JEE Mathematics Continuity and Differentiability MCQs Set 09

Answer: (a) (1) and (2) only

Explanation:
1. For \( x > 1 \), \( x^p \) dominates and \( f(x) = \dfrac{g(x)}{7} \). For \( |x| < 1 \), \( x^p \to 0 \) and \( f(x) = \dfrac{h(x) + 7}{3x + 1} \).
2. Continuity at 1 from the right: \( \dfrac{g(1)}{7} = 7 \), so \( g(1) = 49 \). From the left: \( \dfrac{h(1) + 7}{4} = 7 \), so \( h(1) = 21 \).
3. Then \( g(1) + h(1) = 70 \) and \( g(1) - h(1) = 28 \).

Teacher's Note:
1. Split the limit by the size of x relative to 1.
2. Continuity requires both one-sided limits to equal f(1).

Answer: (a) (2) and (4) only

Explanation:
1. If \( h > 0 \), \( g = \dfrac{3h}{h} = 3 \); if \( h < 0 \), \( g = \dfrac{h}{3h} = \dfrac{1}{3} \).
2. For \( n > 1 \) and \( 0 < \sin x < 1 \), \( \sin^n x < \sin x \), so \( h > 0 \) and \( f = [3] = 3 \) near \( \dfrac{\pi}{2} \), which equals \( f\left(\tfrac{\pi}{2}\right) \): f is continuous and (being constant) differentiable.
3. For \( 0 < n < 1 \), \( \sin^n x > \sin x \), so \( h < 0 \), \( g = \tfrac{1}{3} \) and \( f = 0 \) near \( \tfrac{\pi}{2} \), while \( f\left(\tfrac{\pi}{2}\right) = 3 \): f is discontinuous, hence not differentiable.

Teacher's Note:
1. A power with exponent greater than 1 makes a number between 0 and 1 smaller.
2. The ratio g only takes the values 3 and \( \tfrac{1}{3} \).

Answer: (a) (1), (3) and (4) only

Explanation:
1. \( \sin(\sin x) - \sin x \approx -\dfrac{\sin^3 x}{6} \approx -\dfrac{x^3}{6} \).
2. For a finite non-zero limit the denominator must also behave like \( x^3 \), so \( c = 0 \) and \( a \ne 0 \), and b does not affect the limit.
3. \( \dfrac{-1/6}{a} = -\dfrac{1}{12} \) gives \( a = 2 \).

Teacher's Note:
1. Match the lowest power of x in the numerator and denominator.
2. A non-zero constant in the denominator would make the limit zero.

Answer: (a) \( a = 3,\ b = 5 \)

Explanation:
1. Differentiability needs continuity at 1: \( 1 + 3 + a = b + 2 \).
2. The derivatives must match: \( 2x + 3 = 5 \) at \( x = 1 \), so \( b = 5 \).
3. Then \( 4 + a = 7 \) gives \( a = 3 \).

Teacher's Note:
1. Two conditions (continuity and equal slopes) fix two constants.
2. Substitute the slope condition into the continuity equation.

Answer: (a) \( a = -\dfrac{1}{2},\ b = \dfrac{3}{2} \)

Explanation:
1. The function is even, so check \( x = 1 \). Continuity: \( a + b = 1 \).
2. Slopes at 1: for \( |x| \ge 1 \), \( f = \dfrac{1}{x} \) with slope \( -1 \); inside, the slope is \( 2a \). So \( 2a = -1 \) and \( a = -\dfrac{1}{2} \).
3. Then \( b = 1 - a = \dfrac{3}{2} \). By symmetry the same holds at \( x = -1 \).

Teacher's Note:
1. Use the symmetry of an even function to check only one join.
2. Match both the value and the slope at the join.

Answer: (a) \( a = \dfrac{35}{9},\ b = \dfrac{10}{3} \)

Explanation:
1. Continuity at 3: \( 9a - 3b + 2 = 9b - 3 \), that is \( 9a - 12b + 5 = 0 \).
2. Equal slopes at 3: \( 6a - b = 6b \), so \( a = \dfrac{7b}{6} \).
3. Substituting, \( \dfrac{21b}{2} - 12b + 5 = 0 \Rightarrow b = \dfrac{10}{3} \) and \( a = \dfrac{35}{9} \).

Teacher's Note:
1. Two equations from continuity and equal slopes can be solved together.
2. Express one constant in terms of the other from the slope equation.

Answer: (d) \( \dfrac{1}{4} \)

Explanation:
1. The form is 0/0. By L'Hospital's rule the limit is \( \dfrac{1/x}{4x^3} \) at \( x = 1 \).
2. That equals \( \dfrac{1}{4} \).

Teacher's Note:
1. Differentiate numerator and denominator separately.
2. Substitute \( x = 1 \) after simplifying.

Answer: (a) 2

Explanation:
1. \( \pi - 2\tan^{-1}x = 2\cot^{-1}x \approx \dfrac{2}{x} \) for large x.
2. \( \ell n\left(1 + \dfrac{1}{x}\right) \approx \dfrac{1}{x} \), so the limit is 2.

