JEE Mathematics Continuity and Differentiability MCQs Set 03

Answer: (a) -2

Explanation:
1. For \( x<1 \), \( f(x)=\log(1-x) \), so \( f'(x)=\frac{1}{x-1} \) (the same formula holds for \( x>1 \)).
2. At \( x=\tfrac12 \): \( f'=\frac{1}{-1/2}=-2 \).

Teacher's Note:
1. The derivative of \( \log|u| \) is \( u'/u \) on both sides of the zero of \( u \).
2. Substitute carefully, the denominator is negative here.

Answer: (c) does not exist

Explanation:
1. Put \( t=\frac{\pi}{4}-x \); then \( y=|\tan t| \) and \( t=0 \) at \( x=\frac{\pi}{4} \).
2. Near \( t=0 \), \( \tan t\approx t \), so \( y \) behaves like \( |t| \).
3. Left derivative is \( +1 \) and right derivative is \( -1 \) (in terms of \( x \): \( -1 \) and \( +1 \)), so the derivative does not exist.

Teacher's Note:
1. A modulus of a function that vanishes with non-zero slope gives a corner.
2. Compare the two one-sided derivatives.

Answer: (c) does not exist

Explanation:
1. Near \( x=2 \), \( |x|=x \), so \( y=x+|x-2| \).
2. For \( x>2 \): \( y=2x-2 \), slope \( 2 \). For \( x<2 \): \( y=2 \), slope \( 0 \).
3. The slopes differ, so the derivative does not exist.

Teacher's Note:
1. Only the term that changes behaviour at the point matters.
2. Different one-sided slopes mean no derivative.

Answer: (a) \( \mu=0 \)

Explanation:
1. \( |x|^2=x^2 \) is smooth and the constant \( \lambda \) is smooth.
2. The term \( \mu|x| \) has right slope \( \mu \) and left slope \( -\mu \) at 0.
3. They agree only if \( \mu=-\mu \), i.e. \( \mu=0 \).

Teacher's Note:
1. Separate the smooth part from the modulus part.
2. Only \( |x| \) (an odd power of the modulus) spoils differentiability at 0.

Answer: (c) non-existent

Explanation:
1. For \( 02. So \( f(1^-)=0 \) but \( f(1)=1 \): \( f \) is discontinuous at 1.
3. A function that is not continuous at a point cannot be differentiable there, so \( f'(1) \) does not exist.

Teacher's Note:
1. Check continuity first; it is the quickest test.
2. Greatest-integer functions jump at integers.

Answer: (a) 2

Explanation:
1. Just right of \( x=\frac{\pi}{4} \), \( 2x \) is slightly more than \( \frac{\pi}{2} \), so \( \cos 2x<0 \) and \( f(x)=-\cos 2x \).
2. \( f'(x)=2\sin 2x \), and at \( x=\frac{\pi}{4} \) this is \( 2\sin\frac{\pi}{2}=2 \).

Teacher's Note:
1. First decide the sign inside the modulus on the required side.
2. Then differentiate the resulting expression.

Answer: (b) 0

Explanation:
1. For every real \( x \), \( [x] \) is an integer, so \( \sin\pi[x]=0 \).
2. So \( f \) is the zero function and its derivative on the left of 1 is 0.

Teacher's Note:
1. Look for a simplification before differentiating.
2. The sine of any integer multiple of pi is zero.

Answer: (c) continuous only at \( x=1 \) among the integers

Explanation:
1. At an integer \( n \): \( f(n)=n^2-n^2=0 \), and just right of \( n \) also \( f=0 \).
2. Just left of \( n \) (\( n\ne 1 \)): \( f=2n-2 \) for \( n>0 \) (and a non-zero value for \( n\le0 \)), which is not 0 unless \( n=1 \).
3. At \( n=1 \) the left value \( 2(1)-2=0 \) agrees, so \( f \) is continuous only at \( x=1 \) among integers.

Teacher's Note:
1. Check left and right limits at each integer separately.
2. The special value \( n=1 \) makes \( 2n-2 \) vanish.

Answer: (c) 0

Explanation:
1. Both \( x \) and \( |x-x^2| \) are continuous functions on \( [-1,1] \).
2. Their difference is continuous, so there is no point of discontinuity.

Teacher's Note:
1. Sums and differences of continuous functions are continuous.
2. The modulus of a continuous function is continuous.

