JEE Mathematics Continuity and Differentiability MCQs Set 17

Answer: (d) is discontinuous at \( x=1 \) since neither \( f(1^+) \) nor \( f(1^-) \) exists

Explanation:
1. \( f(x)=\dfrac{\ln(x^2-2x+5)}{\ln(4x-3)} \). As \( x\to1 \) the numerator tends to \( \ln4 \) and the denominator to 0.
2. For \( x\to1^+ \), \( 4x-3\gt 1 \) so the denominator is a small positive number and \( f\to+\infty \). For \( x\to1^- \) the denominator is a small negative number and \( f\to-\infty \).
3. Neither one-sided limit is finite, so f is discontinuous at 1.

Teacher's Note:
1. Change the base of the logarithm.
2. The base 4x - 3 passes through 1 at x = 1.

Answer: (c) only right continuous

Explanation:
1. Right: \( \{x\}=h \): \( \cos^{-1}(1-h^2)\approx\sqrt2h \), \( \sin^{-1}(1-h)\to\tfrac\pi2 \), denominator \( \approx\sqrt2h \): limit \( \tfrac\pi2=f(0) \).
2. Left: \( \{x\}=1-h \): the numerator is \( \approx\tfrac\pi2\cdot h \) and the denominator \( \approx2\sqrt2h \), so the limit is \( \tfrac{\pi}{4\sqrt2}\ne f(0) \).
3. So f is only right continuous.

Teacher's Note:
1. Fractional part is x on the right of 0 and 1 + x on the left.
2. Compute both one-sided limits.

Answer: (c) \( [-2,3]-\{0\} \)

Explanation:
1. On \( [-2,-1] \): \( g=t^2+2t \); on \( [-1,0) \): \( g=-1 \) (slope 0 on both sides of \( -1 \)).
2. At 0: \( g(0)=f(0)=0 \) but \( g(0^-)=-1 \): a jump. On \( [0,1] \): \( g=x^2-2x \) and on \( [1,3] \): \( g=-1 \), slope 0 on both sides of 1.
3. So g fails to be differentiable only at 0.

Teacher's Note:
1. Find g on each interval from the graph of f.
2. Check the junctions at -1, 0 and 1.

Answer: (b) 5

Explanation:
1. \( \sin^{-1}x\in[-\tfrac\pi2,\tfrac\pi2] \) takes the integer values -1, 0, 1 at \( x=-\sin1,0,\sin1 \): the greatest integer jumps at these three points, so \( a=3 \).
2. \( \dfrac2{1+x^2} \) for \( x\ge0 \) lies in \( (0,2] \): it equals 1 at \( x=1 \) (a jump) and equals 2 only at \( x=0 \), where the value 2 differs from the right limit 1 (not differentiable): \( b=2 \).
3. \( a+b=5 \).

Teacher's Note:
1. The greatest integer function jumps where its argument crosses an integer.
2. Count crossings in the given domains.

Answer: (c) 3

Explanation:
1. On \( [-2,0) \): \( g\le0 \). On \( [0,2] \): \( g=0 \). On \( (2,3] \): \( g=x^2-2x \), which increases to \( 9-6=3 \) at \( x=3 \).
2. The maximum value is 3.

Teacher's Note:
1. The running maximum is non-decreasing on [0, 3].
2. Evaluate it at the right end.

Answer: (c) \( \tfrac32 \)

Explanation:
1. Split: \( \dfrac{e^x(e^{\tan x-x}-1)}{\tan x-x}\to1 \).
2. \( \ln(\sec x+\tan x)-x\approx\dfrac{x^3}6 \) (derivative \( \sec x-1 \) integrates to this) and \( \tan x-x\approx\dfrac{x^3}3 \), so this part tends to \( \tfrac12 \).
3. Total \( 1+\tfrac12=\tfrac32 \).

Teacher's Note:
1. Separate the exponential part from the logarithmic part.
2. Both leading terms are of order x cubed.

Answer: (d) \( \left(-\infty,\tfrac38\right] \)

Explanation:
1. \( f(x)=3x \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \); \( g(x)=x/3 \) for \( x\ge0 \) and \( x \) for \( x\lt 0 \). So \( h(x)=x \) for all x.
2. \( f-g=\tfrac{8x}3 \) for \( x\ge0 \) and \( 0 \) for \( x\lt 0 \).
3. We need \( -1\le f-g\le1 \): \( \tfrac{8x}3\le1 \) gives \( x\le\tfrac38 \); all negative x are fine.

Teacher's Note:
1. Compute f(x) - g(x) on each side of 0.
2. The argument of arcsine must lie in [-1, 1].

