Given that for each a ∈ (0, 1), lim(h→0⁺) ∫(h to 1-h) t⁻ᵃ(1-t)ᵃ⁻¹ dt exists. Let this limit be g(a).

Q: Why can't we just differentiate under the integral sign? It looks like we could just bring the derivative inside.

Great question! You *could* use the Leibniz Integral Rule, but the integral is improper (it blows up at the endpoints). Differentiating under the integral sign requires uniform convergence, which is tricky here. The substitution method is much safer and more elegant!

Q: How did you know to use the substitution $t = \frac{1}{1+z}$?

That's the mark of a seasoned problem solver! When you see an integral from $0$ to $1$ with powers of $t$ and $(1-t)$, it's a classic signature of the Beta function. This specific substitution is the standard 'key' to unlock the integral into the form $\int_0^{\infty} \frac{z^{a-1}}{1+z} dz$.

Q: Why is the derivative $g'(1/2)$ equal to zero?

Think about the symmetry! We found $g(1/2) = \pi$. The function $g(a) = \pi \csc(a\pi)$ is symmetric about $a = 1/2$. At the point of symmetry for a smooth function, the slope (derivative) must be zero. It's a beautiful geometric confirmation of our algebra!

Question

Passage:
Given that for each $a \in (0, 1)$, $\lim_{h 	o 0^+} \int_h^{1-h} t^{-a}(1-t)^{a-1} dt$ exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$.

Question 1:
The value of $g\left(\frac{1}{2}ight)$ is
(A) $\pi$
(B) $2\pi$
(C) $\frac{\pi}{2}$
(D) $\frac{\pi}{4}$

Question 2:
The value of $g'\left(\frac{1}{2}ight)$ is
(A) $\frac{\pi}{2}$
(B) $\pi$
(C) $-\frac{\pi}{2}$
(D) 0

Sigma Percentile · Accepted Answer

### The Beauty of the Improper Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful mathematical structure.

We are looking at the function:

$$g(a) = \lim_{h 	o 0^+} \int_h^{1-h} t^{-a}(1-t)^{a-1} dt$$

At first glance, this looks like a daunting improper integral. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to find $g(1/2)$ and its derivative $g'(1/2)$.

### Phase 1: The Symmetry of $a = 1/2$

Let us first tackle the specific case where $a = 1/2$. Substituting this into our integral, we get:

$$g(1/2) = \int_0^1 t^{-1/2}(1-t)^{-1/2} dt$$

Notice the symmetry here? Both terms are raised to the power of $-1/2$. This means we can rewrite the integrand as:

$$\frac{1}{\sqrt{t(1-t)}}$$

Now, we are looking at the integral of $\frac{1}{\sqrt{t - t^2}} dt$. This is a classic quadratic form. To solve it, we complete the square in the denominator:

$$t - t^2 = -(t^2 - t) = -(t^2 - t + 1/4 - 1/4) = 1/4 - (t - 1/2)^2$$

Our integral becomes:

$$\int_0^1 \frac{dt}{\sqrt{(1/2)^2 - (t - 1/2)^2}}$$

This is the standard form for $\sin^{-1}(x)$, specifically $\sin^{-1}(\frac{t - 1/2}{1/2}) = \sin^{-1}(2t - 1)$. Evaluating this from $0$ to $1$, we get:

$$\sin^{-1}(1) - \sin^{-1}(-1) = \pi/2 - (-\pi/2) = \pi$$

The area under this curve is exactly **$\pi$**. A beautiful, clean result.

### Phase 2: The Generalization

Now, we must find $g'(1/2)$. Differentiating the integral directly is a trap. Instead, let us find a general expression for $g(a)$.

We use the substitution $t = \frac{1}{1+z}$. This transforms our limits: as $t 	o 0, z 	o \infty$, and as $t 	o 1, z 	o 0$. The differential $dt$ becomes $-\frac{1}{(1+z)^2} dz$.

