JEE Mathematics Limits Continuity and Differentiability MCQs Set 14

Answer: (c) \((1-\sin1)x+1\)

Explanation:
1. Continuity at \(0\): \(\lim_{x\to0^+}x^{\sin x}=1\), so \(g(0)=1\) and \(g(x)=mx+1\).
2. For \(x>0\), \(f'(x)=x^{\sin x}\left(\cos x\ln x+\dfrac{\sin x}{x}\right)\), so \(f'(1)=\sin1\).
3. \(f(-1)=g(-1)=1-m\). Then \(1-m=\sin1\), so \(m=1-\sin1\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(\left(\dfrac1e,e\right)\)

Explanation:
1. Left limit: the form is \(1^\infty\), exponent \(\dfrac{\cos x-\sin x-1}{\sin x}\to-1\), so \(a=e^{-1}\).
2. Right limit: divide by \(e^{3/x}\); as \(x\to0^+\) the limit is \(\dfrac1b\).
3. Continuity gives \(\dfrac1b=\dfrac1e\), so \(b=e\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) discontinuous at all integer values of \(x\)

Explanation:
1. If \(\sin\pi x>0\) the limit is \(1\); if \(\sin\pi x<0\) the base lies in \([0,1)\) and the limit is \(-1\); if \(\sin\pi x=0\) the limit is \(0\).
2. At every integer, \(f=0\) but \(\sin\pi x\) takes both signs close to it, so the one-sided limits are \(1\) and \(-1\).
3. Hence \(f\) is discontinuous at every integer.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

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

Explanation:
1. \(p=-1\), the slope of \(|x-1|\) to the left of \(1\).
2. With \(t=x-1\): \(\ln\cos^mt=m\ln\cos t\approx-\dfrac{mt^2}2\), so \(g\approx-\dfrac{2t^{n-2}}m\).
3. A finite non-zero limit needs \(n=2\), and then \(-\dfrac2m=-1\) gives \(m=2\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) continuous as well as differentiable

Explanation:
1. \(|f(0)|\le0\), so \(f(0)=0\), and \(|f(x)|\to0\), hence \(f\) is continuous.
2. \(\left|\dfrac{f(h)-f(0)}h\right|\le|h|\to0\), so \(f'(0)=0\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(-1\)

Explanation:
1. Let \(g(x)=\dfrac{x^x-x^{-x}}2=\sinh(x\ln x)\); \(g(1)=0\) and \(g'(1)=\cosh0\cdot(\ln1+1)=1\).
2. \(f'(x)=-\dfrac{g'(x)}{1+g^2}\), so \(f'(1)=-1\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(1+[g(x)]^3\)

Explanation:
1. \(g'(x)=\dfrac1{f'(g(x))}\).
2. \(f'(g(x))=\dfrac1{1+[g(x)]^3}\), hence \(g'(x)=1+[g(x)]^3\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(\dfrac{x^{n-1}}{y^{n-1}}\)

Explanation:
1. Put \(x^n=\sin A\), \(y^n=\sin B\): \(\cos A+\cos B=a(\sin A-\sin B)\), so \(a=\cot\dfrac{A-B}2\).
2. Thus \(A-B\) is constant and \(dA=dB\), i.e. \(\dfrac{x^{n-1}dx}{\sqrt{1-x^{2n}}}=\dfrac{y^{n-1}dy}{\sqrt{1-y^{2n}}}\).
3. Rearranging gives \(\dfrac{x^{n-1}}{y^{n-1}}\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(f(x,y)=\dfrac{x^2}{y^2}\)

Explanation:
1. Put \(x^3=\sin A\), \(y^3=\sin B\); then \(a=\cot\dfrac{A-B}2\) is constant, so \(dA=dB\).
2. \(\dfrac{3x^2dx}{\sqrt{1-x^6}}=\dfrac{3y^2dy}{\sqrt{1-y^6}}\), so \(\dfrac{dy}{dx}=\dfrac{x^2}{y^2}\sqrt{\dfrac{1-y^6}{1-x^6}}\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(2^n\)

Explanation:
1. The sum is the Maclaurin expansion of \(f\) evaluated at \(x=1\).
2. It equals \(f(1)=2^n\). (Equivalently, \(\binom n0+\binom n1+\dots+\binom nn=2^n\).)

