JEE Mathematics Continuity and Differentiability MCQs Set 24

Answer: (a) 1

Explanation:
1. For 0 < \( x \) < \( 1, [x] = 0 \), so \( f(x) = \alpha \). Continuity needs \( \alpha = f(0) = 2 \).
2. For \( x \) < 0 near 0: \( (\sin x - x)/x^{3} = -1/6 + x^{2}/120 \) − ..., which is slightly greater than −1/6 and less than 0, so its integer part is −1. Hence the left limit is \( \beta - 1 = 2 \), so \( \beta = 3 \).
3. \( \beta - \alpha = 3 - 2 = 1 \).

Teacher's Note:
1. Find the one-sided limits; the integer part of a number just above −1/6 is −1.

Answer: (a) a = −1, b = 2

Explanation:
1. Continuity at 1: left value 1, right limit \( a + b \), so \( a + b = 1 \).
2. Derivatives at 1: left derivative = \( 2x \) e^2(x−1) \( + 2x^{2} \) e^2(x−1) = \( 2 + 2 = 4 \); right derivative = \( -2a \sin(2x - 2) \) + 2bx = \( 2b \).
3. So \( 2b = 4, b = 2 \), and then \( a = -1 \).

Teacher's Note:
1. A differentiable function must be continuous, so use both continuity and equality of the one-sided derivatives.

Answer: (a) \( \dfrac1{12} \)

Explanation:
1. Put \( s = \sin x \), so \( s \to 1 \). The bracket is \( (s^{2} - 3s + 2)/(\sqrt{2s^{2}+3s+4} + \sqrt{s^{2}+6s+2}) = (s - 1)(s - 2)/(...) \), and the denominator tends to \( 3 + 3 = 6 \).
2. \( \tan^{2}x = s^{2}/(1 - s^{2}) = s^{2}/((1 - s)(1 + s)) \).
3. The product is \( -s^{2}(s - 2)/(6(1 + s)) \), which tends to \( -(1)(-1)/(6 \cdot 2) = 1/12 \) as \( s \to 1 \). So \( k = 1/12 \).

Teacher's Note:
1. Rationalise the difference; the factor \( 1 - s \) cancels the \( \tan^{2} x \) singularity.

Answer: (a) \( |P|\le1 \)

Explanation:
1. Let \( f(x) \) = a₁ \( \sin x \) + ... + a₈ \( \sin 8x \). Then \( f(0) = 0 \) and \( f'(0) \) = a₁ + 2a₂ + ... + 8a₈ = \( P \).
2. From \( |f(x)| \le |\sin x|, |f(x)/x| \le |\sin x/x| \) for \( x \ne 0 \). Letting \( x \to 0, |f'(0)| \le 1 \).
3. So \( |P| \le 1 \) (equality is possible, for example \( f(x) = \sin x \)).

Teacher's Note:
1. \( P \) is the derivative of the sum at \( x = 0 \); use the squeeze with \( |\sin x/x| \).

Answer: (d) 7

Explanation:
1. \( f \) is constant between jumps of \( [x^{2}] \), and it jumps wherever \( x^{2} \) is an integer. For 1 < \( x \) < \( 3, x^{2} \) takes values in (1, 9).
2. The integers 2, 3, 4, 5, 6, 7, 8 are crossed, giving \( x = \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8} \), all inside (1, 3).
3. At each such point \( f \) jumps (aⁿ to aⁿ⁺¹), so \( f \) is not differentiable at 7 points.

Teacher's Note:
1. A step function is non-differentiable exactly at its jump points.

Answer: (b) has a point of discontinuity

Explanation:
1. \( \sqrt{4 - x^{2}} = \sqrt{1 + x^{2}} \) at \( x^{2} = 3/2 \).
2. At \( x = 0 \): \( f(0) = \max(2, 1) = 2 \), but as \( x \to 0^{+} f(x) = \min(\sqrt{4 - x^{2}}, \sqrt{1 + x^{2}}) \to \min(2, 1) = 1 \).
3. The right limit 1 differs from \( f(0) = 2 \), so \( f \) is discontinuous at \( x = 0 \) (and it also has corners at \( x = \pm \sqrt{3/2} \)), so only option \( (b) \) is true.

