JEE Mathematics Continuity and Differentiability MCQs Set 13

Answer: (d) -1

Explanation:
1. Putting \( y=0 \): \( f(x/2)=\dfrac{f(x)+1}2 \).
2. This (Jensen's equation) forces f to be linear: \( f(x)=1+mx \) with \( m=f'(0)=-1 \).
3. So \( f(2)=1-2=-1 \).

Teacher's Note:
1. The midpoint property is characteristic of linear functions.
2. Use f(0) and f'(0) to write the line.

Answer: (b) \( \dfrac16 \)

Explanation:
1. The equation is satisfied by \( f(x)=c\tan^{-1}x \) (addition formula of the inverse tangent).
2. \( f'(0)=\lim f(x)/x=c=\tfrac13 \).
3. \( f'(x)=\dfrac{1}{3(1+x^2)} \), so \( f'(1)=\dfrac16 \).

Teacher's Note:
1. Recognise the tangent addition formula.
2. Use the given limit as f'(0), which gives the constant.

Answer: (b) \( A=1,\ B=-1 \)

Explanation:
1. Continuity: \( A+\sin^{-1}(1+B)=1 \).
2. Slope: \( \dfrac1{\sqrt{1-(1+B)^2}}=1 \) gives \( 1+B=0 \), so \( B=-1 \).
3. Then \( A=1 \).

Teacher's Note:
1. The slope of the second piece at 1 must equal 1.
2. Solve for B first, then use continuity for A.

Answer: (c) f is differentiable only when \( k=2 \)

Explanation:
1. Both pieces equal 0 at 1, so f is continuous for every k.
2. \( f'(1^-)=2 \) and \( f'(1^+)=k \).
3. Differentiability needs \( k=2 \).

Teacher's Note:
1. Continuity holds for all k, so the slope condition alone decides.
2. Left slope is 2x = 2 at x = 1.

Answer: (b) f is continuous but not differentiable at \( x=2 \)

Explanation:
1. \( \int_0^2(1+|1-t|)dt=2+1=3=f(2) \), so f is continuous at 2.
2. For \( x>2 \), \( f'(x)=1+|1-x|=x \), so \( f'(2^+)=2 \), while \( f'(2^-)=5 \).
3. The slopes differ.

Teacher's Note:
1. Evaluate the integral on [0, 2] to check continuity.
2. The derivative of the integral is the integrand.

Answer: (d) continuous and differentiable at \( x=0 \)

Explanation:
1. Near 0, \( \sin x^2\approx x^2 \), so \( f(x)\approx x^5 \).
2. \( \dfrac{f(x)-f(0)}{x}\approx x^4\to0 \).
3. f is continuous and differentiable at 0 with \( f'(0)=0 \).

Teacher's Note:
1. Use sin t is approximately t for small t.
2. The quotient has a very high power of x so it tends to 0.

Answer: (c) \( a=1,\ b=\dfrac{\sqrt3}2-\dfrac\pi6 \)

Explanation:
1. Slope match: \( 2\cos\tfrac\pi3=1=a \).
2. Value match: \( \sin\tfrac\pi3=\tfrac{\sqrt3}2=a\tfrac\pi6+b \), so \( b=\tfrac{\sqrt3}2-\tfrac\pi6 \).

Teacher's Note:
1. Match the value and the derivative at x = pi/6.
2. The derivative of sin 2x is 2 cos 2x.

Answer: (a) \( a=1 \) and \( 64b^2+c^2=4 \)

Explanation:
1. Right limit: \( \lim\dfrac{e^{ax/2}-1}{x}=\tfrac a2=\tfrac12 \), so \( a=1 \).
2. Right slope: \( \dfrac{e^{x/2}-1}{x}\approx\tfrac12+\tfrac x8 \), so \( f'(0^+)=\tfrac18 \).
3. Left slope: \( b\cdot\dfrac{1/2}{\sqrt{1-c^2/4}}=\dfrac{b}{\sqrt{4-c^2}} \).
4. Equating to \( \tfrac18 \) and squaring: \( 64b^2=4-c^2 \), i.e. \( 64b^2+c^2=4 \).

Teacher's Note:
1. Continuity fixes a; the slope condition links b and c.
2. Square the slope equation to remove the root.

Answer: (b) not continuous at \( x=-2 \)

Explanation:
1. Put \( t=x+2 \). \( \dfrac{|t|}{\tan^{-1}t}\to+1 \) as \( t\to0^+ \) and \( \to-1 \) as \( t\to0^- \).
2. The limit does not exist (and \( f(-2)=2 \) matches neither), so f is not continuous at \( -2 \).

Teacher's Note:
1. Use tan^(-1) t is approximately t.
2. The modulus gives opposite signs on the two sides.

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

Explanation:
1. \( \dfrac{x}{\sqrt{x^2}}=\dfrac{x}{|x|}=\operatorname{sgn}x \).
2. The left limit is \( -1 \), the right limit is \( 1 \) and \( f(0)=0 \).
3. So f is not continuous at 0.

