JEE Mathematics Continuity and Differentiability MCQs Set 25

Answer: (c) for all real values of x

Explanation:
1. Each piece is a polynomial, so it is continuous on its own interval.
2. At \(x=2\): left limit \(=2-1=1\), right limit \(=2(2)-3=1\), and \(f(2)=1\).
3. So \(f\) is continuous for all real x.

Teacher's Note:
1. Check continuity at the point where the definition changes.

Answer: (d) \(\frac{\log\frac12}{2^x}\)

Explanation:
1. \(\frac{d}{dx}a^{u}=a^u\ln a\cdot u'\). With \(a=2\) and \(u=-x\): \(\frac{dy}{dx}=-2^{-x}\ln2\).
2. \(-\ln2=\ln\frac12\), so \(\frac{dy}{dx}=\frac{\log\frac12}{2^x}\).

Teacher's Note:
1. Do not forget the chain-rule factor \(-1\).

Answer: (c) \(\frac{\pi}{180}\sec x^\circ\tan x^\circ\)

Explanation:
1. Write \(x^\circ=\frac{\pi x}{180}\) radians. Then \(y=\sec\frac{\pi x}{180}\).
2. \(\frac{dy}{dx}=\sec\frac{\pi x}{180}\tan\frac{\pi x}{180}\cdot\frac{\pi}{180}\).

Teacher's Note:
1. Formulas for derivatives of trigonometric functions need radians, so convert first.

Answer: (b) \(\frac1x\)

Explanation:
1. \(\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac1x=\frac{1}{x\log x}\).
2. \(e^y=\log x\), so \(e^y\frac{dy}{dx}=\log x\cdot\frac{1}{x\log x}=\frac1x\).

Teacher's Note:
1. Use \(e^{\log(\log x)}=\log x\).

Answer: (c) \(\frac{-2x}{1+x^4}\)

Explanation:
1. \(\frac{d}{du}\cot^{-1}u=-\frac{1}{1+u^2}\).
2. With \(u=x^2\), \(u'=2x\): \(\frac{dy}{dx}=-\frac{2x}{1+x^4}\).

Teacher's Note:
1. Remember the negative sign of the derivative of \(\cot^{-1}\).

Answer: (d) \(\frac{\sec^2\sqrt x}{2\sqrt x\tan\sqrt x}\)

Explanation:
1. \(\frac{dy}{dx}=\frac{1}{\tan\sqrt x}\cdot\sec^2\sqrt x\cdot\frac{1}{2\sqrt x}\).
2. \(=\frac{\sec^2\sqrt x}{2\sqrt x\tan\sqrt x}\).

Teacher's Note:
1. Apply the chain rule through the log, the tangent and the root.

Answer: (a) e

Explanation:
1. \(y=e\cdot e^{\log_ex}=ex\).
2. So \(\frac{dy}{dx}=e\).

Teacher's Note:
1. Simplify \(e^{\log x}=x\) before differentiating.

Answer: (c) \(-2x\sin2x^2\)

Explanation:
1. \(\frac{dy}{dx}=2\cos x^2\cdot(-\sin x^2)\cdot2x=-2x\cdot2\sin x^2\cos x^2\).
2. Using \(2\sin x^2\cos x^2=\sin2x^2\): \(\frac{dy}{dx}=-2x\sin2x^2\).

Teacher's Note:
1. Apply the chain rule three times and combine using the double angle formula.

Answer: (d) \((2y-1)\frac{dy}{dx}-\cos x=0\)

Explanation:
1. The expression repeats itself, so \(y=\sqrt{\sin x+y}\), i.e. \(y^2=\sin x+y\).
2. Differentiating, \(2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}\).
3. So \((2y-1)\frac{dy}{dx}-\cos x=0\).

Teacher's Note:
1. For an infinite nested expression, square once and use self-similarity.

