3. The general solution of the differential equation √(1+x²+y²+x²y²)+xy(dy/dx) = 0

(where C is a constant of integration)

1. 1)
   \(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}= \frac{1}{2}log_{e}\left ( \frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1} \right )+c\)
2. 2)
   \(\sqrt{1+y^{2}}-\sqrt{1+x^{2}}= \frac{1}{2}log_{e}\left ( \frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1} \right )+c\)
3. 3)
   \(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}= \frac{1}{2}log_{e}\left ( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right )+c\)
4. 4)
   \(\sqrt{1+y^{2}}-\sqrt{1+x^{2}}= \frac{1}{2}log_{e}\left ( \frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1} \right )+c\)

4. Let L₁ be a tangent to the parabola y² = 4(x+1) and L₂ be a tangent to the parabola y² = 8(x+2) such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line:

1. 1) x+2y = 0
2. 2) x+2 = 0
3. 3) 2x+1 = 0
4. 4) x+3 = 0

Solution:

1. Let t₁ tangent of y² = 4(x+1)

   L₁ : t₁y = (x+1)+t₁² .......(i)

   and t₂ tangent of y² = 8(x+2)

   L₂ : t₂y = (x+2)+2t₂² ….(ii)

   L₁ perpendicular to L₂

   (1/t₁)(1/t₂) = -1

   t₁t₂ = -1

   t₂(i) -t₁(ii)

   t₁t₂y = t₂(x+1)+ t₂. t₁²

   t₁t₂y = t₁(x+2)+2t₂². t₁

   Subtracting above equations

   (t₂-t₁) x+(t₂-2t₁) + t₂t₁(t₁-2t₂) = 0

   (t₂- t₁)x +(t₂-2t₁)-(t₁-2t₂) = 0

   (t₂ -t₁) x + 3t₂-3t₁ = 0

   ⇒ x+3 = 0

   Answer: (4)

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5. The area (in sq. units) of the region A = {(x, y): |x|+|y| ≤ 1, 2y²|x|}

1. 1) 1/6
2. 2) 5/6
3. 3) 1/3
4. 4) 7/6

Solution:

1. 

   Total area =

   \(4 \int_{0}^{1 / 2}\left[(1-x)-\left(\sqrt{\frac{x}{2}}\right)\right] d x\)

   =

   \(4\left[x-\frac{x^{2}}{2}-\frac{1}{\sqrt{2}} \frac{x^{3 / 2}}{3 / 2} \mid \begin{array}{l} 1 / 2 \\ 0 \end{array}\right]\)

   = 4[(1/2)-(1/8)-(√2/3)(1/2)^3/2]

   = 4×5/24

   = 5/6

   Answer: (2)

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6. The shortest distance between the lines (x-1)/0 = (y+1)/-1 = z/1 and x+y+z+1 = 0, 2x-y+z+3 = 0 is:

1. 1) 1
2. 2) 1/√2
3. 3) 1/√3
4. 4) ½

Solution:

1. Plane through line of intersection is

   x+y+z+1+λ(2x-y+z+3) = 0

   It should be parallel to given line

   0(1+2λ)-1(1-λ)+1(1 +λ) = 0

   ⇒ λ = 0

   Plane: x+y+z+1 = 0

   Shortest distance of (1,-1,0) from this plane

   = 1-1+0+1/√(1²+1²+1²)

   = 1/√3

   Answer: (3)

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7. Let a, b, c, d and p be any non zero distinct real numbers such that (a²+b²+c²)p²-2(ab+bc+cd)p +(b² +c²+d²) = 0. Then:

1. 1) a, c, p are in G.P.
2. 2) a, b, c, d are in G.P.
3. 3) a, b, c, d are in A.P.
4. 4) a, c, p are in A.P.

Solution:

1. (a²+b²+c²)p²-2(ab+bc+cd)p +(b² +c²+d²) = 0

   (a²p²-2abp+b²]+[b²p²-2bcp+c²]+[ c²p²-2cdp+d²] = 0

   (ap-b)²+(bp -c)²+(cp-d)² = 0

   ap = b

   bp = c

   cp = d

   b/a = c/b = d/c = p

   a, b, c, d are in G.P.

   Answer: (2)

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8. Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

1. 1) 2! 3! 4!
2. 2) (3!)³(4!)
3. 3) 3!(4!)³
4. 4) (3!)²(4!)

Solution:

1. Total numbers in three families = 3+3+4 = 10

   So total arrangement = 10!

   

   Favourable cases

   = Arrangement of 3 families × Interval arrangement of family members

   = 3!×3!×3!×4!

