JEE Mathematics Continuity and Differentiability MCQs Set 18

Answer: (c) \( \sqrt2 \)

Explanation:
1. For large x the terms with the largest power dominate: \( x^{1/2} \) in the numerator and \( (2x-3)^{1/2} \) in the denominator.
2. The ratio tends to \( \dfrac{2x^{1/2}}{(2x)^{1/2}}=\sqrt2 \).

Teacher's Note:
1. Compare the powers of x.
2. Only the square root terms matter in the limit.

Answer: (b) \( \tfrac2\pi \)

Explanation:
1. Put \( a_n=\cot\theta_n \). Since \( \cot\theta+\csc\theta=\cot\tfrac\theta2 \), \( a_{n+1}=\cot\tfrac{\theta_n}2 \).
2. With \( a_1=0=\cot\tfrac\pi2 \) we get \( a_n=\cot\dfrac\pi{2^n} \).
3. \( \dfrac{a_n}{2^{n-1}}\approx\dfrac{2^n/\pi}{2^{n-1}}=\dfrac2\pi \).

Teacher's Note:
1. Use a cotangent substitution to solve the recurrence.
2. For small angles cot is about the reciprocal of the angle.

Answer: (b) 1

Explanation:
1. As \( n\to\infty \), \( \tfrac1{4^n}\to0 \) so \( f(0)=\lim f(4^{-n}) \).
2. \( |\sin e^n|e^{-n^2}\le e^{-n^2}\to0 \) and \( \tfrac{n^2}{n^2+1}\to1 \).
3. Hence \( f(0)=1 \).

Teacher's Note:
1. Use continuity of f along the sequence 4 to the minus n.
2. The first term is squeezed to 0.

Answer: (b) f is continuous for all x and \( f' \) is continuous for all x

Explanation:
1. Additivity and differentiability at 0 give \( f'(x)=\lim\dfrac{f(h)}h=f'(0) \) for all x.
2. So f is linear, \( f(x)=kx \): f is continuous, differentiable everywhere, and \( f'=k \) is constant (continuous).

Teacher's Note:
1. Write f(x + h) = f(x) + f(h).
2. The derivative is the same at every point.

Answer: (a) A, B, C and D

Explanation:
1. (A) both sides give 0 at \( -\tfrac\pi2 \). (B) at 0 the left slope is \( \sin0=0 \) and the right slope is 1, with equal values -1.
2. (C) at 1: values 0 and 0, slopes 1 and \( \tfrac1x=1 \). (D) \( -\tfrac32\lt -\tfrac\pi2 \) so f is linear there. All four are true.

Teacher's Note:
1. Check values and slopes at each junction.
2. The point -3/2 lies on a linear piece.

Answer: (b) 4

Explanation:
1. Differentiating with respect to x: \( \tfrac13f'\!\left(\tfrac{x+2y}3\right)=\tfrac13f'(x) \), so f' is constant and equals \( f'(0)=1 \).
2. \( f(x)=x+c \) with \( c=f(0)=2 \), so \( f(2)=4 \).

Teacher's Note:
1. Differentiate with respect to x.
2. A constant derivative gives a linear function.

Answer: (d) \( \mathbb R-\{-1,1\} \)

Explanation:
1. At \( x=1 \): \( f(1)=\tfrac\pi4 \) but the right limit is 0: f is discontinuous at 1, hence not differentiable.
2. Similarly at \( x=-1 \) (\( -\tfrac\pi4 \) against 0).
3. Elsewhere f is differentiable, so the domain of f' is \( \mathbb R-\{-1,1\} \).

Teacher's Note:
1. Check continuity at the joins first.
2. Discontinuity rules out a derivative.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( f=1 \) for \( x\lt 0 \), \( (2x-1)^2 \) on \( [0,1) \) and \( 1 \) for \( x\ge1 \). The values match at 0 and 1: continuous.
2. One-sided slopes: \( 0 \) and \( -4 \) at 0; \( 4 \) and \( 0 \) at 1. Both statements are true and the reason gives exactly the non-differentiability.

Teacher's Note:
1. Write f on the three intervals.
2. Differentiate the middle piece.

