JEE Mathematics Continuity and Differentiability MCQs Set 07

Answer: (b) \( -\dfrac{\pi}{2} \)

Explanation:
1. The form is \( 1^\infty \). Put \( x = a + h \): \( \ln\left(2 - \dfrac{a}{a + h}\right) = \ln\dfrac{a + 2h}{a + h} \approx \dfrac{h}{a} \).
2. \( \tan\left(\dfrac{\pi}{2} + \dfrac{\pi h}{2a}\right) = -\cot\dfrac{\pi h}{2a} \approx -\dfrac{2a}{\pi h} \).
3. The exponent of e is \( a \cdot \left(-\dfrac{2a}{\pi h}\right)\cdot\dfrac{h}{a} = -\dfrac{2a}{\pi} \). Setting \( -\dfrac{2a}{\pi} = 1 \) gives \( a = -\dfrac{\pi}{2} \).

Teacher's Note:
1. Shift the variable so that the limit point becomes 0.
2. Use \( \tan\left(\dfrac{\pi}{2} + u\right) = -\cot u \).

Answer: (a) \( y(x) \) is discontinuous at \( x = 0 \)

Explanation:
1. If \( y \ge 0 \): \( y = 5x - 2y \Rightarrow y = \dfrac{5x}{3} \), valid for \( x \ge 0 \). If \( y < 0 \): \( -y = 5x - 2y \Rightarrow y = 5x \), valid for \( x < 0 \).
2. So \( y = \dfrac{5x}{3} \) for \( x \ge 0 \) and \( y = 5x \) for \( x < 0 \). Both pieces vanish at 0, so y is continuous at 0 and strictly increasing on R, with domain R.
3. Hence only statement (a) is incorrect.

Teacher's Note:
1. Solve the modulus equation separately for each sign of y.
2. Check that each solution satisfies its own sign condition.

Answer: (b) 1

Explanation:
1. Using \( \sin 2\theta = 2\sin\theta\cos\theta \) repeatedly, \( P_n = \dfrac{\sin x}{2^n\sin\left(x/2^n\right)} \).
2. As \( n \to \infty \), \( 2^n\sin\dfrac{x}{2^n} \to x \), so \( g(x) = \dfrac{\sin x}{x} \).
3. Then \( \lim_{x \to 0}g(x) = 1 \).

Teacher's Note:
1. A product of cosines with halving angles telescopes through the double angle formula.
2. This is Viete's product formula.

Answer: (b) 1

Explanation:
1. Let \( g(x) = ||x| - 2| \). It has corners at \( x = 0 \) (a peak of height 2) and \( x = \pm 2 \) (zeros). These three corners stay corners of f.
2. Extra corners occur where \( g(x) = -a \) with \( -a > 0 \). For \( 0 < -a < 2 \) there are 4 such points \( x = \pm(2 - c),\ \pm(2 + c) \) with \( c = -a \), giving \( 3 + 4 = 7 \) points.
3. For \( -a = 2 \) the points are \( 0, \pm 4 \), giving 5 points in all. For \( -a > 2 \) there are 2 extra points, giving 5, and for \( a \ge 0 \) there are only 3.
4. Hence \( a \in (-2, 0) \), and the only integer is \( a = -1 \).

Teacher's Note:
1. Count the corners of the inner function and the new corners created by the outer modulus.
2. Check the borderline values of a separately.

Answer: (d) \( \displaystyle\lim_{x \to b}f(x) \ge f(b) \)

Explanation:
1. For \( x < b \), a decreasing function satisfies \( f(x) \ge f(b) \).
2. Taking the limit as \( x \to b^- \) gives \( \lim_{x \to b}f(x) \ge f(b) \). A decreasing function may have a jump at b, so (a) fails, and \( f'(b) \) may be 0 or not exist, so (b) fails.

Teacher's Note:
1. Decreasing means the function value falls as x increases.
2. Limits preserve non-strict inequalities.

