JEE Mathematics Continuity and Differentiability MCQs Set 04

Answer: (c) \( -4 \)

Explanation:
1. Differentiability needs continuity: \( e^0 + a = 0 - 3 \), so \( a = -4 \).
2. Check derivatives: left derivative \( e^0 = 1 \), right derivative \( 1 \). They are equal, so \( f \) is differentiable at 0 for \( a = -4 \).

Teacher's Note:
1. Always test continuity first and then compare the two one-sided derivatives.
2. Here the derivatives match automatically.

Answer: (b) \( 0 \)

Explanation:
1. Put \( t = \dfrac{x^n}{e^x} \). As \( x \to \infty \), \( t \to 0 \) because the exponential grows faster than any power.
2. The numerator is \( 2^t - 3^t \approx t \ln\dfrac{2}{3} \).
3. The expression is \( \dfrac{t \ln(2/3)}{x^n} = \dfrac{\ln(2/3)}{e^x} \to 0 \).

Teacher's Note:
1. An exponential dominates every power of x as x tends to infinity.
2. Reduce the limit to a form like \( \dfrac{a^t - b^t}{t} \) when t tends to 0.

Answer: (c) equal to \( -\sqrt{\pi} \)

Explanation:
1. The function is defined only for \( x \le 2 \), so take \( x = 2 - h \) with \( h \to 0^+ \).
2. \( \sqrt{4 - x^2} = \sqrt{h(4-h)} \approx 2\sqrt{h} \) and \( \cos\dfrac{\pi(2-h)}{4} = \sin\dfrac{\pi h}{4} \approx \dfrac{\pi h}{4} \), so the second root is \( \dfrac{\sqrt{\pi h}}{2} \).
3. Numerator \( \approx \sqrt{\pi}\,h \) and denominator \( \sin(-h) \approx -h \). The limit is \( -\sqrt{\pi} \).

Teacher's Note:
1. When a function exists on one side only, the limit is the one-sided limit from that side.
2. Use \( \sin(x-2) \approx x-2 \) near 2.

Answer: (c) \( \dfrac{1}{128 r} \)

Explanation:
1. Let BC be at distance \( h \) from A along the diameter through A. Then half of BC is \( \sqrt{h(2r - h)} \) and \( AB = AC = \sqrt{2rh} \).
2. \( \Delta = h\sqrt{h(2r-h)} \approx \sqrt{2r}\,h^{3/2} \) and \( P \approx 4\sqrt{2rh} \), so \( P^3 \approx 64(2r)^{3/2} h^{3/2} \).
3. \( \dfrac{\Delta}{P^3} \to \dfrac{1}{64 \cdot 2r} = \dfrac{1}{128 r} \).

Teacher's Note:
1. Use the chord-height relation \( (\text{half chord})^2 = h(2r - h) \) for a chord at distance h from the end of a diameter.
2. Keep only the leading powers of h.

Answer: (c) \( g \) and \( h \) only

Explanation:
1. \( f'(0) = \lim_{x \to 0} \sin\dfrac{1}{x} \) does not exist, so \( f \) is not differentiable at 0.
2. \( g'(0) = \lim_{x \to 0} x \sin\dfrac{1}{x} = 0 \) and \( h'(0) = \lim_{x \to 0} x|x| = 0 \), so \( g \) and \( h \) are differentiable at 0.

Teacher's Note:
1. Use the definition of the derivative at 0 for functions defined in pieces.
2. A bounded oscillating factor is killed only if it is multiplied by something that tends to zero.

Answer: (b) \( h \) has an irremovable discontinuity at \( x = \dfrac{\pi}{2} \)

Explanation:
1. Put \( x = \dfrac{\pi}{2} + t \). The numerator of \( f \) is \( -2\sin t + \sin 2t \approx -t^3 \) and the denominator is \( 4t^2 \), so the left limit of \( h \) is \( \lim -\dfrac{t}{4} = 0 \).
2. For \( g \): \( -\cos x = \sin t \), so \( g = \dfrac{e^{\sin t} - 1}{8t} \approx \dfrac{t}{8t} = \dfrac{1}{8} \), so the right limit of \( h \) is \( \dfrac{1}{8} \).
3. The one-sided limits (0 and \( \tfrac18 \)) differ, so the discontinuity is a jump and cannot be removed.

Teacher's Note:
1. Compute both one-sided limits separately.
2. Different finite one-sided limits mean a non-removable (jump) discontinuity.

