JEE Mathematics Limits Continuity and Differentiability MCQs Set 05

Answer: (c) 1

Explanation:
1. \( x\sin\frac1x=\frac{\sin(1/x)}{1/x}\to1 \) as \( x\to\infty \).
2. \( \sin\frac1{x^2}\to\sin0=0 \).
3. So the limit is \( 1+0=1 \).

Teacher's Note:
1. Substitute \( t=1/x \) in the first term to use the standard limit \( \frac{\sin t}{t}\to1 \).

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

Explanation:
1. As \( x\to a^- \), \( \frac xa\to1^- \), so \( \left[\frac xa\right]=0 \).
2. Also \( |x|^3\to a^3 \), so the first term tends to \( \frac{a^3}{a}=a^2 \).
3. The limit is \( a^2-0=a^2 \).

Teacher's Note:
1. For a left-hand limit the greatest integer of a number just below 1 is 0.

Answer: (b) \( \ell=0 \), \( m \) is undefined

Explanation:
1. \( \sin\sqrt{x+1}-\sin\sqrt x=2\cos\frac{\sqrt{x+1}+\sqrt x}2\sin\frac{\sqrt{x+1}-\sqrt x}2 \). As \( x\to\infty \), \( \sqrt{x+1}-\sqrt x\to0 \) and the cosine is bounded, so \( \ell=0 \).
2. For \( x\to-\infty \), \( \sqrt x \) and \( \sqrt{x+1} \) are not real numbers, so the function is not defined and \( m \) is undefined.

Teacher's Note:
1. Always check the domain before evaluating a limit.

Answer: (c) \( \cot^{-1}\left(\left(\lim_{x\to0^-}f(x)\right)^2\right)=1 \)

Explanation:
1. For \( 02. For \( x\to0^- \), \( \{x\}\to1 \), so \( \lim_{x\to0^-}f=\sqrt{\cot1} \). This is not \( \cot1 \), so (b) is false.
3. The two one-sided limits differ, so \( \lim_{x\to0}f \) does not exist and (a) is false.
4. The square of the left limit is \( \cot1 \), and \( \cot^{-1}(\cot1)=1 \), so (c) is true.

Teacher's Note:
1. Work out both one-sided limits separately, using the value of \([x]\) and \(\{x\}\) near 0.

Answer: (c) \( 0 \) for \( x\ge0 \) and \( 2x \) for \( x<0 \)

Explanation:
1. For \( x>0 \), \( \frac x{t^2}\to+\infty \) and \( \cot^{-1}\to0 \), so \( f=0 \).
2. For \( x<0 \), \( \frac x{t^2}\to-\infty \) and \( \cot^{-1}\to\pi \), so \( f=\frac{2x}\pi\cdot\pi=2x \).
3. For \( x=0 \), \( f=0 \). So \( f=0 \) for \( x\ge0 \) and \( 2x \) for \( x<0 \).

Teacher's Note:
1. The range of the principal \( \cot^{-1} \) is \( (0,\pi) \), so it tends to \( \pi \) as its argument tends to \( -\infty \).

Answer: (a) 1

Explanation:
1. Let the limit be \( L \). Then \( L+\frac{3L-1}{L^2}=3 \), so \( L^3+3L-1=3L^2 \).
2. This gives \( L^3-3L^2+3L-1=0 \), i.e. \( (L-1)^3=0 \).
3. Hence \( L=1 \).

Teacher's Note:
1. Recognise the expansion of \( (L-1)^3 \) after clearing the denominator.

Answer: (b) \( \ln\frac23 \)

Explanation:
1. Write the terms as \( 2^u \) and \( 3^u \) with \( u=\frac{x^n}{e^x}\to0 \).
2. For small u, \( 2^u-3^u\approx u(\ln2-\ln3) \).
3. The expression becomes \( \frac{e^x\cdot\frac{x^n}{e^x}\ln\frac23}{x^n}=\ln\frac23 \).

Teacher's Note:
1. Since \( x^n/e^x\to0 \), use \( a^u\approx1+u\ln a \).

Answer: (b) is discontinuous at \( x=1 \), since \( f(1^+)=+\infty \) and \( f(1^-)=-\infty \)

Explanation:
1. \( f(x)=\frac{\ln(x^2-2x+5)}{\ln(4x-3)} \). At \( x=1 \) the numerator tends to \( \ln4>0 \) and the denominator tends to 0.
2. As \( x\to1^+ \), \( 4x-3>1 \) so \( \ln(4x-3)\to0^+ \) and \( f\to+\infty \).
3. As \( x\to1^- \), \( 4x-3<1 \) so \( \ln(4x-3)\to0^- \) and \( f\to-\infty \).
4. Neither one-sided limit is finite, so \( f \) is discontinuous at 1.

