JEE Mathematics Continuity and Differentiability MCQs Set 12

Answer: (c) 1

Explanation:
1. f even gives \( f' \) odd, and hence \( f'' \) even.
2. So \( f''(-\pi)=f''(\pi)=1 \).

Teacher's Note:
1. Differentiating an even function gives an odd function, and again gives an even one.
2. Parity alternates with each derivative.

Answer: (a) continuous and differentiable in \( [-1,1] \)

Explanation:
1. For \( t\ge0 \): \( x=t \), \( y=2t^2 \), so \( y=2x^2 \) for \( x\ge0 \).
2. For \( t\lt 0 \): \( x=3t \), \( y=0 \), so \( y=0 \) for \( x\lt 0 \).
3. \( f'(0^+)=0=f'(0^-) \), so f is continuous and differentiable everywhere.

Teacher's Note:
1. Eliminate t separately for t >= 0 and t < 0.
2. Compare the values and slopes of the two pieces at x = 0.

Answer: (c) not differentiable at \( x=\tfrac12 \)

Explanation:
1. \( [\cos\pi x]=0 \) for \( 0\lt |x|\le\tfrac12 \) and \( -1 \) for \( \tfrac12\lt |x|\lt 1 \), but \( 1 \) at \( x=0 \).
2. At \( x=\tfrac12 \) it drops from 0 to -1 (a jump), so f is discontinuous, hence not differentiable.
3. At 0 the value 1 differs from the limit 0, so (b) and (d) fail; at -1 f is continuous, so (a) fails.

Teacher's Note:
1. Tabulate the greatest integer value of cos(pi x) on each sub-interval.
2. A jump in the integer part happens exactly where cos(pi x) crosses 0 or 1.

Answer: (d) \( ab\,f(x) \)

Explanation:
1. \( f'(x)=f(x)\lim_{h\to0}\dfrac{f(h)-1}{h} \).
2. \( f(h)-1=hg(h)G(h) \), so the limit is \( \lim g(h)G(h)=ab \).
3. Hence \( f'(x)=ab\,f(x) \).

Teacher's Note:
1. Use f(x+h) = f(x)f(h) and factor out f(x).
2. The remaining limit is the derivative of f at zero.

Answer: (d) \( \dfrac{2f(x)}{x} \)

Explanation:
1. Putting \( x=y=1 \): \( f(1)=1 \).
2. \( f'(x)=\lim_{h\to0}\dfrac{f\left(x(1+h/x)\right)-f(x)}{h}=\dfrac{f(x)}{x}\lim\dfrac{f(1+h/x)-f(1)}{h/x} \)
3. This equals \( \dfrac{f(x)}{x}f'(1)=\dfrac{2f(x)}{x} \).

Teacher's Note:
1. Write x+h as x(1+h/x) and use the functional equation.
2. The limit that remains is f'(1).

Answer: (c) -2

Explanation:
1. For \( -\tfrac\pi2\lt x\lt 0 \), \( |x|=-x \) and \( |\sin x|=-\sin x \).
2. So \( f(x)=-x-\sin x \) and \( f'(0^-)=-1-\cos0=-2 \).

Teacher's Note:
1. Remove the moduli using the sign on the left side of 0.
2. Differentiate the resulting smooth expression.

Answer: (d) \( (-\infty,\infty)-\{2,3\} \)

Explanation:
1. Put \( t=x^2-5x+6=(x-2)(x-3) \). Then \( f=\dfrac{\sin|t|}{t} \), which tends to \( +1 \) when \( t\to0^+ \) and to \( -1 \) when \( t\to0^- \).
2. Near 2 and 3, t changes sign, so the limit of f does not exist there and f is discontinuous (so not differentiable).
3. Elsewhere f is smooth.

Teacher's Note:
1. Replace the quadratic by t and use sin t / t tends to 1.
2. The modulus makes the two one-sided limits +1 and -1.

Answer: (c) \( (2n+1)\dfrac\pi2,\ n\in I \)

Explanation:
1. The modulus gives a corner wherever \( \cos x=0 \).
2. \( \cos x=0 \) for \( x=(2n+1)\tfrac\pi2 \).

Teacher's Note:
1. A modulus of a smooth function is non-differentiable at its simple zeros.
2. List all the zeros with the integer n.

Answer: (b) 1

Explanation:
1. Continuity needs \( k=\lim_{x\to0}(\cos x)^{1/\sin x}=e^{\lim\frac{\ln\cos x}{\sin x}}=e^{0}=1 \).
2. With \( k=1 \): \( \ln f\approx\dfrac{-x^2/2}{x}=-\dfrac x2 \) gives \( f(x)\approx1-\tfrac x2 \), so both one-sided derivatives equal \( -\tfrac12 \) and f is differentiable.

Teacher's Note:
1. Differentiability needs continuity, which fixes k.
2. Check that the two one-sided derivatives then agree.

Answer: (c) f is not differentiable at \( x=0 \)

Explanation:
1. \( f'(0^+)=\lim_{x\to0^+}\dfrac{x^a-0}{x}=\lim x^{a-1} \).
2. Since \( a-1\lt 0 \), the limit is infinite, so f is not differentiable at 0.
3. f is continuous at 0 and unbounded as \( x\to\infty \).

Teacher's Note:
1. Form the difference quotient and look at the sign of the exponent a - 1.
2. A vertical tangent at 0 means the derivative does not exist.