Teacher's Note:
1. Simplify the root to a modulus first.
2. The function is the sign function.

Answer: (c) f is differentiable at \( x=0 \) but not at \( x=1 \)

Explanation:
1. Near \( x=0 \) the formula is a product of smooth functions, so f is differentiable there.
2. At 1: \( \dfrac{f(x)-f(1)}{x-1}=\sin\dfrac1{x-1} \), which oscillates between -1 and 1 and has no limit.
3. So f is not differentiable at 1.

Teacher's Note:
1. Check x = 0 and x = 1 separately.
2. The difference quotient at 1 oscillates, the function is continuous there but has no derivative.

Answer: (a) \( \lim_{x\to c}f(x)=f(c) \)

Explanation:
1. \( f(x)-f(c)=\dfrac{f(x)-f(c)}{x-c}\cdot(x-c) \).
2. As \( x\to c \) the first factor tends to a finite number and the second to 0, so \( f(x)\to f(c) \).

Teacher's Note:
1. Differentiability implies continuity.
2. Multiply and divide by x - c.

Answer: (a) continuous at \( x=0 \)

Explanation:
1. If \( y\ge0 \): \( x=2y \), so \( y=\tfrac x2 \) for \( x\ge0 \).
2. If \( y\lt 0 \): \( x=10y \), so \( y=\tfrac x{10} \) for \( x\lt 0 \).
3. Both pieces are 0 at 0 so y is continuous, but the slopes \( \tfrac12 \) and \( \tfrac1{10} \) differ.

Teacher's Note:
1. Split on the sign of y and solve for y in terms of x.
2. Compare the slopes at x = 0.

Answer: (c) \( [0,\infty) \)

Explanation:
1. \( f'(x)=\ln x-1+1=\ln x \).
2. For \( x\in[1,\infty) \), \( \ln x \) takes all values in \( [0,\infty) \).

Teacher's Note:
1. Differentiate with the product rule.
2. Then find the range of ln x over the given set.

Answer: (c) \( A\in[64,\infty) \)

Explanation:
1. On \( (7,9) \), \( (x-5)^3 \) lies between 8 and 64.
2. If \( A\ge64 \) then \( 0\lt \dfrac{(x-5)^3}{A}\lt 1 \), so the integer part is the constant 0 and f is smooth.
3. If A is smaller, the integer part jumps somewhere in the interval and the jump multiplies \( \sin(x-5) \), which is non-zero at all but isolated points, creating a break in the derivative.

Teacher's Note:
1. Make the greatest integer factor constant on the interval.
2. The cubic increases from 8 to 64 on (7, 9).

Answer: (b) \( [2,\infty)-\{4\} \)

Explanation:
1. Put \( u=\sqrt{2x-4} \) so \( x=2+\tfrac{u^2}2 \). Then \( x\pm2u=\dfrac{(u\pm2)^2}{2} \).
2. \( f=\dfrac{(u+2)+|u-2|}{\sqrt2} \): a constant \( 2\sqrt2 \) for \( 2\le x\lt 4 \) and \( \sqrt2\,u \) for \( x\ge4 \).
3. There is a corner where \( u=2 \), i.e. at \( x=4 \).

Teacher's Note:
1. Complete the squares under each root using u.
2. The modulus |u - 2| gives the only corner.

Answer: (c) 3

Explanation:
1. The moduli give corners at \( x=0.5 \) and \( x=1 \).
2. \( \tan x \) is undefined at \( x=\tfrac\pi2\approx1.57 \), which lies in \( (0,2) \).
3. So there are 3 points.

Teacher's Note:
1. List the corners of the modulus terms and the points where tan x is undefined.
2. Check each lies inside (0, 2).

Answer: (c) 5

Explanation:
1. Since \( \dfrac{f(1+h)}{h} \) has a finite limit, \( f(1+h)\to0 \), so \( f(1)=0 \) by continuity.
2. \( f'(1)=\lim\dfrac{f(1+h)-f(1)}{h}=\lim\dfrac{f(1+h)}{h}=5 \).

Teacher's Note:
1. Use continuity to get f(1) = 0.
2. Then the given limit is the definition of f'(1).

Answer: (a) \( f(x) \) is differentiable everywhere

Explanation:
1. For \( x\ge0 \), both are \( x+1 \). For \( x\lt 0 \), \( x+1\lt -x+1 \) so the minimum is \( x+1 \).
2. Hence \( f(x)=x+1 \) for all x, a straight line.

Teacher's Note:
1. Compare the two expressions on each side of 0.
2. The minimum is the same line on both sides.

Answer: (c) \( 2af(a)-a^2f'(a) \)

Explanation:
1. Add and subtract \( a^2f(a) \): \( \dfrac{f(a)(x^2-a^2)-a^2(f(x)-f(a))}{x-a} \).
2. Limit \( =2af(a)-a^2f'(a) \).

Teacher's Note:
1. Insert a^2 f(a) to build two difference quotients.
2. The first quotient gives 2a f(a).