Answer: (b) 2

Explanation:
1. Put \( t=x-1 \): numerator \( -at+\sin t\approx(1-a)t \), denominator \( t+\sin t\approx2t \), so the base tends to \( \tfrac{1-a}2 \).
2. The exponent \( \dfrac{1-x}{1-\sqrt x}=1+\sqrt x\to2 \).
3. \( \left(\dfrac{1-a}2\right)^2=\dfrac14 \) gives \( a=0 \) or \( a=2 \); the largest is 2.

Teacher's Note:
1. Evaluate the base and the exponent separately.
2. Solve the resulting quadratic equation in a.

Answer: (c) 6

Explanation:
1. \( (f-g)''=0 \), so \( f-g \) is linear: \( f(x)-g(x)=mx+c \) with \( m=f'(1)-g'(1)=-2 \).
2. At 2: \( f(2)-g(2)=3-9=-6=-4+c \), so \( c=-2 \).
3. Then \( g(2)-f(2)=6 \) (and the formula is \( f(x)-g(x)=-2x-2 \)).

Teacher's Note:
1. Equal second derivatives mean the difference is linear.
2. Use the data at x = 1 and x = 2.

Answer: (c) 3

Explanation:
1. From the pieces \( g=2 \) for \( x\le-1 \), \( 2-x^2-x \) for \( -1\lt x\lt 0 \), \( 2 \) for \( 0\le x\le1 \), \( x+1 \) for \( x\gt 1 \):
2. slopes at -1: 0 and \( -2(-1)-1=1 \); at 0: \( -1 \) and 0; at 1: 0 and 1.
3. All three are corners: 3 points.

Teacher's Note:
1. Differentiate each piece and compare slopes at the three joins.
2. The constant pieces have slope 0.

Answer: (b) 8

Explanation:
1. a: continuity needs \( x-1=x^2-x-2 \), i.e. \( x^2-2x-1=0 \): 2 points.
2. b: \( x^3-3x+1 \) has three distinct real roots with sign changes: 3 points.
3. c: at 1 the factor \( \log x \) vanishes so \( (\log x)|x-1| \) is smooth; the corner at 3 survives, and \( (x-2)^{1/3} \) has a vertical tangent at 2: 2 points.
4. d: only the corner at 3 is a sharp turn: 1.
5. Sum \( 2+3+2+1=8 \).

Teacher's Note:
1. Handle each of the four sub-parts separately.
2. A cube root gives a vertical tangent, not a sharp turn.

Answer: (c) 6

Explanation:
1. The exponent sum over odd indices is about \( \tfrac{n^2}4 \) for the base 2 and about \( \tfrac{n^2}4 \) for the base 3 over even indices.
2. So the product is about \( (2\cdot3)^{n^2/4}=6^{n^2/4} \) and the limit of its power \( \tfrac1{n^2+1} \) is \( 6^{1/4} \).
3. The fourth power is 6.

Teacher's Note:
1. Add the exponents of 2 and of 3 separately.
2. Then compare with the denominator n squared plus 1.

Answer: (c) \( 1-\tfrac1{2^n} \)

Explanation:
1. Row 4 has common difference \( \tfrac3{16}-\tfrac18=\tfrac1{16} \), so \( a_{4j}=\tfrac j{16} \). Column 4: \( a_{24}=1 \), \( a_{44}=\tfrac4{16}=\tfrac14 \) so \( r^2=\tfrac14 \), \( r=\tfrac12 \). Then \( a_{ij}=\dfrac j{2^i} \).
2. Row i has \( a_{ij}=j/2^i \), so \( d_i=\dfrac1{2^i} \).
3. \( \sum_{i=1}^n\dfrac1{2^i}=1-\dfrac1{2^n} \).

Teacher's Note:
1. Read the common difference from consecutive entries of a row.
2. Sum the geometric series.

Answer: (c) 5

Explanation:
1. Divide numerator and denominator by \( n\cdot3^n \): \( \dfrac1{\left(\frac{x-2}3\right)^n+3-\frac1n} \).
2. For the limit to be \( \tfrac13 \) we need \( \left(\frac{x-2}3\right)^n\to0 \), i.e. \( -1\lt \tfrac{x-2}3\lt 1 \), so \( -1\lt x\lt 5 \).
3. The integers are 0, 1, 2, 3, 4: five of them.

Teacher's Note:
1. Divide through by n times 3 to the n.
2. A geometric term tends to 0 only when its ratio is less than 1 in modulus.

