JEE Mathematics Continuity and Differentiability MCQs Set 08

Answer: (a) 16

Explanation:
1. \( x_0 = 0 \) because \( |x| \le [x] \le x \) forces x to be a non-negative integer. So \( f(0) = \dfrac{0 + 5}{0 + 1} = 5 \).
2. Let k(x) be the common ratio: \( a = k e^x \), \( b = k^2 e^x \), \( c = k^3 e^x \), \( f = k^4 e^x \). The terms \( e^x, p, q, r, f \) form an A.P. with common difference d, so \( f = e^x + 4d \).
3. Then \( k^4 = 1 + 4d\,e^{-x} \), and \( f(0) = 5 \) gives \( d = 1 \). So \( a^4 = k^4 e^{4x} = e^{4x} + 4e^{3x} \).
4. \( \lim \dfrac{a^4(x) - 5}{x} = \dfrac{d}{dx}\left(e^{4x} + 4e^{3x}\right)\Big|_0 = 4 + 12 = 16 \).

Teacher's Note:
1. Express every function in terms of the common ratio and the common difference.
2. The limit is the derivative of \( a^4 \) at 0 because \( a^4(0) = 5 \).

Answer: (b) \( \left(-\infty, \ell n\left(\sqrt{5} - 2\right)\right) \)

Explanation:
1. \( \dfrac{b}{a} = k(x) > 0 \) with \( k^4 = 1 + 4e^{-x} \). The inequality \( k < e^{-x/2} \) becomes \( 1 + 4e^{-x} < e^{-2x} \).
2. Put \( u = e^{-x} > 0 \): \( u^2 - 4u - 1 > 0 \), so \( u > 2 + \sqrt{5} \).
3. \( e^{-x} > 2 + \sqrt{5} \Rightarrow x < -\ln\left(2 + \sqrt{5}\right) = \ln\left(\sqrt{5} - 2\right) \), because \( (\sqrt{5} + 2)(\sqrt{5} - 2) = 1 \).

Teacher's Note:
1. Raise both sides to the fourth power since both are positive.
2. Use the product \( (\sqrt{5} + 2)(\sqrt{5} - 2) = 1 \) to rewrite the logarithm.

Answer: (b) 6

Explanation:
1. \( f(x) = \tan x \). For b: with \( y = \dfrac{4}{\pi}\tan^{-1}x \), \( \ln b = \lim\dfrac{\ln\left(\frac{4}{\pi}\tan^{-1}x\right)}{x^2 - 1} = \dfrac{\frac{1}{1 + x^2}\cdot\frac{1}{\tan^{-1}x}}{2x}\Big|_{x=1} = \dfrac{1}{\pi} \), so \( b = e^{1/\pi} \approx 1.375 > 1 \).
2. For \( b > 1 \), \( \tan x = bx \) has one root in \( \left(0, \dfrac{\pi}{2}\right) \) and one in each \( \left(k\pi, k\pi + \dfrac{\pi}{2}\right) \), \( k \ge 1 \).
3. So \( (0, m\pi) \) contains \( m \) roots, and \( m = 6 \).

Teacher's Note:
1. When the slope of the line exceeds 1, the line also meets the first branch of tan x.
2. The branches in which tan x is negative never meet a line with positive slope for x greater than 0.

Answer: (c) 8

Explanation:
1. Here \( f(x) = \tan x \) and \( b = e^{1/\pi} \approx 1.375 > 0 \).
2. In each interval of length \( \pi \), \( |\tan x| \) takes every positive value exactly twice.
3. In \( (0, 4\pi) \) there are 4 such intervals, so \( 4 \times 2 = 8 \) points.

Teacher's Note:
1. The graph of |tan x| repeats every \( \pi \).
2. Count the intersections of a horizontal line with each period.

