Practice MCQs for JEE Physics Mathematical Tools
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Multiple Choice Questions
(a) \( \dfrac{2xy+y^2}{x^2+2xy} \)
(b) \( -\dfrac{2xy+y^2}{x^2+2xy} \)
(c) \( -\dfrac{x^2+2xy}{2xy+y^2} \)
(d) \( -\dfrac{2x+2y}{x^2+y^2} \)
Show Answer & Explanation
Answer: (b) \( -\dfrac{2xy+y^2}{x^2+2xy} \)
Explanation:
1. Differentiate: \( 2xy+x^2y'+y^2+2xyy'=0 \).
2. \( y'(x^2+2xy)=-(2xy+y^2) \).
Teacher's Note:
1. Treat y as a function of x and use the product rule on both terms.
2. Group the \( y' \) terms on one side.
(a) \( \dfrac{18y-3x^2}{3y^2-18x} \)
(b) \( \dfrac{3x^2-18y}{3y^2-18x} \)
(c) \( \dfrac{18y+3x^2}{3y^2+18x} \)
(d) \( \dfrac{x^2-6y}{y^2-6x} \cdot(-1) \)
Show Answer & Explanation
Answer: (a) \( \dfrac{18y-3x^2}{3y^2-18x} \)
Explanation:
1. Differentiate: \( 3x^2+3y^2y'=18y+18xy' \).
2. \( y'(3y^2-18x)=18y-3x^2 \).
Teacher's Note:
1. The two forms (18y − 3x²)/(3y² − 18x) and (6y − x²)/(y² − 6x) are the same.
2. The folium of Descartes has this equation.
(a) \( \pi r^2\dfrac{dr}{dt} \)
(b) \( 2\pi\dfrac{dr}{dt} \)
(c) \( \pi r\dfrac{dr}{dt} \)
(d) \( \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt} \)
Explanation:
1. Chain rule: \( \dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt} \).
2. \( \dfrac{dA}{dr}=2\pi r \).
Teacher's Note:
1. The factor 2πr is the circumference.
2. Rate of area = circumference × rate of radius.
(a) \( 4\pi r^2\dfrac{dr}{dt} \)
(b) \( 4\pi r\dfrac{dr}{dt} \)
(c) \( 8\pi r\dfrac{dr}{dt} \)
(d) \( \dfrac{4}{3}\pi r^3\dfrac{dr}{dt} \)
Show Answer & Explanation
Answer: (c) \( 8\pi r\dfrac{dr}{dt} \)
Explanation:
1. \( \dfrac{dS}{dr}=8\pi r \).
2. \( \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt} \).
Teacher's Note:
1. Chain rule again.
2. The volume would give \( 4\pi r^2 \) instead.
(a) 4 m
(b) 8 m
(c) 12 m
(d) 2 m
Show Answer & Explanation
Answer: (b) 8 m
Explanation:
1. \( \dfrac{dx}{dt}=-2t+4=0 \) gives t = 2 s.
2. \( x_{max}=-4+8+4=8 \) m (the second derivative is −2, so it is a maximum).
Teacher's Note:
1. Set the velocity to zero.
2. Check the second derivative.
(a) 39 and 38
(b) 38 and 39
(c) 40 and 36
(d) 11 and 39
Show Answer & Explanation
Answer: (a) 39 and 38
Explanation:
1. \( y'=6x^2-30x+36=6(x-2)(x-3) \).
2. \( y''=12x-30 \): negative at x = 2 (maximum) and positive at x = 3 (minimum).
3. \( y(2)=16-60+72+11=39 \), \( y(3)=54-135+108+11=38 \).
Teacher's Note:
1. Here the local maximum (at x = 2) happens to exceed the local minimum (at x = 3), but this is not true for every cubic.
2. Use the sign of the second derivative.
(a) \( 6(8x-1)^2 \)
(b) \( 16(8x-1)^2 \)
(c) \( 48(8x-1) \)
(d) \( 48(8x-1)^2 \)
Show Answer & Explanation
Answer: (d) \( 48(8x-1)^2 \)
Explanation:
1. \( \dfrac{dy}{du}=6u^2 \) and \( \dfrac{du}{dx}=8 \).
2. \( \dfrac{dy}{dx}=48u^2=48(8x-1)^2 \).
Teacher's Note:
1. Chain rule.
2. Replace u by 8x − 1 in the answer.
(a) \( \dfrac13\sin\dfrac x3 \)
(b) \( -\dfrac13\cos\dfrac x3 \)
(c) \( -\dfrac13\sin\dfrac x3 \)
(d) \( -3\sin\dfrac x3 \)
Show Answer & Explanation
Answer: (c) \( -\dfrac13\sin\dfrac x3 \)
Explanation:
1. \( \dfrac{dy}{du}=-\sin u \) and \( \dfrac{du}{dx}=-\dfrac13 \).
