JEE Mathematics Continuity and Differentiability MCQs Set 15

Answer: (b) 3

Explanation:
1. For \( x\le0 \): \( 1-x\ge x^2+1 \) iff \( -1\le x\le0 \), so \( h=x^2+1 \) for \( x\lt -1 \) and \( 1-x \) on \( [-1,0] \).
2. For \( x\gt 0 \): \( h=x^2+1 \) on \( (0,1] \) and \( x+1 \) for \( x\gt 1 \).
3. Slopes: -2 and -1 at -1; -1 and 0 at 0; 2 and 1 at 1. So 3 points.

Teacher's Note:
1. Compare the two graphs on each side of 0.
2. Check the slopes at -1, 0 and 1.

Answer: (a) A and D only

Explanation:
1. \( \int_0^2(1+|1-t|)dt=2+1=3 \), so \( f(2^+)=3 \) while \( f(2)=5(2)+1=11 \).
2. Hence f is not continuous at 2: (A) true, (B) and (C) false.
3. The right difference quotient is \( \dfrac{f(x)-11}{x-2}\to-\infty \), so the right derivative does not exist: (D) true.

Teacher's Note:
1. Evaluate the integral at x = 2 and compare with 5x + 1.
2. A jump kills differentiability.

Answer: (b) B and D only

Explanation:
1. The sum is \( \dfrac1n\sum\dfrac{1}{(a+k/n)(b+k/n)}\to\int_0^1\dfrac{dx}{(x+a)(x+b)} \).
2. For \( a\ne b \): \( \dfrac1{b-a}\left[\ln\dfrac{x+a}{x+b}\right]_0^1=\dfrac1{a-b}\ln\dfrac{a(b+1)}{b(a+1)} \) so (B).
3. For \( a=b \): \( \int_0^1\dfrac{dx}{(x+a)^2}=\dfrac1a-\dfrac1{a+1}=\dfrac1{a(a+1)} \) so (D).

Teacher's Note:
1. Recognise a Riemann sum with step 1/n.
2. Use partial fractions for a different from b.

Answer: (c) x = 0 and x = 1

Explanation:
1. For \( -2\lt x\le0 \): \( f(|x|)=-x-1 \) and \( |f(x)|=1 \), so \( g=-x \).
2. For \( 0\lt x\le1 \): \( f(|x|)=x-1 \) and \( |f|=1-x \), so \( g=0 \). For \( 1\lt x\lt 2 \): \( g=2(x-1) \).
3. Slopes: -1 and 0 at 0; 0 and 2 at 1. Both are corners.

Teacher's Note:
1. Write g separately on each interval.
2. Compare slopes at the joins.

Answer: (b) A, C and D only

Explanation:
1. (A) By continuity f keeps the sign of f(c) near c: true.
2. (B) The limit is \( \int_0^1\ln(1+x)dx=2\ln2-1 \), not \( 1+2\ln2 \): false.
3. (C) A continuous non-negative function with zero integral vanishes identically: true.
4. (D) If f had no zero it would keep one sign on [a, b] and the integral could not be 0: true.

Teacher's Note:
1. Convert the sum to an integral for (B).
2. A continuous function with no zero cannot change sign.

Answer: (c) \( \sqrt3+\sqrt4+\sqrt6+\sqrt8 \)

Explanation:
1. Put \( t=\sqrt x \) and \( F(t)=f(t^2) \): \( F'(t)=2\sqrt x\,f'(x) \). The condition is \( \lim[\tan\tfrac\pi8F+F']=\cot\tfrac\pi{12} \).
2. As in the previous type, \( \lim F=\dfrac{\cot(\pi/12)}{\tan(\pi/8)}=\dfrac{2+\sqrt3}{\sqrt2-1} \).
3. Rationalising: \( (2+\sqrt3)(\sqrt2+1)=2\sqrt2+2+\sqrt6+\sqrt3=\sqrt8+\sqrt4+\sqrt6+\sqrt3 \).