Substituting these into our integral $g(a) = \int_0^1 t^{-a}(1-t)^{a-1} dt$, we get:

$$\int_{\infty}^0 \left(\frac{1}{1+z}ight)^{-a} \left(1 - \frac{1}{1+z}ight)^{a-1} \left(-\frac{1}{(1+z)^2}ight) dz$$

After simplifying the algebra, the powers of $(1+z)$ cancel out perfectly, leaving us with the elegant identity:

$$g(a) = \int_0^{\infty} \frac{z^{a-1}}{1+z} dz$$

This is a famous result in analysis, known to be equal to:

$$g(a) = \frac{\pi}{\sin(a\pi)}$$

### Phase 3: The Final Derivative

We have reduced our complex integral to $g(a) = \pi \csc(a\pi)$. Now, finding the derivative is straightforward.

Using the chain rule:

$$g'(a) = \pi \cdot (-\csc(a\pi) \cot(a\pi)) \cdot \pi = -\pi^2 \csc(a\pi) \cot(a\pi)$$

Or, written in terms of sine and cosine:

$$g'(a) = -\frac{\pi^2 \cos(a\pi)}{\sin^2(a\pi)}$$

Finally, we evaluate at $a = 1/2$. Since $\cos(\pi/2) = 0$, the entire expression collapses to **$0$**.

We have navigated the complexity and arrived at a simple, elegant truth. Keep practicing, keep visualizing, and the math will always reveal its secrets to you.

Animated Solution for Mathematics - Definite Integration: Comprehension Passage

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

Select Answer:

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

Select Answer:

Visualized Solution

Defining the Function $g (a)$

Substituting $a = \frac{1}{2}$

Simplifying the Integrand

Completing the Square

Applying the Integration Formula

Evaluating the Limits for $g (\frac{1}{2})$

Generalizing $g (a)$ via Substitution

Transforming the Integral

The Beta Function Identity

Differentiating $g (a)$

Evaluating $g^{'} (\frac{1}{2})$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of the Improper Integral

Phase 1

Phase 2

Phase 3

Similar Questions

Let $f : R \to R$ be a differentiable function such that $f (0) = 0$ , $f^{'} (\frac{π}{2}) = 3$ and $f^{'} (0) = 1$ . If $g (x) = \int_{x}^{π /2} [f^{'} (t) cosec t - cot t cosec t f (t)] d t$ for $x \in (0, \frac{π}{2})$ , then $lim_{x \to 0} g (x) =$

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

The value of the integral $\int_{0}^{1/2} \frac{1 + 3}{(( x + 1 ) ^{2} ( 1 - x ) ) ^{1/4}} d x$ 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?

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

Let $f : (0, 1) \to R$ be the functions defined as $f (x) = n$ if $x \in [\frac{1}{n + 1}, \frac{1}{n})$ where $n \in N$ . Let $g : (0, 1) \to R$ be a function such that $\int_{x^{2}}^{x} \frac{1 - t}{t} d t < g (x) < 2 x$ for all $x \in (0, 1)$ . Then $lim_{x \to 0} f (x) g (x)$

Let the function $f : [0, 2] \to R$ be defined as $f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l n x]}, x \in [0, 1) x \in [1, 2]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is

Let $f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t$ . Then the value of $\int_{0}^{π /2} f (θ) d θ$ is ______.

Animated Solution for Mathematics - Definite Integration: 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

Select Answer:

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

Select Answer:

Visualized Solution

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=21​

Simplifying the Integrand

Completing the Square

Applying the Integration Formula

Evaluating the Limits for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(21​)

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

Transforming the Integral

The Beta Function Identity

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

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

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of the Improper Integral

Phase 1

Phase 2

Phase 3

Similar Questions

.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; } The value of the integral ∫01/2​((x+1)2(1−x))1/41+3​​dx is ________.

.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?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫1a​[x]f′(x)dx,a>1 where [x] denotes the greatest integer not exceeding x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(θ)=sinθ+∫−π/2π/2​(sinθ+tcosθ)f(t)dt. Then the value of ∫0π/2​f(θ)dθ is ______.

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

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

Defining the Function $g (a)$

Substituting $a = \frac{1}{2}$

Evaluating the Limits for $g (\frac{1}{2})$

Generalizing $g (a)$ via Substitution

Differentiating $g (a)$

Evaluating $g^{'} (\frac{1}{2})$

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

The value of the integral $\int_{0}^{1/2} \frac{1 + 3}{(( x + 1 ) ^{2} ( 1 - x ) ) ^{1/4}} d x$ 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?

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

Let $f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t$ . Then the value of $\int_{0}^{π /2} f (θ) d θ$ is ______.