Answer: (c) 3

Explanation:
1. \( \cos x-1\approx-\tfrac{x^2}2 \) and \( \cos x-e^x\approx-x \).
2. The product behaves like \( \tfrac{x^3}2 \).
3. So the limit is finite and non-zero for \( n=3 \).

Teacher's Note:
1. Use the leading terms of each factor.
2. Add the orders of vanishing.

Answer: (d) B and C only

Explanation:
1. Put \( x=y=0 \): \( f(0)=1 \). Then \( f'(x)=f'(0)+2x \) and \( f(x)=x^2+x\sin\varphi+1 \).
2. The minimum value is \( 1-\tfrac{\sin^2\varphi}4\ge\tfrac34 \): so (B) and (C) are true.
3. f is unbounded above, so (D) is false, and (A) is false.

Teacher's Note:
1. Find f(x) from the functional equation.
2. Complete the square to find the minimum.

Answer: (b) 2

Explanation:
1. \( \sin x=\cos x \) at \( x=\tfrac\pi4 \) gives \( \tfrac1{\sqrt2}\lt \tfrac34 \), so the constant 3/4 is the maximum there.
2. Corners occur where \( \sin x=\tfrac34 \) and where \( \cos x=\tfrac34 \): two points.

Teacher's Note:
1. Check which of the three is largest near pi/4.
2. Corners are where the constant meets the sine or the cosine.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The quadratic \( 4x^2-12x+5=(2x-1)(2x-5) \) vanishes at \( \tfrac12 \) and \( \tfrac52 \), which would create corners in the modulus.
2. But \( \cos\pi x \) also vanishes at these points (reason: \( \cos\tfrac\pi2=\cos\tfrac{5\pi}2=0 \)), and a factor that vanishes cancels the corner, so f is differentiable there. The reason explains the assertion.

Teacher's Note:
1. Identify where the modulus argument is zero.
2. A vanishing smooth factor removes the corner.

Answer: (a) A and B only

Explanation:
1. f is continuous at 1 iff \( -2=1+a \), i.e. \( a=-3 \). g is continuous at 2 iff \( 0=1-b \), i.e. \( b=1 \).
2. Exactly one discontinuity means exactly one of these conditions fails: (A) and (B).
3. (C) fails both (two points) and (D) fails neither.

Teacher's Note:
1. Find the value of a that makes f continuous at 1 and of b that makes g continuous at 2.
2. Count the failures.

Answer: (b) 1

Explanation:
1. Put \( x=y=0 \): \( f(0)=1 \). Then \( f'(x)=f'(0)-2x \) and \( f(x)=1-x\sin\alpha-x^2 \).
2. \( f(-\sin\alpha)=1+\sin^2\alpha-\sin^2\alpha=1 \).

Teacher's Note:
1. Find f(0) and then f(x).
2. Substitute x = f'(0).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion: \( g(\cot x-\csc x)=g\cdot\dfrac{\cos x-1}{\sin x}\to0 \), and \( f''=g''\sin x+2g'\cos x-g\sin x\to0 \) at 0: both are 0, true.
2. Reason: \( f'(0)=g'(0)\sin0+g(0)\cos0=g(0) \): true.
3. The reason does not explain why the limit equals \( f''(0) \), so it is not a correct explanation.

Teacher's Note:
1. Compute both sides of the assertion.
2. The reason concerns f' not f''.

Answer: (c) continuous but not differentiable at \( x=1 \)

Explanation:
1. \( f(1^-)=\sin\tfrac\pi2=1 \) and \( f(1)=|2-3|\cdot1=1 \): continuous.
2. Left slope \( \tfrac\pi2\cos\tfrac\pi2=0 \); right slope: on \( [1,\tfrac32) \) \( f=(3-2x)\cdot1 \) has slope \( -2 \).
3. The slopes differ, so f is continuous but not differentiable at 1.

Teacher's Note:
1. Compare values then slopes.
2. Use [x] = 1 just to the right of 1.

Answer: (d) 1

Explanation:
1. Equal roots: \( 4f'^2-4ff''=0 \), so \( ff''=f'^2 \), which gives \( f=e^{kx} \); \( f(0)=1 \), \( f'(0)=2 \) give \( f=e^{2x} \). The repeated root is \( t=\dfrac{f'}f=2 \).
2. \( \lim\dfrac{e^{2x}-1}x=2 \), so the value is \( 2-\tfrac22=1 \).

