JEE Mathematics Continuity and Differentiability MCQs Set 06

Answer: (d) non existent

Explanation:
1. Write the numerator as \( (2x + \sin 2x) + 2 \). The ratio is \( \left(1 + \dfrac{2}{2x + \sin 2x}\right)e^{-\sin x} \).
2. The first factor tends to 1 but \( e^{-\sin x} \) keeps oscillating between \( e^{-1} \) and \( e \).
3. So the limit does not exist.

Teacher's Note:
1. A bounded oscillating factor prevents a limit unless it is multiplied by something tending to zero.
2. Separate the part that tends to 1.

Answer: (c) \( a < 0 \) and \( \alpha < m < \beta \)

Explanation:
1. The limit equals 1 when the quadratic is positive near m.
2. If \( a < 0 \), the parabola opens downward and is positive between the roots, so \( \alpha < m < \beta \) works.
3. In (a) and (d) m can lie between the roots where the quadratic is negative, and in (b) m lies left of both roots where it is negative for \( a < 0 \).

Teacher's Note:
1. The sign of a quadratic depends on the sign of a and on the position relative to the roots.
2. The ratio \( \dfrac{|u|}{u} \) is the sign of u.

Answer: (c) \( e^{-a^2/(2b^2)} \)

Explanation:
1. The form is \( 1^\infty \), so the limit is \( e^{L} \) with \( L = \lim \dfrac{\cos ax - 1}{\sin^2 bx} \).
2. \( \cos ax - 1 \approx -\dfrac{a^2x^2}{2} \) and \( \sin^2 bx \approx b^2x^2 \), so \( L = -\dfrac{a^2}{2b^2} \).

Teacher's Note:
1. Use \( 1 - \cos u \approx \dfrac{u^2}{2} \) and \( \sin u \approx u \).
2. Keep a and b in the right places.

Answer: (d) 1

Explanation:
1. Putting \( x = y = 0 \) gives \( f(0) = -c \).
2. \( f'(x) = \lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h} = \lim\dfrac{f(h) + c}{h} = \lim\dfrac{f(h) - f(0)}{h} = f'(0) = 1 \).

Teacher's Note:
1. Find f(0) first from the functional equation.
2. The derivative is the same at every x because f is linear.

Answer: (b) 2 or 1 according as \( x \) is rational or irrational

Explanation:
1. If x is rational, \( n!\,x \) is an integer for large n, so \( \cos(n!\,\pi x) = \pm 1 \), \( \cos^{2m} = 1 \) and the bracket is \( [2] = 2 \).
2. If x is irrational, \( |\cos(n!\,\pi x)| < 1 \), so \( 0 \le \cos^{2m}(n!\,\pi x) < 1 \) and \( [1 + \cos^{2m}] = 1 \).

Teacher's Note:
1. For rational x the factorial eventually clears the denominator.
2. The greatest integer function turns values just below 2 into 1.

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

Explanation:
1. Each term lies between \( \dfrac{r}{n^2 + 2n} \) and \( \dfrac{r}{n^2 + n + 1} \).
2. The sums are \( \dfrac{n(n+1)}{2(n^2 + 2n)} \) and \( \dfrac{n(n+1)}{2(n^2 + n + 1)} \), and both tend to \( \dfrac{1}{2} \).
3. By the sandwich theorem the limit is \( \dfrac{1}{2} \).

Teacher's Note:
1. Bound the denominator by its smallest and largest values.
2. Use \( \sum r = \dfrac{n(n+1)}{2} \).

Answer: (a) \( \dfrac{19}{45} \)

Explanation:
1. Numerator: \( \sin x - x \approx -\dfrac{x^3}{6} \), so \( (\sin x - x)^2 \approx \dfrac{x^6}{36} \), and \( 1 - \cos x^3 \approx \dfrac{x^6}{2} \). The numerator is \( \dfrac{19}{36}x^6 \).
2. Denominator: \( x^5\sin x \approx x^6 \) and \( (1 - \cos x)^2(2x^2 - \sin^2 x) \approx \dfrac{x^4}{4}\cdot x^2 = \dfrac{x^6}{4} \). The denominator is \( \dfrac{5}{4}x^6 \).
3. The limit is \( \dfrac{19/36}{5/4} = \dfrac{19}{45} \).

Teacher's Note:
1. Expand every term to the same power of x.
2. Only the lowest surviving power matters.

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

Explanation:
1. \( n^2 < n^2 + n + 1 < (n+1)^2 \) for \( n \ge 1 \), so \( n < \sqrt{n^2 + n + 1} < n + 1 \) and the greatest integer is n.
2. \( \sqrt{n^2 + n + 1} - n = \dfrac{n + 1}{\sqrt{n^2 + n + 1} + n} \to \dfrac{1}{2} \).

Teacher's Note:
1. Locate the square root between two consecutive integers.
2. Rationalise the difference \( \sqrt{\ } - n \).