Answer: (a) \( f(x) \) is continuous at all \( x \in I \) but not differentiable

Explanation:
1. At an integer \( n \le 0 \): left limit \( (n - 1) + 1 = n \), right limit \( n + 0 = n \), and \( f(n) = n \). At \( n \ge 1 \): left limit \( \dfrac{n}{(n - 1) + 1} = 1 \), right limit \( \dfrac{n}{n} = 1 \), \( f(n) = 1 \). So f is continuous at every integer.
2. For \( n \le 0 \), \( \sqrt{\{x\}} \) has an infinite slope just to the right of n. For \( n \ge 1 \) the one-sided slopes are \( -\dfrac{1}{n} \) (left) and \( \dfrac{1}{n} \) (right), and at \( n = 1 \) they are \( \dfrac{1}{2} \) and 1.
3. So f is continuous but not differentiable at integers.

Teacher's Note:
1. Compare left and right limits first and then one-sided derivatives.
2. The square root of a fractional part gives a vertical tangent at integers.

Answer: (b) 2

Explanation:
1. The integers in \( (-4, 4) \) are \( -3, -2, -1, 0, 1, 2, 3 \).
2. At \( x = -3 \), \( f = x^2 + 2x \) is positive and smooth, so f is differentiable. At \( x = -2 \), \( x^2 + 2x \) changes sign, giving a corner.
3. At \( x = -1 \): the left derivative is \( -(2x + 2) = 0 \) and the right side is the constant 1 with \( f(-1) = 1 \), so f is continuous and differentiable.
4. At \( x = 0, 1, 2, 3 \): f jumps (for example \( f(0^-) = 1 \) and \( f(0) = 0 \)), so f is not differentiable.
5. So f is differentiable at 2 integers, \( -3 \) and \( -1 \).

Teacher's Note:
1. Greatest integer functions jump at integers, so check each integer separately.
2. Differentiability needs continuity first.

Answer: (c) \( e^{f'(0)/f(0)} \)

Explanation:
1. Here \( f(0) = 1\cdot 2\cdot 4\cdots 2^{n-1} \ne 0 \). The limit has the form \( 1^\infty \).
2. \( \ln\left(\dfrac{f(x + 1/n)}{f(x)}\right)^n = n\ln\dfrac{f(x + 1/n)}{f(x)} \to \dfrac{f'(x)}{f(x)} \).
3. So \( g(0) = e^{f'(0)/f(0)} \).

Teacher's Note:
1. The limit of \( n\left[\ln f(x + 1/n) - \ln f(x)\right] \) is the derivative of \( \ln f \).
2. This gives the logarithmic derivative in the exponent.

Answer: (b) \( -4 \)

Explanation:
1. The zeros are \( -1, -2, -4, \dots \), so \( x_r = -2^{r-1} \).
2. \( \sum_{r=1}^{\infty}\dfrac{r}{x_r} = -\sum_{r=1}^{\infty}\dfrac{r}{2^{r-1}} = -\dfrac{1}{(1 - 1/2)^2} = -4 \).

Teacher's Note:
1. Use \( \sum r x^{r-1} = \dfrac{1}{(1 - x)^2} \) for \( |x| < 1 \).
2. Arrange the roots in the given decreasing order first.

Answer: (c) 7

Explanation:
1. \( g(x) = x\sin x\,(\sin x + x\cos x) \), which is an odd function, so its roots are symmetric about 0.
2. Positive roots in \( (0, 2\pi) \): \( x = \pi \) from \( \sin x = 0 \); and the roots of \( \tan x = -x \), one in \( \left(\dfrac{\pi}{2}, \pi\right) \) and one in \( \left(\dfrac{3\pi}{2}, 2\pi\right) \) (no root where tan is positive). This gives 3 positive roots.
3. Total: 3 positive, 3 negative and \( x = 0 \), that is 7 roots.

Teacher's Note:
1. Break g into factors and count the roots of each.
2. Use a sketch of \( \tan x \) and \( -x \) to count the roots of \( \tan x = -x \).

Answer: (c) \( (0, 3] \)

Explanation:
1. For \( |x| \le 1 \), \( x^2 \le |x| \), and for \( |x| > 1 \), \( x^2 > |x| \). So \( u(x) = x^2 \) and \( v(x) = |x| \) for all x, giving \( f(x) = x^2 - 2|x| \).
2. f falls from 3 at \( t = -3 \) to \( -1 \) at \( t = -1 \), rises to 0 at \( t = 0 \), falls to \( -1 \) at \( t = 1 \), and rises to 0 at \( t = 2 \) and 3 at \( t = 3 \).
3. So \( g(x) = x^2 + 2x \) on \( [-3, -1] \), \( g = -1 \) on \( [-1, 0) \), \( g = 0 \) on \( [0, 2] \), and \( g = x^2 - 2x \) on \( [2, 3] \).
4. A value k with \( 0 < k \le 3 \) is taken once on \( [-3, -1) \) and once on \( (2, 3] \), giving exactly two solutions. Values in \( (-1, 0) \) are taken once, and \( -1 \) and 0 are taken infinitely often.

