JEE Mathematics Continuity and Differentiability MCQs Set 10

Answer: (a) f(x) is continuous at x = 2 but not at x = −2

Explanation:
1. At x = 2: for x slightly above 2, \( \{x^2\}\to 0 \) and \( \{x\}\to0 \); for x slightly below 2, \( \{x^2\}\to1 \) and \( (\{x\})^2\to1 \). Both limits are 0, and f(2) = 0, so f is continuous at 2.
2. At x = −2: for x slightly above −2, \( x^2 \) is just below 4, so \( \{x^2\}\to1 \) while \( \{x\}\to0 \), giving limit 1. But f(−2) = 0.
3. Hence f is not continuous at x = −2.

Teacher's Note:
1. Fractional-part functions jump at integers, so check each one-sided limit carefully.
2. Near −2 the square x² approaches 4 from below, which is why {x²} tends to 1.

Answer: (c) \( -\frac52,\ -\frac32 \)

Explanation:
1. Expand: \( x+ax\cos x-b\sin x=(1+a-b)x+\left(\frac b6-\frac a2\right)x^3+\dots \)
2. For a finite limit we need \( 1+a-b=0 \), and then the limit is \( \frac b6-\frac a2=1 \).
3. Substituting \( b=1+a \): \( \frac{1+a}{6}-\frac a2=1\Rightarrow a=-\frac52,\ b=-\frac32 \).

Teacher's Note:
1. Use Maclaurin series of sin x and cos x to read off the coefficients of x and x³.
2. Continuity at 0 means the limit equals f(0) = 1.

Answer: (d) discontinuous at the origin because |x|/x is discontinuous there

Explanation:
1. \( |x| \) is continuous at 0, so the question depends on \( \frac{|x|}{x} \).
2. \( \frac{|x|}{x}=-1 \) for x < 0 and \( +1 \) for x > 0 (and is undefined at 0): it has a jump at the origin.
3. A continuous function plus a discontinuous one is discontinuous, so f is discontinuous at the origin because |x|/x is.

Teacher's Note:
1. Test each component separately; |x| is continuous everywhere.
2. The jump of |x|/x from −1 to +1 is transferred to f.

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

Explanation:
1. Divide by \( |x-y| \): \( \left|\frac{f(x)-f(y)}{x-y}\right|\le|x-y|^2 \).
2. Let y → x: \( |f'(x)|\le0 \), so f′(x) = 0 for all x.
3. Hence f is constant, and this is exactly what Reason R states, so R explains A.

Teacher's Note:
1. This is the standard 'higher power of |x−y|' argument showing the derivative vanishes.
2. R is the fact used in the last step of A.

Answer: (a) \( \frac{x^{n-1}}{y^{n-1}} \)

Explanation:
1. Put \( x^n=\sin A,\ y^n=\sin B \): \( \cos A+\cos B=a(\sin A-\sin B) \).
2. This gives \( \cot\frac{A-B}{2}=a \), so A − B is constant, and \( \frac{dA}{dx}=\frac{dB}{dx} \).
3. Differentiating: \( nx^{n-1}dx=\cos A\,dA \) and \( ny^{n-1}dy=\cos B\,dB \). Since dA = dB, \( \frac{dy}{dx}=\frac{x^{n-1}\cos B}{y^{n-1}\cos A} \).
4. With \( \cos A=\sqrt{1-x^{2n}} \) and \( \cos B=\sqrt{1-y^{2n}} \), \( \sqrt{\frac{1-x^{2n}}{1-y^{2n}}}\,\frac{dy}{dx}=\frac{x^{n-1}}{y^{n-1}} \).

Teacher's Note:
1. The trigonometric substitution x^n = sin A turns the relation into a constant difference of angles.
2. Use dA/dx = dB/dx to relate the two derivatives.

Answer: (c) \( 3\left(\frac{d^2y}{dx^2}\right)^2 \)

Explanation:
1. Write \( y=\frac ac+\frac{k}{cx+d} \) with \( k=\frac{bc-ad}{c} \). Then \( y'=-\frac{kc}{(cx+d)^2},\ y''=\frac{2kc^2}{(cx+d)^3},\ y'''=-\frac{6kc^3}{(cx+d)^4} \).
2. \( 2y'y'''=\frac{12k^2c^4}{(cx+d)^6} \) and \( 3(y'')^2=\frac{12k^2c^4}{(cx+d)^6} \).

Teacher's Note:
1. A linear fractional function is a constant plus k/(cx+d); differentiate that form.
2. The relation 2y′y‴ = 3(y″)² is a standard identity for such functions.