Teacher's Note:
1. The factor 2 sqrt(x) f'(x) is the derivative with respect to sqrt(x).
2. Use cot 15 = 2 + sqrt 3 and tan 22.5 = sqrt 2 - 1.

Answer: (c) \( e \)

Explanation:
1. Use \( \sin^3\theta=\tfrac14(3\sin\theta-\sin3\theta) \): \( 3^{r-1}\sin^3\tfrac x{3^r}=\tfrac14\left(3^r\sin\tfrac x{3^r}-3^{r-1}\sin\tfrac x{3^{r-1}}\right) \).
2. The sum telescopes to \( \tfrac14\left(3^n\sin\tfrac x{3^n}-\sin x\right)\to\tfrac14(x-\sin x) \).
3. So \( g(x)=\sin x \) and \( \lim(1+\sin x)^{\cot x}=e^{\lim\cos x}=e \).

Teacher's Note:
1. Convert the cube of sine into a telescoping difference.
2. Then the limit is of the form 1 to infinity.

Answer: (d) \( \pm2 \)

Explanation:
1. Let \( m=5|n| \). Then \( f=2+[m\sin x] \). The greatest integer function jumps where \( m\sin x \) equals an integer.
2. For each integer value \( 1\le j\le m-1 \) there are two such points in \( (0,\pi) \), and the maximum \( m \) is reached at the single point \( x=\tfrac\pi2 \).
3. So the count is \( 2(m-1)+1=2m-1=10|n|-1 \). Setting this to 19 gives \( |n|=2 \).

Teacher's Note:
1. Count the points where m sin x crosses each integer value.
2. The peak at pi/2 adds one more point.

Answer: (a) 5

Explanation:
1. Write the sum as \( \sum_{r=1}^{2n}\dfrac1{2n+r}-\sum_{r=1}^{n}\dfrac1{2n+2r} \).
2. As integrals: \( \int_0^2\dfrac{dx}{2+x}-\dfrac12\int_0^1\dfrac{dx}{1+x}=\ln2-\dfrac12\ln2=\dfrac12\ln2 \).
3. With \( c=2 \), \( a/b=1/2 \) the least sum is \( 1+2+2=5 \) (taking \( c=4 \) would give a larger sum).

Teacher's Note:
1. Subtract the even denominators from all denominators.
2. Convert both sums to definite integrals.

Answer: (a) 0

Explanation:
1. Putting \( x=y \) gives \( f(0)=0 \). With \( y=0 \): \( f(x)=f(x)g(0) \), so \( g(0)=1 \).
2. Putting \( x=0 \) in the second relation: \( g(-y)=g(0)g(y)+f(0)f(y)=g(y) \), so g is even.
3. From \( g^2+f^2=1 \) and the existence of \( f'(0) \) (left and right derivatives agree), \( g'(0)=-f(0)f'(0)/g(0)=0 \).

Teacher's Note:
1. Evaluate at x = y and at y = 0 first.
2. An even differentiable function has zero derivative at 0.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For \( x\lt \tfrac\pi4 \): \( f=0 \), and \( f(\tfrac\pi4)=1 \). Then \( \dfrac{f(x)-f(\pi/4)}{x-\pi/4}=\dfrac{-1}{x-\pi/4}\to+\infty \).
2. So the left hand derivative is infinite (not finite), which means the derivative does not exist, and the function is also discontinuous there. Both statements are true and the reason accounts for the assertion.

Teacher's Note:
1. Compute the left difference quotient.
2. An infinite one-sided derivative means no derivative.

Answer: (b) continuous, and differentiable except at \( x=1 \)

Explanation:
1. \( f'(t)=3t^2-2t+1\gt 0 \) (discriminant \( 4-12\lt 0 \)), so f is increasing and \( g(x)=f(x) \) on \( [0,1] \).
2. \( g(1^-)=f(1)=2=3-1 \): continuous at 1.
3. \( g'(1^-)=f'(1)=2 \) while \( g'(1^+)=-1 \), so g is not differentiable at 1 only.

Teacher's Note:
1. Show that f is increasing to remove the maximum.
2. Compare the slopes at x = 1.

Answer: (d) \( 1+x^2 \)

Explanation:
1. Let \( g=f-1 \). The equation becomes \( g(x)g(y)=g(xy) \), with \( g(0)=0 \).
2. The polynomial solutions are \( g=x^n \), so \( f=1+x^n \) and \( f'(1)=n=2 \).
3. Hence \( f(x)=1+x^2 \) (one-one on \( [0,\infty) \)).

Teacher's Note:
1. Substitute g = f - 1 to get a multiplicative equation.
2. Use f'(1) to fix the power.

Answer: (b) 2

Explanation:
1. \( [x^3+1]=[x^3]+1 \) jumps when \( x^3 \) crosses an integer. For \( x\in(1,\tfrac32) \), \( x^3\in(1,3.375) \) so the jumps are at \( x^3=2 \) and \( x^3=3 \), i.e. \( x=2^{1/3} \) and \( 3^{1/3} \).
2. The base jumps there while the exponent is positive, so f is discontinuous, hence not differentiable, at exactly these two points.

Teacher's Note:
1. Find where the greatest integer part changes inside the interval.
2. A jump in the base gives a jump in f.

Answer: (a) 0

Explanation:
1. Since f is odd, \( f(-a+h)=-f(a-h) \) and \( f(-a)=-f(a) \), so \( f'(-a^-)=\lim_{h\to0^-}\dfrac{f(a-h)-f(a)}{-h}=f'(a^+) \).
2. By the symmetry \( f(x)=f(2a-x) \) about \( x=a \): \( f'(a^+)=-f'(a^-)=0 \).
3. Hence the left hand derivative at -a is 0.

Teacher's Note:
1. Use oddness to move the point from -a to a.
2. The second condition is a symmetry about x = a.

Answer: (b) 6

Explanation:
1. Odd powers of x multiplied by \( |x| \) give odd integrands, which integrate to 0 over \( [-1,1] \).
2. \( I_n=2\int_0^1x\left(1+\dfrac{x^2}2+\dfrac{x^4}4+\cdots+\dfrac{x^{2n}}{2n}\right)dx=1+\dfrac12\sum_{k=1}^n\left(\dfrac1k-\dfrac1{k+1}\right) \).
3. The limit is \( 1+\tfrac12=\tfrac32 \), so \( p=3,\ q=2 \) and \( pq=6 \).

Teacher's Note:
1. Discard the odd integrands first.
2. The remaining sum telescopes.

Answer: (c) \( 1+\tfrac12+\tfrac13+\cdots+\tfrac1k \)

Explanation:
1. \( \lim_{x\to0}\dfrac{f(x/m)}{x}=\dfrac1m\lim\dfrac{f(x/m)-f(0)}{x/m}=\dfrac{f'(0)}m=\dfrac1m \).
2. Adding for \( m=1,\dots,k \) gives \( 1+\tfrac12+\cdots+\tfrac1k \).

Teacher's Note:
1. Divide and multiply by m to form a difference quotient.
2. Use f(0) = 0.

Answer: (b) 4

Explanation:
1. Write \( f(x)=\dfrac{e^{sx}f(x)}{e^{sx}} \) with \( s=\sin\tfrac\pi{10} \) and apply L'Hospital's rule: \( \lim f=\lim\dfrac{s\,f+f'}{s}=\dfrac{\sec(\pi/5)}{\sin(\pi/10)} \).
2. \( \sec36^\circ=\dfrac4{\sqrt5+1} \) and \( \sin18^\circ=\dfrac{\sqrt5-1}4 \).
3. The ratio is \( \dfrac{16}{(\sqrt5+1)(\sqrt5-1)}=4 \).

Teacher's Note:
1. Multiply and divide by e raised to s x.
2. Use the exact values of sin 18 and cos 36 degrees.

