JEE Mathematics Limits Continuity and Differentiability MCQs Set 06

Answer: (a) FTFT

Explanation:
1. S1: for \( x\to0^- \), \( [x]=-1 \), so the expression is \( -1/x\to+\infty \), a definite infinite limit and not an indeterminate form. So S1 is false.
2. S2: \( |\sin(3^x)|\le1 \) while \( 3^x\to\infty \), so the limit is 0. S2 is true.
3. S3: dividing by x, the quantity under the root tends to 1, so the limit exists and equals 1. S3 is false.
4. S4: \( (n+2)!+(n+1)!=(n+1)!(n+3) \) and \( (n+3)!=(n+1)!(n+2)(n+3) \), so the ratio is \( \frac1{n+2}\to0 \). S4 is true.
5. The pattern is FTFT.

Teacher's Note:
1. Check each statement separately. An infinite limit is not an indeterminate form, and a bounded function divided by a quantity tending to infinity tends to 0.

Answer: (b) 1

Explanation:
1. Near \( x=\pi/4\approx0.785 \) we have \( [x]=0 \), so the base is exactly 1.
2. The exponent \( \frac1{\ln\tan x}\) is large but \( 1^{\text{anything}}=1 \).
3. So the limit is 1.

Teacher's Note:
1. The base is exactly 1 near the point, so this is not a true 1 to the power infinity form; the value is simply 1.

Answer: (d) \( \frac x4 \)

Explanation:
1. Since \( k^3x-1<[k^3x]\le k^3x \), the sum lies between \( x\sum k^3-n \) and \( x\sum k^3 \).
2. \( \sum k^3=\frac{n^2(n+1)^2}{4} \), so dividing by \( n^4 \) gives \( \frac x4 \) in the limit, and the correction \( n/n^4\to0 \).
3. Hence the limit is \( \frac x4 \).

Teacher's Note:
1. Sandwich the greatest integer function between \( t-1 \) and \( t \), then use the sum of cubes.

Answer: (c) \( e^2 \)

Explanation:
1. The base is \( 1+\frac{2x-1}{x^2-4x+2} \), a \( 1^\infty \) form.
2. The limit is \( e^{L} \) with \( L=\lim x\cdot\frac{2x-1}{x^2-4x+2}=2 \).
3. So the value is \( e^2 \).

Teacher's Note:
1. For a \( 1^\infty \) form use \( e^{\lim g(x)(f(x)-1)} \).

Answer: (b) Continuous but not differentiable at \( x=0 \)

Explanation:
1. As \( x\to0^+ \), \( e^{1/x}\to\infty \) and \( f\approx x\cdot\frac{3e^{1/x}}{-e^{1/x}}=-3x\to0 \). As \( x\to0^- \), \( e^{1/x}\to0 \) and \( f\approx\frac{4x}{2}\to0 \). So \( f \) is continuous at 0.
2. \( \frac{f(h)}{h} \) tends to \(-3\) as \( h\to0^+ \) and to \( 2 \) as \( h\to0^- \).
3. The one-sided derivatives differ, so \( f \) is not differentiable at 0.

Teacher's Note:
1. Compare left and right derivatives; here \( e^{1/x} \) behaves very differently on the two sides.

Answer: (d) None of these

Explanation:
1. As \( x\to0^+ \), \( \{x\}=x \) and \( f\to0 \). As \( x\to0^- \), \( \{x\}=1+x \) and \( f\to 0+1+0=1 \). So \( f \) is discontinuous at 0, and (a), (b) are false.
2. At \( x=2 \): \( f(2)=2 \), but as \( x\to2^- \), \( \{x\}=x-1\to1 \) and \( f\to 2+1+2\sin1\ne2 \). So \( f \) is discontinuous at 2 and (c) is false.
3. Hence none of the given statements is correct.

Teacher's Note:
1. The fractional part function jumps from nearly 1 to 0 at every integer, so test both sides of each integer.

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

Explanation:
1. \( |f|^2=f^2 \), which is a product of differentiable functions, so it is differentiable everywhere.
2. \( |f| \) fails where \( f \) changes sign through zero (for example \( f=x \)).
3. \( f|f| \) has derivative \( 2|f|f' \), so it is differentiable.
4. \( f+|f|=2\max(f,0) \) fails to be differentiable for \( f=x \) at 0. So only (b) is always true.

Teacher's Note:
1. Test statements with the simple choice \( f(x)=x \) to find counterexamples.

Answer: (c) \( g(x) \) is differentiable everywhere except at \( x=1 \)

Explanation:
1. \( f \) increases up to \( x=\frac12 \) where it has maximum \( \frac14 \). So \( g(x)=x-x^2 \) for \( 0\le x\le\frac12 \) and \( g(x)=\frac14 \) for \( \frac12\le x\le1 \).
2. At \( x=\frac12 \) the left derivative \( 1-2x=0 \) equals the right derivative 0, so \( g \) is smooth there.
3. At \( x=1 \): \( g(1)=\frac14 \) but \( g(1^+)=\sin\pi=0 \), so \( g \) is discontinuous (hence not differentiable) at 1 only.
4. Hence g is differentiable everywhere except at \( x=1 \).