Answer: (b) \( \dfrac{-e^4 + 4e^2 + 1}{2} \)

Explanation:
1. From the functional equation, \( f(x) = \ln(1 + x) - x \).
2. \( g(x) = \int_0^x\left(\ln(1 + t) - t\right)dt = (1 + x)\ln(1 + x) - x - \dfrac{x^2}{2} \).
3. At \( x = e^2 - 1 \): \( g = 2e^2 - (e^2 - 1) - \dfrac{(e^2 - 1)^2}{2} = \dfrac{-e^4 + 4e^2 + 1}{2} \).

Teacher's Note:
1. Use \( \int\ln(1 + t)\,dt = (1 + t)\ln(1 + t) - t \).
2. Substitute \( x = e^2 - 1 \) so that \( \ln(1 + x) = 2 \).

Answer: (b) \( -9 \)

Explanation:
1. \( f'(x) = -\dfrac{x}{1 + x} = -1 + \dfrac{1}{1 + x} \).
2. \( f''(x) = -\dfrac{1}{(1 + x)^2} \), so \( f''(2) = -\dfrac{1}{9} \).
3. Therefore \( \dfrac{1}{f''(2)} = -9 \).

Teacher's Note:
1. Derive f' from the functional equation first.
2. Differentiate \( \dfrac{1}{1 + x} \) carefully.

Answer: (a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Explanation:
1. \( f(0) = 0 \) and \( f(1) = 0 \), so \( f(0) = f(1) \), and \( g(x) = \{x\} \) has period 1. Both statements are true.
2. At an integer n, \( g(n^-) = 1 \) and \( g(n) = 0 \), but \( f(1) = f(0) \), so \( f(g(x)) \) has no jump. Statement-2 is exactly the reason, so it explains Statement-1.

Teacher's Note:
1. A jump of the inner function disappears if the outer function has equal values at the two ends.
2. The periodicity of g makes the same check work at every integer.

Answer: (b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1.

Explanation:
1. \( [x - 1] \) jumps at \( x = 2 \) (value 0 just left and 1 at 2), while \( |x - 2| \) is continuous, so Statement-1 is true.
2. At \( x = 2 \) the function is not even continuous, hence not derivable, so Statement-2 is true too. But non-derivability is a consequence of discontinuity, not its explanation, so Statement-2 does not explain Statement-1.

Teacher's Note:
1. Discontinuity implies non-differentiability, not the other way round.
2. Check whether the second statement is the cause of the first.

Answer: (c) Statement-1 is true, Statement-2 is false.

Explanation:
1. Both \( x\sin x \) and \( x\cos x + \sin x \) have amplitudes that grow with x, so neither is periodic: Statement-1 is true.
2. Statement-2 is false: \( f(x) = x + \sin x \) is aperiodic, but \( f'(x) = 1 + \cos x \) is periodic.

Teacher's Note:
1. Test a general claim with a counter-example.
2. A linear term makes a function aperiodic but disappears in the derivative.

Answer: (c) Statement-1 is true, Statement-2 is false.

Explanation:
1. \( \sin(\cos^{-1}x) = \sqrt{1 - x^2} \), which is differentiable on \( (-1, 1) \) and takes values in \( [0, 1] \), so Statement-1 is true.
2. Statement-2 is false: \( f(x) = \dfrac{1}{x} \) is differentiable on \( (0, 1) \) but is not bounded.

Teacher's Note:
1. Rewrite the composite function in a simple form.
2. Differentiability on an open interval does not guarantee boundedness.

Answer: (b) \( f(x) = \dfrac{1}{1 + x^2} \) and \( f(x) = \dfrac{1}{1 + \sqrt{x}} \) only

Explanation:
1. The range of \( \dfrac{1}{1 + x} \) is \( R - \{0\} \).
2. \( \dfrac{1}{1 + x^2} \) has range \( (0, 1] \) and \( \dfrac{1}{1 + \sqrt{x}} \) (defined for \( x \ge 0 \)) also has range \( (0, 1] \).
3. \( \dfrac{1}{\sqrt{3 - x}} \) has range \( (0, \infty) \). Only the second and third functions have the same range.