Teacher's Note:
1. Simplify u and v first; they collapse to simple functions.
2. Track the running minimum and maximum of f to build g.

Answer: (b) 5

Explanation:
1. \( g(x) = x^2 + 2x \) on \( [-3, -1] \) (values from 3 down to \( -1 \)), \( g = -1 \) on \( [-1, 0) \), \( g = 0 \) on \( [0, 2] \) and \( g = x^2 - 2x \) on \( [2, 3] \) (values from 0 to 3).
2. The range is \( [-1, 3] \), which contains the integers \( -1, 0, 1, 2, 3 \): 5 integers.

Teacher's Note:
1. The range is the union of the ranges of each piece.
2. Include the end values that are actually attained.

Answer: (c) \( (1, 2) \)

Explanation:
1. \( g(x) = x^2 + 2x \) on \( [-3, -1] \), \( -1 \) on \( [-1, 0) \), 0 on \( [0, 2] \), \( x^2 - 2x \) on \( [2, 3] \).
2. At \( x = 0 \) the left value is \( -1 \) and the value is 0, so g is discontinuous: \( a = 1 \).
3. At \( x = -1 \) both slopes are 0 (smooth). At \( x = 2 \) the slope changes from 0 to 2, a corner. So the non-differentiable points are \( x = 0 \) and \( x = 2 \): \( b = 2 \).

Teacher's Note:
1. A discontinuity is also a point of non-differentiability.
2. Check the slopes on both sides of every joining point.

Answer: (b) 683

Explanation:
1. From \( a_k = (k - 1)(2k + 3) \approx 2k^2 \), \( \sqrt{a_k} \approx \sqrt{2}\,k \) for large k.
2. The numerator is \( \sqrt{2}\,n\left(1 + 4 + 4^2 + \dots + 4^{10}\right) = \sqrt{2}\,n\,\dfrac{4^{11} - 1}{3} \). The denominator is \( \sqrt{2}\,n\left(2^{11} - 1\right) \).
3. The ratio is \( \dfrac{(2^{11} - 1)(2^{11} + 1)}{3(2^{11} - 1)} = \dfrac{2049}{3} = 683 \).

Teacher's Note:
1. For large k, only the leading term of \( a_k \) matters.
2. Use geometric series sums and factor \( 4^{11} - 1 \) as a difference of squares.

Answer: (d) 1

Explanation:
1. Since \( a_k \approx 2k^2 \), \( \sqrt{a_k} \approx \sqrt{2}\,k \).
2. The numerator is proportional to \( \dfrac{9^{11} - 1}{8} \) and the denominator to \( \dfrac{3^{11} - 1}{2} \), so \( L = \dfrac{9^{11} - 1}{4\left(3^{11} - 1\right)} = \dfrac{3^{11} + 1}{4} \).
3. \( \dfrac{4L}{3^{11}} = 1 + \dfrac{1}{3^{11}} \), which lies just above 1, so its greatest integer value is 1.

Teacher's Note:
1. Factor \( 9^{11} - 1 = \left(3^{11} - 1\right)\left(3^{11} + 1\right) \).
2. The extra term \( \dfrac{1}{3^{11}} \) is tiny and does not change the integer part.

Answer: (a) 4

Explanation:
1. Putting \( y = 0 \): \( f^2(x) = 4 + 4x \). Since \( f \ge 0 \), \( f(x) = 2\sqrt{1 + x} \), with domain \( [-1, \infty) \).
2. \( f(3) = 2\sqrt{4} = 4 \).

Teacher's Note:
1. Substitute y = 0 to remove one variable.
2. Choose the positive root because f is non-negative.

