JEE Mathematics Limits Continuity and Differentiability MCQs Set 13

Answer: (a) \( \frac\pi2 \)

Explanation:
1. \( \frac{dy}{dx}=2x-5 \), so the slopes are \( m_1=-1 \) at \( (2,0) \) and \( m_2=1 \) at \( (3,0) \).
2. \( m_1m_2=-1 \), so the tangents are perpendicular and the angle is \( \frac\pi2 \). Answer: (a)

Teacher's Note:
1. Find the slope of the tangent at each point.
2. Check the product of the slopes.

Answer: (a) 41

Explanation:
1. The expression equals \( 1+\frac{10}{3x^2+9x+7} \), which is largest when the denominator is smallest (and positive).
2. \( 3x^2+9x+7=3\left(x+\frac32\right)^2+\frac14 \) has minimum value \( \frac14 \) at \( x=-\frac32 \).
3. Maximum value \( =1+\frac{10}{1/4}=41 \). Answer: (a)

Teacher's Note:
1. Divide to write the fraction as \( 1+\frac{10}{\cdots} \).
2. Complete the square in the denominator.

Answer: (b) Interval \( \left(-\infty,\frac13\right] \); function \( 3x^2-2x+1 \)

Explanation:
1. (a) \( f'=3x(x+4)\gt0 \) for \( x\lt-4 \): correct. (c) \( f'=6(x-2)(x+1)\ge0 \) for \( x\ge2 \): correct. (d) \( f'=3(x-1)^2\ge0 \) everywhere: correct.
2. (b) \( f'(x)=6x-2\gt0 \) only for \( x\gt\frac13 \), so \( 3x^2-2x+1 \) is decreasing, not increasing, on \( \left(-\infty,\frac13\right] \). The incorrectly matched pair is (b). Answer: (b)

Teacher's Note:
1. Differentiate each function.
2. Check the sign of the derivative on the given interval.

Answer: (d) \( f(6)\ge8 \)

Explanation:
1. By the mean value theorem, \( \frac{f(6)-f(1)}{6-1}=f'(c)\ge2 \) for some \( c\in(1,6) \).
2. So \( f(6)-f(1)\ge10 \), i.e. \( f(6)\ge10-2=8 \). Answer: (d)

Teacher's Note:
1. Apply the mean value theorem on \( [1,6] \).
2. Use the lower bound on \( f' \).

Answer: (a) it is at a constant distance from the origin

Explanation:
1. \( \frac{dx}{d\theta}=a\theta\cos\theta \) and \( \frac{dy}{d\theta}=a\theta\sin\theta \), so \( \frac{dy}{dx}=\tan\theta \) and the slope of the normal is \( -\cot\theta \).
2. The normal is \( y-a(\sin\theta-\theta\cos\theta)=-\frac{\cos\theta}{\sin\theta}(x-a\cos\theta-a\theta\sin\theta) \), which simplifies to \( x\cos\theta+y\sin\theta=a \).
3. Its distance from the origin is \( \frac{a}{\sqrt{\cos^2\theta+\sin^2\theta}}=a \), a constant. Answer: (a)

Teacher's Note:
1. Find \( dy/dx \) parametrically.
2. Simplify the equation of the normal and compute its distance from the origin.

Answer: (b) 21 s

Explanation:
1. Let the lizard catch the insect after \( t \) seconds. The insect covers \( 20t \) cm and the lizard covers \( \frac12(2)t^2=t^2 \) cm.
2. The lizard covers 21 cm more: \( t^2=21+20t\Rightarrow t^2-20t-21=0\Rightarrow(t-21)(t+1)=0 \).
3. Rejecting \( t=-1 \), \( t=21 \) s. Answer: (b)

Teacher's Note:
1. Write the distance covered by each animal.
2. Equate lizard's distance with 21 plus the insect's distance.