Teacher's Note:
1. Find the range of each function by studying its denominator.
2. The domain restriction matters in the third function.

Answer: (a) (1) and (4) only

Explanation:
1. Expanding the determinant gives \( F(x) = -\sin^2 x - \cos^2 x + x\left(\cos^2 x + \sin^2 x\right) = x - 1 \).
2. The range of \( x - 1 \) is R, so (1) is true. \( F'(x) = 1 \), so (2) is false. F is unbounded, so (3) is false. F is a polynomial, so (4) is true.

Teacher's Note:
1. Expand the determinant before testing the statements.
2. The trigonometric terms cancel completely.

Answer: (a) (1) and (3) only

Explanation:
1. For small x, \( \ell n(\cos x) \approx -\dfrac{x^2}{2} \) and \( \ell n(1 + x^2) \approx x^2 \), so \( f(x) \approx -\dfrac{x}{2} \to 0 = f(0) \): f is continuous.
2. \( \dfrac{f(x) - f(0)}{x} \to -\dfrac{1}{2} \) from both sides, so f is differentiable at 0 with \( f'(0) = -\dfrac{1}{2} \).

Teacher's Note:
1. A differentiable function is always continuous.
2. Use series for both logarithms.

Answer: (b) (3) and (4) only

Explanation:
1. \( f(x) = \dfrac{x\left(2^x - 1\right)}{1 - \cos x} \approx \dfrac{x\cdot x\ln 2}{x^2/2} = 2\ln 2 = \ln 4 \), so (3) is true and (1) is false.
2. As \( x \to \infty \), \( \dfrac{\ln 2}{2^x} \to 0 \), so \( g(x) \approx 2^x\cdot\dfrac{\ln 2}{2^x} = \ln 2 \). So (4) is true and (2) is false.

Teacher's Note:
1. Use \( 2^x - 1 \approx x\ln 2 \) and \( 1 - \cos x \approx \dfrac{x^2}{2} \).
2. For g, use \( \sin u \approx u \) with \( u = \dfrac{\ln 2}{2^x} \).

Answer: (a) (2), (3) and (4) only

Explanation:
1. (1) As \( x \to 0^+ \), \( 2^{\cot x} \to \infty \) and \( f \to 0 \); as \( x \to 0^- \), \( 2^{\cot x} \to 0 \) and \( f \to 1 \). The limits differ, so the discontinuity is not removable.
2. (2) \( \dfrac{|\sin x|}{x} \to \pm 1 \) from the two sides, and \( \cos(\pm 1) = \cos 1 \), so the limit exists. (3) \( x\sin\dfrac{\pi}{x} \to 0 \). (4) \( \ell n|x| \to -\infty \), so \( f \to 0 \).
3. In (2), (3) and (4) the limit exists, so the discontinuity is removable.

Teacher's Note:
1. A discontinuity is removable if the limit exists and is finite.
2. Compare the left and right limits in every case.

Answer: (a) (1), (3) and (4) only

Explanation:
1. \( |\sin x| \) is continuous everywhere, so (1) is true.
2. It has a corner wherever \( \sin x = 0 \), that is at \( x = n\pi \) for all integers n, so (3) and (4) are true. Between these points it is smooth, so (2) is false.

Teacher's Note:
1. The modulus creates a corner at every zero of the inside function when it changes sign.
2. The zeros of sin x are infinitely many.

Answer: (a) all four, (1), (2), (3) and (4)

Explanation:
1. For \( x \ge 0 \), \( [|x|] = [x] = |[x]| \), so \( f = 0 \) and f is continuous at all positive integers.
2. For negative non-integers, \( [|x|] = -[x] - 1 \) and \( |[x]| = -[x] \), so \( f = -1 \). At every non-positive integer f equals 0 while its neighbours on the left equal \( -1 \), so f is discontinuous there.
3. The range is \( \{0, -1\} \), which is finite, and \( f \le 0 \) everywhere.

Teacher's Note:
1. Treat non-negative and negative x separately.
2. The function only takes the values 0 and -1.