Teacher's Note:
1. Use \( \tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2} \).
2. For large x, \( \cot^{-1}x \approx \dfrac{1}{x} \).

Answer: (a) 2

Explanation:
1. \( e^x - e^{-x} = 2x + \dfrac{x^3}{3} + \dots \), so the numerator is \( \dfrac{x^3}{3} \).
2. \( x - \sin x \approx \dfrac{x^3}{6} \). The limit is \( \dfrac{1/3}{1/6} = 2 \).

Teacher's Note:
1. Use the series of the hyperbolic sine.
2. The leading terms cancel, so go to the cubic power.

Answer: (a) \( \dfrac{11}{2} \)

Explanation:
1. \( \sqrt{1 - x^2} \approx 1 - \dfrac{x^2}{2} \) and \( \cos x\cos 3x \approx 1 - \dfrac{x^2}{2} - \dfrac{9x^2}{2} = 1 - 5x^2 \).
2. The product is \( 1 - \dfrac{x^2}{2} - 5x^2 = 1 - \dfrac{11x^2}{2} \).
3. So the numerator is \( \dfrac{11x^2}{2} \) and the limit is \( \dfrac{11}{2} \).

Teacher's Note:
1. Multiply the series and keep terms up to \( x^2 \).
2. The corrections from each factor add.

Answer: (a) (2) and (4) only

Explanation:
1. As \( x \to \dfrac{\pi}{2}^- \), put \( x = \dfrac{\pi}{2} - h \): \( \cot 3x = \tan 3h \approx 3h \) and \( \cot 2x = -\cot 2h \approx -\dfrac{1}{2h} \), so the exponent \( \approx -6h^2 \to 0 \) and the left limit is 1.
2. As \( x \to \dfrac{\pi}{2}^+ \), \( |\cos x| \approx h \) and \( |\tan x| \approx \dfrac{1}{h} \), so the limit is \( e^{a/b} \).
3. Continuity: \( b + 3 = 1 \Rightarrow b = -2 \) and \( e^{a/b} = 1 \Rightarrow a = 0 \). Then \( |a - b| = 2 \) and \( |a + 2b| = 4 \), so (2) and (4) are true.

Teacher's Note:
1. Both limits at \( \dfrac{\pi}{2} \) must equal the value of the function.
2. Shift the variable so that the point becomes 0.

Answer: (b) \( 2\pi \)

Explanation:
1. For \( a \ne n\pi \), the limit is \( \dfrac{-\sin a}{-\operatorname{cosec}^2 a} = \sin^3 a \).
2. So f follows \( \sin^3 a \) with the values 2 and \( -2 \) at even and odd multiples of \( \pi \). Both parts repeat with period \( 2\pi \), and no smaller period works because \( f(a + \pi) \ne f(a) \).
3. The fundamental period is \( 2\pi \).

Teacher's Note:
1. Simplify the limit as a derivative ratio.
2. Check whether a half period repeats the pattern.

Answer: (a) 0 and every value in \( [1, 4] \)

Explanation:
1. With \( b = 2 \), for \( |x| \le 2 \), \( g(x) = 2^{|x|} \) takes every value in \( [1, 4] \).
2. For \( |x| > 2 \), \( g(x) = \operatorname{sgn}|4 - x| \) equals 0 at \( x = 4 \) and 1 elsewhere.
3. So g takes the value 0 and all the values in \( [1, 4] \).

Teacher's Note:
1. Evaluate each piece of g over its own interval.
2. The sign function only takes the values 0 and 1 for a modulus.

Answer: (a) \( \dfrac{1}{2} \) and \( \dfrac{\sqrt{3}}{2} \)

Explanation:
1. The smaller of \( |x| \) and \( \sqrt{1 - x^2} \) is \( |x| \) for \( |x| \le \dfrac{1}{\sqrt{2}} \) and \( \sqrt{1 - x^2} \) for \( |x| \ge \dfrac{1}{\sqrt{2}} \). So \( f = \sin^{-1}|x| \) in the first range and \( \cos^{-1}|x| \) in the second.
2. \( \sin^{-1}|x| = \dfrac{\pi}{6} \Rightarrow |x| = \dfrac{1}{2} \) (valid since \( \tfrac{1}{2} < \tfrac{1}{\sqrt{2}} \)).
3. \( \cos^{-1}|x| = \dfrac{\pi}{6} \Rightarrow |x| = \dfrac{\sqrt{3}}{2} \) (valid since \( \tfrac{\sqrt{3}}{2} > \tfrac{1}{\sqrt{2}} \)).

Teacher's Note:
1. Find which of the two quantities is smaller on each range of x.
2. Check that each solution lies in the correct range.

