JEE Mathematics Continuity and Differentiability MCQs Set 05

Answer: (c) is continuous on R

Explanation:
1. \( g \) is continuous except at integers, where it jumps from 1 (left limit) to 0 (value).
2. At an integer \( n \), \( h(n^-) = f(1) \) and \( h(n) = h(n^+) = f(0) \). Because \( f(0) = f(1) \), the one-sided limits and the value agree.
3. Elsewhere \( h \) is a composition of continuous functions, so h is continuous on R.

Teacher's Note:
1. A jump in the inner function is hidden when the outer function takes equal values at the two ends.
2. This is how a periodic extension of a continuous function is built.

Answer: (d) \( f'(0^+) = f'(0^-) = -\dfrac{1}{2} \)

Explanation:
1. Continuity gives \( f(0) = \lim \dfrac{1 - e^x}{x} = -1 \).
2. Expand \( \dfrac{1 - e^x}{x} = -1 - \dfrac{x}{2} - \dfrac{x^2}{6} - \dots \), so \( \dfrac{f(x) - f(0)}{x} \to -\dfrac{1}{2} \) from both sides.

Teacher's Note:
1. Use the series of \( e^x \) to find f(0) and the slope at once.
2. The derivative is the same from both sides, so f is differentiable at 0.

Answer: (a) 1

Explanation:
1. \( x^{-a}\log_a x \to 0 \), so \( \cot^{-1}(0^+) \to \dfrac{\pi}{2} \).
2. \( a^x \log_x a = \dfrac{a^x \ln a}{\ln x} \to \infty \), so \( \sec^{-1}(\infty) = \dfrac{\pi}{2} \).
3. The ratio is \( \dfrac{\pi/2}{\pi/2} = 1 \).

Teacher's Note:
1. Powers beat logarithms and exponentials beat powers as x tends to infinity.
2. Remember the limiting values of the inverse trigonometric functions.

Answer: (b) \( -16 \)

Explanation:
1. For \( 1 < x < 2 \), \( |4x - 8| = 4(2 - x) \), so the upper bound is \( \dfrac{4x^2(2 - x)}{x - 2} = -4x^2 \to -16 \).
2. The lower bound \( \dfrac{\cos(2x - 4) - 33}{2} \to \dfrac{1 - 33}{2} = -16 \).
3. By the sandwich theorem, \( \lim_{x \to 2^-} f(x) = -16 \).

Teacher's Note:
1. Evaluate both bounds as x approaches the point.
2. Remove the modulus using the sign of the expression on the given side.

Answer: (c) 5

Explanation:
1. \( \sin x + \cos x = \sqrt{2}\sin\left(x + \dfrac{\pi}{4}\right) \) takes values in \( [-\sqrt{2}, \sqrt{2}] \), so the integer values that matter are \( -1, 0, 1 \).
2. The function equals 1 at \( x = \dfrac{\pi}{2} \); 0 at \( x = \dfrac{3\pi}{4},\ \dfrac{7\pi}{4} \); and \( -1 \) at \( x = \pi,\ \dfrac{3\pi}{2} \). The value crosses an integer level at each of these points.
3. The points \( x = 0 \) and \( x = 2\pi \) are excluded, so there are 5 points of discontinuity.

Teacher's Note:
1. A greatest-integer function is discontinuous where the inside expression crosses an integer.
2. Do not count end points that lie outside the interval.

Answer: (d) \( \dfrac{4^{n+1} - 1}{3 \cdot 2^n} \)

Explanation:
1. \( x^2 + 2x + 3 = (x+1)^2 + 2 \ge 2 \), so \( a = 2 \). \( b = \lim \dfrac{x}{2x} = \dfrac{1}{2} \).
2. \( \sum_{r=0}^{n} 2^r \left(\dfrac{1}{2}\right)^{n-r} = \dfrac{1}{2^n}\sum_{r=0}^{n} 4^r = \dfrac{1}{2^n} \cdot \dfrac{4^{n+1} - 1}{3} \).

Teacher's Note:
1. Complete the square to find the minimum of a quadratic.
2. The sum is a geometric series with ratio 4.

Answer: (c) \( \dfrac{3}{2} \)

Explanation:
1. Continuity and \( f(a) = 0 \) give \( f(x) \to 0 \) as \( x \to a \).
2. Using \( \dfrac{\ln(1 + t)}{t} \to 1 \) with \( t = 3f(x) \), the expression is \( \dfrac{3}{2} \cdot \dfrac{\ln(1 + 3f)}{3f} \to \dfrac{3}{2} \).

Teacher's Note:
1. Identify the variable that tends to zero and use the standard logarithmic limit.
2. Adjust the constants outside the limit.