Answer: (d) \( [f(0)] \cdot \{f(0)\} = -1.5 \)

Explanation:
1. Expand: \( x - (1 + x + \tfrac{x^2}{2}) + (1 - 2x^2) = -\tfrac{5}{2}x^2 + \dots \), so \( f(0) = -\dfrac{5}{2} \).
2. \( [-2.5] = -3 \) and \( \{-2.5\} = -2.5 - (-3) = 0.5 \), so options (b) and (c) are wrong.
3. \( [f(0)]\cdot\{f(0)\} = -3 \times 0.5 = -1.5 \).

Teacher's Note:
1. Continuity at 0 means f(0) equals the limit.
2. For negative numbers the greatest integer is below the number and the fractional part is always non-negative.

Answer: (d) for no value of \( b \)

Explanation:
1. Continuity at 0 needs \( b = \cos 0 = 1 \).
2. Left derivative \( = 1 \), right derivative \( = -\sin 0 = 0 \). They are never equal, whatever the value of b.

Teacher's Note:
1. The slopes on the two sides are fixed and do not depend on b.
2. Adjusting b can fix continuity but not a difference in slopes.

Answer: (a) 3

Explanation:
1. Split the quotient as \( \dfrac{f(h) - f(0)}{h} + 2 \cdot \dfrac{f(0) - f(-2h)}{2h} \).
2. Each part tends to \( f'(0) = 1 \), so the limit is \( 1 + 2 \times 1 = 3 \).

Teacher's Note:
1. Add and subtract f(0) to form difference quotients.
2. Match the denominator to the change in x inside each function.

Answer: (b) \( \dfrac{1}{2} \)

Explanation:
1. Let \( u = 1 - \cos x \) and let the nested root be \( y \). Then \( y^2 = u + y \), so \( y = \dfrac{1 + \sqrt{1 + 4u}}{2} \).
2. \( y - 1 = \dfrac{\sqrt{1+4u} - 1}{2} \approx u \) for small u, and \( u \approx \dfrac{x^2}{2} \).
3. The limit is \( \dfrac{1}{2} \).

Teacher's Note:
1. An infinite nested radical y satisfies an equation of the form \( y^2 = u + y \).
2. Use the positive root only.

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

Explanation:
1. The limit is of the form 0/0, so differentiate numerator and denominator.
2. The derivative of the numerator is \( \dfrac{1}{3 + x} + \dfrac{1}{3 - x} \), which equals \( \dfrac{2}{3} \) at \( x = 0 \).

Teacher's Note:
1. This is the derivative of \( \ln(3+x) - \ln(3-x) \) at 0.
2. Do not forget the sign from the chain rule on \( \ln(3-x) \).

Answer: (c) \( u(x) = \operatorname{sgn}(|x| - 1) \)

Explanation:
1. For \( |x| > 1 \), \( x^{2n} \to \infty \) and \( f = 1 \). For \( |x| < 1 \), \( x^{2n} \to 0 \) and \( f = -1 \). For \( |x| = 1 \), \( f = 0 \).
2. This is exactly \( \operatorname{sgn}(|x| - 1) \). The others have the wrong sign pattern or the wrong zeros.

Teacher's Note:
1. Treat the cases \( |x| > 1 \), \( |x| = 1 \), \( |x| < 1 \) separately.
2. Compare the zeros of each candidate with those of f.

Answer: (c) If \( \lim_{x \to c} (f(x) + g(x)) \) and \( \lim_{x \to c} f(x) \) exist, then \( \lim_{x \to c} g(x) \) exists.

Explanation:
1. In (c), \( g = (f + g) - f \) is a difference of two functions with limits, so its limit exists.
2. (a) fails for \( f(x) = x \), \( g(x) = \dfrac{1}{x} \) at \( c = 0 \). (b) fails for \( f = \sin\dfrac{1}{x} \), \( g = 0 \). (d) fails for \( f = \sin\dfrac{1}{x} \), \( g = -\sin\dfrac{1}{x} \).

Teacher's Note:
1. A limit of a sum or difference needs both limits to exist, but the converse is not true.
2. Give a counter-example for each wrong option.

Answer: (a) equals \( \dfrac{\pi}{4} \)

Explanation:
1. The denominator is \( (x + 6)(x - 2) \) and the numerator is 0 at \( x = 2 \), so apply L'Hospital's rule.
2. Derivative of the numerator at 2: \( 2\pi\sec^2 4\pi + \dfrac{\pi}{2}\cos\pi + \dfrac{\pi}{2}\sec^2\pi = 2\pi - \dfrac{\pi}{2} + \dfrac{\pi}{2} = 2\pi \).
3. Derivative of the denominator at 2 is \( 2(2) + 4 = 8 \), so \( f(2) = \dfrac{2\pi}{8} = \dfrac{\pi}{4} \).