Answer: (b) \( \sqrt{1+9y^2} \)

Explanation:
1. Differentiate both sides with respect to x using the Leibniz rule: \( 1=\frac{1}{\sqrt{1+9y^2}}\cdot\frac{dy}{dx} \).
2. Hence \( \frac{dy}{dx}=\sqrt{1+9y^2} \).

Teacher's Note:
1. The derivative of an integral with variable upper limit y is the integrand at y times dy/dx.
2. Do not forget the chain-rule factor dy/dx.

Answer: (c) \( -\frac2ae^{-\pi/2} \)

Explanation:
1. In polar form, \( r=ae^{\theta} \), so \( x=ae^\theta\cos\theta,\ y=ae^\theta\sin\theta \).
2. \( \frac{dy}{dx}=\frac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta} \) and \( \frac{d^2y}{dx^2}=\frac{2}{ae^{\theta}(\cos\theta-\sin\theta)^3} \).
3. x = 0 corresponds to \( \theta=\pi/2 \): \( y''=\frac{2}{ae^{\pi/2}(-1)^3}=-\frac2ae^{-\pi/2} \).

Teacher's Note:
1. Polar substitution makes this curve a simple exponential spiral.
2. x = 0 with y > 0 means θ = π/2.

Answer: (c) 26

Explanation:
1. Taylor expansion about 2: \( P(x)=-1+(x-2)^2-2(x-2)^3+(x-2)^4 \).
2. Differentiate twice: \( P''(x)=2-12(x-2)+12(x-2)^2 \).
3. At x = 1: \( P''(1)=2+12+12=26 \).

Teacher's Note:
1. A degree-4 polynomial is fully determined by its Taylor data at one point.
2. Coefficients are P⁽ᵏ⁾(2)/k!.

Answer: (c) \( \left(\frac{-3b}{a^3}\right)\csc^4\theta\cot\theta \)

Explanation:
1. \( \frac{dy}{dx}=-\frac ba\cot\theta \).
2. \( \frac{d^2y}{dx^2}=\frac{(b/a)\csc^2\theta}{-a\sin\theta}=-\frac{b}{a^2}\csc^3\theta \).
3. \( \frac{d^3y}{dx^3}=\frac{-\frac b{a^2}(-3\csc^3\theta\cot\theta)}{-a\sin\theta}=-\frac{3b}{a^3}\csc^4\theta\cot\theta \).

Teacher's Note:
1. Differentiate with respect to θ and divide by dx/dθ = −a sin θ each time.
2. Keep track of the sign at each step.

Answer: (a) \( 32/9 \)

Explanation:
1. \( x^2F(x)=\int_4^x\{4t^2-2F'(t)\}dt \), so F(4) = 0.
2. Differentiate: \( 2xF+x^2F'=4x^2-2F' \), i.e. \( (x^2+2)F'=4x^2-2xF \).
3. At x = 4: \( 18F'(4)=64\Rightarrow F'(4)=\frac{32}{9} \).

Teacher's Note:
1. Multiply out by x² first so the integral stands alone, then differentiate.
2. The lower limit gives F(4) = 0.

Answer: (a) \( \frac{-2(1+y^2)}{y^5} \)

Explanation:
1. Differentiate: \( \frac{y'}{1+y^2}-y'+1=0\Rightarrow y'=\frac{1+y^2}{y^2} \).
2. \( y''=\frac{d}{dy}\left(y^{-2}+1\right)\cdot y'=-\frac{2}{y^3}\cdot\frac{1+y^2}{y^2}=-\frac{2(1+y^2)}{y^5} \).

Teacher's Note:
1. Express y′ as a function of y, then differentiate through y with the chain rule.
2. d/dx = (d/dy)·y′ for functions of y.

Answer: (b) \( -\operatorname{sgn}(\sin x) \)

Explanation:
1. \( y'=\frac{-\sin x}{\sqrt{1-\cos^2x}}=\frac{-\sin x}{|\sin x|} \).
2. This equals \( -\operatorname{sgn}(\sin x) \), defined wherever sin x ≠ 0.

Teacher's Note:
1. √(1 − cos²x) = |sin x|, not sin x.
2. The ratio sin x/|sin x| is sgn(sin x).

Answer: (c) \( \frac{1}{2+(g(x)-x)^2} \)

Explanation:
1. For inverse functions, \( g'(x)=\frac{1}{f'(g(x))} \) and \( f'(t)=1+\sec^2t=2+\tan^2t \).
2. Since f(g(x)) = x, \( g+\tan g=x\Rightarrow\tan g=x-g \).
3. So \( g'(x)=\frac{1}{2+(x-g(x))^2}=\frac{1}{2+(g(x)-x)^2} \).

Teacher's Note:
1. Use the inverse-function derivative rule and eliminate tan g with f(g(x)) = x.
2. (x − g)² and (g − x)² are the same.