Answer: (a) (1), (2) and (4) only

Explanation:
1. \( 1 - x^2 \ge 0 \) gives \( -1 \le x \le 1 \), and \( 1 - \sqrt{1 - x^2} \ge 0 \) there, so the domain is \( [-1, 1] \): (1) is true.
2. Near 0, \( 1 - \sqrt{1 - x^2} \approx \dfrac{x^2}{2} \), so \( f(x) \approx \dfrac{|x|}{\sqrt{2}} \). The one-sided derivatives are \( \pm\dfrac{1}{\sqrt{2}} \), which are finite but different: (2) and (4) are true, and (3) is false.

Teacher's Note:
1. Expand the inner square root to find the behaviour near 0.
2. The result looks like a modulus function.

Answer: (a) (1), (2) and (3) only

Explanation:
1. (1) For \( x \to 0^+ \), \( [2x] = 0 \), and for \( x \to 0^- \), \( [0] = 0 \), so the limit is 0.
2. (2) Both sides give a quantity tending to 0. (3) For \( x \to 0^+ \), \( \dfrac{e^{1/x}}{1 + e^{1/x}} \to 1 \), so \( f \to 0 \); for \( x \to 0^- \), \( e^{1/x} \to 0 \), so \( f \to 0 \).
3. (4) \( \dfrac{\tan^{-1}|x|}{x} \to 1 \) from the right and \( -1 \) from the left, so the limit does not exist.

Teacher's Note:
1. Split each case into the left and right limit.
2. Odd extensions of modulus functions usually give different one-sided limits.

Answer: (d) f is discontinuous at both \( x = 1 \) and \( x = 0 \)

Explanation:
1. For \( x < 0 \) near 0, \( |\cos x| < 1 \), so \( \cos^{2n}x \to 0 \) and \( f(0^-) = 0 \). For \( 0 \le x \le 1 \), \( \sqrt[n]{1 + x^n} \to 1 \), so \( f(0) = 1 \) and \( f(1) = 1 \).
2. At 0, the left limit 0 differs from \( f(0) = 1 \). For \( x > 1 \), \( x^n \to \infty \) and \( f \to 0 \), while \( f(1) = 1 \).
3. So f is discontinuous at both points.

Teacher's Note:
1. Evaluate the limit in n separately in each interval.
2. Compare one-sided limits with the value of f at the join.

Answer: (d) is equal to \( -1 \)

Explanation:
1. At \( \theta = -1 \), the left bound is \( \dfrac{1 - 1 - 2}{2} = -1 \) and the right bound is \( \dfrac{1 - 2 - 1}{2} = -1 \).
2. By the sandwich theorem \( \dfrac{f(\theta)}{\theta^2} \to -1 \), so \( f(\theta) \to -1 \times 1 = -1 \). The value \( f(-1) \) is not given, so (a) cannot be concluded.

Teacher's Note:
1. Both bounds meet at the same value, so the middle expression is squeezed.
2. Multiply back by \( \theta^2 \) at the end.

Answer: (a) \( f(x) \) takes every value between \( f(-2) \) and \( f(2) \)

Explanation:
1. On \( [-2, 0) \), \( f = 4 - (x - 1)^2 \) is increasing from \( f(-2) = -5 \) towards 3. On \( [0, 2] \), \( f = x^3 + 2 \) takes all values from 2 to 10.
2. The range is \( [-5, 3) \cup [2, 10] = [-5, 10] \), a closed interval, so every value between \( f(-2) = -5 \) and \( f(2) = 10 \) is taken (a is correct) and \( -6 \) is not taken.
3. The only root is \( x = -1 \) (from \( x^2 - 2x - 3 = 0 \)); \( x^3 + 2 = 0 \) gives \( x = -\sqrt[3]{2} \notin [0, 2] \). So (b) is wrong, and (c) and (d) are wrong.

Teacher's Note:
1. A jump discontinuity can still leave overlapping value ranges.
2. Find the range by taking the union of the ranges of both pieces.