JEE Mathematics Continuity and Differentiability MCQs Set 16

Answer: (b) \( \left[\tfrac54,\tfrac{11}4\right] \)

Explanation:
1. \( f(x)=3x \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \); \( g(x)=x/3 \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \). So \( h(x)=x \) for all x.
2. The domain requires \( -1\le x\le1 \). Both \( \cos^{-1}x \) and \( \cot^{-1}x \) decrease, so the sum goes from \( \pi+\tfrac{3\pi}4=\tfrac{7\pi}4 \) at \( x=-1 \) to \( 0+\tfrac\pi4 \) at \( x=1 \).
3. So \( k\in\left[1+\tfrac14,1+\tfrac74\right]=\left[\tfrac54,\tfrac{11}4\right] \).

Teacher's Note:
1. First simplify h.
2. Evaluate the decreasing sum at the end points.

Answer: (b) B, C and D only

Explanation:
1. (A) the limits from the two sides are 0 and 1: a jump, not removable.
2. (B) \( |\sin x|/x\to\pm1 \) and cosine is even, so both limits equal \( \cos1 \): removable.
3. (C) \( |x\sin(\pi/x)|\le|x|\to0 \): removable.
4. (D) \( \ln|x|\to-\infty \) so the function tends to 0: removable.

Teacher's Note:
1. A removable discontinuity means the two-sided limit exists and is finite.
2. Evaluate each limit.

Answer: (d) at least three real roots

Explanation:
1. f is odd, so \( f(0)=0 \), \( f(3)=-2 \) and \( f(-5)=-4 \).
2. \( f(3)=-2\lt 0\lt f(5)=4 \) gives a root in \( (3,5) \) by the intermediate value theorem.
3. Similarly \( f(-5)=-4\lt 0\lt f(-3)=2 \) gives a root in \( (-5,-3) \). With the root 0 there are at least three roots.

Teacher's Note:
1. An odd function vanishes at 0.
2. Apply the intermediate value theorem on both sides.

Answer: (b) 3

Explanation:
1. \( 1-\cos^mx\approx\dfrac{m x^2}2 \), so the argument is \( \approx\dfrac{m\pi}{2x^{n-2}} \).
2. For \( n=1 \) it tends to 0 (limit 0); for \( n=2 \) it is the constant \( \tfrac{m\pi}2 \) (limit exists).
3. For \( n\ge3 \) the argument tends to infinity and the sine oscillates.
4. The possible n are 1 and 2: sum 3.

Teacher's Note:
1. Find how the argument behaves as x tends to 0.
2. A sine of an unbounded argument has no limit.

Answer: (b) \( 3\pi \)

Explanation:
1. Near 4, \( f(x)=9\tan\dfrac{\pi x}{12} \).
2. \( f'(x)=\dfrac{9\pi}{12}\sec^2\dfrac{\pi x}{12} \).
3. At \( x=4 \): \( \sec^2\tfrac\pi3=4 \), so \( f'(4)=\tfrac{9\pi}{12}\cdot4=3\pi \).

Teacher's Note:
1. Use the tangent branch because 4 lies in [3, 6).
2. Differentiate with the chain rule.

Answer: (c) A, C and D only

Explanation:
1. For \( 0\lt x\lt 1 \): \( \operatorname{sgn}[x]=0 \), \( \{x\}=x \), so \( f=x^3 \). For \( 1\le x\lt 2 \): \( 1+(x-1)=x \), so again \( f=x^3 \). Hence f is smooth at 1: (A) true, (B) false.
2. At 2: \( f(2^-)=8 \) but \( f(2)=\sin2+1\approx1.91 \): discontinuous, hence not differentiable: (C) and (D) true.

Teacher's Note:
1. Simplify the formula on each side of 1.
2. Compare f(2) with the left limit.

Answer: (b) 20

Explanation:
1. \( \sin x/x\lt 1 \), \( x/\sin x\gt 1 \), \( \tan x/x\gt 1 \) and \( x/\tan x\lt 1 \) near 0.
2. \( 4\sin x/x\to4^- \) so \( L=3 \); \( 5x/\sin x\to5^+ \) so \( M=5 \); \( 6\tan x/x\to6^+ \) so \( N=6 \); \( 7x/\tan x\to7^- \) so \( P=6 \).
3. Sum \( 3+5+6+6=20 \).

Teacher's Note:
1. Decide whether each ratio approaches its limit from below or above.
2. Then take the greatest integer.

