JEE Mathematics Continuity and Differentiability MCQs Set 11

Answer: (c) differentiable at \( x_0 \) when \( f(x_0)\ne g(x_0) \)

Explanation:
1. If \( f(x_0)\ne g(x_0) \), then near \( x_0 \) h equals the larger of the two functions throughout a neighbourhood, so it is differentiable.
2. If \( f(x_0)=g(x_0) \), h may or may not be differentiable (for equal slopes it is differentiable, for unequal slopes there is a corner).

Teacher's Note:
1. The max of two differentiable functions can only have a corner where the graphs cross.
2. Away from crossing points the max simply copies one of the functions.

Answer: (b) \( |f|^2 \) is differentiable everywhere

Explanation:
1. \( |f|^2=f^2 \) is a product of differentiable functions, so it is differentiable everywhere.
2. \( |f| \) fails where f crosses zero (for example \( f(x)=x \)).
3. \( [f] \) and \( \operatorname{sgn}f \) are discontinuous where f passes through an integer or zero.

Teacher's Note:
1. Squaring removes the sign change that makes the modulus non-differentiable.
2. Use the counter-example \( f(x)=x \) to test the other options.

Answer: (d) does not exist

Explanation:
1. \( f(0)=0 \) but \( f(0^+)=1 \) and \( f(0^-)=-1 \).
2. The function is not continuous at 0, hence not differentiable.

Teacher's Note:
1. Differentiability requires continuity first.
2. A jump discontinuity rules out any derivative.

Answer: (c) 4

Explanation:
1. \( \dfrac{dy}{dx}=f'(f(x))\,f'(x) \)
2. At x = 0: \( f'(f(0))f'(0)=f'(0)\cdot f'(0)=2\cdot2=4 \).

Teacher's Note:
1. Apply the chain rule and use f(0)=0 to bring the inner value back to 0.
2. The answer is simply the square of f'(0).

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

Explanation:
1. For a > 1, \( f'(0^+)=\lim_{h\to0^+}\dfrac{a^{1/h}-a^{-1/h}}{a^{1/h}+a^{-1/h}}=1 \) and \( f'(0^-)=-1 \).
2. f is continuous at 0 (limit 0) but the one-sided derivatives differ, so it is not differentiable.

Teacher's Note:
1. Divide numerator and denominator by the large exponential.
2. Equal limits of f only give continuity; derivatives need equal one-sided slopes.

Answer: (b) \( a=2x_0,\ b=-x_0^2 \)

Explanation:
1. Continuity: \( x_0^2=ax_0+b \).
2. Equal derivatives: \( 2x_0=a \).
3. So \( b=x_0^2-2x_0^2=-x_0^2 \).

Teacher's Note:
1. Match the value and the slope of the two pieces at the joining point.
2. Solve the two equations for a and b.

Answer: (a) \( (-1)^k(k-1)\pi \)

Explanation:
1. Just left of k, \( [x]=k-1 \) and \( f(k)=0 \).
2. \( f'(k^-)=\lim_{h\to0^-}\dfrac{(k-1)\sin(k\pi+h\pi)}{h}=(k-1)(-1)^k\pi \).

Teacher's Note:
1. Near k from the left the greatest integer function is the constant k-1.
2. Use sin(k pi + t) = (-1)^k sin t.

Answer: (d) \( f \) is not continuous

Explanation:
1. \( f(0^-)=1 \) while \( f(0)=f(0^+)=0 \).
2. The function has a jump, so it is not continuous at 0.

Teacher's Note:
1. Compare left limit, right limit and the value.
2. A jump immediately rules out differentiability as well.

Answer: (b) differentiable at \( x=2 \)

Explanation:
1. At 1: \( f'(1^-)=1 \), \( f'(1^+)=-1 \) so not differentiable.
2. At 2: values agree (0 and 0), \( f'(2^-)=-1 \) and \( f'(2^+)=3-2(2)=-1 \), so differentiable.

Teacher's Note:
1. Check continuity first, then compare the slopes of adjacent pieces.
2. At a joining point, differentiability needs equal slopes.

Answer: (c) \( f \) is discontinuous at \( x=1 \)

Explanation:
1. \( f(1)=f(1^-)=\tfrac12 \) but \( f(1^+)=2-3+3=2 \).
2. The one-sided limits differ, so f is discontinuous at \( x=1 \).
3. The function is not even defined at 0 or 2 in the given pieces, so the first two options also fail.

Teacher's Note:
1. Evaluate the second formula at x = 1 and compare with the first.
2. A jump rules out continuity, and therefore differentiability.

