JEE Mathematics Continuity and Differentiability MCQs Set 14

Answer: (d) \( \frac{a+b}{ab} \)

Explanation:
1. \( \lim_{x\to0}\frac{\log(1+x/a)}{x}=\frac1a \) and \( \lim_{x\to0}\frac{-\log(1-x/b)}{x}=\frac1b \).
2. Adding, \( f(0)=\frac1a+\frac1b=\frac{a+b}{ab} \).

Teacher's Note:
1. The standard limit log(1 + u)/u = 1 gives each part.
2. Combine the two fractions.

Answer: (a) \( R-\{-1\} \)

Explanation:
1. For \( x\ne-1,-2 \): \( f(x)=\frac{x+2}{(x+1)(x+2)}=\frac1{x+1} \).
2. At \( x=-2 \): the limit is \( \frac1{-1}=-1=f(-2) \), so f is continuous.
3. At \( x=-1 \): \( \frac1{x+1} \) is unbounded, so f is discontinuous.
4. So f is continuous on \( R-\{-1\} \).

Teacher's Note:
1. Cancel the common factor (x + 2) before taking limits.
2. A pole cannot be removed by redefining the value.

Answer: (a) 3

Explanation:
1. Right side: \( f(0)=\frac{3}{1}=3 \).
2. Left limit: \( \lim_{x\to0^-}\frac{\sqrt{1+cx}-\sqrt{1-cx}}{x}=\lim\frac{2cx}{x(\sqrt{1+cx}+\sqrt{1-cx})}=\frac{2c}2=c \).
3. Continuity at 0 gives c = 3.

Teacher's Note:
1. Rationalise the numerator.
2. Compare the left limit with f(0).

Answer: (d) \( \frac2{15} \)

Explanation:
1. \( \frac{x\tan2x}{\sin3x\sin5x}=\frac{2x^2\cdot\frac{\tan2x}{2x}}{15x^2\cdot\frac{\sin3x}{3x}\cdot\frac{\sin5x}{5x}} \).
2. As \( x\to0 \) each ratio tends to 1, so the limit is \( \frac2{15} \).

Teacher's Note:
1. Normalise each factor with the standard limits.
2. Count the powers of x: numerator and denominator both give x^2.

Answer: (a) 1/2

Explanation:
1. Put \( t=2^x \): \( \frac{4t-16}{t^2-16}=\frac{4(t-4)}{(t-4)(t+4)}=\frac4{t+4} \).
2. As \( x\to2 \), \( t\to4 \) and the limit is \( \frac48=\frac12 \).

Teacher's Note:
1. Substitute t = 2^x to turn the expression into a rational function.
2. Cancel the common factor t - 4.

Answer: (b) 4

Explanation:
1. \( \frac{(e^x-1)^2}{\sin(x/a)\log(1+x/a)}\to\frac{x^2}{(x/a)(x/a)}=a^2 \).
2. So \( f(x)\to(a^2)^n \).
3. \( a^{2n}=16^n \) gives \( a^2=16 \) and (among the given options) \( a=4 \).

Teacher's Note:
1. Replace each factor by its first-order approximation near 0.
2. The nth power just carries through.

Answer: (b) 1

Explanation:
1. As \( n\to\infty \), \( \frac1{4n}\to0 \), so \( f(0)=\lim f\left(\frac1{4n}\right) \).
2. \( |\sin(e^n)e^{-n^2}|\le e^{-n^2}\to0 \) and \( \frac{n^2}{n^2+1}\to1 \).
3. So f(0) = 1.

Teacher's Note:
1. Continuity lets us take the limit along the sequence 1/(4n).
2. The sine term is bounded and multiplied by a vanishing factor.

Answer: (a) \( x\ge0 \)

Explanation:
1. \( \sqrt x \) is defined and continuous for \( x\ge0 \) and \( |x-1| \) is continuous for all x.
2. The difference is continuous wherever both are defined, i.e. for \( x\ge0 \).

Teacher's Note:
1. Sum and difference of continuous functions are continuous on the common domain.
2. Check where the square root exists.