Answer: (c) \( \tfrac7{52} \)

Explanation:
1. \( g'(0)=\lim\dfrac{g(x)}{x}=\lim\dfrac{x+\tan x-\tan2x}{ax+\tan x-\tan3x} \).
2. Numerator \( \approx-\tfrac73x^3 \). Denominator \( =(a-2)x-\tfrac{26}3x^3+\cdots \).
3. A finite non-zero limit needs \( a=2 \), then \( b=\dfrac{-7/3}{-26/3}=\dfrac7{26} \) and \( \dfrac ba=\dfrac7{52} \).

Teacher's Note:
1. Expand tan x, tan 2x and tan 3x to the cubic term.
2. The linear term of the denominator must vanish.

Answer: (c) 1

Explanation:
1. \( f(g(x))=x \) gives \( f'(g(a))g'(a)=1 \), so \( f'(b)=\tfrac12 \).
2. Hence \( 2f'(b)=1 \).

Teacher's Note:
1. Differentiate the identity f(g(x)) = x.
2. Evaluate at x = a.

Answer: (d) 1

Explanation:
1. \( \sum_{r=0}^\infty(\tfrac12)^r=2 \), so the denominator is \( [x+2]=[x]+2 \).
2. The ratio is \( \dfrac{[x]+2+\log_{10}(x+2)}{[x]+2}=1+\dfrac{\log_{10}(x+2)}{[x]+2} \).
3. The second term tends to 0, so the limit is 1.

Teacher's Note:
1. Sum the geometric series first.
2. A logarithm grows slower than a linear term.

Answer: (d) A, C and D only

Explanation:
1. For \( x\to0^+ \): \( \{x\}=x \), \( [x]=0 \), so \( f=\dfrac{\tan^2x}{x^2}\to1 \): (A) true, and (D) true since \( \tan^{-1}1=\tfrac\pi4 \).
2. For \( x\to0^- \): \( \{x\}\to1 \), so \( f\to\sqrt{\cot1} \), not \( \cot1 \): (B) false.
3. (C): \( \cot^{-1}(\cot1)=1 \) is true.

Teacher's Note:
1. Fractional part tends to 1 as x approaches 0 from the left.
2. The limit from the left carries a square root.

Answer: (d) is continuous but not differentiable

Explanation:
1. \( f(3)=9\tan\tfrac\pi4=9 \).
2. Left limits: rationals \( 9-3+3=9 \), irrationals \( 2^3+1=9 \); right: \( 9\tan\tfrac\pi4=9 \). So f is continuous at 3.
3. Left derivative along rationals: \( 2x-1=5 \); along irrationals: \( 2^x\ln2=8\ln2\ne5 \). The left derivative does not exist, so f is not differentiable.

Teacher's Note:
1. Check the three branches meeting at 3.
2. The two left branches have different slopes.

Answer: (b) continuous but not differentiable for all x

Explanation:
1. \( f(x)=3x \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \); \( g(x)=x/3 \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \). So \( h(x)=x \) for all x.
2. \( g(f(x))=x \) so \( f(g(f(x)))=f(x) \), and \( g(f(g(x)))=g(h(x))=g(x) \).
3. \( T=f+g \) equals \( \tfrac{10x}3 \) for \( x\ge0 \) and \( 2x \) for \( x\lt 0 \).
4. Continuous at 0, with slopes \( \tfrac{10}3 \) and 2 on the two sides, so not differentiable at 0. (It is not odd either.)

Teacher's Note:
1. Simplify the compositions using the earlier results.
2. Compare slopes at 0.

Answer: (c) 0 and 2

Explanation:
1. Left of 0: \( g=t^2+2t \) then \( -1 \). At 0: \( g(0)=f(0)=0 \) but \( g(0^-)=-1 \) (jump).
2. For \( x\ge0 \): the maximum of f on \( [0,x] \) is 0 for \( x\le2 \) (since \( f(2)=0 \)), then \( f(x)=x^2-2x \) for \( 2\lt x\le3 \).
3. At 2 the slopes are 0 and \( 2x-2=2 \): a corner. No other points.

Teacher's Note:
1. Find the running maximum of f on the positive side.
2. It stays 0 until f climbs back to 0 at x = 2.

Answer: (b) 4

Explanation:
1. The line \( y=x/n \) crosses the arches of \( |\sin\pi x| \) and creates corners. At integers the line lies above the zero of the arch, so the original corners there disappear.
2. The arch on \( (0,1) \) gives 1 crossing; each of the arches on \( (1,2),(2,3),(3,4) \) gives 2 crossings provided the line is below the peak, i.e. \( \tfrac{3.5}n\lt 1 \).
3. That gives 7 points, the maximum, for \( n\gt 3.5 \), so the least integer is 4. (For \( n=3 \) only 5 points arise.)

