JEE Mathematics Continuity and Differentiability MCQs Set 19

Answer: (d) 3

Explanation:
1. On the interval \( 0\lt \cos x\le1 \), so \( 3+2\cos x\in(3,5] \).
2. The value jumps when it crosses 4 (\( \cos x=\tfrac12 \), \( x=\pm\tfrac\pi3 \)) and at the peak value 5 when \( x=0 \) (isolated).
3. 3 points: \( -\tfrac\pi3,\ 0,\ \tfrac\pi3 \).

Teacher's Note:
1. Find where 3 + 2 cos x is an integer.
2. The maximum value 5 occurs at only one point.

Answer: (c) A, B and D only

Explanation:
1. On \( (-1,1) \) the product \( x\sin\pi x \) lies in \( [0,0.6) \) (maximum about 0.57), so \( f=0 \) there: (A), (B), (D) true.
2. At 1: left values are small positive (floor 0), right values small negative (floor -1): a jump, so (C) is false.

Teacher's Note:
1. Bound x sin(pi x) on (-1, 1).
2. Look at the sign just right of 1.

Answer: (b) \( a=-\tfrac12,\ b=\tfrac32 \)

Explanation:
1. At \( x=1 \): continuity \( a+b=1 \); slope \( 2a=-1 \) (derivative of \( 1/x \) is \( -1 \)).
2. So \( a=-\tfrac12 \) and \( b=\tfrac32 \) (the same holds at \( x=-1 \) by symmetry).

Teacher's Note:
1. Match values and slopes at x = 1.
2. The function is even.

Answer: (d) -1

Explanation:
1. \( \dfrac{f(x^2)-f(x)}{f(x)-f(0)}=\dfrac{f(x^2)-f(0)}{f(x)-f(0)}-1 \).
2. \( \dfrac{f(x^2)-f(0)}{f(x)-f(0)}=\dfrac{x\cdot\frac{f(x^2)-f(0)}{x^2}}{\frac{f(x)-f(0)}{x}}\to\dfrac{0\cdot f'(0)}{f'(0)}=0 \).
3. The limit is \( 0-1=-1 \).

Teacher's Note:
1. Add and subtract f(0) in the numerator.
2. The first ratio tends to 0.

Answer: (b) 4

Explanation:
1. The numerator is \( (\cos x-1)(\cos x-e^x)-\tfrac{x^3}2 \).
2. \( (\cos x-1)(\cos x-e^x)=\left(-\tfrac{x^2}2+\cdots\right)\left(-x-\tfrac{x^2}2\cdots\right)=\tfrac{x^3}2+\tfrac{x^4}2+\cdots \).
3. After subtracting \( \tfrac{x^3}2 \) the leading term is \( \tfrac{x^4}2 \), so the limit is finite and non-zero for \( n=4 \).

Teacher's Note:
1. Expand cos x - 1 and cos x - e^x separately.
2. The cubic term cancels with the subtraction.

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

Explanation:
1. \( f'=6x+12\gt 0 \) on \( [-1,2] \): (A).
2. \( f(2^-)=35=f(2) \) and \( f(2^+)=35 \): (B).
3. Slopes at 2: 24 on the left and -1 on the right: (C).
4. f increases up to 35 at 2 and then decreases: (D). All four are true.

Teacher's Note:
1. Check the value and slopes at 2.
2. The right piece is decreasing.

Answer: (c) B and D only

Explanation:
1. (A) \( \ln|x|\to0 \), so the function tends to infinity: not removable.
2. (B) \( \dfrac{x+1}{x^2+x+1}\to\tfrac23 \): removable.
3. (C) the limits from left and right are 0 and 1: not removable.
4. (D) rationalising, \( \dfrac{1-x}{x(x-1)(\sqrt{x+1}+\sqrt{2x})}\to-\dfrac1{2\sqrt2} \): removable.

Teacher's Note:
1. Test whether the two-sided limit exists and is finite.
2. Cancel common factors where possible.

Answer: (d) A is false, but R is true.

Explanation:
1. Assertion: \( f(1)=2=3-1 \), but the right slope is \( -2 \) and the left slope is \( 3 \): not differentiable, so the assertion is false.
2. Reason: this is the correct definition of a (two-sided) derivative: true.

Teacher's Note:
1. Compare left and right slopes at 1.
2. The reason is simply the definition.