Answer: (b) \( \sqrt{pq} \)

Explanation:
1. The form is \( 1^\infty \). The limit is \( e^{L} \) with \( L = \lim n\left(\dfrac{p^{1/n} + q^{1/n}}{2} - 1\right) \).
2. Using \( p^{1/n} - 1 \approx \dfrac{\ln p}{n} \), \( L = \dfrac{\ln p + \ln q}{2} \).
3. So the limit is \( e^{\frac{1}{2}\ln(pq)} = \sqrt{pq} \).

Teacher's Note:
1. This is the geometric mean limit.
2. Use \( a^{1/n} - 1 \approx \dfrac{\ln a}{n} \).

Answer: (d) 11

Explanation:
1. If x is not an integer, \( \{-x\} > 0 \) so \( \operatorname{sgn}\{-x\} = 1 \), and since \( \{x\} < 1 \) we get \( f(x) = 1 \).
2. If x is an integer, \( \{x\} = 0 \) and \( \{-x\} = 0 \), so \( f(x) = 0 \).
3. So f equals 1 except at integers where it is 0. It is discontinuous, hence not differentiable, at the 11 integers from \( -5 \) to 5.

Teacher's Note:
1. Compute the sign of the fractional part carefully at integers and non-integers.
2. Discontinuity implies non-differentiability.

Answer: (a) 19

Explanation:
1. For \( f''(1) \) to exist, f, f' and the second derivative must match at 1.
2. \( f(1) \): \( a + b + c = 1 \). \( f'(1) \): \( 2a + b = 3 \). \( f''(1) \): \( 2a = 6 \).
3. So \( a = 3,\ b = -3,\ c = 1 \) and \( a^2 + b^2 + c^2 = 9 + 9 + 1 = 19 \).

Teacher's Note:
1. Existence of the second derivative forces continuity of the function and of its first derivative as well.
2. Match three conditions for three unknowns.

Answer: (a) \( -2n \)

Explanation:
1. For \( 0 < x < \dfrac{\pi}{2} \), \( \tan x > \sin x \) and \( \tan x > x \), so \( 1 - \tan^n x \) is the smallest of the three. Hence \( f(x) = 1 - \tan^n x \) near \( \dfrac{\pi}{4} \).
2. The derivative is \( -n\tan^{n-1}x\,\sec^2 x \), which at \( \dfrac{\pi}{4} \) equals \( -n \cdot 1 \cdot 2 = -2n \).

Teacher's Note:
1. Decide which branch of a minimum is active near the point.
2. Compare tan x with sin x and x for positive x.

Answer: (a) 0

Explanation:
1. At \( x = 0 \) the first and second rows are the same, so \( g(0) = 0 \) and the limit is \( g'(0) \).
2. Differentiating the determinant row by row, only the first row changes and it becomes \( f'(c), f'(2c), f'(3c) \) at \( x = 0 \), the same as the third row. So \( g'(0) = 0 \).

Teacher's Note:
1. A determinant with two equal rows is zero.
2. The limit of \( \dfrac{g(x)}{x} \) with \( g(0) = 0 \) is the derivative at 0.

Answer: (a) \( a f(x) \)

Explanation:
1. \( f'(0) = \lim\dfrac{f(x) - 1}{x} = \lim\left(p(x) + x\,q(x)\right) = a \).
2. From \( f(x + h) = f(x)f(h) \): \( f'(x) = f(x)\lim\dfrac{f(h) - 1}{h} = f(x)f'(0) = a f(x) \).

Teacher's Note:
1. For \( f(x+y) = f(x)f(y) \), the derivative is f'(0) times f(x).
2. Remember the exponential nature of such a function.

Answer: (a) \( \dfrac{8}{3} \)

Explanation:
1. \( 4n^2\left(1 - \cos\dfrac{x}{n}\right) \to 2x^2 \), so \( f(x) = \tan^{-1}(2x^2) \). \( \dfrac{n^2}{2}\ln\cos\dfrac{2x}{n} \to \dfrac{n^2}{2}\cdot\left(-\dfrac{2x^2}{n^2}\right) = -x^2 \), so \( g(x) = -x^2 \).
2. Then \( e^{-2g} = e^{2x^2} = 1 + 2x^2 + 2x^4 + \dfrac{4}{3}x^6 + \dots \).
3. \( \tan^{-1}(2x^2) = 2x^2 - \dfrac{8}{3}x^6 + \dots \), so \( e^{f} = 1 + 2x^2 + 2x^4 + \left(-\dfrac{8}{3} + \dfrac{4}{3}\right)x^6 + \dots = 1 + 2x^2 + 2x^4 - \dfrac{4}{3}x^6 + \dots \).
4. The difference is \( \left(\dfrac{4}{3} + \dfrac{4}{3}\right)x^6 = \dfrac{8}{3}x^6 \), so the limit is \( \dfrac{8}{3} \).

Teacher's Note:
1. Find f and g first, then expand both exponentials to the sixth power.
2. Series expansion is faster than repeated L'Hospital.