Answer: (d) none of these

Explanation:
1. At \( x=\sqrt3 \): left \( \sqrt{1+3}=2 \), right \( \sqrt3\cdot\sqrt3-1=2 \). Continuous.
2. At \( x=4 \): left \( 4\sqrt3-1\approx5.93 \), right \( [4]=4 \). Discontinuous.
3. At \( x=5 \): left limit \( [x]\to4 \), \( f(5)=|1-5|=4 \). Continuous. On \( 4\le x<5 \), \( [x]=4 \) is constant.
4. So there is exactly one point of discontinuity (\( x=4 \)), which is none of the listed numbers.

Teacher's Note:
1. Test only the break points of the definition.
2. Remember \( [x] \) is constant between consecutive integers.

Answer: (a) 0

Explanation:
1. The integrand \( t\sin\frac1t \) is bounded near 0 (it tends to 0) and continuous elsewhere, so it is integrable.
2. An integral function \( \int_0^x g(t)dt \) of an integrable bounded function is continuous.
3. So there is no point of discontinuity.

Teacher's Note:
1. Integrals of bounded functions are continuous in the upper limit.
2. No need to evaluate the integral.

Answer: (a) \( f(x)=x \) for at least one \( 0\le x\le1 \)

Explanation:
1. Let \( g(x)=f(x)-x \). Then \( g(0)=f(0)\ge0 \) and \( g(1)=f(1)-1\le0 \).
2. By the intermediate value theorem \( g \) has a zero in \( [0,1] \), so \( f(x)=x \) for some \( x \).
3. Option (b) is false (continuity does not give differentiability). In (c), \( f(x)+x=0 \) would need \( f(x)=-x \), impossible for \( x>0 \) unless \( f(0)=0 \); it is not guaranteed.

Teacher's Note:
1. This is the fixed point idea for a continuous self map of [0, 1].
2. Use \( g(x)=f(x)-x \) at the end points.

Answer: (b) 10

Explanation:
1. A continuous function on an interval takes every value between any two of its values.
2. If it took two different rational values it would take an irrational value in between, which is not allowed.
3. So \( f \) is constant, \( f(x)=10 \) and \( f(1.5)=10 \).

Teacher's Note:
1. Rational-valued and continuous forces a constant function.
2. Use the intermediate value property.

Answer: (a) 0

Explanation:
1. \( f'(0)=\lim_{h\to0}\frac{e^{-1/h^2}-0}{h}=\lim_{h\to0}\frac{1/h}{e^{1/h^2}} \).
2. As \( h\to0 \), \( e^{1/h^2} \) grows much faster than \( 1/h \), so the limit is 0 from both sides.
3. Hence \( f'(0)=0 \). (The source key gave (c); the correct value is 0.)

Teacher's Note:
1. Use the definition of the derivative at the point.
2. Exponentials dominate powers.

Answer: (a) non-existent

Explanation:
1. \( h(x)=[\sin x+1] \).
2. Just near \( x=\frac{\pi}{2} \) (both sides) \( \sin x<1 \), so \( h=1 \); but \( h(\pi/2)=[2]=2 \).
3. So \( h \) is discontinuous at \( \frac{\pi}{2} \) and the derivative does not exist.

Teacher's Note:
1. Compose first, then test the point.
2. A jump at the point means no derivative.

Answer: (c) non-existent

Explanation:
1. \( h(x)=[|x|] \). For \( x<-1 \) (near \( -1 \)): \( |x|>1 \) so \( h=1 \). For \( -12. The two sides differ, so \( h \) is discontinuous at \( -1 \) and \( h'(-1) \) does not exist.

Teacher's Note:
1. Check the values on each side of the point.
2. A discontinuity rules out differentiability.

Answer: (b) 3

Explanation:
1. \( \left|x-\frac12\right| \) is not differentiable at \( x=\frac12 \); \( |x-1| \) at \( x=1 \).
2. \( \tan x \) is not defined (so not differentiable) at \( x=\frac{\pi}{2}\approx1.57 \), which lies in \( [0,2] \).
3. Total \( 3 \) points.

Teacher's Note:
1. Remember points where a function is undefined.
2. Count each kink separately.

Answer: (d) \( 2p-1 \)

Explanation:
1. \( f(x)=n+[p\sin x] \) and \( p\sin x \) takes values in \( (0,p] \) on \( (0,\pi) \).
2. Each integer \( k=1,2,\dots,p-1 \) is reached at two points (one on each side of \( \frac{\pi}{2} \)): \( 2(p-1) \) points of jump.
3. The maximum value \( p \) is reached only at \( x=\frac{\pi}{2} \): 1 more point.
4. Total \( 2(p-1)+1=2p-1 \).