   = (3!)³(4!)

   Answer: (2)

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9. The values of λ and μ for which the system of linear equations

x+y+z = 2

x+2y+3z = 5

x+3y+λz = μ

has infinitely many solutions are, respectively:

1. 1) 6 and 8
2. 2) 5 and 8
3. 3) 5 and 7
4. 4) 4 and 9

Solution:

1. x+y+z = 2

   x+2y+3z = 5

   x+3y+λz = μ

   has infinitely many solutions

   Δ =

   \(\begin{vmatrix} 1 & 1 &1 \\ 1 &2 & 3\\ 1& 3 &\lambda \end{vmatrix}=0\)

   R₂ → R₂-R₁

   R₃ → R₃-R₁

   ⇒

   \(\begin{vmatrix} 1 & 1 &1 \\ 0 &1 & 2\\ 0& 2 &\lambda -1 \end{vmatrix}=0\)

   (λ-1-4) = 0

   λ = 5

   Δ₃ =

   \(\begin{vmatrix} 1 & 1 &2 \\ 1 &2 & 5\\ 1& 3&\mu \end{vmatrix}=0\)

   R₂ → R₂-R₁

   R₃ → R₃-R₁

   ⇒

   \(\begin{vmatrix} 1 & 1 &2 \\ 0 &1 & 3\\ 0& 2&\mu -2 \end{vmatrix}=0\)

   (μ-2-6) = 0

   ⇒ μ = 8

   λ = 5, μ = 8

   Answer: (2)

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10. Let m and M be respectively the minimum and maximum values of

\(\begin{vmatrix} \cos ^{2}x& 1+\sin ^{2}x& \sin 2x\\ 1+\cos ^{2}x & \sin ^{2} x&\sin 2x \\ \cos ^{2}x& \sin ^{2} x& 1+\sin 2x \end{vmatrix}\)

Then the ordered pair (m, M) is equal to:

1. 1) (-3, -1)
2. 2) (-4, -1)
3. 3) (1, 3)
4. 4) (-3, 3)

Solution:

1. \(\begin{vmatrix} \cos ^{2}x& 1+\sin ^{2}x& \sin 2x\\ 1+\cos ^{2}x & \sin ^{2} x&\sin 2x \\ \cos ^{2}x& \sin ^{2} x& 1+\sin 2x \end{vmatrix}\)

   R₁ → R₁-R₂

   R₃→ R₃-R₂

   \(\begin{vmatrix} -1 &1 &0 \\ 1+\cos ^{2}x& \sin ^{2}x &\sin 2x \\ -1&0 & 1 \end{vmatrix}\)

   ⇒ -1(sin²x) -1(1+cos²x +sin2x)

   ⇒ -sin²x –cos²x -1- sin2x

   = -2-sin2x

   So minimum value when sin2x = 1

   m = -2-1 = -3

   So maximum value when sin2x = -1

   (m, M) = (-3, -1)

   Answer: (1)

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11. A ray of light coming from the point (2, 2√3) is incident at an angle 30⁰ on the line x = 1 at the point A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB passes through the point:

1. 1) (4,-√3)
2. 2) (3, -1/√3)
3. 3) (3,-√3)
4. 4) (4, -√3/2)

Solution:

1. 

   Equation of line P’B passing through (0, 2√3)

   (y-y₁) = m (x-x₁)

   (y-2√3) = tan 120⁰ (x- 0)

   y-2√ 3 = -√3x

   √3 x+y = 2√ 3

   Check options

   (3,-√3) satisfy the line.

   Answer: (3)

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12. Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:

1. 1) 10/99
2. 2) 5/33
3. 3) 15/101
4. 4) 5/101

Solution:

1. Case-1

   E, O, E, O, E, O, E, O, E, O, E

   2b = a+c → Even

   ⇒ Both a and c should be either even or odd.

   P = (⁶C₂+⁵C₂ )/ ¹¹C₃ = 5/33

   Case -2

   O, E, O, E, O, E, O, E, O, E, O

   P = (⁵C₂+⁶C₂ )/ ¹¹C₃ = 5/33

   Total probability = (1/2)×(5/33)+(1/2)×(5/33)

   = 5/33

   Answer: (2)

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13. If f(x + y) = f(x) f(y) and

\(\sum_{x=1}^{\infty }f(x)=2\)

x, y ∈N, where N is the set of all natural number, then the value of f(4)/f(2) is:

1. 1) 2/3
2. 2) 1/9
3. 3) 1/3
4. 4) 4/9

Solution:

1. f(x + y) = f(x) f(y)