Answer: (a) 1

Explanation:
1. In an A.P. \( B = 60^\circ \), so \( A + C = 120^\circ \).
2. \( 4\sin A\sin C = 2\left[\cos(A - C) - \cos(A + C)\right] = 2\cos(A - C) + 1 \), so \( 3 - 4\sin A\sin C = 2 - 2\cos(A - C) = 4\sin^2\dfrac{A - C}{2} \).
3. The expression is \( \dfrac{2\left|\sin\frac{A - C}{2}\right|}{|A - C|} \to 1 \).

Teacher's Note:
1. Use \( 2\sin A\sin C = \cos(A - C) - \cos(A + C) \).
2. The condition A.P. fixes the middle angle at 60 degrees.

Answer: (b) 6

Explanation:
1. \( f(x) = \pm 2 \) at every point. Continuity on \( (-\infty, 1) \) and on \( (1, \infty) \) forces f to be constant on each interval (a continuous function cannot jump between 2 and \( -2 \)).
2. If the left and right constants differ (2 choices), f is discontinuous at 1 for either value of \( f(1) \): \( 2 \times 2 = 4 \) functions.
3. If both constants equal c, f is discontinuous only when \( f(1) = -c \): 2 functions. Total \( 4 + 2 = 6 \).

Teacher's Note:
1. A continuous function with only two possible values is constant on an interval.
2. Count cases by whether the two sides agree.

Answer: (b) 2

Explanation:
1. Numerator: \( \ln(1 + x^3\tan^2 x) \approx x^3\cdot x^2 = x^5 \).
2. Denominator: \( e^{\sin^3 x}\left(e^{x^3 - \sin^3 x} - 1\right) \approx x^3 - \sin^3 x \). Since \( \sin x \approx x - \dfrac{x^3}{6} \), \( \sin^3 x \approx x^3 - \dfrac{x^5}{2} \), so the denominator is \( \approx \dfrac{x^5}{2} \).
3. The limit is \( \dfrac{x^5}{x^5/2} = 2 \).

Teacher's Note:
1. Factor out one exponential to get the form \( e^u - 1 \approx u \).
2. Keep the next term of the expansion of sin x.

Answer: (c) \( \dfrac{1}{e} \)

Explanation:
1. \( \ln(1 + x) - x \approx -\dfrac{x^2}{2} \).
2. \( (1 + x)^{1/x} = e^{\frac{\ln(1+x)}{x}} = e^{1 - \frac{x}{2} + \dots} \approx e\left(1 - \dfrac{x}{2}\right) \), so \( (1 + x)^{1/x} - e \approx -\dfrac{ex}{2} \).
3. The ratio is \( \dfrac{-x^2/2}{-e x^2/2} = \dfrac{1}{e} \).

Teacher's Note:
1. Expand \( \ln(1+x) \) to the second term before exponentiating.
2. The limit \( (1+x)^{1/x} \to e \) hides a first-order correction of \( -\dfrac{ex}{2} \).

Answer: (a) equal to 1

Explanation:
1. \( x + \dfrac{1}{x} \ge 2 \) with equality only at \( x = 1 \), so near 1 the value of \( x + \dfrac{1}{x} \) is \( 2 + \varepsilon \) with small \( \varepsilon > 0 \) on both sides.
2. Then \( -x - \dfrac{1}{x} = -2 - \varepsilon \), which lies just below the integer \( -2 \), so its fractional part is \( 1 - \varepsilon \to 1 \) from both sides.
3. The limit is 1 (the value at \( x = 1 \) would be 0).

Teacher's Note:
1. Use the AM-GM inequality to see on which side of \( -2 \) the expression lies.
2. The fractional part of a number just below an integer is close to 1.

Answer: (a) \( \dfrac{7}{4} \)

Explanation:
1. \( 1 - \cos x\cos 2x\cos 3x \approx \dfrac{x^2}{2}\left(1^2 + 2^2 + 3^2\right) = 7x^2 \) because the leading terms of the product of cosines add up.
2. \( \sin^2 2x \approx 4x^2 \).
3. The limit is \( \dfrac{7}{4} \).

Teacher's Note:
1. For a product of cosines, \( 1 - \prod\cos(k_i x) \approx \dfrac{x^2}{2}\sum k_i^2 \).
2. Match this with the denominator \( 4x^2 \).

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

Explanation:
1. Subtracting the sums for k and k - 1 gives \( \cos^{-1}\beta_k = \dfrac{\pi}{2} \) for every k, so every \( \beta_r = 0 \) and \( A = 0 \).
2. As \( x \to 0 \), \( (1 + x^2)^{1/3} \approx 1 \) and \( (1 - 2x)^{1/4} \approx 1 - \dfrac{x}{2} \), so the numerator is \( \approx \dfrac{x}{2} \).
3. The limit is \( \dfrac{x/2}{x} = \dfrac{1}{2} \).

Teacher's Note:
1. If a sum formula holds for every k, the general term is the difference of consecutive sums.
2. Then evaluate the limit at \( A = 0 \) by a binomial expansion.