Answer: (d) \( \left(\dfrac{b^b}{a^a}\right)^{\frac{1}{b-a}} \)

Explanation:
1. The base tends to 1 as \( t \to 0 \), so the form is \( 1^\infty \). The limit is \( e^{L} \) with \( L = \lim_{t \to 0}\dfrac{1}{t}\left(\dfrac{b^{t+1} - a^{t+1}}{b - a} - 1\right) \).
2. By L'Hospital, \( L = \dfrac{b\ln b - a\ln a}{b - a} \).
3. So the limit is \( e^{\frac{b\ln b - a\ln a}{b - a}} = \left(\dfrac{b^b}{a^a}\right)^{\frac{1}{b-a}} \). Options (a), (b) and (c) are the exponent only.

Teacher's Note:
1. A limit of the form \( 1^\infty \) equals \( e \) raised to the limit of (base minus 1) times the power.
2. Do not forget the exponential at the end.

Answer: (d) \( k(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2 - x, & 1 < x \le 2 \end{cases} \) at \( x = 1 \)

Explanation:
1. (a) both one-sided derivatives are 0. (b) both are 1. (c) both are 2, and the function is continuous at 0 in each case.
2. In (d), the left derivative at 1 is 1 and the right derivative is \( -1 \), so k has a corner and is not differentiable at \( x = 1 \).

Teacher's Note:
1. Continuity alone is not enough; the one-sided slopes must agree.
2. A corner point appears where the slopes differ.

Answer: (a) \( \beta = 1 \) and \( \gamma = \alpha + 5 \)

Explanation:
1. Expanding along the first row, the determinant equals \( \alpha x \cdot \dfrac{1}{x^2} + \beta - \dfrac{\gamma}{x} = \beta + \dfrac{\alpha - \gamma}{x} \).
2. For \( x \ln\left(\beta + \dfrac{\alpha - \gamma}{x}\right) \) to have a finite limit, \( \beta \) must be 1; then the limit is \( \alpha - \gamma \).
3. So \( \alpha - \gamma = -5 \), that is \( \gamma = \alpha + 5 \).

Teacher's Note:
1. A finite limit of \( x \ln D \) requires \( D \to 1 \).
2. Then \( x\ln(1 + u/x) \to u \).

Answer: (b) \( \displaystyle\lim_{x \to 0}\frac{\sin(\cos x)}{\cos x} \)

Explanation:
1. (a) \( \dfrac{\sin(\tan t)}{\sin t} \approx \dfrac{\tan t}{t} \to 1 \). (c) Rationalising gives \( \dfrac{2}{\sqrt{1+x} + \sqrt{1-x}} \to 1 \). (d) \( \dfrac{|x|}{x} = 1 \) for \( x > 0 \).
2. In (b), \( \cos x \to 1 \), so the expression tends to \( \sin 1 \approx 0.84 \), which is not 1.

Teacher's Note:
1. The standard limit \( \dfrac{\sin u}{u} \to 1 \) needs u to tend to 0.
2. Here the angle tends to 1 radian, not 0.

Answer: (a) \( e^{-1/12} \)

Explanation:
1. The base tends to \( \dfrac{3}{1 + 2} = 1 \) and the power tends to infinity, so the form is \( 1^\infty \) and the limit is \( e^{L} \) where \( L = \lim \dfrac{1}{\sin x}\left(\dfrac{3}{1 + \sqrt{4 + x}} - 1\right) \).
2. \( \dfrac{3}{1 + \sqrt{4+x}} - 1 \approx -\dfrac{x}{12} \) because \( \sqrt{4 + x} \approx 2 + \dfrac{x}{4} \), so \( \dfrac{3}{3 + x/4} - 1 \approx -\dfrac{x}{12} \).
3. Hence \( L = -\dfrac{1}{12} \).

Teacher's Note:
1. For a \( 1^\infty \) form use \( e^{\lim (\text{base} - 1)\times \text{power}} \).
2. Expand the square root to first order.

Answer: (a) 1

Explanation:
1. \( \sqrt{x+1} - \sqrt{x} = \dfrac{1}{\sqrt{x+1} + \sqrt{x}} \to 0^+ \), so the numerator tends to \( \cot^{-1}0 = \dfrac{\pi}{2} \).
2. The base \( \dfrac{2x+1}{x-1} \to 2 \), so \( \left(\dfrac{2x+1}{x-1}\right)^x \to \infty \) and the denominator tends to \( \sec^{-1}\infty = \dfrac{\pi}{2} \).
3. The limit is 1.

Teacher's Note:
1. Rationalise a difference of square roots.
2. \( \sec^{-1} \) of a very large number is close to \( \dfrac{\pi}{2} \).

Answer: (c) \( \dfrac{\tan x - x}{x\tan x} \)

Explanation:
1. Use \( \tan\theta = \cot\theta - 2\cot 2\theta \). Then \( \dfrac{1}{2^k}\tan\dfrac{x}{2^k} = \dfrac{1}{2^k}\cot\dfrac{x}{2^k} - \dfrac{1}{2^{k-1}}\cot\dfrac{x}{2^{k-1}} \).
2. The sum telescopes to \( \dfrac{1}{2^n}\cot\dfrac{x}{2^n} - \cot x \).
3. As \( n \to \infty \), \( \dfrac{1}{2^n}\cot\dfrac{x}{2^n} \to \dfrac{1}{x} \), so the limit is \( \dfrac{1}{x} - \cot x = \dfrac{\tan x - x}{x\tan x} \).