2. \( \dfrac{dy}{dx}=\left(-\sin u\right)\left(-\dfrac13\right)=\dfrac13\sin u=\dfrac13\sin\left(-\dfrac x3\right)=-\dfrac13\sin\dfrac x3 \).
Teacher's Note:
1. Two negative signs from the chain rule give a positive factor, and then sin(−x/3) = −sin(x/3) brings the sign back.
2. Check by writing y = cos(x/3), which is the same function: its derivative is −(1/3) sin(x/3).
(a) \( \dfrac2{x^2}-\sin x \)
(b) \( -\dfrac2{x^2}-\sin x \)
(c) \( -\dfrac1{x^2}-\sin x \)
(d) \( -\dfrac2x-\cos x \)
Show Answer & Explanation
Answer: (b) \( -\dfrac2{x^2}-\sin x \)
Explanation:
1. \( y'=\dfrac2x+\cos x \).
2. \( y''=-\dfrac2{x^2}-\sin x \).
Teacher's Note:
1. \( \ln x^2=2\ln x \).
2. Differentiate twice.
(a) \( \tfrac17x^{-6/7}+\sec^2x \) and \( -\tfrac6{49}x^{-13/7}+2\tan x\sec^2x \)
(b) \( \tfrac17x^{-6/7}+\sec^2x \) and \( \tfrac6{49}x^{-13/7}+2\tan x\sec^2x \)
(c) \( 7x^{6}+\sec^2x \) and \( 42x^5+2\sec^2x \)
(d) \( \tfrac17x^{-6/7}+\sec x\tan x \) and \( -\tfrac6{49}x^{-13/7}+\sec x \)
Show Answer & Explanation
Answer: (a) \( \tfrac17x^{-6/7}+\sec^2x \) and \( -\tfrac6{49}x^{-13/7}+2\tan x\sec^2x \)
Explanation:
1. \( y'=\tfrac17x^{1/7-1}+\sec^2x \).
2. \( y''=\tfrac17\cdot\left(-\tfrac67\right)x^{-13/7}+2\sec x\cdot\sec x\tan x \).
Teacher's Note:
1. Subtract 1 from the exponent: 1/7 − 1 = −6/7.
2. The derivative of sec²x is 2 sec²x tan x.
(a) \( \cos\theta+\sec\theta\tan\theta \)
(b) \( \cos\theta-\sec^2\theta \)
(c) \( \sec^2\theta \)
(d) \( \cos\theta+\sec^2\theta \)
Show Answer & Explanation
Answer: (d) \( \cos\theta+\sec^2\theta \)
Explanation:
1. \( r=\sin\theta+\tan\theta \) since \( \sec\theta\sin\theta=\tan\theta \).
2. \( \dfrac{dr}{d\theta}=\cos\theta+\sec^2\theta \).
Teacher's Note:
1. Simplify before differentiating.
2. secθ sinθ = tanθ.
(a) \( \dfrac{-\csc^2x}{1+\cot x} \)
(b) \( \dfrac{\csc^2x}{(1+\cot x)^2} \)
(c) \( \dfrac{-\csc^2x}{(1+\cot x)^2} \)
(d) \( \dfrac{-\csc^2x}{(1+\cot x)^2}\cot x \)
Show Answer & Explanation
Answer: (c) \( \dfrac{-\csc^2x}{(1+\cot x)^2} \)
Explanation:
1. Quotient rule: \( \dfrac{-\csc^2x(1+\cot x)+\cot x\csc^2x}{(1+\cot x)^2} \).
2. The numerator simplifies to \( -\csc^2x \).
Teacher's Note:
1. The cot x · csc²x terms cancel in the numerator.
2. The denominator is squared.
(a) \( 3\sin^2x+3\cos3x \)
(b) \( 3\sin^2x\cos x+3\cos3x \)
(c) \( 3\sin^2x\cos x+\cos3x \)
(d) \( 3\cos^2x\sin x+3\cos3x \)
Show Answer & Explanation
Answer: (b) \( 3\sin^2x\cos x+3\cos3x \)
Explanation:
1. Chain rule on \( \sin^3x \): \( 3\sin^2x\cos x \).
2. Chain rule on \( \sin3x \): \( 3\cos3x \).
Teacher's Note:
1. The two terms need different applications of the chain rule.
2. Do not drop the inner derivative.
(a) \( 4x\sin(x^2+1)\cos(x^2+1) \)
(b) \( 2\sin(x^2+1)\cos(x^2+1) \)
(c) \( 2x\sin(x^2+1) \)
(d) \( 4x\sin^2(x^2+1) \)
Show Answer & Explanation
Answer: (a) \( 4x\sin(x^2+1)\cos(x^2+1) \)
Explanation:
