JEE Physics Mathematical Tools MCQs Set 01

Answer: (a) \( \frac{1}{\cos t-1} \)

Explanation:
1. Quotient rule: numerator \( \cos t(1-\cos t)-\sin t\cdot\sin t=\cos t-1 \).
2. So \( \frac{ds}{dt}=\frac{\cos t-1}{(1-\cos t)^2}=\frac{-1}{1-\cos t}=\frac{1}{\cos t-1} \).

Teacher's Note:
1. Write the rule you are using (product, quotient or chain) before substituting.
2. Simplify the result with trigonometric identities to match the standard form.

Answer: (a) \( 2r^2 \)

Explanation:
1. Sides \( 2r\cos\phi \) and \( 2r\sin\phi \) (diagonal is the diameter).
2. Area \( =4r^2\sin\phi\cos\phi=2r^2\sin2\phi \).
3. Maximum when \( \sin2\phi=1 \): \( 2r^2 \) (a square).

Teacher's Note:
1. For a maximum, set the derivative to zero or use the largest value of the trigonometric factor.
2. The rectangle of largest area in a circle is always a square.

Answer: (b) \( 113^\circ \)

Explanation:
1. The resultant of 5 N and 13 N must be 12 N: \( 144=25+169+2(5)(13)\cos\theta \).
2. So \( \cos\theta=-\frac{50}{130}=-\frac5{13} \).
3. Since \( \cos67^\circ=\sin23^\circ=\frac5{13} \), \( \theta=180^\circ-67^\circ=113^\circ \).

Teacher's Note:
1. Resolve every vector along two perpendicular axes, add components, then recombine.
2. Use the dot product for angles and projections, and the cross product for areas and perpendicular directions.

Answer: (b) \( \frac{\csc\theta}{\cot\theta+\csc\theta} \)

Explanation:
1. Chain rule: \( \frac{dr}{d\theta}=-(\csc\theta+\cot\theta)^{-2}(-\csc\theta\cot\theta-\csc^2\theta) \).
2. Factor: \( \frac{\csc\theta(\cot\theta+\csc\theta)}{(\csc\theta+\cot\theta)^2}=\frac{\csc\theta}{\cot\theta+\csc\theta} \).

Teacher's Note:
1. Write the rule you are using (product, quotient or chain) before substituting.
2. Simplify the result with trigonometric identities to match the standard form.

Answer: (a) \( \hat i+\pi\hat j+2\hat k \)

Explanation:
1. \( \frac{d\vec A}{dt}=\hat i-\pi\cos(\pi t)\hat j+2t\hat k \).
2. At \( t=1 \): \( \cos\pi=-1 \) so \( \hat i+\pi\hat j+2\hat k \).

Teacher's Note:
1. Write the rule you are using (product, quotient or chain) before substituting.
2. Simplify the result with trigonometric identities to match the standard form.

Answer: (c) \( \frac25(5s+4)^{1/2}+C \)

Explanation:
1. Put \( u=5s+4,\ du=5\,ds \).
2. Integral \( \frac15\int u^{-1/2}du=\frac15\cdot2u^{1/2} \).
3. Result: \( \frac25(5s+4)^{1/2}+C \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (c) \( -\frac34\cos(x^{4/3}-8)+C \)

Explanation:
1. Put \( u=x^{4/3}-8,\ du=\frac43x^{1/3}dx \).
2. Integral \( \frac34\int\sin u\,du=-\frac34\cos u \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (c) \( 4y^2-\frac{8}{3}y^{3/4}+C \)

Explanation:
1. Write as \( 8y-2y^{-1/4} \).
2. Integrate: \( 4y^2-2\cdot\frac{y^{3/4}}{3/4} \).
3. Result: \( 4y^2-\frac83y^{3/4}+C \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (c) \( 150^\circ \)

Explanation:
1. Perpendicular resultant: \( u+v\cos\theta=0 \) so \( \cos\theta=-u/v \).
2. Magnitude: \( v^2\sin^2\theta=v^2/4 \) so \( \sin\theta=\tfrac12 \).
3. Since \( \cos\theta \) is negative, \( \theta=150^\circ \).

Teacher's Note:
1. Resolve every vector along two perpendicular axes, add components, then recombine.
2. Use the dot product for angles and projections, and the cross product for areas and perpendicular directions.

Answer: (d) \( 60^\circ \)

Explanation:
1. \( |A-B|^2=2A^2(1-\cos\theta)=A^2 \) gives \( \cos\theta=\tfrac12 \).
2. So \( \theta=60^\circ \).

Teacher's Note:
1. Resolve every vector along two perpendicular axes, add components, then recombine.
2. Use the dot product for angles and projections, and the cross product for areas and perpendicular directions.

Answer: (b) \( x^2+\frac{2}{x}+C \)

Explanation:
1. Expand: \( 2x-2x^{-2} \).
2. Integrate: \( x^2-2\cdot\frac{x^{-1}}{-1}=x^2+\frac2x \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (b) \( \frac{x^2+x+4}{(x+0.5)^2} \)

Explanation:
1. Quotient rule: \( g'=\frac{(x+0.5)(2x)-(x^2-4)(1)}{(x+0.5)^2} \).
2. Numerator: \( 2x^2+x-x^2+4=x^2+x+4 \).
3. So \( g'(x)=\frac{x^2+x+4}{(x+0.5)^2} \).

