0$ (at $x=3$) as the negative area cancels out the positive area.

### The Beauty of Symmetry and $\alpha$

Because the function $f(t)$ repeats this pattern of positive and negative triangles, the integral $g(x)$ will return to zero at every odd integer.

The integral from $3$ to $5$ is a shifted version of the integral from $1$ to $3$. Since the net area from $1$ to $3$ is $0$, the net area from $3$ to $5$ is also $0$. Therefore:

$$g(5) = g(3) + \int_3^5 f(t) dt = 0 + 0 = 0$$

This pattern continues indefinitely. The roots of $g(x) = 0$ in the interval $(1, 8]$ are $x = 3, 5, 7$. Counting these, we find **$\alpha = 3$**.

### The Elegance of the Limit and $\beta$

Finally, we tackle $\beta = \lim_{x 	o 1^+} \frac{g(x)}{x-1}$. This is a $0/0$ indeterminate form, so we apply L'Hopital's Rule:

$$\beta = \lim_{x 	o 1^+} \frac{\frac{d}{dx} \int_1^x f(t) dt}{\frac{d}{dx} (x-1)}$$

By the Fundamental Theorem of Calculus, the derivative of the integral is simply the integrand $f(x)$, and the derivative of the denominator is $1$. Thus:

$$\beta = \lim_{x 	o 1^+} f(x) = 4 - 2(1) = 2$$

### Final Calculation

We have navigated the geometry, exploited the symmetry, and utilized the Fundamental Theorem of Calculus. We found $\alpha = 3$ and $\beta = 2$.

Adding them together, we arrive at our final answer:

$$\alpha + \beta = 5$$

Remember, in JEE Advanced, you are not just a calculator; you are an architect of logic. When you see a complex function, look for the symmetry, trust the theorems, and let the math tell its story.

Enter Numerical Value:

Visualized Solution

Analyzing $f (t)$

Equation of $f (t)$ for $1 < t < 3$

Defining $g (x)$

Integral for $1 < x < 3$

Evaluating $g (x)$

Roots of $g (x)$ in $(1, 3]$

Geometric Area Intuition

Finding Other Roots

Value of $α$

Setting up $β$

Applying L'Hopital's Rule

Evaluating $β$

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Visualizing the Rhythm of $f (t)$

The Integral as a Net Signed Area

The Beauty of Symmetry and $α$

The Elegance of the Limit and $β$

Final Calculation

Similar Questions

Comprehension Passage

The value of $g (\frac{1}{2})$ is

The value of $g^{'} (\frac{1}{2})$ is

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is:

Let $f (x) = \int_{0}^{x} g (t) lo g_{e} (\frac{1 - t}{1 + t}) d t$ , where $g$ is a continuous odd function. If $\int_{- π /2}^{π /2} (f (x) + \frac{x ^{2} c o s x}{1 + e ^{x}}) d x = (\frac{π}{α})^{2} - α$ , then $α$ is equal to

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

If the value of the integral $\int_{0}^{5} \frac{x + [ x ]}{e ^{x - [x]}} d x = α e^{- 1} + β$ , where $α, β \in R$ , $5 α + 6 β = 0$ , and $[x]$ denotes the greatest integer less than or equal to $x$ ; then the value of $(α + β)^{2}$ is equal to :

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}, x \in R - {- 1}, n \in N, n > 2$ . If $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$ , then $lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f_{n} (x)) d x$ is equal to

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

Let $g (t) = \int_{- π /2}^{π /2} cos (\frac{π}{4} t + f (x)) d x$ , where $f (x) = lo g_{e} (x + x^{2} + 1), x \in R$ . Then which one of the following is correct?

Enter Numerical Value:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1<t<3

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1<x<3

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,3]

Geometric Area Intuition

Finding Other Roots

Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Applying L'Hopital's Rule

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Visualizing the Rhythm of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

The Integral as a Net Signed Area

The Beauty of Symmetry and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Elegance of the Limit and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of g(21​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of g′(21​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)+2f(x1​)=x2+5 and 2g(x)−3g(21​)=x,x>0. If α=∫12​f(x)dx, and β=∫12​g(x)dx, then the value of 9α+β is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫0x​g(t)loge​(1+t1−t​)dt, where g is a continuous odd function. If ∫−π/2π/2​(f(x)+1+exx2cosx​)dx=(απ​)2−α, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0x​f(t)dt, where f is such that 21​≤f(t)≤1 for t∈[0,1] and 0≤f(t)≤21​ for t∈[1,2]. Then g(2) satisfies the inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=(1+xn)1/nx​,x∈R−{−1},n∈N,n>2. If fn​(x)=(f∘f∘… upto n times)(x), then limn→∞​∫01​xn−2(fn​(x))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(t)=∫−π/2π/2​cos(4π​t+f(x))dx, where f(x)=loge​(x+x2+1​),x∈R. Then which one of the following is correct?

Analyzing $f (t)$

Equation of $f (t)$ for $1 < t < 3$

Defining $g (x)$

Integral for $1 < x < 3$

Evaluating $g (x)$

Roots of $g (x)$ in $(1, 3]$

Value of $α$

Setting up $β$

Evaluating $β$

Visualizing the Rhythm of $f (t)$

The Beauty of Symmetry and $α$

The Elegance of the Limit and $β$

The value of $g (\frac{1}{2})$ is

The value of $g^{'} (\frac{1}{2})$ is

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is:

Let $f (x) = \int_{0}^{x} g (t) lo g_{e} (\frac{1 - t}{1 + t}) d t$ , where $g$ is a continuous odd function. If $\int_{- π /2}^{π /2} (f (x) + \frac{x ^{2} c o s x}{1 + e ^{x}}) d x = (\frac{π}{α})^{2} - α$ , then $α$ is equal to

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}, x \in R - {- 1}, n \in N, n > 2$ . If $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$ , then $lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f_{n} (x)) d x$ is equal to

Let $g (t) = \int_{- π /2}^{π /2} cos (\frac{π}{4} t + f (x)) d x$ , where $f (x) = lo g_{e} (x + x^{2} + 1), x \in R$ . Then which one of the following is correct?