Teacher's Note:
1. For continuity at a point, f(point) must equal the limit.
2. Factor the denominator to see why the limit is of the form 0/0.

Answer: (c) 0

Explanation:
1. At \( x = 3 \) the numerator is \( 3f(3) - 3f(3) = 0 \), so apply L'Hospital's rule.
2. The limit is \( f(3) - 3f'(3) = 6 - 6 = 0 \).

Teacher's Note:
1. Check that the form is 0/0 before differentiating.
2. Treat f(3) as a constant when differentiating the numerator.

Answer: (a) \( f(x) = x^{1/3} \)

Explanation:
1. \( x^{1/3} \) is continuous on R, but \( f'(x) = \dfrac{1}{3}x^{-2/3} \) tends to infinity at 0, so the graph has a vertical tangent there and f is not differentiable at 0.
2. The domain of \( \dfrac{|x|}{x} \) excludes 0, and \( e^{-x} \) and \( \tan x \) are differentiable at every point of their domains.

Teacher's Note:
1. A vertical tangent means the derivative does not exist even though the function is continuous.
2. Always test only the points inside the domain.

Answer: (b) exists and is \( \dfrac{3}{4} \)

Explanation:
1. Adding and subtracting the two limits gives \( \lim f = \dfrac{3}{2} \) and \( \lim g = \dfrac{1}{2} \).
2. So \( \lim f g = \dfrac{3}{2} \cdot \dfrac{1}{2} = \dfrac{3}{4} \).

Teacher's Note:
1. When the sum and the difference have limits, each function has a limit.
2. Then the product rule for limits applies.

Answer: (b) has an isolated point removable discontinuity

Explanation:
1. For \( 1 \le x < 2 \), \( [x] = 1 \), so \( f(x) = x \to 2 \) as \( x \to 2^- \). For \( x > 2 \), \( \sqrt{6 - x} \to 2 \).
2. Both one-sided limits equal 2, but \( f(2) = 1 \). The limit exists and differs from the defined value, so the discontinuity is removable and f(2) is an isolated displaced point.

Teacher's Note:
1. Missing point: f(a) is undefined. Isolated point: f(a) is defined but differs from the limit.
2. Redefining f(2) = 2 would remove the discontinuity.

Answer: (c) 4

Explanation:
1. Simplify the slope: \( \dfrac{s^2 + 2s - 3}{s - 1} = \dfrac{(s+3)(s-1)}{s - 1} = s + 3 \).
2. The slope of the secant PQ tends to \( f'(1) \) as \( Q \to P \), that is \( s \to 1 \). So \( f'(1) = 1 + 3 = 4 \).

Teacher's Note:
1. The derivative is the limit of the secant slope as the second point approaches the first.
2. Cancel the common factor before taking the limit.

Answer: (b) \( b \)

Explanation:
1. Write \( (a^n + b^n)^{1/n} = b\left[1 + \left(\dfrac{a}{b}\right)^n\right]^{1/n} \).
2. Since \( 0 < \dfrac{a}{b} < 1 \), \( \left(\dfrac{a}{b}\right)^n \to 0 \) and the bracket tends to \( 1^0 = 1 \). The limit is \( b \).

Teacher's Note:
1. The larger base dominates the sum.
2. Factor out the largest term before taking the limit.

Answer: (c) 3

Explanation:
1. \( |x - 0.5| \) has a corner at \( x = 0.5 \) and \( |x - 1| \) has a corner at \( x = 1 \), while \( \tan x \) is smooth there, so f is not differentiable at both points.
2. \( \tan x \) is not defined at \( x = \dfrac{\pi}{2} \approx 1.57 \), which lies in \( (0, 2) \), so f has no derivative there either.
3. Total points: \( 0.5,\ 1,\ \dfrac{\pi}{2} \), that is 3.

Teacher's Note:
1. A function has no derivative at points where it is undefined as well as at corners.
2. List the corners of every modulus term separately.

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

Explanation:
1. (a) jumps from \( -1 \) to 0 at \( x = 0 \). (b) \( \dfrac{1}{x} \to \infty \) as \( x \to 0^+ \). (c) jumps from 0 to \( \tfrac{1}{2} \) just after \( x = \tfrac{1}{2} \).
2. In (d), both pieces tend to 0 at \( x = 0 \) and \( f(0) = 0 \), and each piece is continuous elsewhere, so f is continuous on \( [-1, 1] \).

Teacher's Note:
1. Look at the join points of each piece.
2. Jumps and unbounded growth both destroy continuity.