Answer: (a) the whole interval \( [-1, 1] \)

Explanation:
1. \( f \) equals \( \sin^{-1}|x| \) for \( |x| \le \dfrac{1}{\sqrt{2}} \) and \( \cos^{-1}|x| \) for \( |x| \ge \dfrac{1}{\sqrt{2}} \). In both ranges \( f \le \sin^{-1}\dfrac{1}{\sqrt{2}} = \dfrac{\pi}{4} \).
2. Since \( \dfrac{\pi}{4} < \dfrac{\pi}{3} \), the inequality holds for every x in \( [-1, 1] \).

Teacher's Note:
1. The largest value of f occurs where the two quantities are equal, at \( |x| = \dfrac{1}{\sqrt{2}} \).
2. Compare that maximum with the bound.

Answer: (a) \( \dfrac{1}{2} \)

Explanation:
1. \( f(x) = \sin^{-1}|x| \) for \( |x| \le \dfrac{1}{\sqrt{2}} \) and \( \cos^{-1}|x| \) for \( |x| \ge \dfrac{1}{\sqrt{2}} \), with maximum \( \dfrac{\pi}{4} \) at \( |x| = \dfrac{1}{\sqrt{2}} \).
2. For \( 0 < k < \dfrac{\pi}{4} \) there are 4 roots \( \pm\sin k \) and \( \pm\cos k \). For \( k = \dfrac{\pi}{4} \) there are only 2 roots, and \( \dfrac{\sqrt{3}}{2} \) and 1 exceed \( \dfrac{\pi}{4} \approx 0.785 \).
3. So \( k = \dfrac{1}{2} \) is the valid choice.

Teacher's Note:
1. The graph rises from 0 to \( \dfrac{\pi}{4} \) and falls back to 0 on each side of the y-axis.
2. A value strictly between 0 and the maximum is met four times.

Answer: (b) 1

Explanation:
1. Let \( h(x) = (x + 1)e^{2x} + 5xe^x - 25 \). As \( x \to -\infty \), \( h \to -25 \) from below, and h decreases until \( x = -\dfrac{3}{2} \) and then increases (its derivative \( e^{2x}(2x + 3) + 5e^x(x + 1) \) is positive for \( x > -1 \)).
2. So h stays negative up to \( x = 0 \) where \( h(0) = -24 \), and it is increasing and unbounded afterwards, for example \( h(2) = 3e^4 + 10e^2 - 25 > 0 \).
3. Hence h has exactly one real zero.

Teacher's Note:
1. Study the sign and the monotonicity of the function.
2. Use the values at the ends and at a convenient point.

Answer: (a) 0

Explanation:
1. With \( u = \dfrac{1}{h} \to 0 \), \( h\ln\dfrac{\sin(x + u)}{\sin x} = \dfrac{\ln\sin(x + u) - \ln\sin x}{u} \to \dfrac{d}{dx}\ln\sin x = \cot x \).
2. \( \cot x \) is differentiable at every point of \( \left(0, \dfrac{\pi}{2}\right) \), so there are no points of non-derivability.

Teacher's Note:
1. The limit is the derivative of \( \ln\sin x \).
2. The cotangent is smooth wherever it is defined.

Answer: (b) 2

Explanation:
1. \( \sum_{k=1}^{r}k = \dfrac{r(r + 1)}{2} \) and \( \sum_{k=1}^{r}k^3 = \dfrac{r^2(r + 1)^2}{4} \), so the ratio is \( \dfrac{2}{r(r + 1)} \).
2. \( \sum_{r=1}^{n}\dfrac{2}{r(r + 1)} = 2\left(1 - \dfrac{1}{n + 1}\right) \to 2 \).

Teacher's Note:
1. Use partial fractions \( \dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1} \).
2. The sum telescopes.

Answer: (b) 1

Explanation:
1. \( \sqrt{\dfrac{2 - \sqrt{3}}{4}} = \sin 15^\circ \), so \( \sin^{-1}(\cdot) = \dfrac{\pi}{12} \). \( \cos^{-1}\dfrac{\sqrt{12}}{4} = \cos^{-1}\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6} \) and \( \sec^{-1}\sqrt{2} = \dfrac{\pi}{4} \).
2. The sum is \( \dfrac{\pi + 2\pi + 3\pi}{12} = \dfrac{\pi}{2} \).
3. \( \cot\dfrac{\pi}{2} = 0 \) and \( \cos 0 = 1 \).

Teacher's Note:
1. Identify each inverse trigonometric value from a standard angle.
2. Use \( \sin^2 15^\circ = \dfrac{1 - \cos 30^\circ}{2} \).

Answer: (b) 2

Explanation:
1. As \( x \to 0^+ \), \( e^{1/x} \to \infty \), so the limit is \( \dfrac{8}{2 + b + b^2} = 2 \), giving \( b^2 + b - 2 = 0 \) and \( b = 1 \) (positive).
2. As \( x \to 0^- \), \( e^{1/x} \to 0 \), so the limit is \( 1 + a^3 = 2 \), giving \( a = 1 \).
3. \( a^2 + b^2 = 2 \).

Teacher's Note:
1. Evaluate the limit separately from the right and from the left.
2. \( e^{1/x} \) tends to infinity on the right and to zero on the left.