Answer: (d) all of these

Explanation:
1. \( f'(x)=0 \) for \( x\lt 0 \) and \( 2x \) for \( x\ge0 \), with \( f'(0^-)=f'(0^+)=0 \).
2. So f is differentiable and f' is continuous, which also means f is continuous.

Teacher's Note:
1. Check the derivative from both sides of 0.
2. Options 1, 2, 3 are all true, so option 4 is correct.

Answer: (a) LHD of f at \( x=0 \) is 1

Explanation:
1. \( f'(0^-)=\lim_{h\to0^-}\dfrac{1}{1+2^{1/h}}=\dfrac{1}{1+0}=1 \).
2. \( f'(0^+)=\lim_{h\to0^+}\dfrac1{1+2^{1/h}}=0 \).
3. So LHD = 1, RHD = 0 and f is not differentiable.

Teacher's Note:
1. As h tends to 0 from the left, 1/h goes to minus infinity and 2^(1/h) goes to 0.
2. The derivative from the right goes the other way.

Answer: (a) f is continuous at \( x=1 \) and \( x=4 \)

Explanation:
1. At 1: left value \( \tfrac12-1+3+\tfrac12=3 \) equals \( |1-4|=3 \), so f is continuous at 1; at 4 the formula \( |x-4| \) is continuous.
2. \( f'(1^-)=\tfrac32-2+3=\tfrac52 \) and \( f'(1^+)=-1 \), so f is not differentiable at 1.
3. \( |x-4| \) has a corner at 4, so f is not differentiable there. Only option (a) holds.

Teacher's Note:
1. First check the values at the joining point, then compare the slopes.
2. A modulus corner can sit inside one piece, so check it separately.

Answer: (b) 2

Explanation:
1. Split: \( \dfrac12\left[\dfrac{f(3+h^2)-f(3)}{h^2}+\dfrac{f(3)-f(3-h^2)}{h^2}\right] \).
2. As \( h\to0 \), \( h^2\to0^+ \) and each fraction tends to \( f'(3) \).
3. The limit is \( \tfrac12(2+2)=2 \).

Teacher's Note:
1. Insert f(3) in the numerator and recognise two difference quotients.
2. Here the increment is h squared, so both fractions are one-sided derivative quotients.

Answer: (b) f is continuous but not differentiable

Explanation:
1. \( f'(0^+)=\lim_{h\to0^+}\dfrac{he^h}{h}=1 \) and \( f'(0^-)=\lim_{h\to0^-}\dfrac{-he^h}{h}=-1 \).
2. f is continuous but has a corner at 0.

Teacher's Note:
1. The factor e^x is smooth, so the corner comes from |x|.
2. Option (a) is true but weaker than option (b).

Answer: (c) \( (-\infty,\infty)-\{0\} \)

Explanation:
1. Near 0, \( 1-e^{-x^2}\approx x^2 \), so \( f(x)\approx|x| \) which has a corner at 0.
2. For \( x\ne0 \) the root is of a positive differentiable function, so f is differentiable.

Teacher's Note:
1. Expand e^(-x^2) for small x to see the behaviour at 0.
2. The square root of x squared is the modulus.

Answer: (a) 0

Explanation:
1. \( f'(0^+)=\lim_{x\to0^+}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\dfrac{\sqrt x}{x+2} \).
2. The limit is \( 0 \).

Teacher's Note:
1. Cancel x from numerator and denominator before taking the limit.
2. The domain is x >= 0, so only the right hand derivative makes sense.

Answer: (a) 1/2

Explanation:
1. \( f(g(x))=x \), so \( f'(g(x))\,g'(x)=1 \).
2. At x = a: \( f'(b)\cdot2=1 \), so \( f'(b)=\tfrac12 \).

Teacher's Note:
1. Differentiate the identity f(g(x)) = x using the chain rule.
2. Then put x = a so that g(a) is replaced by b.

Answer: (b) \( p+q=0 \); r is any real number

Explanation:
1. \( f'(0^+)=p+q \) and \( f'(0^-)=-p-q \) (the term \( r|x|^3 \) has zero derivative at 0).
2. Equality gives \( p+q=0 \) with r free.

Teacher's Note:
1. Compute each term's one-sided derivative at 0.
2. The cubic modulus is smooth enough not to add a corner.

Answer: (c) LHD of f at \( x=0 \) is 0 and RHD at \( x=1 \) is 0

Explanation:
1. \( 4\pi[x] \) is an integral multiple of \( \pi \), so \( \sin4\pi[x]=0 \) for all x.
2. Hence \( f(x)=0 \) for all x: f is the zero function, differentiable everywhere with derivative 0.

Teacher's Note:
1. Test the numerator for integer values of the greatest integer function.
2. A constant function has zero derivative everywhere, one-sided or not.