Let f(x) = ^(-1)(√((1- x)/(1+ x))); x (0,). A normal to y = f(x) at x = ()/3 passes through the point: Option 1: (()/3, ()/2) Option 2: (()/3, 0) Option 3: (0, 0) Option 4: (()/3, ()/6)

Answer: (()/3, ()/6). Solution: Simplify the expression inside the inverse tangent using half-angle trigonometric identities: 1 - x = 2^2(x/2) 1 + x = 2^2(x/2) Substitute these into the function: f(x) = ^(-1)(√((2^2(x/2))/(2^2(x/2)))) f(x) = ^(-1)(√(^2(x/2))) f(x) = ^(-1)(, (x/2), ) Since x (0, ), the half-angle x/2 (0, ()/2). In this interval, (x/2) is positive, so, (x/2), = (x/2). f(x) = ^(-1)((x/2)) f(x) = x/2 The normal is at x = ()/3. The y-coordinate at this point: y = f(()/3) y = (/3)/2 y = ()/6 The point of tangency is (()/3, ()/6). The derivative of the simplified function f(x) = x/2: f'(x) = 1/2 This is the slope of the tangent. The slope of the normal is the negative reciprocal of the tangent slope: m_n = -1/(1/2) m_n = -2 Using the point-slope form with point (()/3, ()/6) and slope -2: y - ()/6 = -2(x - ()/3) y - ()/6 = -2x + (2)/3 2x + y = (2)/3 + ()/6 2x + y = (4 + )/6 2x + y = (5)/6 Check which of the given points satisfies the equation 2x + y = (5)/6: Option 1: (()/3, ()/2) 2(()/3) + ()/2 = (7)/6 ≠ (5)/6 Option 2: (()/3, 0) 2(()/3) + 0 = (4)/6 ≠ (5)/6 Option 3: (0, 0) 0 + 0 = 0 ≠ (5)/6 Option 4: (()/3, ()/6) 2(()/3) + ()/6 = (4)/6 + ()/6 = (5)/6 ✓ Therefore, the normal passes through the point (()/3, ()/6).

Calculus

Differentiation

Hard

Let $f(x) = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ ; $x \in (0,\pi)$ . A normal to $y = f(x)$ at $x = \frac{\pi}{3}$ passes through the point:

$\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$

$\left(\frac{\pi}{3}, 0\right)$

$(0, 0)$

$\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$

✅ Correct Option: 4

Simplify the expression inside the inverse tangent using half-angle trigonometric identities:

$1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$

$1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$

Substitute these into the function:

$f(x) = \tan^{-1}\left(\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}}\right)$

$f(x) = \tan^{-1}\left(\sqrt{\tan^2\left(\frac{x}{2}\right)}\right)$

$f(x) = \tan^{-1}\left(\left|\tan\left(\frac{x}{2}\right)\right|\right)$

Since $x \in (0, \pi)$ , the half-angle $\frac{x}{2} \in (0, \frac{\pi}{2})$ . In this interval, $\tan(\frac{x}{2})$ is positive, so $|\tan(\frac{x}{2})| = \tan(\frac{x}{2})$ .

$f(x) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right)$

$f(x) = \frac{x}{2}$

The normal is at $x = \frac{\pi}{3}$ . The $y$ -coordinate at this point:

$y = f\left(\frac{\pi}{3}\right)$

$y = \frac{\pi/3}{2}$

$y = \frac{\pi}{6}$

The point of tangency is $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$ .

The derivative of the simplified function $f(x) = \frac{x}{2}$ :

$f'(x) = \frac{1}{2}$

This is the slope of the tangent. The slope of the normal is the negative reciprocal of the tangent slope:

$m_n = -\frac{1}{1/2}$

$m_n = -2$

Using the point-slope form with point $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$ and slope $-2$ :

$y - \frac{\pi}{6} = -2\left(x - \frac{\pi}{3}\right)$

$y - \frac{\pi}{6} = -2x + \frac{2\pi}{3}$

$2x + y = \frac{2\pi}{3} + \frac{\pi}{6}$

$2x + y = \frac{4\pi + \pi}{6}$

$2x + y = \frac{5\pi}{6}$

Check which of the given points satisfies the equation $2x + y = \frac{5\pi}{6}$ :

Option 1: $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$

$2\left(\frac{\pi}{3}\right) + \frac{\pi}{2} = \frac{7\pi}{6} \neq \frac{5\pi}{6}$

Option 2: $\left(\frac{\pi}{3}, 0\right)$

$2\left(\frac{\pi}{3}\right) + 0 = \frac{4\pi}{6} \neq \frac{5\pi}{6}$

Option 3: $(0, 0)$

$0 + 0 = 0 \neq \frac{5\pi}{6}$

Option 4: $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$

$2\left(\frac{\pi}{3}\right) + \frac{\pi}{6} = \frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6}$ ✓

Therefore, the normal passes through the point $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$ .

2025: Sample Paper

The slope of the tangent to the curve $y = x^2 - 4\cos x$ , $x \in [0, \pi]$ is maximum when $x$ is equal to

2025: Actual Paper

Let $f: (0, \frac{6}{5}) \to \mathbb{R}$ & $g: (0, \frac{6}{5}) \to \mathbb{R}$ be functions defined by

$f(x) = [x^2]$ and $g(x) = (|x - 1| + |x - 2|)f(x)$

Here $[a] =$ the highest integer $\leq a$ . Then

2025: Actual Paper

Let $h(x) = min[\sin x], [\cos x]]$ , for all real numbers $x$ . Let S be the set of points in $(0, \frac{\pi}{2})$ where $h(x)$ is not differentiable. Then the cardinality of S is:

Let f(x)=tan⁡−1(1−cos⁡x1+cos⁡x)f(x) = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)f(x)=tan−1(1+cosx1−cosx​​); x∈(0,π)x \in (0,\pi)x∈(0,π). A normal to y=f(x)y = f(x)y=f(x) at x=π3x = \frac{\pi}{3}x=3π​ passes through the point:

Solution

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UGAT Bangalore Past Year Questions (Topic-Wise)

Let $f(x) = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ ; $x \in (0,\pi)$ . A normal to $y = f(x)$ at $x = \frac{\pi}{3}$ passes through the point: