Let g(t) = ∫(-π/2 to π/2) cos(π/4 t + f(x)) dx, where f(x) = log_e(x + √(x² + 1)), x ∈ R. Then which

Q: Wait, why do we reject the negative root when rationalizing $f(-x)$?

We don't reject it; we use the property of the logarithm. When you rationalize $f(-x)$, you get $\log_e(\sqrt{x^2+1}-x)$. Since $(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x) = (x^2+1) - x^2 = 1$, it follows that $\sqrt{x^2+1}-x = \frac{1}{\sqrt{x^2+1}+x}$. Thus, $\log_e(\sqrt{x^2+1}-x) = \log_e((x+\sqrt{x^2+1})^{-1}) = -\log_e(x+\sqrt{x^2+1}) = -f(x)$. It is a beautiful algebraic identity, not a rejection!

Q: Why does the integral of $\sin(f(x))$ vanish?

Because $f(x)$ is an odd function, and the sine of an odd function is also odd. When you integrate an odd function over a symmetric interval $[-a, a]$, the area on the negative side perfectly cancels the area on the positive side, resulting in zero. It is one of the most powerful shortcuts in calculus!

Q: Can we evaluate the constant $K$ without knowing the integral?

Yes! In many JEE problems, you don't need to compute the exact value of the integral if it appears as a common factor in both $g(1)$ and $g(0)$. By defining the integral as $K$, we can express both $g(1)$ and $g(0)$ in terms of $K$ and then find their ratio, which eliminates $K$ entirely.

Question

Let $g(t) = \int_{-\pi/2}^{\pi/2} \cos\left(\frac{\pi}{4} t + f(x)\right) dx$, where $f(x) = \log_e(x + \sqrt{x^2 + 1}), x \in \mathbf{R}$. Then which one of the following is correct?

Options:

(A) $g(1) = g(0)$

(B) $\sqrt{2} g(1) = g(0)$

(C) $g(1) = \sqrt{2} g(0)$

(D) $g(1) + g(0) = 0$

Sigma Percentile · Accepted Answer

### The Elegance of Symmetry: A Journey Through Calculus

My dear student, welcome to a problem that might look like a daunting mountain of logarithms and trigonometric functions, but I promise you, it is actually a beautifully choreographed dance of symmetry.

When you first encounter an integral like

$$g(t) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\left(\frac{\pi}{4} t + f(x)\right) dx$$

your heart might skip a beat. But take a deep breath. The secret to solving this lies not in brute-force integration, but in understanding the soul of the function $f(x) = \log_e(x + \sqrt{x^2 + 1})$.

### The Mystery of $f(x)$

First, let us investigate the parity of $f(x)$. In the world of JEE, whenever you see symmetric limits like $[-\frac{\pi}{2}, \frac{\pi}{2}]$, your intuition should immediately scream, "Check for even or odd functions!"

Let us test $f(-x)$:

$$f(-x) = \log_e(-x + \sqrt{(-x)^2 + 1}) = \log_e(\sqrt{x^2 + 1} - x)$$

Now, here is the trick. Multiply and divide by the conjugate, $\sqrt{x^2 + 1} + x$. The numerator becomes:

$$(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x) = (x^2 + 1) - x^2 = 1$$

Thus, $f(-x) = \log_e\left(\frac{1}{x + \sqrt{x^2 + 1}}\right)$. Using the laws of logarithms, this is simply $-\log_e(x + \sqrt{x^2 + 1})$, which is exactly $-f(x)$.

We have discovered that **$f(x)$ is an odd function**. This is a massive breakthrough!

### The Expansion

Now, let us look at our integral $g(t)$. We have a cosine of a sum. Let us use the compound angle identity: $\cos(A + B) = \cos A \cos B - \sin A \sin B$.

Here, $A = \frac{\pi}{4}t$ and $B = f(x)$. So:

$$g(t) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[\cos\left(\frac{\pi}{4}t\right) \cos(f(x)) - \sin\left(\frac{\pi}{4}t\right) \sin(f(x))\right] dx$$

Because we are integrating with respect to $x$, the terms involving $t$ are constants. We can pull them out:

$$g(t) = \cos\left(\frac{\pi}{4}t\right) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(f(x)) dx - \sin\left(\frac{\pi}{4}t\right) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(f(x)) dx$$

### The Power of Symmetry

Look at the second integral. We are integrating $\sin(f(x))$. Since $f(x)$ is odd, $\sin(f(x))$ is also odd.

The integral of an odd function over symmetric limits $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is zero. The entire second term vanishes!

We are left with $g(t) = K \cdot \cos(\frac{\pi}{4}t)$, where:

$$K = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(f(x)) dx$$

We do not even need to calculate $K$!

### The Final Relation

Now, we simply evaluate $g(0)$ and $g(1)$. For $t=0$:

$$g(0) = K \cdot \cos(0) = K$$

For $t=1$:

$$g(1) = K \cdot \cos\left(\frac{\pi}{4}\right) = K \cdot \frac{1}{\sqrt{2}}$$

Therefore, $g(1) = \frac{g(0)}{\sqrt{2}}$, which rearranges to **$\sqrt{2} g(1) = g(0)$**.

See? The complexity melted away because we respected the symmetry of the problem. Keep this mindset, and you will conquer any JEE problem.

Animated Solution for Mathematics - Definite Integration: 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?

Select Answer:

Visualized Solution

Understanding the function $g (t)$

Analyzing the nature of $f (x)$

Rationalizing to find $f (- x)$

Conclusion on $f (x)$ parity

Expanding the Cosine Term

Applying Linearity of Integration

Parity of $sin (f (x))$ and $cos (f (x))$

Evaluating the odd integral

Simplifying $g (t)$

Evaluating $g (0)$

Evaluating $g (1)$

Final Comparison and Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Elegance of Symmetry

The Mystery of $f (x)$

The Expansion

The Power of Symmetry

The Final Relation

Similar Questions

If $g (x) = \int_{0}^{x} cos 4 t d t$ , then $g (x + π)$ equals

If $g (x) = \int_{0}^{x} cos^{4} t d t$ , then $g (x + π)$ equals

If $f (x) = \frac{2 - x c o s x}{2 + x c o s x}$ and $g (x) = lo g_{e} x, (x > 0)$ then the value of integral $\int_{- π /4}^{π /4} g (f (x)) d x$ 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 ______.

Comprehension Passage

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

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

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π$ , then the value of $\int_{0}^{π /2} \frac{π ^{2} d t}{f ( t )}$ equals

For $x > 0$ , let $f (x) = \int_{e}^{x} \frac{l n t}{1 + t} d t$ . Find the function $f (x) + f (\frac{1}{x})$ and show that $f (e) + f (\frac{1}{e}) = \frac{1}{2}$ .

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y$ , then $e + f (0)$ is equal to .........

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

Select Answer:

Visualized Solution

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

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

Rationalizing to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−x)

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

Expanding the Cosine Term

Applying Linearity of Integration

Parity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(f(x)) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(f(x))

Evaluating the odd integral

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

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

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