Teacher's Note:
1. Write the rule you are using (product, quotient or chain) before substituting.
2. Simplify the result with trigonometric identities to match the standard form.

Answer: (b) \( \frac{32\pi Rv}{15} \)

Explanation:
1. The water rises at \( v/15 \) while the cylinder goes down at \( v \), so the wet length grows at \( v+\frac v{15}=\frac{16v}{15} \).
2. Wet curved area \( =2\pi R\times\text{length} \).
3. Rate \( 2\pi R\cdot\frac{16v}{15}=\frac{32\pi Rv}{15} \).

Teacher's Note:
1. Relative speed between cylinder and water surface decides how fast the wet length grows.
2. Multiply by the circumference to get the area rate.

Answer: (d) \( -\frac12\cos2x+\cot x+C \)

Explanation:
1. \( \int\sin2x\,dx=-\frac12\cos2x \) and \( \int\csc^2x\,dx=-\cot x \).
2. So the integral is \( -\frac12\cos2x+\cot x+C \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (d) \( \frac{x^{3/2}}{3}+4x^{1/2}+C \)

Explanation:
1. Write as \( \frac12x^{1/2}+2x^{-1/2} \).
2. Integrate: \( \frac12\cdot\frac{x^{3/2}}{3/2}+2\cdot\frac{x^{1/2}}{1/2} \).
3. Result: \( \frac{x^{3/2}}{3}+4x^{1/2}+C \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (a) \( 2 \)

Explanation:
1. Antiderivative: \( -x^2+4x \).
2. At \( 3/2 \): \( -\tfrac94+6=\tfrac{15}{4} \); at \( 1/2 \): \( -\tfrac14+2=\tfrac74 \).
3. Difference: \( \frac{15}{4}-\frac74=2 \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (d) \( -5\sin t,\ -5\cos t \)

Explanation:
1. \( v=\frac{ds}{dt}=-5\sin t \).
2. \( a=\frac{dv}{dt}=-5\cos t \).

Teacher's Note:
1. Write the rule you are using (product, quotient or chain) before substituting.
2. Simplify the result with trigonometric identities to match the standard form.

Answer: (d) 3 km due North

Explanation:
1. Second displacement: \( 5\cos37^\circ=4 \) km East and \( 5\sin37^\circ=3 \) km South.
2. Required total is \( 6 \) km East, 0 North, so the third has East part \( 6-2-4=0 \) and North part \( 0+3=3 \).
3. The third displacement is 3 km due North.

Teacher's Note:
1. Resolve every vector along two perpendicular axes, add components, then recombine.
2. Use the dot product for angles and projections, and the cross product for areas and perpendicular directions.

Answer: (c) \( \sqrt2:\sqrt3:\sqrt2 \)

Explanation:
1. Let \( R^2=P^2+Q^2+2PQc \) with \( c=\cos\theta \).
2. Conditions: \( P^2+4Q^2+4PQc=4R^2 \) and \( P^2+Q^2-2PQc=4R^2 \).
3. Solving: \( Q=\frac{\sqrt6}{2}P,\ c=-\frac{\sqrt6}{4} \), which gives \( R=P \).
4. Hence \( P:Q:R=1:\tfrac{\sqrt6}{2}:1=\sqrt2:\sqrt3:\sqrt2 \).

Teacher's Note:
1. Resolve every vector along two perpendicular axes, add components, then recombine.
2. Use the dot product for angles and projections, and the cross product for areas and perpendicular directions.

Answer: (a) \( \frac12\sin^6\frac{x}{3}+C \)

Explanation:
1. Put \( u=\sin\frac x3,\ du=\frac13\cos\frac x3\,dx \).
2. Integral \( 3\int u^5du=\frac{u^6}{2} \).
3. Result: \( \frac12\sin^6\frac x3+C \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (d) \( 2.5 \)

Explanation:
1. Split at \( x=0 \): \( \int_{-2}^{0}(-x)dx+\int_0^1x\,dx \).
2. The first part is \( 2 \) and the second part is \( 0.5 \).
3. Total area \( 2.5 \) square units.

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.

Answer: (c) \( 20\sin x\cos^{-5}x \)

Explanation:
1. Chain rule: \( 5(-4)\cos^{-5}x\cdot(-\sin x) \).
2. This is \( 20\sin x\cos^{-5}x \).

Teacher's Note:
1. Differentiate the outer power first and then multiply by the derivative of the inner function.
2. Count the two negative signs carefully.

Answer: (c) \( \frac{v}{15} \)

Explanation:
1. The water surface is an annulus of area \( \pi(16R^2-R^2)=15\pi R^2 \).
2. Volume pushed aside per second \( =\pi R^2v \), and this equals \( 15\pi R^2\frac{dh}{dt} \).
3. Hence \( \frac{dh}{dt}=\frac{v}{15} \).

Teacher's Note:
1. Equate the rate of volume displaced to the rate of volume gained by the water.
2. Use the area of the free surface only, not the whole vessel.

Answer: (b) \( 9b^3 \)

Explanation:
1. \( \left[\frac{x^3}{3}\right]_0^{3b}=\frac{27b^3}{3}=9b^3 \).

Teacher's Note:
1. For a substitution, choose the inner function whose derivative also appears in the integrand.
2. After integrating, differentiate your answer to confirm it.