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(x=n\pi\), \(n=\pm1,\pm2,\pm3,\dots\)

Explanation:
1. Only the zeros of \(\sin x\) can be trouble. At \(x=n\pi\) the one-sided derivatives are \(\pm n\pi\cdot1\) (opposite signs), differing by \(2|n\pi|\).
2. For \(n\ne0\) they differ. At \(x=0\), \(f(x)\approx x|x|\) is differentiable with \(f'(0)=0\).
3. So the exceptions are the non-zero integer multiples of \(\pi\), positive and negative.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(\left\{x:x=\dfrac2{2n+1},\ n\in\mathbb Z\right\}\)

Explanation:
1. At \(x=0\): \(\dfrac{h^2|\cos(\pi/h)|}h\to0\), so \(f\) is differentiable there.
2. Elsewhere \(|\cos(\pi/x)|\) has corners where \(\cos\dfrac\pi x=0\), i.e. \(x=\dfrac2{2n+1}\), and the factor \(x^2\ne0\) keeps the corner.
3. These points (for \(n\ne0,-1\), those lying in \([-1,1]\)) are the non-differentiable points; \(0\) is not one.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) II and III are true, I is false

Explanation:
1. For \(t\in(0,1)\), \(f(t)=f(t^{3/2})=f(t^{(3/2)^k})\to f(0)\); for \(t>1\), \(f(t)=f(t^{(2/3)^k})\to f(1)=f(0)\). So \(f\) is constant on \([0,\infty)\).
2. For \(x<0\), \(f(x^3)=f(x^2)=c\), so \(f\) is constant on \(\mathbb R\).
3. A constant function is even and differentiable, but not odd unless it is zero.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (c) differentiable and the derivative is continuous

Explanation:
1. \(f(x)\approx x\) near \(0\), so \(f\) is continuous and \(f'(0)=\lim\dfrac{\sin h^2}{h^2}=1\).
2. For \(x\ne0\), \(f'(x)=2\cos x^2-\dfrac{\sin x^2}{x^2}\to2-1=1=f'(0)\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(4\)

Explanation:
1. Put \(\operatorname{cosec}\theta=e^u\): \(x=2\sinh u\) and \(y=2\sinh nu\).
2. \(\dfrac{dy}{dx}=\dfrac{n\cosh nu}{\cosh u}\) and \(x^2+4=4\cosh^2u\).
3. \((x^2+4)\left(\dfrac{dy}{dx}\right)^2-n^2y^2=4n^2(\cosh^2nu-\sinh^2nu)=4n^2\), so \(k=4\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (c) \(2\)

Explanation:
1. The condition makes \(f\) an isometry of \([0,1]\) into itself, so \(f(0)\in\{0,1\}\) (extremes must go to extremes since \(|f(1)-f(0)|=1\)).
2. This gives exactly \(f(x)=x\) and \(f(x)=1-x\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(2\)

Explanation:
1. \(\cot x=\dfrac1x-\dfrac x3-\dots\), so \(f(x)=\dfrac{\alpha+\beta}{x^2}-\dfrac\alpha3+\dots\).
2. A finite limit needs \(\alpha+\beta=0\), and then \(-\dfrac\alpha3=\dfrac13\), so \(\alpha=-1\), \(\beta=1\).
3. \(\alpha^2+\beta^2=2\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(4\)

Explanation:
1. The only possible trouble spots are \(x=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4\) where the definition switches (the inner absolute values vanish only at these points, which belong to the second region, or are not interior to a region).
2. At each of them the left and right values differ (for example at \(\dfrac\pi4\): \(3\) on the left, \(2+\sqrt2\) on the right), so \(f\) is discontinuous and hence non-differentiable.
3. There are \(4\) such points.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(5\)

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. Since \(g'=f''=-f\) and \(g=f'\), this is \(fg-gf=0\).
3. So \(F\) is constant: \(F(10)=F(5)=5\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(0\)

Explanation:
1. Oddness gives \(f'_-(-a)=f'_+(a)\).
2. For \(x>a\), \(f(x)=f(2a-x)\), so \(f'_+(a)=-f'_-(a)=0\).
3. Hence the required derivative is \(0\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (c) \(1\)

Explanation:
1. At \(x=\dfrac\pi2\) the first factor \(\cos x\) vanishes, so only the term in which it is differentiated survives.
2. \(f'\left(\tfrac\pi2\right)=-\sin\tfrac\pi2\cdot\cos\pi\cos2\pi\cdots=(-1)(-1)(1)\cdots=1\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) II and IV only

Explanation:
1. Differentiating \(n\) times replaces the first row by \(n!\), \(\sin\left(x+\frac{n\pi}2\right)\), \(\cos\left(x+\frac{n\pi}2\right)\).
2. At \(x=0\) the first two rows coincide, so the determinant is \(0\) for every \(a\).
3. So II and IV are correct.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) I and III only