Teacher's Note:
1. Use the discriminant being zero.
2. Then solve the resulting differential equation.

Answer: (c) 6

Explanation:
1. \( 2\cos x \) is an integer at \( 2\cos x=1,0,-1 \) (the values \( \pm2 \) occur only at 0, \( 2\pi \) or at the tangent point \( x=\pi \), where no jump occurs).
2. \( x=\tfrac\pi3,\tfrac\pi2,\tfrac{2\pi}3,\tfrac{4\pi}3,\tfrac{3\pi}2,\tfrac{5\pi}3 \): 6 points.

Teacher's Note:
1. List where 2 cos x takes integer values.
2. At the minimum -2 the floor does not jump.

Answer: (b) \( f'(0)=0=f(0) \)

Explanation:
1. Since \( \tfrac1n\to0 \) and f is continuous, \( f(0)=\lim f(1/n)=0 \).
2. \( f'(0)=\lim_{n\to\infty}\dfrac{f(1/n)-f(0)}{1/n}=0 \).
3. The other options need not hold (f need not vanish identically on (0, 1]).

Teacher's Note:
1. Use continuity and then the definition of the derivative along 1/n.
2. Look for an f with zeros only at 1/n for the other options.

Answer: (b) 1

Explanation:
1. Put \( x=y=0 \): \( f(0)=0 \). Then \( f(y^{2n+1})=\{f(y)\}^{2n+1} \), so f is additive: \( f(x)=cx \).
2. The relation gives \( c=c^{2n+1} \), so \( c\in\{0,1,-1\} \); not identically zero and \( f'(0)=c\ge0 \) give \( c=1 \).
3. Hence \( f'(6)=1 \).

Teacher's Note:
1. Show that f(x) = cx.
2. Use c = c^(2n+1) and the sign condition.

Answer: (c) 6

Explanation:
1. \( (\cos kx)^{1/k}\approx1-\dfrac{kx^2}2 \).
2. The product is \( \approx1-\dfrac{x^2}2(2+3+\cdots+n) \), so the limit is \( \tfrac12\left(\tfrac{n(n+1)}2-1\right) \).
3. Setting it equal to 10: \( \tfrac{n(n+1)}2=21 \), so \( n=6 \).

Teacher's Note:
1. Use the first order approximation of each root of a cosine.
2. Sum the coefficients.

Answer: (c) -2

Explanation:
1. \( x^2+(k-1)|x|-k=(|x|-1)(|x|+k) \).
2. For \( k=-2 \) the zeros are \( |x|=1 \) and \( |x|=2 \) (four simple zeros, each a corner of the modulus) and the inner \( |x| \) gives a corner at 0 where the inner value \( -k=2\ne0 \): 5 points.
3. For \( k=3,4,5 \) there are only 3 points (\( 0,\pm1 \)).

Teacher's Note:
1. Factor the quadratic in |x|.
2. Count the simple zeros plus the corner at 0.

Answer: (d) \( \pm\tfrac\pi2 \)

Explanation:
1. The limit is \( e^{\ln(1+b^2)}=1+b^2 \).
2. So \( \sin^2\theta=\dfrac{1+b^2}{2b}\ge1 \) (AM-GM), which forces \( \sin^2\theta=1 \) and \( b=1 \).
3. Hence \( \theta=\pm\tfrac\pi2 \).

Teacher's Note:
1. Evaluate the limit of the form 1 to infinity.
2. Use (1 + b squared)/(2b) is at least 1.

Answer: (b) B and C only

Explanation:
1. (A) false: \( f(x)=0 \) with any g gives \( fg=0 \).
2. (C) true: if \( f(a)\ne0 \) then \( g=\dfrac{fg}f \) near a is a quotient of differentiable functions.
3. (B) is the contrapositive of the fact that non-zero \( f(a) \) forces g to be differentiable, hence continuous.

Teacher's Note:
1. Test (A) with f = 0.
2. Divide by f when f(a) is not zero.