JEE Mathematics Limits Continuity and Differentiability MCQs Set 07

Answer: (a) F, T, F, T

Explanation:
1. S1: for small negative x, [x] = -1, so the quotient is -1/x, which tends to +infinity: not an indeterminate form (false).
2. S2: the numerator is bounded and the denominator tends to infinity, so the limit is 0 (true).
3. S3: dividing by x, the quantity tends to \( \sqrt{1}=1 \), so the limit exists (false).
4. S4: \( \frac{(n+1)!\,(n+3)}{(n+3)(n+2)(n+1)!}=\frac1{n+2}\to0 \) (true).

Teacher's Note:
1. Evaluate each statement separately.
2. An indeterminate form needs both parts to tend to 0 or infinity in a way that does not settle the limit.

Answer: (b) \( -1 \)

Explanation:
1. As \( x\to1^+ \): \( [x-1]=0 \), \( [1-x]=-1 \), and \( 1-x\to0 \), so the value tends to -1.
2. As \( x\to1^- \): \( [x-1]=-1 \), \( [1-x]=0 \), giving again -1.
3. Both one-sided limits are -1.

Teacher's Note:
1. Evaluate the greatest integer parts on each side of 1.
2. Check both left and right limits.

Answer: (a) \( -1 \)

Explanation:
1. Near 0, \( \cos^{-1}(\cos x)=|x| \) and \( \sin^{-1}(\sin x)=x \).
2. For x < 0 the ratio is \( |x|/x=-1 \).

Teacher's Note:
1. The inverse trigonometric functions return principal values.
2. For x < 0, |x| = -x.

Answer: (c) \( 9 \)

Explanation:
1. As \( x\to3 \): \( \ln(x-2)\sim x-3 \) and \( x^3+27\to54 \).
2. The limit is \( \frac{54(x-3)}{(x-3)(x+3)}=\frac{54}{6}=9 \).

Teacher's Note:
1. Use ln(1 + u) ~ u with u = x - 3.
2. Cancel the common factor x - 3.

Answer: (c) \( 3p(\ln4)^3 \)

Explanation:
1. Near 0: \( 4^x-1\sim x\ln4 \), \( \sin(x/p)\sim x/p \), \( \ln(1+x^2/3)\sim x^2/3 \).
2. So the limit is \( \frac{x^3(\ln4)^3}{(x/p)(x^2/3)}=3p(\ln4)^3 \).

Teacher's Note:
1. Replace each factor by its leading term.
2. The powers of x cancel.

Answer: (d) \( 1 \)

Explanation:
1. Put \( h=x-2 \): \( \sin(e^h-1)\sim h \) and \( \ln(1+h)\sim h \).
2. The ratio tends to 1.

Teacher's Note:
1. Both numerator and denominator behave like h.
2. Use standard limits.

Answer: (b) \( 1 \)

Explanation:
1. Near 0, \( \ln(1+x)\sim x \) so \( \sin(\ln(1+x))\sim x \).
2. Also \( \ln(1+\sin x)\sim\sin x\sim x \).
3. The limit is 1.

Teacher's Note:
1. Use sin u ~ u and ln(1 + u) ~ u for small u.
2. Both parts are asymptotic to x.

Answer: (d) does not exist because the left-hand limit is not equal to the right-hand limit

Explanation:
1. Since \( 1-\cos2t=2\sin^2t \), the expression is \( \frac{\sqrt2|\sin(x-1)|}{x-1} \).
2. For x > 1 it tends to \( \sqrt2 \) and for x < 1 it tends to \( -\sqrt2 \).

Teacher's Note:
1. Use 1 - cos 2t = 2 sin^2 t.
2. Remember sqrt(sin^2 t) = |sin t|.

Answer: (d) \( \frac12 \)

Explanation:
1. Expand: \( (1+x^2)^{1/3}\approx1+\frac{x^2}3 \) and \( (1-2x)^{1/4}\approx1-\frac x2 \).
2. The numerator is about \( \frac x2 \) and the denominator about x.
3. The limit is \( \frac12 \).

Teacher's Note:
1. Use binomial expansions to the first order.
2. The leading term of the numerator is x/2.

Answer: (b) \( \frac{\sqrt2}{8} \)

Explanation:
1. Rationalise: \( \frac{2-(1+\cos x)}{\sin^2x(\sqrt2+\sqrt{1+\cos x})}=\frac{1-\cos x}{\sin^2x(\sqrt2+\sqrt{1+\cos x})} \).
2. Since \( \frac{1-\cos x}{\sin^2x}=\frac1{1+\cos x}\to\frac12 \), the limit is \( \frac12\cdot\frac1{2\sqrt2}=\frac{\sqrt2}8 \).

Teacher's Note:
1. Multiply numerator and denominator by the conjugate.
2. sin^2 x = (1 - cos x)(1 + cos x).