Teacher's Note:
1. Convert each term into a difference of two consecutive terms to telescope.
2. The identity \( \tan\theta = \cot\theta - 2\cot 2\theta \) is worth remembering.

Answer: (b) it is not derivable at the origin

Explanation:
1. Near 0, \( f(x) = x(1 + |x| + \dots) = x + x|x| + \dots \), so \( f'(0) = 1 \): f is derivable at the origin and (b) is incorrect.
2. Each piece \( \dfrac{x}{1 - x} \) and \( \dfrac{x}{1 + x} \) is smooth on its domain, and the values \( (0, \infty),\ (-\infty, 0),\ (-\infty, -1),\ (1, \infty) \) together with \( f(0) = 0 \) cover all of R.

Teacher's Note:
1. Test differentiability at 0 using the definition when a modulus is present.
2. The term \( x|x| \) has zero slope at 0.

Answer: (a) \( \dfrac{f'(1)}{x} \)

Explanation:
1. Differentiate with respect to y and put \( y = 1 \): \( x f'(x) = f'(1) \).
2. So \( f'(x) = \dfrac{f'(1)}{x} \).

Teacher's Note:
1. This is the logarithmic functional equation.
2. Differentiate partially with respect to one variable and substitute a convenient value.

Answer: (a) 1

Explanation:
1. As \( x \to 0 \) with \( x \ne 0 \), \( f(x) = \sin x \) is close to 0 but not equal to 0 (and not equal to 2).
2. So \( g(f(x)) = \sin^2 x + 1 \to 1 \). The value \( g(0) = 4 \) is irrelevant for the limit.

Teacher's Note:
1. A limit ignores the value of the function at the limit point.
2. Check that the inner function never takes the special values near the point.

Answer: (d) \( a + b = 0 \)

Explanation:
1. For small \( x > 0 \), \( f'(0^+) = a + b \). For small \( x < 0 \), \( f'(0^-) = -a - b \) because \( |\sin x| \) and \( e^{|x|} \) both have slopes that change sign.
2. \( |x|^3 \) is differentiable at 0 with derivative 0.
3. Equality of the one-sided derivatives needs \( a + b = -(a + b) \), that is \( a + b = 0 \), with c arbitrary.

Teacher's Note:
1. Terms with a modulus have slopes that change sign at 0.
2. Compare the one-sided derivatives of the whole sum.

Answer: (a) \( \ell = m \)

Explanation:
1. As \( x \to \infty \), \( e^x \) dominates \( x^2 \), so \( f \approx \dfrac{x}{2x} = \dfrac{1}{2} \).
2. As \( x \to -\infty \), \( e^x \to 0 \), so \( f \approx \dfrac{\ln x^2}{\ln x^4} = \dfrac{2\ln|x|}{4\ln|x|} = \dfrac{1}{2} \).
3. So \( \ell = m = \dfrac{1}{2} \).

Teacher's Note:
1. Decide which term dominates at each end before simplifying.
2. Exponentials dominate powers at plus infinity but vanish at minus infinity.

Answer: (c) \( l \) and \( m \) both exist

Explanation:
1. For \( 0 < |x| < 1 \), \( 0 \le x^2 < 1 \), so \( [x^2] = 0 \).
2. Both quotients are 0 for all small non-zero x, so \( l = 0 \) and \( m = 0 \) both exist.

Teacher's Note:
1. The greatest integer of a small positive number is 0.
2. The limit is found by values near the point, not at the point.

Answer: (c) is equal to 289

Explanation:
1. \( \log_{\cos(x/2)}\cos x = \dfrac{\ln\cos x}{\ln\cos(x/2)} \approx \dfrac{-x^2/2}{-x^2/8} = 4 \).
2. The expression is \( (1 + 4^2)^2 = 17^2 = 289 \).

Teacher's Note:
1. Change the base of the logarithm to natural logarithms.
2. Use \( \ln\cos u \approx -\dfrac{u^2}{2} \) for small u.

Answer: (d) 2

Explanation:
1. The function is defined for \( x \ge 0 \), so use \( x \to 0^+ \).
2. \( \ln\left(1 + x^2 + 2\sqrt{x} + \dots\right) \approx 2\sqrt{x} \) and \( \tan\sqrt{x} \approx \sqrt{x} \).
3. The limit is 2, so \( f(0) = 2 \).

Teacher's Note:
1. Continuity needs f(0) to equal the limit.
2. Use \( \ln(1 + u) \approx u \) and \( \tan u \approx u \) for small u.