Answer: (b) 3

Explanation:
1. \( f(x) = 2\sqrt{1 + x} \) with domain \( [-1, \infty) \), so \( a = -1 \).
2. On \( [-1, 0] \), \( 7 - |x| = 7 + x \ge 6 \) is the largest. On \( [0, 1] \), \( 7 - x \ge 6 \). On \( [1, 8] \), the constant 6 is the largest (f reaches 6 at \( x = 8 \)). On \( [8, 10] \), \( f(x) \) is the largest.
3. The slope changes at \( x = 0 \) (from 1 to \( -1 \)), at \( x = 1 \) (from \( -1 \) to 0) and at \( x = 8 \) (from 0 to \( \tfrac{1}{3} \)). That gives 3 points.

Teacher's Note:
1. Find where each branch of the maximum takes over.
2. The point \( x = -1 \) is an end point of the domain and is not counted.

Answer: (b) \( \dfrac{4}{3} \)

Explanation:
1. \( f(x) = 2\sqrt{1 + x} \) with domain \( [-1, \infty) \), so \( a = -1 \).
2. \( \int_{-1}^{0}2\sqrt{1 + x}\,dx = 2\cdot\dfrac{2}{3}(1 + x)^{3/2}\Big|_{-1}^{0} = \dfrac{4}{3} \).

Teacher's Note:
1. The domain starts where the square root becomes zero.
2. Integrate with the substitution \( u = 1 + x \).

Answer: (c) 3

Explanation:
1. \( f \) falls from \( \tfrac{3}{2} \) at \( t = -\tfrac{3}{2} \) to 1 at \( t = -1 \), rises as \( 2 + t \) to 2 at \( t = 0 \), falls as \( 2 - t \) to 1 at \( t = 1 \), and then rises as t.
2. The running maximum is \( g_1 = \tfrac{3}{2} \) up to \( x = -\tfrac{1}{2} \), then \( 2 + x \) up to \( x = 0 \), then 2 up to \( x = 2 \), then x.
3. The slope changes at \( x = -\tfrac{1}{2},\ 0,\ 2 \), so there are 3 points of non-differentiability.

Teacher's Note:
1. A running maximum is constant while the function is below its previous maximum.
2. Corners appear where the maximum starts or stops following the function.

Answer: (a) 1

Explanation:
1. \( g_1 \) is non-decreasing from \( g_1\left(-\tfrac{3}{2}\right) = \tfrac{3}{2} \), so \( \min g_1 = \tfrac{3}{2} \).
2. \( g_2(x) = f(x) \) while f falls from \( \tfrac{3}{2} \) to 1 and then equals 1, so the largest value of \( g_2 \) is \( \tfrac{3}{2} \).
3. \( g_3 \) takes the values 2, \( 2 - x \) and 1, so \( \min g_3 = 1 \).
4. The smallest of \( \tfrac{3}{2},\ \tfrac{3}{2},\ 1 \) is 1.

Teacher's Note:
1. Find the extreme value of each function separately and then compare.
2. The running minimum eventually stays at the global minimum of f over the range.

Answer: (d) \( \left(\dfrac{1}{2}, -\dfrac{1}{2}, -2\right) \)

Explanation:
1. Expand the numerator: constant term \( a + b \); coefficient of x: \( 1 + a - b + c \); coefficient of \( x^2 \): \( \dfrac{a + b - c}{2} \).
2. For a finite limit the constant and x terms must vanish: \( a + b = 0 \) and \( 1 + a - b + c = 0 \). The limit equals the \( x^2 \) coefficient, so \( \dfrac{a + b - c}{2} = 1 \), which gives \( c = -2 \).
3. Then \( a - b = 1 \) and \( a + b = 0 \), so \( a = \dfrac{1}{2} \), \( b = -\dfrac{1}{2} \).

Teacher's Note:
1. Expand each function in a series and equate coefficients.
2. The limit equals the coefficient of the first power that survives.

Answer: (c) one positive and one negative

Explanation:
1. The limit condition gives \( a = \dfrac{1}{2} \), \( b = -\dfrac{1}{2} \), \( c = -2 \).
2. The equation becomes \( \dfrac{1}{2}x^2 - \dfrac{1}{2}x - 2 = 0 \), that is \( x^2 - x - 4 = 0 \).
3. The product of the roots is \( -4 < 0 \), so the roots are real with opposite signs.

Teacher's Note:
1. A negative product of roots means one positive and one negative root.
2. The discriminant \( 1 + 16 \) is positive.