Answer: (b) \( 24\sqrt2 \)

Explanation:
1. Numerator: \( 72^x-9^x-8^x+1=(8^x-1)(9^x-1)\approx(x\log8)(x\log9)=6x^2\log2\log3 \).
2. Denominator: \( \sqrt2-\sqrt{1+\cos x}=\frac{1-\cos x}{\sqrt2+\sqrt{1+\cos x}}\approx\frac{x^2/2}{2\sqrt2}=\frac{x^2}{4\sqrt2} \).
3. Limit \( =24\sqrt2\log2\log3 \), so \( K=24\sqrt2 \).

Teacher's Note:
1. Factorise the numerator as (8^x - 1)(9^x - 1).
2. Rationalise the denominator.

Answer: (a) \( a=\frac79,\ b=\frac13 \)

Explanation:
1. Expand \( f(kx)\approx f(0)+kxf'(0) \) in the numerator. The denominator is of order \( x^2 \), so the constant and the x terms must vanish.
2. Constant term: \( (2-3a+b)f(0)=0 \) so \( b=3a-2 \).
3. x term: \( (2-6a+8b)f'(0)=0 \) so \( 2-6a+8b=0 \).
4. Substituting: \( 2-6a+24a-16=0 \) gives \( a=\frac79,\ b=\frac13 \).

Teacher's Note:
1. A finite limit needs the numerator to vanish to second order.
2. Use the first-order Taylor expansion of f.

Answer: (a) \( x=-1,1 \) only

Explanation:
1. For \( |x|<1 \): \( x^{2n}\to0 \), so \( f(x)=\log(2+x) \). For \( |x|>1 \): \( x^{2n}\to\infty \), so \( f(x)=-\sin x \).
2. At \( x=1 \): left limit \( \log3 \), right limit \( -\sin1 \): they differ.
3. At \( x=-1 \): left limit \( -\sin(-1)=\sin1 \), right limit \( \log1=0 \): they differ.
4. So f is discontinuous at x = 1 and x = -1 only.

Teacher's Note:
1. Evaluate the limit separately for |x| < 1, |x| = 1 and |x| > 1.
2. Compare one-sided limits at the boundary points.

Answer: (c) f is continuous at x = 2 and x = 3

Explanation:
1. Near any point both rational and irrational numbers occur, so f can be continuous only where \( 5x=x^2+6 \).
2. \( x^2-5x+6=0 \) gives \( x=2,\ 3 \).
3. At these two points both branches agree with the limit, so f is continuous exactly at x = 2 and x = 3.

Teacher's Note:
1. The two branches must agree at the point.
2. Solve the resulting quadratic.

Answer: (a) \( a=\frac12,\ b=1 \)

Explanation:
1. For \( x\to9^- \): \( [x]=8 \), so \( \frac{x-8}{x-8}=1 \) and \( b=1 \).
2. For \( x\to9^+ \): \( x^2-15x+56=(x-7)(x-8)>0 \), so \( \frac{a(x-7)(x-8)}{x-8}=a(x-7)\to2a \).
3. \( 2a=1 \) gives \( a=\frac12 \).

Teacher's Note:
1. Factorise the quadratic as (x - 7)(x - 8).
2. Find the greatest integer on each side of 9.

Answer: (d) \( x^2-3x+2=0 \)

Explanation:
1. Continuity at \( x=-1 \): \( b-a=a+b-2 \) gives \( a=1 \).
2. Derivatives at \( x=-1 \): \( 2bx=2ax-b \) gives \( -2b=-2a-b \), so \( b=2a=2 \).
3. The roots 1 and 2 satisfy \( x^2-3x+2=0 \).

Teacher's Note:
1. Equate function values and derivatives at the joining point.
2. Sum 3 and product 2 give the quadratic.

Answer: (a) \( (-\infty,-1)\cup(3,\infty) \)

Explanation:
1. A function of this type is continuous only where the two branches are equal: \( x^2-ax+3=2-x \), i.e. \( x^2+(1-a)x+1=0 \).
2. Exactly two points require two distinct real roots: \( (1-a)^2-4>0 \).
3. \( (a-1)^2>4 \) gives \( a<-1 \) or \( a>3 \).

Teacher's Note:
1. Rational and irrational branches can match in a limit only where they are equal.
2. Use the discriminant condition for two distinct roots.