Answer: (b) \(2^{x-y}\frac{2^y-1}{1-2^x}\)

Explanation:
1. Differentiating: \(2^x\ln2+2^y\ln2\,y'=2^{x+y}\ln2\,(1+y')\).
2. \(2^x-2^{x+y}=y'(2^{x+y}-2^y)\), so \(y'=\frac{2^x(1-2^y)}{2^y(2^x-1)}\).
3. This equals \(2^{x-y}\cdot\frac{2^y-1}{1-2^x}\).

Teacher's Note:
1. Cancel \(\ln2\) and collect the terms in \(y'\).

Answer: (d) has infinitely many discontinuities on \((-\infty,\infty)\)

Explanation:
1. \(f\) is undefined where \(\tan x\) is undefined (\(x=\frac\pi2+n\pi\)) and where \(1+\tan x=0\) (\(x=\frac{3\pi}{4}+n\pi\)).
2. These points repeat for every integer n, so there are infinitely many discontinuities.

Teacher's Note:
1. Periodic functions have repeating discontinuities.

Answer: (c) \(f(x)\) is discontinuous at \(x=0\)

Explanation:
1. \(\sqrt{x^2}=|x|\), so \(f(x)=\frac{x}{|x|}\), which is 1 for \(x>0\) and \(-1\) for \(x<0\).
2. The left limit is \(-1\) and the right limit is 1, so the limit does not exist at 0.
3. Hence \(f\) is discontinuous at \(x=0\).

Teacher's Note:
1. The signum function has a jump at 0.

Answer: (d) None of these

Explanation:
1. Left limit at 0: \(\lambda(0)=0\). Right limit: \(4(0)+1=1\).
2. The two limits are different for every value of \(\lambda\), so no \(\lambda\) makes f continuous.

Teacher's Note:
1. If the left value is 0 for every \(\lambda\), it cannot match 1.

Answer: (b) 2

Explanation:
1. \(\frac{1+\cos2\theta}{1-\cos2\theta}=\cot^2\theta\), so \(y=|\cot\theta|\).
2. At \(\theta=\frac{3\pi}{4}\), \(\cot\theta<0\), so \(y=-\cot\theta\).
3. \(\frac{dy}{d\theta}=\csc^2\theta=\csc^2\frac{3\pi}{4}=2\).

Teacher's Note:
1. Take the sign of \(\cot heta\) at the given point when removing the modulus.

Answer: (c) \(\frac12\)

Explanation:
1. \(\frac{dx}{dt}=\cos t\cos2t-2\sin t\sin2t\), which at \(t=\frac\pi4\) equals \(0-\sqrt2=-\sqrt2\).
2. \(\frac{dy}{dt}=-\sin t\sin2t+2\cos t\cos2t=-\frac{\sqrt2}{2}+0\).
3. \(\frac{dy}{dx}=\frac{-\sqrt2/2}{-\sqrt2}=\frac12\).

Teacher's Note:
1. Differentiate x and y with respect to t and divide.

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

Explanation:
1. \(\tan^{-1}\frac{a-x}{1+ax}=\tan^{-1}a-\tan^{-1}x\).
2. Its derivative is \(-\frac{1}{1+x^2}\).

Teacher's Note:
1. Use \( an^{-1}A- an^{-1}B= an^{-1}rac{A-B}{1+AB}\) to simplify first.

Answer: (c) \(-\frac xy\)

Explanation:
1. With \(t=\tan\frac\theta2\): \(x=\cos\theta\), \(y=\sin\theta\).
2. \(\frac{dy}{dx}=\frac{\cos\theta}{-\sin\theta}=-\frac xy\).

Teacher's Note:
1. Recognise the half-angle tangent substitution. Alternatively, note \(x^2+y^2=1\).

Answer: (b) \(\frac14\)

Explanation:
1. \(f(0)=\lim_{x\to0}\frac{\sqrt{4+x}-2}{x}\).
2. Rationalising: \(\frac{(4+x)-4}{x(\sqrt{4+x}+2)}=\frac{1}{\sqrt{4+x}+2}\to\frac14\).

Teacher's Note:
1. Multiply by the conjugate to remove the 0/0 form.