Answer: (c) 5

Explanation:
1. f(|x|) is even. For \( x\ge0 \): \( 2x+3 \) on \( [0,1] \) and \( 6-x^2 \) on \( (1,\infty) \).
2. Corners: at 0 (slopes \( \mp2 \)), at \( \pm1 \) (slopes 2 and -2), and where \( 6-x^2=0 \) i.e. \( x=\pm\sqrt6 \) (sign change, giving a corner in the modulus).
3. Total: 0, \( \pm1 \), \( \pm\sqrt6 \) = 5 points.

Teacher's Note:
1. Write f(|x|) on each interval.
2. The outer modulus adds corners where f(|x|) changes sign.

Answer: (d) \( \left(\tfrac{24}{25},\tfrac{-7}{25}\right) \)

Explanation:
1. \( \cos^3x-\cos^2x=\cos^2x(\cos x-1)\to0^- \), so the first limit is \( f(0^-)=4 \).
2. \( \sin^2x-\sin^3x=\sin^2x(1-\sin x)\to0^+ \), so the second limit is \( f(0^+)=3 \).
3. The line is \( 4y=3x \), i.e. \( 3x-4y=0 \).
4. Reflecting \( (0,1) \): \( d=\dfrac{3(0)-4(1)}{25}=-\dfrac4{25} \), image \( \left(0-2\cdot3d,\ 1+2\cdot4d\right)=\left(\tfrac{24}{25},-\tfrac7{25}\right) \).

Teacher's Note:
1. Decide from which side each argument approaches 0.
2. Use the reflection formula for a point in a line.

Answer: (c) 5

Explanation:
1. On \( [1,3] \): \( [\ln x]=0 \) for \( x\lt e \) and 1 for \( x\ge e \). So f is 0 up to \( e \) (the sign change at 2 is harmless) and jumps to \( \operatorname{sgn}(e-2)\cdot1=1 \) at \( x=e \): one discontinuity.
2. At 3: \( f(3)=1 \) but \( f(3^+)=\{9^+\}=0 \): second.
3. On \( (3,3.5] \): \( \{x^2\} \) jumps when \( x^2=10,11,12 \): three more.
4. Total 5.

Teacher's Note:
1. Check the greatest integer jump at x = e.
2. Do not forget the join at x = 3.

Answer: (b) A and C only

Explanation:
1. (A) true: split the quotient into two derivatives, each tending to \( f'(c)/2 \)... giving \( f'(c) \).
2. (B) false: \( f(x)=|x| \) at 0 gives a symmetric quotient 0 but f is not differentiable.
3. (C) \( \dfrac{g(h)}{h}=h\sin\tfrac1{h^2}\to0 \), so \( g'(0)=0 \) and g' exists everywhere: true.
4. (D) \( \dfrac{g(h)}h=\sin\tfrac1{h^2} \) has no limit: false.

Teacher's Note:
1. Test (B) with the modulus function.
2. Compute g'(0) from the definition in (C) and (D).

Answer: (c) 3

Explanation:
1. Along rationals near 2: \( f\to4-2+3=5 \). Irrationals from the left: \( 2+a \). Irrationals from the right: \( 2^2+1=5 \).
2. Continuity needs \( 2+a=5 \), so \( a=3 \).

Teacher's Note:
1. Compare the limits along rational and irrational sequences.
2. All three must equal f(2) = 5.

Answer: (d) 2

Explanation:
1. Row 4 has common difference \( \tfrac3{16}-\tfrac18=\tfrac1{16} \), so \( a_{4j}=\tfrac j{16} \). Column 4: \( a_{24}=1 \), \( a_{44}=\tfrac4{16}=\tfrac14 \) so \( r^2=\tfrac14 \), \( r=\tfrac12 \). Then \( a_{ij}=\dfrac j{2^i} \).
2. \( a_{ii}=\dfrac i{2^i} \).
3. \( \sum_{i=1}^\infty\dfrac i{2^i}=\dfrac{1/2}{(1-1/2)^2}=2 \).

Teacher's Note:
1. Write the diagonal entries.
2. Use the sum of i x^i with x = 1/2.

Answer: (d) 6

Explanation:
1. A continuous integer valued function on R is constant, so \( f(x)/c \) stays inside one interval \( [k,k+1) \).
2. For \( k\ge1 \) we would need \( c\le2/k \) and \( c\ge\sqrt{26}/(k+1) \), which no positive integer satisfies.
3. For \( k=0 \): \( \sqrt{26}/c\le1 \), so \( c\ge\sqrt{26}\approx5.1 \): the least integer is 6.

Teacher's Note:
1. g must be constant.
2. Compare the interval for f over c with a single unit interval.