Answer: (c) 3

Explanation:
1. With \( f=1+x^2 \): \( h=\min\{\tfrac2{1+x^2},x^2,|1-x|\} \).
2. The corners occur where two of the branches meet: \( x^2=1-x \) at \( x=\tfrac{\sqrt5-1}2 \), at \( x=1 \) where \( |1-x| \) vanishes, and where \( \tfrac2{1+x^2}=x-1 \) (about \( x=1.57 \)).
3. That gives exactly 3 points.

Teacher's Note:
1. Sketch the three curves and follow the lower envelope.
2. Count the points where the minimum switches branch.

Answer: (d) g is continuous on R but not differentiable at a and not differentiable at b

Explanation:
1. g is continuous at a and b (the integral over an interval of length 0 is 0, and g is constant outside).
2. \( g'(a^-)=0 \) but \( g'(a^+)=f(a)\ge1 \); also \( g'(b^-)=f(b)\ge1 \) but \( g'(b^+)=0 \).
3. So g is not differentiable at a and not differentiable at b.

Teacher's Note:
1. Use the fundamental theorem of calculus for the one-sided derivatives.
2. The range [1, infinity) keeps f away from zero.

Answer: (c) A, B, C and D

Explanation:
1. \( f'(x)=-f'(1-x) \) gives \( f'(\tfrac12)=0 \) (B) and \( f'(\tfrac34)=0 \). Since \( f'(\tfrac14)=f'(\tfrac12)=f'(\tfrac34)=0 \), Rolle's theorem gives two zeros of f'' in \( (\tfrac14,\tfrac34) \) (A).
2. \( f(\tfrac12+x)=f(\tfrac12-x) \) makes \( f(x+\tfrac12) \) even, so its product with the odd \( \sin x \) integrates to 0 (C).
3. In (D) put \( u=1-t \) on the right side: it becomes the left side.

Teacher's Note:
1. f(x) = f(1 - x) means the graph is symmetric about x = 1/2.
2. Use Rolle's theorem on f'.

Answer: (b) B and C only

Explanation:
1. \( f'(x)=\dfrac1x+\sqrt{1+\sin x} \) and \( \sqrt{1+\sin x}=\left|\sin\tfrac x2+\cos\tfrac x2\right| \) is continuous with a corner where \( \sin x=-1 \). So (B) is true and (A) is false.
2. \( f'(x)\le\tfrac1x+\sqrt2 \) is bounded while \( f(x)\to\infty \) (it grows linearly), so eventually \( f(x)\gt f'(x)\gt 0 \): (C) is true.
3. Since f is unbounded, (D) is false.

Teacher's Note:
1. Rewrite sqrt(1 + sin x) as a modulus to see the corners.
2. Compare the growth of f (unbounded) with the bounded derivative.

Answer: (b) 2

Explanation:
1. Put \( x=y=1 \): \( f(1)=0 \). Write \( f(x)=e^xg(x) \): the equation becomes \( g(xy)=g(x)+g(y) \).
2. The continuous (differentiable) solution is \( g(x)=k\ln x \), so \( f(x)=ke^x\ln x \); \( f'(1)=ke=e \) gives \( k=1 \).
3. \( f(e)=e^e \), \( \ln f(e)=e\approx2.718 \) and its greatest integer is 2.

Teacher's Note:
1. Look for the function g(x) = f(x) e^(-x).
2. The logarithmic Cauchy equation gives a logarithm.

Answer: (a) A, B and C

Explanation:
1. The integral is a constant c, so \( f'-f=c \) (B).
2. Differentiating, \( f''=f' \), so \( f''-f=f'-f=c \) (C).
3. Solving, \( f=Ae^x-c \); if \( f\to0 \) as \( x\to\infty \) then \( A=0 \), which forces \( c=0 \) and \( f\equiv0 \), so \( f'=f \) holds (A).

Teacher's Note:
1. The definite integral is just a number.
2. Solve the linear equation to test statement A.