1. Outer: \( 2\sin(\cdot) \); middle: \( \cos(x^2+1) \); inner: \( 2x \).
2. Product: \( 4x\sin(x^2+1)\cos(x^2+1) \).
Teacher's Note:
1. A three-layer chain rule.
2. This equals \( 2x\sin(2x^2+2) \).
(a) \( \dfrac{1}{\sqrt{2r-r^2}} \)
(b) \( \dfrac{2-2r}{\sqrt{2r-r^2}} \)
(c) \( \dfrac{r-1}{\sqrt{2r-r^2}} \)
(d) \( \dfrac{1-r}{\sqrt{2r-r^2}} \)
Show Answer & Explanation
Answer: (d) \( \dfrac{1-r}{\sqrt{2r-r^2}} \)
Explanation:
1. \( \dfrac{dq}{dr}=\dfrac{1}{2\sqrt{2r-r^2}}(2-2r) \).
2. \( =\dfrac{1-r}{\sqrt{2r-r^2}} \).
Teacher's Note:
1. The factor 2 in the denominator cancels with the 2 in 2 − 2r.
2. q is the upper half of a circle, so dq/dr = 0 at r = 1.
(a) \( 10\pi rh \)
(b) \( 5\pi r^2+10\pi rh \)
(c) \( 5\pi r^2 \)
(d) \( \pi r^2 \)
Show Answer & Explanation
Answer: (c) \( 5\pi r^2 \)
Explanation:
1. \( \dfrac{dV}{dt}=\pi r^2\dfrac{dh}{dt} \) when r is constant.
2. \( =5\pi r^2 \).
Teacher's Note:
1. Differentiate only the changing quantity.
2. The units are m³/s.
(a) \( 5\pi r^2 \)
(b) \( 10\pi rh \)
(c) \( 5\pi rh \)
(d) \( 5\pi r^2+10\pi rh \)
Show Answer & Explanation
Answer: (b) \( 10\pi rh \)
Explanation:
1. \( \dfrac{dV}{dt}=2\pi rh\dfrac{dr}{dt} \).
2. \( =2\pi rh\times5=10\pi rh \).
Teacher's Note:
1. The power 2 on r gives the factor 2.
2. h is constant here.
(a) \( 5\pi r^2+10\pi rh \)
(b) \( 5\pi r^2 \)
(c) \( 10\pi rh \)
(d) \( 25\pi r^2h \)
Show Answer & Explanation
Answer: (a) \( 5\pi r^2+10\pi rh \)
Explanation:
1. \( \dfrac{dV}{dt}=\pi r^2\dfrac{dh}{dt}+2\pi rh\dfrac{dr}{dt} \).
2. \( =5\pi r^2+10\pi rh \).
Teacher's Note:
1. Add the two contributions.
2. Use the product rule on \( r^2h \).
(a) x = 20, y = 40
(b) x = 10, y = 50
(c) x = 25, y = 35
(d) x = y = 30
Show Answer & Explanation
Answer: (d) x = y = 30
Explanation:
1. \( P=x(60-x) \), \( \dfrac{dP}{dx}=60-2x=0 \) gives x = 30.
2. \( P''=-2\lt 0 \), so this is a maximum: x = y = 30 and P = 900.
Teacher's Note:
1. For a fixed sum the product is maximum when the numbers are equal.
2. Check the second derivative.
(a) \( 40/3 \) m
(b) \( \sqrt{10} \) m
(c) \( \sqrt{40/3} \) m
(d) \( \sqrt{20} \) m
Show Answer & Explanation
Answer: (c) \( \sqrt{40/3} \) m
Explanation:
1. Base side x and height h: \( x^2+4xh=40 \), so \( h=\dfrac{40-x^2}{4x} \).
2. \( V=x^2h=\dfrac{40x-x^3}{4} \), \( \dfrac{dV}{dx}=\dfrac{40-3x^2}{4}=0 \).
3. \( x^2=\dfrac{40}{3} \), so \( x=\sqrt{40/3}\approx3.65 \) m (then h = x/2).
Teacher's Note:
1. The sheet area covers the base and four sides.
2. The second derivative is negative, so it is a maximum.
(a) \( \dfrac1x+\dfrac1{2x^2}+C \)
(b) \( -\dfrac1x-\dfrac1{2x^2}+C \)
(c) \( -\dfrac1x+\dfrac1{2x^2}+C \)
(d) \( \dfrac{x^{-2}}{-2}+x^{-3}+C \)
Show Answer & Explanation
Answer: (b) \( -\dfrac1x-\dfrac1{2x^2}+C \)
Explanation:
1. Expand: \( x^{-2}+x^{-3} \).
2. \( \int x^{-2}dx=-x^{-1} \) and \( \int x^{-3}dx=-\dfrac{x^{-2}}2 \).