Teacher's Note:
1. Compare \( f(0) \) with the limit from the right, where the minimum is used.

Answer: (b) \( b-a<\dfrac{\pi}{24} \)

Explanation:
1. Multiply by \( f \) > 0: \( f f' \) > \( f^{4} + 1 \). With \( u = f^{2}, u' \) = 2ff' > \( 2(u^{2} + 1) \).
2. So \( d(\tan^{-1}u)/dx = u'/(1 + u^{2}) \) > 2. Integrating over \( (a, b) \): \( \tan^{-1} u(b) - \tan^{-1} u(a) \) > \( 2(b - a) \).
3. \( u(a) = 1 \) and \( u(b) = \sqrt{3} \), so the left side is \( \pi/3 - \pi/4 = \pi/12 \). Hence \( 2(b - a) \) < π/12, i.e. \( b - a \) < π/24.

Teacher's Note:
1. Substitute \( u = f^{2} \) and integrate the inequality for \( d(\tan^{-1}u)/dx \).

Answer: (d) 2

Explanation:
1. On \( (0, \pi/2) \): \( |\tan x| = \tan x \) equals \( \cos x \) when \( \sin x = \cos^{2}x = 1 - \sin^{2}x \), i.e. \( \sin x = (\sqrt{5} - 1)/2 \). This gives one crossing with different slopes (\( a \) corner).
2. Near \( x = 0, |\tan x| \approx |x| \) < \( \cos x \), so \( f = \cos|x| = \cos x \) is smooth at 0. On \( (\pi/2, \pi), \cos x \) < \( 0 \le |\tan x| \), so \( f = |\tan x| \) is smooth.
3. By symmetry the crossing also occurs at the negative value, so \( f \) has exactly two points of non-differentiability.

Teacher's Note:
1. Find where the two branches cross; at 0 the larger branch is the smooth \( \cos x \).

Answer: (d) \( \{1,2\} \)

Explanation:
1. \( \sqrt (x(2 - x)) = 2 - x \) gives \( x(2 - x) = (2 - x)^{2} \), so \( x = 1 \) or \( x = 2 \). For \( 0 \le x \) < \( 1, 2 - x \) is larger; for 1 < \( x \) < \( 2, \sqrt (x(2 - x)) \) is larger.
2. At \( x = 1 \) the slope changes from −1 (line) to 0 (the semicircle top), so there is a corner.
3. Near \( x = 2, f(x) = \sqrt (x(2 - x)) \) whose slope tends to −∞, so \( f \) is not differentiable at \( x = 2 \). Near \( x = 0, f = 2 - x \) is smooth.
4. So \( f \) is non-differentiable at \( x = 1 \) and \( x = 2 \).

Teacher's Note:
1. A vertical tangent also counts as non-differentiable.

Answer: (d) 4

Explanation:
1. Sketch the three pieces. For \( 0 \le x \) < π/2: \( f = \cos x \) up to \( x = \pi/4 \) and \( f = \sin x \) from π/4 to π/2 (a corner at π/4).
2. At \( x = \pi/2 \) and \( x = \pi, {\sin x} \) jumps, so \( f \) is discontinuous there (not counted). Between them \( f = \sin x \) on (π/2, 5π/6], and then \( f = 1/2 \): a corner at 5π/6.
3. For π < \( x \) < \( 7\pi/6, f = 1 + \sin x \), and it meets the level 1/2 at 7π/6: a corner. Then \( f = 1/2 \) up to 5π/3, where \( \cos x \) takes over: a corner at 5π/3.
4. The corners are at \( \pi/4, 5\pi/6, 7\pi/6 \) and 5π/3: four points.

Teacher's Note:
1. Treat each region separately; jumps of the fractional part are discontinuities, not corners.

Answer: (a) it is differentiable at x = 0

Explanation:
1. 0 ≤ {e^1/x} < 1, so \( |f(x)| \le x^{2} \to 0 \), hence \( k = 0 \).
2. \( (f(x) - f(0))/x \) = x{e^1/x}, which tends to 0 as \( x \to 0 \), so \( f'(0) = 0 \) exists.
3. Statement \( (d) \) is false: {e^1/x} is discontinuous wherever e^1/x is an integer \( (x = 1/\ln n, n \ge 2) \).