Answer: (a) \( \sqrt2 \)

Explanation:
1. Put \( \theta=\cos^{-1}(1-x) \), so \( x=1-\cos\theta=2\sin^2\frac\theta2 \) and \( \sqrt x=\sqrt2\sin\frac\theta2 \).
2. The ratio is \( \frac{\theta}{\sqrt2\sin(\theta/2)}\to\frac{\theta}{\sqrt2\,\theta/2}=\sqrt2 \).

Teacher's Note:
1. Substitute theta = arccos(1 - x).
2. Use sin(theta/2) ~ theta/2.

Answer: (b) \( 5050 \)

Explanation:
1. Write the numerator as \( \sum_{k=1}^{100}(x^k-1) \).
2. Each \( \frac{x^k-1}{x-1}\to k \).
3. So the limit is \( 1+2+\dots+100=5050 \).

Teacher's Note:
1. Subtract 1 from each of the 100 terms.
2. Use (x^k - 1)/(x - 1) -> k.

Answer: (b) \( 1 \)

Explanation:
1. Since \( x^3\sin\frac1x\approx x^3\cdot\frac1x=x^2 \) for large x, the numerator is about \( x^2+x+1 \).
2. So the ratio tends to 1.

Teacher's Note:
1. For large x, sin(1/x) ~ 1/x.
2. Compare leading powers.

Answer: (d) \( -\frac13 \)

Explanation:
1. For large |x|, \( x^2\sin\frac1x\approx x \) and \( \sqrt{9x^2+x+1}\approx3|x|=-3x \) (x < 0).
2. The ratio tends to \( \frac{x}{-3x}=-\frac13 \).

Teacher's Note:
1. sqrt(x^2) = |x| = -x when x is negative.
2. sin(1/x) ~ 1/x.

Answer: (b) \( 0 \)

Explanation:
1. The denominator is dominated by \( 3^{2n+3}=27\cdot9^n \).
2. The numerator is dominated by \( 5^{n+1} \) (since \( 4^n<5^n \)), which is much smaller than \( 9^n \).
3. So the limit is 0.

Teacher's Note:
1. Compare the fastest-growing terms top and bottom.
2. 9^n grows faster than 5^n.

Answer: (b) \( -2 \)

Explanation:
1. Put \( x=\frac\pi2+t \): \( \frac{t}{\cos(\pi/2+t)}=-\frac{t}{\sin t} \).
2. For small t (either sign), \( \frac{t}{\sin t}>1 \), so the quotient is slightly less than -1, say -1.0001.
3. Its integer part is -2 on both sides.

Teacher's Note:
1. The quotient tends to -1 but from below.
2. The greatest integer of a number just below -1 is -2.

Answer: (c) \( -\frac34 \)

Explanation:
1. Divide numerator and denominator by n.
2. The terms \( \frac{(-1)^n}{n} \) tend to 0 in both cases.
3. So the limit is \( \frac{-3}{4} \).

Teacher's Note:
1. The alternating terms are bounded by 1.
2. Divide by n and let n grow.

Answer: (d) \( \frac12 \)

Explanation:
1. Combine: \( \frac{2-(1+x)}{1-x^2}=\frac{1-x}{(1-x)(1+x)}=\frac1{1+x} \).
2. As x tends to 1 this is \( \frac12 \).

Teacher's Note:
1. Take the common denominator.
2. Cancel the factor 1 - x.

Answer: (b) \( \frac12 \)

Explanation:
1. Put \( h=1/x \): \( \frac{h-\ln(1+h)}{h^2} \).
2. Expand \( \ln(1+h)=h-\frac{h^2}2+\dots \) so the ratio tends to \( \frac12 \).

Teacher's Note:
1. Substitute h = 1/x and use the series of ln(1 + h).
2. The linear terms cancel.

Answer: (d) \( \frac1{12} \)

Explanation:
1. Expand: \( e^{-x^2/2}=1-\frac{x^2}2+\frac{x^4}8-\dots \) and \( \cos x=1-\frac{x^2}2+\frac{x^4}{24}-\dots \).
2. The difference is \( x^4\left(\frac18-\frac1{24}\right)=\frac{x^4}{12} \).
3. Dividing by \( x^3\sin x\sim x^4 \) gives \( \frac1{12} \).

Teacher's Note:
1. Use Maclaurin series up to x^4.
2. x^3 sin x ~ x^4.

Answer: (c) \( -12 \)

Explanation:
1. The numerator is \( \sim6x^2 \).
2. With \( u=2x^2-x \): \( \ln\cos u\approx-\frac{u^2}2\approx-\frac{x^2}{2} \) (since \( u\approx-x \)).
3. The ratio tends to \( \frac{6x^2}{-x^2/2}=-12 \).

Teacher's Note:
1. ln cos u ~ -u^2/2 for small u.
2. u is about -x near 0.

Answer: (a) \( -\cos a \)

Explanation:
1. The numerator is the third forward difference of sin at a, which behaves like \( h^3\,f'''(a) \).
2. With \( f=\sin \): \( f'''(a)=-\cos a \).

Teacher's Note:
1. Third forward difference ~ h^3 f'''(a).
2. The third derivative of sin is -cos.