Answer: (c) 2

Explanation:
1. \( h(x)=x \) for \( x\lt 0 \) and \( x\ge1 \), and \( x^2 \) for \( 0\le x\lt 1 \).
2. At 0: slopes 1 and 0. At 1: slopes \( 2 \) (left) and \( 1 \) (right).
3. So h has corners at exactly two points, 0 and 1.

Teacher's Note:
1. Find where the two curves cross, those are the only candidates.
2. Compare slopes at x = 0 and at x = 1.

Answer: (c) neither continuous in \( [-1,7) \) nor differentiable at \( x=1 \)

Explanation:
1. \( f(1)=f(1^-)=3 \) but \( f(1^+)=6 \).
2. So f has a jump at 1, it is discontinuous there and hence not differentiable at 1.

Teacher's Note:
1. Compare 3 to the power 1 with 7 - 1.
2. A jump discontinuity rules out both continuity and differentiability.

Answer: (b) \( 2f(x)g(0) \)

Explanation:
1. \( f'(x)=f(x)\lim_{h\to0}\dfrac{f(h)-1}{h}=f(x)\lim\dfrac{\sin2h\cdot g(h)}{h} \).
2. \( \lim\dfrac{\sin2h}{h}=2 \) and \( g(h)\to g(0) \).
3. So \( f'(x)=2f(x)g(0) \).

Teacher's Note:
1. Use f(x+h) = f(x)f(h) and factor out f(x).
2. Then rewrite f(h) - 1 as sin 2h times g(h).

Answer: (d) does not exist

Explanation:
1. \( \cos x-\sin x \) is positive just left of \( \tfrac\pi4 \) and negative just right of it.
2. Left: \( f=\cos x-\sin x \), slope \( -\sin\tfrac\pi4-\cos\tfrac\pi4=-\sqrt2 \).
3. Right: \( f=\sin x-\cos x \), slope \( +\sqrt2 \).
4. The slopes differ, so there is a corner.

Teacher's Note:
1. A modulus of a function with a simple zero has a corner there.
2. Compute the slope on each side by removing the modulus with the right sign.

Answer: (b) g is differentiable at 0 if f is continuous at 0

Explanation:
1. \( g'(0)=\lim_{x\to0}\dfrac{xf(x)-0}{x}=\lim_{x\to0}f(x) \).
2. This limit exists and equals \( f(0) \) when f is continuous at 0.

Teacher's Note:
1. Use the definition at 0, the factor x cancels.
2. Only continuity of f is needed, not differentiability.

Answer: (c) f(x) is not differentiable at \( x=2 \)

Explanation:
1. On \( [-1,0) \) f = \( -x \); on \( [0,1) \) f = 0; on \( [1,2) \) f = \( x \); \( f(2)=4 \).
2. At 0: values and limits all 0, so f is continuous at 0 (option a is false).
3. At 1: \( f(1^-)=0\ne1=f(1) \), discontinuous, so not differentiable.
4. At 2: \( f(2^-)=2\ne4 \), so f is not differentiable at 2.

Teacher's Note:
1. Write f piece by piece on each unit interval.
2. Compare the left limit with the value at each integer.

Answer: (d) \( \sin(|x|)-|x| \)

Explanation:
1. \( \cos|x|=\cos x \) is smooth, so options (a) and (b) carry the corner of \( \pm|x| \).
2. \( \sin|x|+|x| \) has slopes \( \pm2 \).
3. \( \sin|x|-|x| \) has slope \( \cos0-1=0 \) on the right and \( -\cos0+1=0 \) on the left, so it is differentiable.

Teacher's Note:
1. Remove the moduli on each side of 0 and compare the slopes.
2. The corner of sin|x| cancels the corner of -|x| exactly.

Answer: (a) \( \{(a,1-2a,a)\mid a\in R,a\ne0\} \)

Explanation:
1. Continuity at 1: \( a+b+c=1 \).
2. Equal slopes at 1: \( 2a+b=1 \), so \( b=1-2a \).
3. Then \( c=1-a-b=a \). The triplets are \( (a,1-2a,a) \).

Teacher's Note:
1. Two conditions (value and slope) at the joining point fix b and c in terms of a.
2. Substitute b into the continuity condition to find c.

Answer: (a) 6

Explanation:
1. The relation is satisfied by \( f(x)=1+x^n \): indeed \( (1+x^n)(1+y^n)=1+x^n+y^n+(xy)^n \).
2. \( f(2)=1+2^n=5 \) gives \( n=2 \), so \( f(x)=1+x^2 \).
3. Hence \( f'(3)=2\cdot3=6 \).

Teacher's Note:
1. Test the polynomial form 1 + x^n, which fits the equation.
2. Use f(2) = 5 to fix n.

Answer: (a) 15

Explanation:
1. \( f(2)=f(2)f(0)f(0) \) and \( f(2)\ne0 \) give \( f(0)^2=1 \); with \( f(x+h)=f(x)f(h)f(0) \) one finds \( f(0)=1 \).
2. \( f'(2)=f(2)\lim\dfrac{f(h)-1}{h}=f(2)f'(0)=5\cdot3=15 \).

Teacher's Note:
1. Show f(0) = 1 first, then use f(2+h) = f(2)f(h).
2. The limit that remains is f'(0).