Teacher's Note:
1. A bounded factor times \( x^{2} \) tends to 0 at a rate fast enough for differentiability.

Answer: (b) φ (the empty set)

Explanation:
1. For \( x \) > 0: \( f = \sin x - x + 4(x - \pi)\cos x \), so \( f'(0^{+}) = \cos 0 - 1 + 4\cos 0 - 0 = 4 \).
2. For \( x \) < 0: \( f = -\sin x + x + 4(x - \pi)\cos x \), so \( f'(0^{-}) = -1 + 1 + 4 = 4 \).
3. The one-sided derivatives agree and \( f \) is smooth elsewhere, so \( K \) is empty.

Teacher's Note:
1. Compare the left and right derivatives at \( x = 0 \).

Answer: (a) \( \left\{\dfrac\pi4\right\} \)

Explanation:
1. On \( (-\pi/2, \pi/2) \cos x \) > 0, and \( \sin x = \cos x \) only at x = π/4 (sin x < 0 < \( \cos x \) for \( x \) < 0).
2. For \( x \) < \( \pi/4, f = \sin x \) (slope \( \cos x \)); for \( x \) > \( \pi/4, f = \cos x \) (slope \( -\sin x \)). At π/4 the slopes are √2/2 and −√2/2, so there is a corner.
3. Hence \( S = {\pi/4} \).

Teacher's Note:
1. A minimum of two smooth functions has corners only where they cross with different slopes.

Answer: (b) f is non-differentiable at 5 points and g at 7 points

Explanation:
1. For \( t = |x| \ge 0 \), compare \( h(t) = |t^{2} - 2t| \) with \( t \). They are equal at \( t = 0, 1 \) and 3.
2. \( f = \max \): \( f = h \) on \( [0, 1], f = t \) on \( [1, 3], f = h \) for \( t \ge 3 \). Corners at \( x = 0 \) (from |x|), ±1 and ±3: 5 points.
3. \( g = \min \): \( g = t \) on \( [0, 1], g = h \) on [1, 3] (with \( h \) touching 0 at \( t = 2, a \) corner), \( g = t \) for \( t \ge 3 \). Corners at \( x = 0, \pm 1, \pm 2, \pm 3 \): 7 points.

Teacher's Note:
1. Reduce to \( t = |x| \) and list the crossings and the zero of \( |t^{2} - 2t| \).

Answer: (b) \( \dfrac13 \)

Explanation:
1. The derivative of the inverse at \( f(c) \) is \( 1/f'(c) \).
2. \( f'(x) \) = e^x \( + 1 \), so \( f'(\ln 2) = 2 + 1 = 3 \), and the required value is 1/3.

Teacher's Note:
1. Use \( (f^{-1})'(f(c)) = 1/f'(c) \).

Answer: (a) only two points

Explanation:
1. On [−1, 0:) \( f = -x - 1 \), and at 0: left limit −1 but \( f(0) = 0 \), so \( f \) is discontinuous at 0.
2. On [0, 1:) \( f = x \), with limit 1 at \( x \to 1^{-} \), but \( f(1) = 1 + 1 = 2 \): discontinuous at 1.
3. On [1, 2:) \( f = 2x \), which tends to 4 as \( x \to 2^{-} \), and \( f(2) = 2 + 2 = 4 \): continuous at 2. On [2, 3:) \( f = x + 2 \), which is smooth inside.
4. So in \( (-1, 3) f \) is discontinuous only at \( x = 0 \) and \( x = 1 \).

Teacher's Note:
1. Check the left limit and the value at each break point.

Answer: (a) differentiable if f'(c) = 0

Explanation:
1. Near \( c, f(x) \approx f'(c)(x - c) \), so \( g(x) \approx |f'(c)| |x - c| \).
2. If \( f'(c) \ne 0 \), this has a corner at \( c \) (slopes \( \pm |f'(c)| \)), so \( g \) is not differentiable.
3. If \( f'(c) = 0 \), then \( |f(x)|/|x - c| \to 0 \), so \( g'(c) = 0 \) exists: \( g \) is differentiable.