Teacher's Note:
1. Expand before integrating.
2. Raise the power by one and divide by the new power.
(a) \( 2x+\cot x+C \)
(b) \( 2x-\cot x+C \)
(c) \( x+\cot x+C \)
(d) \( x-\csc^2x+C \)
Show Answer & Explanation
Answer: (a) \( 2x+\cot x+C \)
Explanation:
1. \( 1-\cot^2x=2-\csc^2x \).
2. \( \int(2-\csc^2x)dx=2x+\cot x+C \).
Teacher's Note:
1. Use \( 1+\cot^2x=\csc^2x \).
2. \( \int\csc^2x\,dx=-\cot x \).
(a) \( \cos\theta+\theta+C \)
(b) \( \sin\theta+\theta+C \)
(c) \( -\cos\theta-\theta+C \)
(d) \( -\cos\theta+\theta+C \)
Show Answer & Explanation
Answer: (d) \( -\cos\theta+\theta+C \)
Explanation:
1. \( \cos\theta\tan\theta=\sin\theta \) and \( \cos\theta\sec\theta=1 \).
2. \( \int(\sin\theta+1)d\theta=-\cos\theta+\theta+C \).
Teacher's Note:
1. Simplify the integrand first.
2. The integral of sin is minus cos.
(a) \( 12(y^4+4y^2+1)^3+C \)
(b) \( 4(y^4+4y^2+1)^3+C \)
(c) \( (y^4+4y^2+1)^3+C \)
(d) \( \dfrac{(y^4+4y^2+1)^3}{3}+C \)
Show Answer & Explanation
Answer: (c) \( (y^4+4y^2+1)^3+C \)
Explanation:
1. \( du=(4y^3+8y)\,dy=4(y^3+2y)\,dy \), so \( 12(y^3+2y)dy=3\,du \).
2. \( \int3u^2du=u^3+C \).
Teacher's Note:
1. Match the factor \( y^3+2y \) with du.
2. Check by differentiating the answer.
(a) \( \dfrac15\sqrt{5x+8}+C \)
(b) \( \dfrac25\sqrt{5x+8}+C \)
(c) \( 2\sqrt{5x+8}+C \)
(d) \( \dfrac25(5x+8)^{3/2}+C \)
Show Answer & Explanation
Answer: (b) \( \dfrac25\sqrt{5x+8}+C \)
Explanation:
1. Put \( u=5x+8 \), \( dx=du/5 \).
2. \( \dfrac15\int u^{-1/2}du=\dfrac15\cdot2u^{1/2}=\dfrac25\sqrt{5x+8} \).
Teacher's Note:
1. Divide by the coefficient of x.
2. \( \int u^{-1/2}du=2\sqrt u \).
(a) \( -\dfrac13(3-2s)^{3/2}+C \)
(b) \( \dfrac13(3-2s)^{3/2}+C \)
(c) \( -\dfrac23(3-2s)^{3/2}+C \)
(d) \( -\dfrac12(3-2s)^{3/2}+C \)
Show Answer & Explanation
Answer: (a) \( -\dfrac13(3-2s)^{3/2}+C \)
Explanation:
1. Put \( u=3-2s \), \( ds=-du/2 \).
2. \( -\dfrac12\cdot\dfrac{u^{3/2}}{3/2}=-\dfrac13u^{3/2} \).
Teacher's Note:
1. The inner derivative is −2, which gives the negative sign.
2. \( \tfrac12\cdot\tfrac23=\tfrac13 \).
(a) \( 3\tan(3x+2)+C \)
(b) \( \tan(3x+2)+C \)
(c) \( \dfrac13\sec(3x+2)\tan(3x+2)+C \)
(d) \( \dfrac13\tan(3x+2)+C \)
Show Answer & Explanation
Answer: (d) \( \dfrac13\tan(3x+2)+C \)
Explanation:
1. Put \( u=3x+2 \), \( dx=du/3 \).
2. \( \dfrac13\int\sec^2u\,du=\dfrac13\tan u+C \).
Teacher's Note:
1. Divide by the coefficient of x.
2. \( \int\sec^2u\,du=\tan u \).
(a) \( \dfrac{3}{(2+\sin t)^2}+C \)
(b) \( -\dfrac{6}{(2+\sin t)^2}+C \)
(c) \( -\dfrac{3}{(2+\sin t)^2}+C \)
(d) \( -\dfrac{3}{2+\sin t}+C \)
Show Answer & Explanation
Answer: (c) \( -\dfrac{3}{(2+\sin t)^2}+C \)
Explanation:
1. Put \( u=2+\sin t \), \( du=\cos t\,dt \).
2. \( 6\int u^{-3}du=6\cdot\dfrac{u^{-2}}{-2}=-\dfrac3{u^2} \).
Teacher's Note:
1. The numerator cos t is exactly du.
2. Differentiate the answer to check.
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FAQs
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