Teacher's Note:
1. Change the base of the logarithm and watch the sign of \( \ln(4x-3) \) on either side of 1.

Answer: (c) (i), (ii), (iii) and (iv)

Explanation:
1. Let \( x=-\pi+h \). Then \( \sin x=-\sin h \). For \( h\to0^+ \): \( f=\frac h{-\sin h}\to-1 \), so (i) is true.
2. For \( h\to0^- \): \( |h|=-h \) so \( f=\frac{-h}{-\sin h}\to1 \), so (ii) is true.
3. The one-sided limits are different, so (iii) is true.
4. At \( x=\pi \), the numerator tends to \( 2\pi \) while \( \sin x \) tends to 0 with opposite signs on the two sides, so \( f\to+\infty \) on one side and \(-\infty\) on the other. The limit does not exist and (iv) is true.

Teacher's Note:
1. Shift the variable to \( h=x+\pi \) and use \( \sin(-\pi+h)=-\sin h \).

Answer: (a) \( a=2,\ L=\frac1{64} \)

Explanation:
1. \( \sqrt{a^2-x^2}=a-\frac{x^2}{2a}-\frac{x^4}{8a^3}-\dots \), so the numerator is \( \left(\frac1{2a}-\frac14\right)x^2+\frac{x^4}{8a^3}+\dots \).
2. For a finite limit the \( x^2 \) term must vanish, so \( a=2 \).
3. Then \( L=\frac1{8a^3}=\frac1{64} \).

Teacher's Note:
1. Expand the square root by the binomial series and cancel the lowest power.

Answer: (d) \( \pm\frac\pi2 \)

Explanation:
1. The limit is \( e^{\ln(1+b^2)}=1+b^2 \).
2. So \( \sin^2\theta=\frac{1+b^2}{2b}\ge1 \) by AM-GM, since \( 1+b^2\ge2b \).
3. As \( \sin^2\theta\le1 \), equality holds: \( b=1 \) and \( \sin^2\theta=1 \), giving \( \theta=\pm\frac\pi2 \).

Teacher's Note:
1. Use \( 1+b^2\ge2b \) to see that the value is forced to the extreme.

Answer: (b) \( a=1,\ b=-4 \)

Explanation:
1. \( \frac{x^2+x+1}{x+1}=x+\frac1{x+1} \).
2. The expression becomes \( (1-a)x-b+\frac1{x+1} \).
3. For a finite limit \( a=1 \), and then the limit is \( -b=4 \), so \( b=-4 \).

Teacher's Note:
1. Divide the numerator by the denominator before taking the limit.

Answer: (b) \(-\frac12\) and \(-1\)

Explanation:
1. For small a, \( (1+a)^{1/n}-1\approx\frac an \).
2. Dividing the equation by a gives \( \frac{x^2}3+\frac x2+\frac16\to0 \), i.e. \( 2x^2+3x+1=0 \).
3. The roots are \( -\frac12 \) and \(-1\).

Teacher's Note:
1. Divide the whole equation by a and let a tend to 0 using the binomial approximation.

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

Explanation:
1. At \( x=0 \) the function is \( (x-1)\sin\frac1{x-1} \), a product of differentiable functions near 0, so it is differentiable there.
2. At \( x=1 \): \( \frac{f(1+h)-f(1)}h=\sin\frac1h \), which oscillates between \(-1\) and 1 and has no limit.
3. So f is not differentiable at 1.

Teacher's Note:
1. Points other than the joining point behave normally; test the joining point by the difference quotient.

Answer: (d) 1

Explanation:
1. Since f is positive and increasing, \( f(x)\le f(2x)\le f(3x) \) for \( x>0 \).
2. Dividing by \( f(x) \): \( 1\le\frac{f(2x)}{f(x)}\le\frac{f(3x)}{f(x)} \).
3. The right side tends to 1, so by the sandwich theorem the limit is 1.

Teacher's Note:
1. Use the monotonic property to sandwich \( f(2x) \) between \( f(x) \) and \( f(3x) \).

Answer: (d) 3

Explanation:
1. The denominator tends to 0, so for the limit to be 0 the numerator must also tend to 0.
2. So \( L^2-9=0 \) where \( L=\lim f(x) \), giving \( L=\pm3 \).
3. Since \( f(x)\ge0 \), \( L=3 \).

Teacher's Note:
1. A fraction with denominator tending to 0 can have a finite limit only if the numerator tends to 0.