Answer: (b) is discontinuous

Explanation:
1. Each term is \( \dfrac1{(k-1)x+1}-\dfrac1{kx+1} \), so the partial sums telescope to \( 1-\dfrac1{nx+1} \).
2. For \( x\ne0 \) the sum is 1; at \( x=0 \) every term is 0, so \( f(0)=0 \).
3. The limit at 0 is 1, not equal to f(0): f is discontinuous at 0.

Teacher's Note:
1. Split each term into partial fractions.
2. Evaluate the sum at x = 0 separately.

Answer: (b) \( \pi \)

Explanation:
1. At \( x=0 \): \( y=\pi \).
2. First derivative: \( y'\cos x-y\sin x+\cos y-xy'\sin y=0 \) gives \( y'(0)+\cos\pi=0 \), so \( y'(0)=1 \).
3. Second derivative at 0 (all terms with \( \sin 0 \) or x vanish): \( y''-y\cos0=0 \), so \( y''(0)=\pi \).

Teacher's Note:
1. Differentiate implicitly twice.
2. Evaluate at x = 0 where y = pi.

Answer: (b) both \( f'(1) \) and \( f'(2) \) do not exist

Explanation:
1. On \( [0,1) \): \( f=0 \); on \( [1,2) \): \( f=x \). At 1: left slope 0, right slope 1.
2. On \( [2,3) \): \( f=2(x-1) \) with slope 2, while the slope on \( [1,2) \) is 1; f is continuous at 2 (value 2). So neither derivative exists.

Teacher's Note:
1. Write f on each unit interval.
2. Compare the slopes on both sides.

Answer: (d) B, C and D only

Explanation:
1. If \( 0\lt a\lt b \) the base tends to \( a/b\lt 1 \), so the limit is 0 (B). If \( a\gt b\gt 0 \) the base tends to \( a/b\gt 1 \), so f tends to infinity: no finite limit (C).
2. If \( a=b \): \( \left(1-\dfrac1{ax+2}\right)^x\to e^{-1/a} \) (D). (A) is false because of (C).

Teacher's Note:
1. Compare a/b with 1.
2. For a = b use the 1 to infinity form.

Answer: (a) 6

Explanation:
1. Let \( s=\sin x \). \( \ln\left(\sqrt{1+s^2}-s\right)=-\sinh^{-1}s=-s+\tfrac{s^3}6-\cdots \).
2. The numerator is \( \tfrac{s^3}6+\cdots \), so \( L=\tfrac16 \) and \( 6L+5=6 \).

Teacher's Note:
1. Recognise the logarithm as an inverse hyperbolic sine.
2. Expand to the cubic term.

Answer: (b) A and C only

Explanation:
1. Near 0: \( \log\cos x\approx-\tfrac{x^2}2 \) and \( \log(1+x^2)\approx x^2 \), so \( f(x)\approx-\dfrac x2 \).
2. Hence \( f(x)\to0=f(0) \) and \( f'(0)=-\tfrac12 \) exists.

Teacher's Note:
1. Use the leading terms of the logarithms.
2. f behaves like a linear function near 0.

Answer: (b) 5

Explanation:
1. \( F'(x)=f\!\left(\tfrac x2\right)f'\!\left(\tfrac x2\right)+g\!\left(\tfrac x2\right)g'\!\left(\tfrac x2\right) \).
2. \( g'=f''=-f \) and \( g=f' \), so \( F'(x)=fg-gf=0 \).
3. F is constant: \( F(10)=F(5)=5 \).

Teacher's Note:
1. Compute F' using f'' = -f.
2. A function with zero derivative is constant.

Answer: (c) B and D only

Explanation:
1. At \( x=2n \): left \( b_n+\cos2n\pi=b_n+1 \), right \( a_n+\sin2n\pi=a_n \), so \( a_n-b_n=1 \) (B).
2. At \( x=2n+1 \): left \( a_n+\sin(2n+1)\pi=a_n \), right \( b_{n+1}+\cos(2n+1)\pi=b_{n+1}-1 \), so \( a_n-b_{n+1}=-1 \), i.e. \( a_{n-1}-b_n=-1 \) (D).

Teacher's Note:
1. Impose continuity at the even and the odd integers.
2. Shift the index for the second relation.