   Put x = 1, y = 1

   f(2) = (f(1))²

   Put x = 2, y = 1

   f(3) = f(2). f(1) = f((1))³

   Put x = 2, y = 2

   f(4) = f((2))² = f((1))⁴

   f(n) = (f(1))ⁿ

   \(\sum_{x=1}^{\infty }f(x)=\)
   = f(1) + f(2) + f(3) + ....... = 2

   ⇒ f(1)+f((1))²+f(1))³+… = 2

   f(1)/(1-f(1)) = 2

   f(1) = 2/3

   f(2) = (2/3)²

   f(4) = (2/3)⁴

   f(4)/f(2) = (2/3)⁴(2/3)² = 4/9

   Answer: (4)

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14. If {p} denotes the fractional part of the number p, then {3²⁰⁰/8} is equal to:

1. 1) 5/8
2. 2) 1/8
3. 3) 7/8
4. 4) 3/8

Solution:

1. {3²⁰⁰/8} = {9¹⁰⁰/8} = {(8+1)¹⁰⁰/8}

   {(¹⁰⁰C₀ 1¹⁰⁰+¹⁰⁰C₁ (8)1⁹⁹+¹⁰⁰C₂ (8²)1⁹⁸+…¹⁰⁰C₁₀₀ 8¹⁰⁰)/8}

   = {(¹⁰⁰C₀ 1¹⁰⁰+8k)/8}

   = {(1+8k)/8}

   = {(1/8)+k} k∈I

   = 1/8

   Answer: (2)

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15. Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse (x²/4)+(y²/2) = 1 from any of its foci ?

1. 1) (-1,√3)
2. 2) (-2, √3)
3. 3) (-1, √2)
4. 4) (1,2)

Solution:

1. Let foot of perpendicular is (h,k)

   

   (x²/4)+(y²/2) = 1 (Given)

   a = 2, b = √2

   e = √(1-2/4)

   = 1/√2

   So focus (ae,0) = (√2,0)

   Equation of tangent

   y = mx+√(a²m²+b²)

   y = mx+√(4m²+2)

   Passes through (h,k) (k -mh)² = 4m²+2

   line perpendicular to tangent will have slope -1/m

   y-0 = -(1/m)(x-√2)

   my = -x+√2

   (h+mk)² = 2

   Add equation (1) and (2)

   k²(1+m²)+h²(1+m²) = 4(1+m²)

   h²+k² = 4

   x²+y² = 4 (auxiliary circle)

   (-1,√3) lies on the locus.

   Answer: (1)

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16.

\(\lim _{x \rightarrow 1}\left(\frac{\int_{0}^{(x-1)^{2}} t \cos \left(t^{2}\right) d t}{(x-1) \sin (x-1)}\right)\)

1. 1) is equal to 1
2. 2) is equal to 1/2
3. 3) does not exist
4. 4) is equal to -1/2

Solution:

1. \(\lim _{x \rightarrow 1}\left(\frac{\int_{0}^{(x-1)^{2}} t \cos \left(t^{2}\right) d t}{(x-1) \sin (x-1)}\right)\)

   Using L hospital rule

   =

   \(\lim _{x \rightarrow 1} \frac{2(x-1) \cdot(x-1)^{2} \cos (x-1)^{4}-0}{(x-1) \cdot \cos (x-1)+\sin (x-1)}\left(\frac{0}{0}\right)\)

   

   On taking limit

   = 0/(1+1)

   = 0

   Answer: Bouns

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17. If

\(\sum_{i=1}^{n}\left(x_{i}-a\right)=n\)

and

\(\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\)

, (n, a>1) then the standard deviation of n observations x₁, x₂, x₃…x_n is:

1. 1) n√(a-1)
2. 2) √(na-1)
3. 3) a-1
4. 4) √(a-1)

Solution:

1. S.D =

   \(\sqrt{\frac{\Sigma\left(x_{i}-a\right)^{2}}{n}-\left(\frac{\sum\left(x_{i}-a\right)}{n}\right)^{2}}\)

   =

   \(\sqrt{\left(\frac{n a}{n}\right)-\left(\frac{n}{n}\right)^{2}}=\sqrt{a-1}\)

   Answer: (4)

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18. If α and β be two roots of the equation x² -64x+256 = 0. Then the value of (α³/β⁵)^1/8+(β³/α⁵)^1/8 is:

1. 1) 1
2. 2) 3
3. 3) 2
4. 4) 4

Solution:

1. x² -64x+256 = 0.