Answer: (d) \( \frac{a^2}2(\alpha-\beta)^2 \)

Explanation:
1. \( ax^2+bx+c=a(x-\alpha)(x-\beta) \), and \( 1-\cos u\approx\frac{u^2}2 \) as \( u\to0 \).
2. The fraction becomes \( \frac{a^2(x-\alpha)^2(x-\beta)^2}{2(x-\alpha)^2}=\frac{a^2(x-\beta)^2}2 \).
3. As \( x\to\alpha \) this is \( \frac{a^2}2(\alpha-\beta)^2 \).

Teacher's Note:
1. Factorise the quadratic and use \( 1-\cos u\approx u^2/2 \).

Answer: (b) \( (-\infty,\infty) \)

Explanation:
1. For \( x\ge0 \), \( f=\frac x{1+x} \) with derivative \( \frac1{(1+x)^2} \), equal to 1 at 0.
2. For \( x<0 \), \( f=\frac x{1-x} \) with derivative \( \frac1{(1-x)^2} \), also 1 at 0.
3. The one-sided derivatives agree and \( 1+|x|\ne0 \), so f is differentiable everywhere.

Teacher's Note:
1. Differentiate each branch separately and compare the derivatives at 0.

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

Explanation:
1. For \( 22. f is continuous on \([2,5]\), differentiable in \( (2,5) \) and \( f(2)=f(5)=3 \). Statement 2 is true.
3. These are the conditions of Rolle's theorem, which only guarantees that \( f'(c)=0 \) for some c in \( (2,5) \), not specifically at 4. So Statement 2 does not explain Statement 1.

Teacher's Note:
1. Rolle's theorem gives the existence of some point; here the real reason is that f is constant on the interval.

Answer: (d) 2

Explanation:
1. \( 1-\cos2x=2\sin^2x\approx2x^2 \), \( 3+\cos x\to4 \) and \( x\tan4x\approx4x^2 \).
2. The limit is \( \frac{2x^2\cdot4}{4x^2}=2 \).

Teacher's Note:
1. Replace each factor by its leading term as x tends to 0.

Answer: (b) 1

Explanation:
1. At \( x=1 \): left of 1, \( [x]=0 \) and \( [x^2]=0 \), so \( f=0 \); \( f(1)=1-1=0 \); right of 1, \( [x]=1 \) and \( [x^2]=1 \), so \( f=0 \). So f is continuous at 1.
2. At \( x=0 \): the left limit is \( 1-0=1 \) but \( f(0)=0 \). So f is discontinuous at 0.
3. At an integer \( n\ge2 \) the left limit is \( (n-1)^2-(n^2-1)=2-2n\ne0=f(n) \).
4. At an integer \( n\le-1 \) the left limit is \( (n-1)^2-n^2=1-2n\ne0=f(n) \). So the only integral point of continuity is \( x=1 \).

Teacher's Note:
1. Compare left limit, value and right limit at each integer; they agree only at 1.

Answer: (a) 0

Explanation:
1. \( 1,x^2 \) and \( x^3 \) are continuous everywhere.
2. The minimum of finitely many continuous functions is continuous.
3. So the number of points of discontinuity is 0.

Teacher's Note:
1. Min and max of continuous functions are continuous (they can fail only to be differentiable).

Answer: (c) The empty set

Explanation:
1. For \( x>0 \), \( f=\frac{\sin x}{1+x} \) and \( f'(0^+)=\frac{\cos0\cdot1-0}{1}=1 \).
2. For \( x<0 \), \( f=\frac{\sin x}{1-x} \) and \( f'(0^-)=1 \) as well.
3. The one-sided derivatives agree and the denominator never vanishes, so f is differentiable everywhere and the set is empty.

Teacher's Note:
1. The only place that could cause trouble is \( x=0 \), where \( |x| \) has a corner; check both one-sided derivatives.

Answer: (a) 0

Explanation:
1. \( \cos\left(\frac{2x-1}2\pi\right)=\sin\pi x \), so \( f(x)=[x]\sin\pi x \).
2. Where \( [x] \) jumps (at integers), \( \sin\pi x=0 \), so there is no jump in f.
3. f is continuous everywhere, so the number of discontinuities is 0.

Teacher's Note:
1. A bounded jump multiplied by a function that is zero at the jump point gives no discontinuity.

Answer: (b) 1

Explanation:
1. Divide the numerator and denominator by x.
2. \( \frac{\sin^2x}x\to0 \) and \( \frac{\cos^2x}x\to0 \) since the sine and cosine are bounded.
3. The fraction tends to 1, and so does its square root.

Teacher's Note:
1. Bounded terms are negligible compared with x as x tends to infinity.