Teacher's Note:
1. Count where \( p\sin x \) crosses each integer.
2. The peak value gives a single point.

Answer: (c) \( \{0\} \)

Explanation:
1. At \( x=0 \): \( |x| \) has a corner and the other term is smooth, so \( f \) is not differentiable at 0.
2. At \( x=1 \): \( \frac{f(x)-f(1)}{x-1}=(x-1)\cos\frac{1}{x-1}-\frac{|x|-1}{x-1} \). Near 1, \( |x|=x \) so the last term is 1.
3. The limit is \( 0-1=-1 \), so \( f'(1)=-1 \) exists. The set is \( \{0\} \).

Teacher's Note:
1. Test each suspect point with the definition.
2. A factor like \( (x-1)^2 \) times a bounded term is differentiable.

Answer: (b) 0

Explanation:
1. Put \( x=\frac{\pi}{2}-u \): \( 1-\sin x=1-\cos u\approx\frac{u^2}{2} \) and \( \sin2x=\sin(\pi-2u)=\sin2u\approx2u \).
2. So \( f(x)\approx\frac{u^2/2}{2u}=\frac u4\to0 \).
3. Continuity requires \( f(\pi/2)=0 \).

Teacher's Note:
1. Use a substitution so the limit is at 0.
2. Compare orders of the small terms.

Answer: (a) 0

Explanation:
1. \( \cos(\sin x)=1-\frac{\sin^2x}{2}+\frac{\sin^4x}{24}-\dots \) and \( \sin^2x=x^2-\frac{x^4}{3}+\dots \).
2. \( \cos(\sin x)=1-\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^4}{24}+\dots \) while \( \cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\dots \).
3. The difference is \( \frac{x^4}{6}+\dots \), so \( f(x)\approx\frac{x^2}{6}\to0 \). Hence \( a=0 \).

Teacher's Note:
1. Expand both cosines to the fourth power of x.
2. The leading terms cancel.

Answer: (c) \( -\frac12 \)

Explanation:
1. Continuity at \( \frac\pi4 \) needs \( \lambda=\lim_{x\to\pi/4}\frac{1-\tan x}{4x-\pi} \).
2. This is of the form 0/0; by L'Hopital: \( \frac{-\sec^2x}{4}=-\frac{2}{4}=-\frac12 \).

Teacher's Note:
1. Recognise the derivative of tan at pi/4.
2. The limit is \( \frac{d}{dx}(1-\tan x) \) divided by 4.

Answer: (b) 1

Explanation:
1. Let \( x=\frac\pi2-u \): \( f=(\cos u)^{1/(2u)} \).
2. \( \ln f=\frac{\ln\cos u}{2u}\approx\frac{-u^2/2}{2u}=-\frac u4\to0 \).
3. So \( f\to e^0=1 \).

Teacher's Note:
1. Take logarithms for a power-type limit.
2. Use \( \ln\cos u\approx-\frac{u^2}{2} \).

Answer: (d) for no value of \( f(0) \)

Explanation:
1. \( \sin\frac1x \) oscillates between \( -1 \) and \( 1 \) infinitely often as \( x\to0 \).
2. The limit does not exist, so no choice of \( f(0) \) gives continuity.

Teacher's Note:
1. Continuity needs the limit to exist first.
2. Oscillation means no limit.

Answer: (b) \( f(x) \) is continuous at \( x=0 \)

Explanation:
1. Near 0, \( 0\le\tan^2x<1 \), so \( f(x)=0 \) in a neighbourhood of 0.
2. So the limit is 0 and equals \( f(0) \): continuous.
3. The function is constant near 0, so \( f'(0)=0 \), which makes (c) and (d) false.

Teacher's Note:
1. Look at the value of the inside near the point.
2. A locally constant function is differentiable.

Answer: (a) is broken at two points

Explanation:
1. On \( [1,2) \): \( f=\log1=0 \); on \( [2,3) \): \( f=\log2 \); on \( [3,4) \): \( f=\ln x \) (from \( \ln3 \)).
2. Jumps occur at \( x=2 \) (0 to \( \ln2 \)) and at \( x=3 \) (\( \ln2 \) to \( \ln3 \)): two breaks.
3. At the breaks there is no tangent; elsewhere the graph is smooth. So the graph is broken at exactly two points.

Teacher's Note:
1. Evaluate the function on each sub-interval.
2. Count the jumps between pieces.