Teacher's Note:
1. The running maximum of a function stays constant after its peak; check the joining points.

Answer: (d) \( f'(0)=f'(1)-2 \)

Explanation:
1. Differentiate both sides with respect to y: \( 2f'(x+2y)=2f'(2y)+4x \).
2. Put \( y=0 \): \( f'(x)=f'(0)+2x \).
3. At \( x=1 \): \( f'(1)=f'(0)+2 \), i.e. \( f'(0)=f'(1)-2 \).

Teacher's Note:
1. Differentiating a functional equation with respect to one variable is a quick way to get a relation between derivatives.

Answer: (a) TTTF

Explanation:
1. S1: \( f=|x| \) except at \( x=\pm1 \) where \( \operatorname{sgn}=0 \) and \( f=0 \). So \( f \) fails at \( 0 \) (corner) and at \( \pm1 \) (jump). That is 3 points, so S1 is true.
2. S2: left limit \( a\sin\frac\pi2=a \) and \( \lim_{x\to0^+}\frac{\tan x-\sin x}{x^3}=\frac12 \). So \( a=\frac12 \) and S2 is true.
3. S3: \( \sqrt[3]{x^2|x|}=|x| \), which is differentiable everywhere except 0, so S3 is true.
4. S4: \( \sin^{-1}(\sin x) \) has corners at \( \frac\pi2,\frac{3\pi}2,\frac{5\pi}2 \) and is zero at \( \pi,2\pi \) where \( f \) is undefined. That gives 5 points, so S4 is false.

Teacher's Note:
1. For each statement, list the points of trouble: sign changes, corners and undefined points.

Answer: (b) Statement 1 is true, Statement 2 is true and Statement 2 is NOT a correct explanation for Statement 1

Explanation:
1. Statement 1: \( \sqrt{\frac{1-\cos2x}2}=|\sin x| \), so the expression is \( \frac{|\sin x|}x \), which tends to \( 1 \) from the right and \(-1\) from the left. The limit does not exist, so it is true.
2. Statement 2: put \( t=\sin x=1+\varepsilon \). The numerator is \( t-t^t=-\varepsilon^2+\dots \) and the denominator is \( -\frac{\varepsilon^2}2+\dots \), so the limit is 2. It is true.
3. The two statements are unrelated, so Statement 2 does not explain Statement 1.

Teacher's Note:
1. Expand around \( t=1 \) up to the second order terms for the second limit.

Answer: (c) Statement 1 is true, Statement 2 is false

Explanation:
1. Statement 1: divide by \( x^4 \); the limit is \( \frac23 \). It is true.
2. Statement 2 holds only when P and Q have the same degree. If the degrees differ, the limit is 0 or infinite, e.g. \( \frac{x^2}{x^3}\to0 \). So Statement 2 as stated is false.

Teacher's Note:
1. A general rule must hold for every case; one counterexample with unequal degrees is enough to disprove it.

Answer: (a) Statement 1 is true, Statement 2 is true and Statement 2 is a correct explanation for Statement 1

Explanation:
1. Each of the three parts is continuous throughout its own domain by Statement 2.
2. A sum of functions that are continuous on the common domain is continuous there, so Statement 1 follows directly from Statement 2.
3. Hence both are true and Statement 2 explains Statement 1.

Teacher's Note:
1. The sum of continuous functions is continuous on the common domain.

Answer: (a) \( a=3,\ b=12,\ c=9 \)

Explanation:
1. The denominator behaves like \( x^3 \), so the coefficients of \( x \) and \( x^2 \) in the numerator must vanish.
2. Coefficient of \( x \): \( a-b+c=0 \). Coefficient of \( x^2 \): \( a+\frac b2-c=0 \). These give \( b=4a \) and \( c=3a \).
3. Coefficient of \( x^3 \): \( \frac a2-\frac b3+\frac c2=\frac{2a}3=2 \), so \( a=3 \), \( b=12 \), \( c=9 \).

Teacher's Note:
1. Use series for \( e^x \), \( e^{-x} \) and \( \ln(1+x) \) up to \( x^3 \).

Answer: (a) \( \frac1{\sqrt2} \)

Explanation:
1. \( \sec(\tan^{-1}x)=\sqrt{1+x^2} \).
2. \( \frac{dy}{dx}=\frac x{\sqrt{1+x^2}} \).
3. At \( x=1 \) this is \( \frac1{\sqrt2} \).

Teacher's Note:
1. Convert the composite trigonometric expression to an algebraic one first.

Answer: (b) 0

Explanation:
1. For \( x\ne0 \), \( h(x)=|x|>0 \), so \( g(h(x))=|x|+1 \).
2. Since \( |x|+1\ge1 \), \( f(|x|+1)=|x|+1-1=|x| \).
3. So the limit as \( x\to0 \) is 0.

Teacher's Note:
1. Work from the inside out and decide which branch of each piecewise function applies near 0.