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JEE Main 2020 Maths Paper With Solutions Sep 4 Shift 1

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September 4 Shift 1 – Maths

Question 1.Let y = y(x) be the solution of the differential equation, xy’-y = x²(xcosx+sinx),x > 0. If y(π) = π, then y’’(π/2)+y(π/2) is equal to:
a) 2+(π/2)+(π²/4)
b) 2+(π/2)
c) 1+(π/2)
d) 1+(π/2)+ (π²/4)

xy’-y = x²(x cosx + sinx) x > 0, y(π) =π

y’-(1/x)y = x(x cosx+sin x)

I.F =

\begin{array}{l} e^{- \int \frac{1}{x} d x} \end{array}

= e^{– ln x}

= 1/x

y(1/x) = ∫(1/x)x(x cos x+sin x) dx

(y/x) = ∫(x cos x+ sin x) dx

(y/x) = ∫ (d/dx)(x sin x) dx

(y/x) = x sin x+C

y = x² sin x+Cx

x = π, y = π

π = πC

C = 1

y = x² sin x+x

y(π/2) = (π²/4)+(π/2)

y’ = 2x sinx+x²cosx+1

y’’ = 2sinx+2xcosx+2x cosx-x²sinx

y’’(π/2) = 2-(π²/4)

y(π/2)+ y’’(π/2) = 2+π/2

Answer:(b)

Question 2. The value of

\begin{array}{l} \sum_{r = 0}^{20}^{50 - r} C_{6} \end{array}

is equal to:
a) ⁵¹C₇–³⁰C₇
b) ⁵¹C₇+³⁰C₇
c) ⁵⁰C₇–³⁰C₇
d) ⁵⁰C₆–³⁰C₆

\begin{array}{l} \sum_{r = 0}^{20}^{50 - r} C_{6} \end{array}

= ⁵⁰C₆+⁴⁹C₆+⁴⁸C₆+…+³¹C₆+³⁰C₆

Add and subtract ³⁰C₇

Using ⁿC_r+ⁿC_r-1 = ⁿ⁺¹C_r

³⁰C₆+³⁰C₇= ³¹C₇

³¹C₆+³¹C₇= ³²C₇

Similarly solving

⁵¹C₇–³⁰C₇

Answer: (a)

Question 3. Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]²+2[x+2]-7 = 0 has:

a) exactly four integral solutions.
b) infinitely many solutions.
c) no integral solution.
d) exactly two solutions.

[x]²+2[x+2]-7 = 0

[x]²+2[x]-3 = 0

let [x] = y

y²+3y-y -3 = 0

(y -1)(y+3) = 0

[x] = 1 or [x] = -3

x ∈ [1,2) or x ∈ [-3,-2)

Answer: (b)

Question 4. Let P(3, 3) be a point on the hyperbola, (x²/a²)-(y²/b²) = 1. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a², e²) is equal to:
a) (9, 3)
b) (9/2, 2)
c) (9/2, 3)
d) (3/2, 2)

(x²/a²)-(y²/b²) = 1

Point P(3, 3) on hyperbola.

(9/a²)-(9/b²) = 1 ..(i)

Equation of normal (a²x/3)+(b²y)/3 = a²e²

At x axis y = 0

a²x/3 = a²e²

x = 3e² = 9

3e² = 9

e² = 3

e = √3

e² = 1+b²/a² = 3

b² = 2a² ..(ii)

Put in equation 1

(9/a²)-(9/2a²) = 1

(9/2a²) = 1

a² = 9/2

(a², e²) = (9/2,3)

Answer: (c)

Question 5. Let (x²/a²)+(y²/b²) = 1 (a>b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, Φ(t) = (5/12)+t-t², then a²+b² is equal to:
a) 135
b) 116
c) 126
d) 145

L.R = 2b²/a = 10 ..(i)

Φ(t) = (5/12)-(t-1/2)²+(1/4)

= (8/12)-(t-1/2)²

Φ(t)_max = 2/3 = e

e² = 1-(b²/a²)

= 4/9

b²/a² = 5/9

From (i)

b²/a.a = 5/9

5/a = 5/9

a = 9

b² = 45

a²+b² = 81+45 = 126

Answer: (c)

Question 6. Let

\begin{array}{l} f (x) = \int \frac{\sqrt{x}}{(1 + x)^{2}} d x \end{array}

, x≥0. Then f(3)-f(1) is equal to:
a) (-π/6)+(1/2)+(√3/4)
b) (π/6)+(1/2)-(√3/4)
c) (-π/12)+(1/2)+(√3/4)
d) (π/12)+(1/2)-(√3/4)

\begin{array}{l} f (x) = \int \frac{\sqrt{x}}{(1 + x)^{2}} d x \end{array}

Substituting x = tan²t

dx = 2tan t sec²t dt

f(x) =

\begin{array}{l} \int \frac{\tan t . 2 \tan t \sec^{2} t}{\sec^{4} t} d t \end{array}

f(x) = 2∫sin²t dt

x = 3 t = π/3

x = 1 t = π/4

JEE Main 2020 Paper With Solution Maths Sept 4 Shift 1

Answer: (d)