Teacher's Note:
1. The modulus creates a corner exactly when the function crosses zero with non-zero slope.

Answer: (d) not continuous for any values of a and b

Explanation:
1. At \( x = 1 \): \( a + b = 5 \). At \( x = 2 \): \( a + 2b = b + 10 \), so \( a + b = 10 \).
2. These two conditions contradict each other (5 ≠ 10), so no choice of a, \( b \) makes \( f \) continuous.

Teacher's Note:
1. Write the continuity condition at each break point and test for consistency.

Answer: (a) into

Explanation:
1. Replacing \( x \) by \( x^{2} \) gives \( f(x^{2}) + f(x^{4}) = 2 \), so \( f(x^{4}) = f(x) \), and repeatedly \( f(x) \) = f(x²ⁿ).
2. For \( |x| \) < 1, x²ⁿ → 0, so by continuity \( f(x) = f(0) \); for \( |x| \) > 1 use roots of \( x \) tending to 1 to get \( f(x) = f(1) \). Also \( f(0) + f(0) = 2 \) gives \( f(0) = 1 \) and \( f(1) = 1 \).
3. So \( f(x) = 1 \) for all \( x \): a constant function, which is into (not onto), not one-one, and periodic.

Teacher's Note:
1. A continuous function with \( f(x) = f(x^{4}) \) is forced to be constant.

Answer: (c) \( \{5,10,15\} \)

Explanation:
1. \( f(x) = x + 5 \) for \( x \le 10 \) and \( 25 - x \) for \( x \ge 10 \), with a corner at 10 and \( f(10) = 15 \).
2. \( g = f(f(x)) \) can fail to be differentiable where \( f \) has a corner \( (x = 10) \), or where \( f(x) = 10 \) (the corner of the outer \( f \)): \( x + 5 = 10 \) gives \( x = 5 \), and \( 25 - x = 10 \) gives \( x = 15 \).
3. At \( x = 10 \): \( g = 20 - x \) for \( x \) < 10 and \( g = x \) for \( x \) > 10, so there is a corner. At \( x = 20, f(20) = 5 \) is not a corner of the outer function, so \( g \) is smooth there.
4. The set is {5, 10, 15}.

Teacher's Note:
1. List where the inner function has a corner or takes the value 10.

Answer: (a) \( \dfrac12 \)

Explanation:
1. At \( x = \pi/4 \) both numerator and denominator vanish \( (\sqrt{2} \cos(\pi/4) = 1, \cot(\pi/4) = 1) \).
2. By L'Hospital's rule: limit = \( (-\sqrt{2} \sin x)/(-\csc^{2}x) \) at \( \pi/4 = (-1)/(-2) = 1/2 \).
3. So \( k = 1/2 \).

Teacher's Note:
1. Apply L'Hospital's rule at the 0/0 point.

Answer: (a) \( \dfrac{2}{5-\pi} \)

Explanation:
1. At \( x = 5, |\pi - x| = |x - \pi| = 5 - \pi \) (since π < 5).
2. Left value: \( a(5 - \pi) + 1 \); right limit: \( b(5 - \pi) + 3 \). Equating, \( (a - b)(5 - \pi) = 2 \).
3. So \( a - b = 2/(5 - \pi) \).

Teacher's Note:
1. Both modulus terms equal 5 − π at \( x = 5 \).

Answer: (d) \( \left(-\dfrac32,\dfrac12\right) \)

Explanation:
1. Right limit: \( (\sqrt{x + x^{2}} - \sqrt{x})/x^{3/2} = \sqrt{x}(\sqrt{1 + x} - 1)/x^{3/2} = (\sqrt{1 + x} - 1)/x \to 1/2 \). So \( q = 1/2 \).
2. Left limit: \( \sin((p + 1)x)/x + \sin x/x \to (p + 1) + 1 = p + 2 \). Setting \( p + 2 = 1/2 \) gives \( p = -3/2 \).
3. So \( (p, q) = (-3/2, 1/2) \).

Teacher's Note:
1. Evaluate the left and right limits separately and equate them to \( f(0) \).