Answer: (d) \( (a-b)^2 \)

Explanation:
1. Let \( x=a^2\cos^2\theta+b^2\sin^2\theta \). Then \( u^2=(a^2+b^2)+2\sqrt{x(a^2+b^2-x)} \), where \( x \) varies between \( a^2 \) and \( b^2 \).
2. \( x(a^2+b^2-x) \) is largest at \( x=\frac{a^2+b^2}2 \): \( u^2_{\max}=2(a^2+b^2) \). It is smallest at the ends \( x=a^2 \) or \( b^2 \), where it equals \( a^2b^2 \): \( u^2_{\min}=a^2+b^2+2ab=(a+b)^2 \).
3. Difference \( =2(a^2+b^2)-(a+b)^2=(a-b)^2 \). Answer: (d)

Teacher's Note:
1. Write \( u^2 \) in terms of \( x=a^2\cos^2\theta+b^2\sin^2\theta \).
2. Find the extreme values of \( x(a^2+b^2-x) \).

Answer: (b) \( a=1,\ b\in\mathbb R \)

Explanation:
1. The form is \( 1^\infty \), so the limit is \( e^{L} \) with \( L=\lim2x\left(\frac ax+\frac b{x^2}\right)=\lim\left(2a+\frac{2b}x\right)=2a \).
2. \( e^{2a}=e^2\Rightarrow a=1 \), and \( b \) does not affect the limit, so \( b \) is any real number. Answer: (b)

Teacher's Note:
1. Recognise the \( 1^\\infty \) form.
2. Find the exponent limit; the \( b/x \) term vanishes.

Answer: (c) \( -\frac12 \)

Explanation:
1. Continuity at \( \frac\pi4 \) requires \( f\left(\frac\pi4\right)=\lim_{x\to\pi/4}\frac{1-\tan x}{4x-\pi} \), a \( \frac00 \) form.
2. By L'Hopital's rule: \( \lim\frac{-\sec^2x}4=-\frac{(\sqrt2)^2}4=-\frac12 \). Answer: (c)

Teacher's Note:
1. Continuity means the value equals the limit.
2. Use L'Hopital's rule.

Answer: (c) \( \frac{1-x}x \)

Explanation:
1. The exponent repeats itself, so \( x=e^{y+x} \). Taking logs: \( \log x=y+x \).
2. Differentiate: \( \frac1x=\frac{dy}{dx}+1 \), so \( \frac{dy}{dx}=\frac{1-x}x \). Answer: (c)

Teacher's Note:
1. Use the self-similar structure of the exponent.
2. Take logarithms and differentiate.

Answer: (d) \( \left(\frac98,\frac92\right) \)

Explanation:
1. Differentiate with respect to time: \( 2y\frac{dy}{dt}=18\frac{dx}{dt} \). Given \( \frac{dy}{dt}=2\frac{dx}{dt} \): \( 4y=18\Rightarrow y=\frac92 \).
2. Then \( \frac{81}4=18x\Rightarrow x=\frac98 \). The point is \( \left(\frac98,\frac92\right) \). Answer: (d)

Teacher's Note:
1. Differentiate the equation of the parabola with respect to time.
2. Substitute the given rate relation.

Answer: (b) \( (x-1)^3 \)

Explanation:
1. Integrate: \( f'(x)=3(x-1)^2+C \). The tangent at \( (2,1) \) has slope 3, so \( f'(2)=3+C=3 \) and \( C=0 \).
2. Integrate again: \( f(x)=(x-1)^3+k \). Since \( f(2)=1 \), \( 1+k=1 \) and \( k=0 \).
3. \( f(x)=(x-1)^3 \). Answer: (b)

Teacher's Note:
1. Integrate \( f'' \) twice.
2. Use the slope and the point to find the constants.

Answer: (a) AP

Explanation:
1. Let \( f(x)=Ax^2+Bx+C \). \( f(1)=f(-1)\Rightarrow2B=0\Rightarrow B=0 \), so \( f(x)=Ax^2+C \) and \( f'(x)=2Ax \).
2. \( f'(a)=2Aa,\ f'(b)=2Ab,\ f'(c)=2Ac \) are constant multiples of \( a,b,c \), which are in AP. So they are in AP. Answer: (a)

Teacher's Note:
1. Use \( f(1)=f(-1) \) to find the form of \( f \).
2. Differentiate.