Answer: (d) does not exist

Explanation:
1. For \( x\to0^+ \): \( \{x\}=x \) and \( \left(\dfrac{(1+x)^{1/x}}e\right)^{1/x}=\exp\dfrac{\ln(1+x)/x-1}{x}\to e^{-1/2} \).
2. For \( x\to0^- \): \( \{x\}\to1 \), the base tends to \( \dfrac2e\lt 1 \) and the exponent \( 1/x\to-\infty \), so the expression tends to infinity.
3. The two sides differ: the limit does not exist.

Teacher's Note:
1. Treat x tending to 0 from each side separately.
2. The fractional part tends to 1 from the left.

Answer: (d) A is false, but R is true.

Explanation:
1. Assertion: the one-sided derivatives at 0 are 1 and 0, which differ, so f is not differentiable at 0: false.
2. Reason: \( \dfrac{f(h)}h=\dfrac1{1+e^{1/h}} \) tends to 1 as \( h\to0^- \) and to 0 as \( h\to0^+ \), and the formula for \( f'(x) \) is the correct quotient rule result: true.

Teacher's Note:
1. Test the assertion at x = 0 using the one-sided derivatives.
2. A true reason can still contradict the assertion.

Answer: (c) \( \tfrac1{128r} \)

Explanation:
1. Half the base is \( \sqrt{r^2-(r-h)^2}=\sqrt{2hr-h^2} \), so \( A=h\sqrt{2hr-h^2} \).
2. For small h: \( A\approx h\sqrt{2hr} \) and \( P\approx4\sqrt{2hr} \).
3. \( \dfrac A{P^3}\approx\dfrac{h\sqrt{2hr}}{64(2hr)^{3/2}}=\dfrac h{64\cdot2hr}=\dfrac1{128r} \).

Teacher's Note:
1. Find the half base from the right triangle through the centre.
2. Keep only the leading terms as h tends to 0.

Answer: (b) \( -\tfrac12 \) and \( -1 \)

Explanation:
1. For small a: \( \sqrt[k]{1+a}-1\approx\dfrac ak \).
2. Divide the equation by a: \( \tfrac13x^2+\tfrac12x+\tfrac16=0 \), i.e. \( 2x^2+3x+1=0 \).
3. The roots are \( -\tfrac12 \) and \( -1 \).

Teacher's Note:
1. Replace each root by its first order expansion.
2. Divide by a and let a tend to 0.

Answer: (c) A, B and C only

Explanation:
1. (A) \( g(x)\to1 \) as \( x\to1 \) but \( g(1)=0 \): true. (B) the one-sided limits of \( [x] \) at 1 are 0 and 1: true.
2. (C) \( [x] \) is an integer, so \( g(f(x))=0 \) for all x: true.
3. (D) \( f(g(x))=[x^2] \) for non-integer x, which jumps near \( x=1 \) (0 on the left, 1 on the right): false.

Teacher's Note:
1. Evaluate g of an integer separately.
2. The composition f of g inherits a jump near 1.

Answer: (b) 0 and 1

Explanation:
1. For \( h\to0^+ \): \( \dfrac{f(h)}h=\dfrac1{1+e^{1/h}}\to0 \).
2. For \( h\to0^- \): \( e^{1/h}\to0 \) so the quotient tends to 1.

Teacher's Note:
1. Divide f(h) by h.
2. Examine the exponential on each side.

Answer: (c) \( n=2,\ m=2 \)

Explanation:
1. For \( x\lt 1 \): \( |x-1|=1-x \), so \( p=-1 \).
2. With \( t=x-1\to0^+ \): \( \log\cos^mt=m\ln\cos t\approx-\dfrac{mt^2}2 \), so \( g\approx-\dfrac{2t^{n-2}}m \).
3. The limit is \( -1 \) only if \( n=2 \) and \( m=2 \).

Teacher's Note:
1. Find p from the left side of the modulus.
2. Expand the logarithm of the cosine power.

Answer: (d) does not exist

Explanation:
1. Near \( a_m \): \( A_i=+1 \) for \( i\lt m \) and \( A_i=-1 \) for \( i\gt m \), giving a factor \( (-1)^{n-m} \).
2. \( A_m=-1 \) for \( x\lt a_m \) and \( +1 \) for \( x\gt a_m \).
3. The left limit is \( -(-1)^{n-m} \) and the right limit is \( (-1)^{n-m} \): they differ, so the limit does not exist.

Teacher's Note:
1. Find the sign of each factor on both sides of a_m.
2. The factor A_m flips sign.