   αβ = 256

   α+β = 64

   (α³/β⁵)^1/8+(β³/α⁵)^1/8

   = (α+β)/(αβ)^5/8

   = 64/(256)^5/8

   = 64/32

   = 2

   Answer: (3)

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19. The position of a moving car at time t is given by f(t) = at²+bt+c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t₁, t₂] is attained at the point :

1. 1) (t₁+t₂)/2
2. 2) 2a(t₁+t₂)+b
3. 3) (t₂-t₁)/2
4. 4) a(t₂-t₁)+b

Solution:

1. f’(t) = V_av = f(t₂)-f(t₁)/(t₂-t₁)

   = [a(t₂²-t₁²)+b(t₂-t₁)]/(t₂-t₁)

   = a(t₁+t₂)+b

   = 2at+b

   t = (t₁+t₂)/2

   Answer: (1)

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20. If

\(I_{1}=\int_{0}^{1}(1-x^{50})^{100}dx\)

and

\(I_{2}=\int_{0}^{1}(1-x^{50})^{101}dx\)

such that I₂ = αI₁ then α equals to:

1. 1) 5050/5049
2. 2) 5050/5051
3. 3) 5051/5050
4. 4) 5049/5050

Solution:

1. 

   I₂ =

   \(I_{1}-0+\int_{0}^{1}\frac{(1-x^{50})^{101}}{-5050}dx\)

   I₂ = I₁-I₂/5050

   (5051/5050)I₂ = I₁

   I₂ = (5050/5051)I₁

   α = 5050/5051

   Answer: (2)

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21. If

\(\vec{a}\)

and

\(\vec{b}\)

are unit vectors, then the greatest value of

\(\sqrt{3}\left | \vec{a}+\vec{b}\right |+\left |\vec{a}-\vec{b} \right |\)

is

Solution:

1. \(\sqrt{3}\left | \vec{a}+\vec{b}\right |+\left |\vec{a}-\vec{b} \right |\)

   = √3(√(2+2cosθ)+√(2-2cosθ)

   = √6(√(1+cosθ)+√2(1-cosθ)

   = √6(√(1+2cos² (θ/2)-1)+√2(√(1-1+2sin² θ/2)

   = √6(√(2cos²θ/2)+√2(√(2sin²θ/2))

   { greatest value of trigonometric function a cosθ+b sinθ ≤√(a²+b²)}

   = 2√3cos(θ/2)+2sinθ/2

   ≤ √(2√3)²+(2)²

   = 4

   Answer: 4

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22. Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point A such that MD²+MC² is minimum is:

Solution:

1. 

   (MD)² = x²+8² = x²+64

   (MC)² = (10-x)² +(11)² = (x-10)²+121

   f(x) = (MD)²+(MC)² = x²+64 +(x-10)²+(121)

   Differentiate

   f'(x) = 0

   2x+2(x-10) = 0

   4x = 20

   ⇒ x = 5

   f’’(x) = 4 > 0

   at x = 5 point of minima

   Answer: 5

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23. Let f : R → R be defined as

The value of λ for which f''(0) exists, is

Solution:

1. 

   

   If f’’(0) exists then

   f’’(0⁺) = f’’(0^-)

   2λ = 10

   ⇒ λ = 5

   Answer: 5

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24. The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45⁰. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30⁰ to the horizontal plane, the angle of elevation of the top of the hill becomes 75⁰. Then the height of the hill (in meters) is:

Solution:

1. 

   x = 80 cos 30⁰ = 40√3

   y = 80 sin 30⁰ = 40

   In ΔADC, tan 45⁰ = h/(x+z)

   ⇒ h = x+z

   ⇒h = 40√3+z ..(i)

   In ΔEDF

   tan 75⁰ = (h-y)/z

   2+√3 = (h-40)/z

   ⇒ z = (h-40)/(2+√3) ..(ii)

   Put the value of z from (i)

   h-40√3 = (h-40)/(2+√3)

   h(1+√3) = 40(2√3+3-1)

   h(1+√3) = 80(1+√3)

   h = 80

   Answer: 80

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25. Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m×n is

Solution:

1. A and B are two sets having m, n element respectively

   No. of subsets of A = 2^m

   No. of subsets of B = 2ⁿ

   2^m = 2ⁿ +112

   2^m -2ⁿ = 112

   2ⁿ (2^m-n-1) = 112

   2ⁿ(2^m-n-1) = 2⁴ (2³-1)

   n = 4

   m-n = 3

   m-4 = 3

   ⇒ m = 7

   m×n = 28

   Answer: 28

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Video Lessons - September 6 Shift 1 Maths

JEE Main 2020 Maths Paper With Solutions Shift 1 September 6

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