Teacher's Note:
1. Draw the arches and the line.
2. The line must stay below the last peak at x = 3.5.

Answer: (c) continuous but not differentiable

Explanation:
1. \( g(\tfrac\pi4)=3-1=2 \), and \( f(2)=4 \). For \( x\lt \tfrac\pi4 \): \( f(g)=3+\tan x\to4 \). For \( x\gt \tfrac\pi4 \): \( g\gt 2 \) so \( f(g)=6-(3-\cot x)=3+\cot x\to4 \). Continuous.
2. Slopes at \( \tfrac\pi4 \): left \( \sec^2\tfrac\pi4=2 \), right \( -\csc^2\tfrac\pi4=-2 \).
3. So it is continuous but not differentiable.

Teacher's Note:
1. g crosses the value 2 at pi/4, where f changes its formula.
2. Compare left and right slopes.

Answer: (d) 9

Explanation:
1. For \( x\lt 0 \), \( e^{-|x|}=e^{x} \) so the left hand derivative at 0 is 1.
2. \( 1-\cos\tfrac x2\approx\tfrac{x^2}8 \) and \( 1-\cos u\approx\tfrac{u^2}2 \), so the numerator \( \approx\dfrac{x^4}{128} \).
3. For a finite non-zero limit \( n=4 \), and \( \dfrac1{128\cdot2^m}=1 \) gives \( 2^m=2^{-7} \), \( m=-7 \).
4. \( n^2+m=16-7=9 \).

Teacher's Note:
1. The numerator behaves like x to the fourth.
2. Match the power of x first, then the constant.

Answer: (b) 6

Explanation:
1. Put \( u=x^4 \). Numerator: \( \sin u-u\cos u+u^5=\tfrac{u^3}3+O(u^5) \).
2. Denominator: \( u\left(2u^2+\tfrac43u^3+\cdots\right)=2u^3+\cdots \).
3. \( L=\tfrac13\cdot\tfrac12=\tfrac16 \), so \( 1/L=6 \).

Teacher's Note:
1. Substitute u = x to the fourth.
2. Expand sine, cosine and the exponential.

Answer: (d) 0

Explanation:
1. If \( 1\in[x,x^2+x+1] \) (i.e. \( x\le-1 \) or \( 0\le x\le1 \)) then \( g=2 \). For \( -1\lt x\lt 0 \) the interval lies left of 1, where f decreases: \( g=f(x^2+x+1)=2-x^2-x \). For \( x\gt 1 \): \( g=f(x)=x+1 \).
2. The pieces agree at \( x=-1,0,1 \) (values 2, 2, 2), so g is continuous everywhere: 0 points.

Teacher's Note:
1. Locate where the minimum of f over the interval occurs.
2. Check the joins at -1, 0 and 1.

Answer: (b) 2

Explanation:
1. From the previous part: \( f(g(x))=3+\tan x \) on \( [0,\tfrac\pi4) \) (positive) and \( 3+\cot x \) on \( [\tfrac\pi4,\pi) \).
2. The corner at \( \tfrac\pi4 \) remains.
3. \( 3+\cot x=0 \) has one solution in \( (\tfrac\pi2,\pi) \), where the sign changes with non-zero slope, giving a second corner. Total 2.

Teacher's Note:
1. Find the formula of f(g(x)) on each part.
2. The modulus adds a corner at each sign change.

Answer: (d) \( \tfrac1{32} \)

Explanation:
1. Row 4 has common difference \( \tfrac3{16}-\tfrac18=\tfrac1{16} \), so \( a_{4j}=\tfrac j{16} \). Column 4: \( a_{24}=1 \), \( a_{44}=\tfrac4{16}=\tfrac14 \) so \( r^2=\tfrac14 \), \( r=\tfrac12 \). Then \( a_{ij}=\dfrac j{2^i} \).
2. \( S_n=\dfrac1{16}\cdot\dfrac{n(n+1)}2=\dfrac{n(n+1)}{32} \).
3. So the limit is \( \tfrac1{32} \).

Teacher's Note:
1. Find the common difference of row 4 and the ratio of the columns.
2. Then sum the arithmetic progression.

Answer: (c) -1

Explanation:
1. For \( x\lt 0 \): \( f(t)=t^2+2t=(t+1)^2-1 \) on \( [-2,0) \), with minimum \( -1 \) at \( t=-1 \).
2. For \( x\ge-1 \) the interval \( [-2,x] \) contains \( -1 \), so \( g(x)=-1 \).

Teacher's Note:
1. Write f on the negative side as a perfect square minus 1.
2. The minimum is reached at t = -1.