Explanation:
1. \([\pi^2]=9\), so \(f(x)=x+|x|+\cos9x\) and \(f\) is continuous but not differentiable at \(0\); \(g\) is smooth.
2. \(f+g\) is continuous everywhere, but not differentiable at \(0\).
3. At \(0\), \(\dfrac{f(h)g(h)}h=f(h)\dfrac{\sin h}h\to f(0)\cdot1=1\) from both sides, so \(fg\) is differentiable. Hence I and III.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(72\)

Explanation:
1. \(f(5)=125+75-165-33=2\), so \(g(2)=5\).
2. \(f'(x)=3x^2+6x-33\), \(f'(5)=72\), so \(g'(2)=\dfrac1{72}\) and \(k=72\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) I and III only

Explanation:
1. \((\sin^{-1}x)^2\) is continuous on \([-1,1]\) and \(\cos\frac1x\) is bounded; at \(0\), \(f\to0=f(0)\), so \(f\) is continuous (also at \(\pm1\)). II is false.
2. At \(0\): \(\dfrac{(\sin^{-1}h)^2\cos(1/h)}h\to0\); elsewhere in \((-1,1)\) \(f\) is a composition of smooth functions. III is true; IV is false.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (c) \(0\)

Explanation:
1. \(\dfrac{d^2x}{dy^2}=-\dfrac{y''}{(y')^3}\) where \(y'=\dfrac{dy}{dx}\).
2. So the expression is \(-y''+y''=0\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(8\)

Explanation:
1. Left limit: \(\dfrac{2\sin^22x}{x^2}\to8\).
2. Right limit: rationalise, \(\dfrac{\sqrt x\left(\sqrt{16+\sqrt x}+4\right)}{\sqrt x}\to8\). So \(a=8\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(3^n\cdot e^7\)

Explanation:
1. Each differentiation multiplies by \(3\), so \(y_n=3^ne^{3x+7}\).
2. At \(x=0\): \(3^ne^7\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (c) continuous but not differentiable

Explanation:
1. \(|x^p\sin\frac1x|\le|x|^p\to0\), so \(f\) is continuous.
2. \(\dfrac{f(h)-f(0)}h=h^{p-1}\sin\dfrac1h\); for \(p\le1\) the factor \(h^{p-1}\) does not tend to \(0\), so the limit does not exist.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(x=2\)

Explanation:
1. Near \(x=2\): from the left \([x^2]=3\), \([-x]=-2\), giving \(3+4=7\); from the right \([x^2]=4\), \([-x]=-3\), giving \(4+9=13\).
2. The one-sided limits differ, so \(f\) is discontinuous at \(2\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) neither continuous nor differentiable

Explanation:
1. \(f(0)=1\). For small \(x>0\): \(f=\cos\pi x\to1\). For small \(x<0\): \([x]=-1\), \(|x|=-x\), \(f=\cos\pi(-x-1)=-\cos\pi x\to-1\).
2. The left limit differs from the right limit, so \(f\) is discontinuous at \(0\) and therefore not differentiable.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(-5\) and \(\dfrac23\)

Explanation:
1. As \(x\to0^+\), \(e^{1/x}\to\infty\): \(f\to\dfrac5{-1}=-5\).
2. As \(x\to0^-\), \(e^{1/x}\to0\): \(f\to\dfrac23\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) continuous but not differentiable

Explanation:
1. \(xf(x)\to0\) from both sides, which equals the value at \(0\): continuous.
2. The difference quotient is \(f(h)\), whose one-sided limits are \(-5\) and \(\dfrac23\): not differentiable.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (d) \(0\)

Explanation:
1. \(|x|^3\) has derivative \(3x|x|\) for all \(x\), and the derivative exists and equals \(0\) at \(x=0\).
2. So there are no points of non-differentiability.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (a) \(1\)

Explanation:
1. For \(x>0\), \(f'=-\dfrac1{\sqrt{1-x^2}}\to-1\) and for \(x<0\), \(f'=+\dfrac1{\sqrt{1-x^2}}\to1\) as \(x\to0\).
2. The one-sided derivatives differ only at \(x=0\), so there is exactly one such point in \((-1,1)\).

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.

Answer: (b) \(1\)

Explanation:
1. For \(x\ne0\) the function is smooth.
2. At \(0\), \(\dfrac{\sqrt{|h|}}h=\pm\dfrac1{\sqrt{|h|}}\to\pm\infty\), so it is not differentiable at \(0\) only.

Teacher's Note:
1. Check the left and right limits separately at the point and compare them with the value of the function.
2. For differentiability test the limit of the difference quotient from both sides; remember that continuity is only necessary, not sufficient.