Answer: (c) 0

Explanation:
1. \( f^{(k)}(1)=n(n-1)\cdots(n-k+1) \), so \( \frac{f^{(k)}(1)}{k!}={}^nC_k \).
2. The sum is \( {}^nC_0-{}^nC_1+{}^nC_2-\cdots+(-1)^n\,{}^nC_n=(1-1)^n=0 \). Answer: (c)

Teacher's Note:
1. Compute \( f^{(k)}(1)/k! \).
2. Recognise the binomial expansion of \( (1-1)^n \).

Answer: (c) \( \frac23 \)

Explanation:
1. The limit is a \( \frac00 \) form. By L'Hopital's rule it equals \( \lim\left(\frac1{3+x}+\frac1{3-x}\right)=\frac13+\frac13 \).
2. So \( k=\frac23 \). Answer: (c)

Teacher's Note:
1. Check the 0/0 form.
2. Differentiate numerator and denominator.

Answer: (b) continuous for all \( x \) but not differentiable at \( x=0 \)

Explanation:
1. For \( x\lt0 \): \( \frac1{|x|}+\frac1x=0 \), so \( f(x)=x \). For \( x\gt0 \): \( f(x)=xe^{-2/x} \). Both tend to 0 as \( x\to0 \), so \( f \) is continuous at 0 (and elsewhere).
2. Left derivative at 0: \( \frac{f(h)-0}h=1 \) for \( h\lt0 \). Right derivative: \( \frac{he^{-2/h}}h=e^{-2/h}\to0 \).
3. The one-sided derivatives are 1 and 0, so \( f \) is not differentiable at 0. Answer: (b)

Teacher's Note:
1. Remove the modulus for each sign of \( x \).
2. Compare one-sided derivatives at 0.

Answer: (c) 2

Explanation:
1. \( f'(x)=6x^2-18ax+12a^2=6(x-a)(x-2a) \), so the critical points are \( x=a \) and \( x=2a \).
2. \( f''(x)=12x-18a \): \( f''(a)=-6a\lt0 \) (maximum at \( a \)) and \( f''(2a)=6a\gt0 \) (minimum at \( 2a \)). So \( p=a \) and \( q=2a \).
3. \( p^2=q\Rightarrow a^2=2a\Rightarrow a=2 \) (since \( a\gt0 \)). Answer: (c)

Teacher's Note:
1. Find the critical points.
2. Use the second derivative to decide max and min.

Answer: (a) \( n^2y \)

Explanation:
1. \( \frac{dy}{dx}=n\left(x+\sqrt{1+x^2}\right)^{n-1}\left(1+\frac x{\sqrt{1+x^2}}\right)=\frac{ny}{\sqrt{1+x^2}} \), so \( (1+x^2)y_1^2=n^2y^2 \).
2. Differentiate again: \( (1+x^2)\,2y_1y_2+2xy_1^2=2n^2yy_1 \). Dividing by \( 2y_1 \): \( (1+x^2)y_2+xy_1=n^2y \). Answer: (a)

Teacher's Note:
1. Show that \( (1+x^2)y_1^2=n^2y^2 \).
2. Differentiate this relation once more.

Answer: (d) does not exist

Explanation:
1. \( 1-\cos2x=2\sin^2x \), so the expression is \( \frac{\sqrt2|\sin x|}{\sqrt2\,x}=\frac{|\sin x|}x \).
2. For \( x\to0^+ \) it tends to \( 1 \), and for \( x\to0^- \) it tends to \( -1 \).
3. The one-sided limits differ, so the limit does not exist. Answer: (d)

Teacher's Note:
1. Use \( 1-\\cos2x=2\\sin^2x \).
2. A square root of a square gives a modulus.

Answer: (a) \( e^4 \)

Explanation:
1. \( \frac{x^2+5x+3}{x^2+x+2}=1+\frac{4x+1}{x^2+x+2} \), so the form is \( 1^\infty \).
2. The limit is \( e^{L} \) with \( L=\lim x\cdot\frac{4x+1}{x^2+x+2}=\lim\frac{4+1/x}{1+1/x+2/x^2}=4 \). So the limit is \( e^4 \). Answer: (a)

Teacher's Note:
1. Write the base as \( 1+\frac{\cdots}{\cdots} \).
2. Compute the limit of the exponent times the added term.

Answer: (b) \( \frac{\sin^2(a+y)}{\sin a} \)

Explanation:
1. \( x=\frac{\sin y}{\sin(a+y)} \), so \( \frac{dx}{dy}=\frac{\sin(a+y)\cos y-\sin y\cos(a+y)}{\sin^2(a+y)} \).
2. The numerator is \( \sin((a+y)-y)=\sin a \), so \( \frac{dx}{dy}=\frac{\sin a}{\sin^2(a+y)} \).
3. Hence \( \frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a} \). Answer: (b)

Teacher's Note:
1. Express \( x \) as a function of \( y \).
2. Use the sine of a difference.

Answer: (d) \( \frac{\log x}{(1+\log x)^2} \)

Explanation:
1. Take logarithms: \( y\log x=x-y \), so \( y(1+\log x)=x \) and \( y=\frac x{1+\log x} \).
2. \( \frac{dy}{dx}=\frac{(1+\log x)-x\cdot\frac1x}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2} \). Answer: (d)

Teacher's Note:
1. Take natural logarithms.
2. Solve for \( y \) and use the quotient rule.

Answer: (a) cut at right angle

Explanation:
1. For the first curve: \( 3x^2-3y^2-6xy\frac{dy}{dx}=0 \), so \( \frac{dy}{dx}=\frac{x^2-y^2}{2xy} \).
2. For the second curve: \( 6xy+3x^2\frac{dy}{dx}-3y^2\frac{dy}{dx}=0 \), so \( \frac{dy}{dx}=\frac{-2xy}{x^2-y^2} \).
3. The product of the slopes is \( \frac{x^2-y^2}{2xy}\cdot\frac{-2xy}{x^2-y^2}=-1 \), so the curves cut at right angles. Answer: (a)

Teacher's Note:
1. Find the slope of the tangent to each curve.
2. Multiply the slopes.

Answer: (c) \( (-\infty,\infty) \)

Explanation:
1. \( f'(x)=-\frac1{1+x^2}+1=\frac{x^2}{1+x^2}\ge0 \) for all real \( x \).
2. So \( f \) is increasing on the whole real line \( (-\infty,\infty) \), which contains the other intervals. Answer: (c)

Teacher's Note:
1. Differentiate \( f \).
2. Check the sign of \( f' \) for all real \( x \).

Answer: (b) 2

Explanation:
1. \( f'(x)=\frac13\left[\frac1{(x+1)^{2/3}}-\frac1{(x-1)^{2/3}}\right] \), which is zero when \( (x-1)^{2/3}=(x+1)^{2/3} \), i.e. \( x=0 \), and does not exist at \( x=\pm1 \).
2. Compare values: \( f(0)=1+1=2 \) and \( f(1)=2^{1/3}\approx1.26 \). The greatest value is 2. Answer: (b)

Teacher's Note:
1. Find the critical points and the end points.
2. Compare the values of \( f \).

Answer: (c) \( e^{-5} \)

Explanation:
1. \( \frac{x-3}{x+2}=1-\frac5{x+2} \), a \( 1^\infty \) form.
2. The limit is \( e^{L} \) with \( L=\lim x\cdot\left(-\frac5{x+2}\right)=-5 \), so the limit is \( e^{-5} \). Answer: (c)

Teacher's Note:
1. Write the base as \( 1+\frac{\cdots}{\cdots} \).
2. Evaluate the limit of the exponent.

Answer: (c) -4

Explanation:
1. Write \( xf(2)-2f(x)=f(2)(x-2)+2f(2)-2f(x) \), so the expression is \( f(2)-2\,\frac{f(x)-f(2)}{x-2} \).
2. As \( x\to2 \) this tends to \( f(2)-2f'(2)=4-8=-4 \). Answer: (c)

Teacher's Note:
1. Add and subtract \( 2f(2) \) in the numerator.
2. Use the definition of \( f'(2) \).