Question 7. If 1+(1-2²×1)+(1-4²×3)+(1-6²×5)+……+(1-20²×19) = α – 220β, then an ordered pair (α, β) is equal to:
a) (10,97)
b) (11,103)
c) (11,97)
d) (10,103)

T_n = 1-(2n)²(2n-1)

= 1-4n²(2n-1)

= 1-8n³+4n²

S_n =

\begin{array}{l} \sum_{n = 1}^{10} T_{n} \end{array}

= n-∑8n³+∑4n²

= n-8×(n²(n+1)²/4)+4n(n+1)(2n+1)/6

S₁₀ = 10-2×100×121 +(2/3)×10×11×21

= 10-24200+1540

= 10-22660

Sum of series = 11-220×103

= α – 220β

α = 11

β = 103

Answer: (b)

Question 8. The integral

\begin{array}{l} \int {(\frac{x}{x \sin x + \cos x})}^{2} d x \end{array}

is equal to: (where C is a constant of integration)
a) tan x-(x sec x/(x sin x+cos x))+C
b) sec x-(x tan x/(x sin x+cos x))+C
c) sec x+(x tan x/(x sin x+cos x))+C
d) tan x+(x sec x/(x sin x+cos x))+C

\begin{array}{l} \int {(\frac{x}{x \sin x + \cos x})}^{2} d x \end{array}

JEE Main 2020 Papers With Solutions Sept 4 Maths Shift 1

= tan x-(x sec x/(x sin x+cos x))+C

Answer: (a)

Question 9. Let f(x)= |x-2| and g(x) = f(f(x)) ,x∈0,4]. Then

\begin{array}{l} \int_{0}^{3} (g (x) - f (x)) d x \end{array}

is equal to:
a) 1/2
b) 0
c) 1
d) 3/2

f(x) = |x-2| =

\begin{array}{l} {\begin{array}{c} x - 2, & x \geq 2 \\ 2 - x, & x < 2 \end{array} \end{array}

g(x) = ||x-2|-2| =

\begin{array}{l} {\begin{array}{c} | x - 4 |, & x \geq 2 \\ | x |, & x < 2 \end{array} \end{array}

=

\begin{array}{l} {\begin{array}{c} 4 - x, & x \in [2, 4) \\ x - 4, & x \geq 4 \\ x, & x \in [0, 2) \end{array} \end{array}

\begin{array}{l} \int_{0}^{3} (g (x) - f (x)) d x \end{array}

=

\begin{array}{l} \int_{0}^{3} g (x) d x - \int_{0}^{3} f (x) d x \end{array}

= [(1/2)×2×2+1+(1/2)×1×1]-[(1/2)×2×2+(1/2)×1×1)]

= (7/2)-(5/2)

= 1

JEE Main 2020 Maths Papers With Solutions Sept 4 Shift 1

Answer: (c)

Question 10. Let x₀ be the point of local maxima of f(x) =

\begin{array}{l} \vec{a} . (\vec{b} \times \vec{c}) \end{array}

where

\begin{array}{l} \vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k} \end{array}

,

\begin{array}{l} \vec{b} = - 2 \hat{i} + x \hat{j} - \hat{k} \end{array}

and

\begin{array}{l} \vec{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k} \end{array}

. Then the value of

\begin{array}{l} \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} \end{array}

at x = x₀ is:
a) -22
b) -4
c) -30
d) 14

\begin{array}{l} \vec{a} . (\vec{b} \times \vec{c}) \end{array}

=

\begin{array}{l} | \begin{array}{c} x & - 2 & 3 \\ - 2 & x & - 1 \\ 7 & - 2 & x \end{array} | \end{array}

= x(x²-2)+2(-2x+7)+ 3(4-7x)

= x³-2x- 4x+14+12-21x

f(x) = x³– 27x+26

f’(x) = 3x²-27 = 0

x = 3

f’’(x) = 6x

at x = 3, f” (3) = 6×3= 18 > 0

at x = -3, f” (-3) = 6×3 = -18 < 0

Max at x₀ = -3

\begin{array}{l} \vec{a} = (- 3, - 2, 3) \end{array}

,

\begin{array}{l} \vec{b} = (- 2, - 3, - 1) \end{array}

,

\begin{array}{l} \vec{c} = (7, - 2, - 3) \end{array}

So

\begin{array}{l} \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} \end{array}

= 6+6-3-14+6+3-21+4-9

= 25-47

= -22

Answer: (a)

Question 11.A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If BAC = 90⁰ and ar(ΔABC) = 5√5 s units, then the abscissa of the vertex C is:
a) 1+√5
b) 1+2√ 5
c) 2√5-1
d) 2+√5

AB = √(4+1) = √5

(1/2)×√5×x = 5√5

x = 10

m_AB = -1/2

m_AC = 2 = tanθ

Since sinθ = 2/√5

cosθ = 1/√5

by parametric co-ordinates

a = 1+ xcosθ = 1+10×1/√5

= 1+2√5

Answer: (b)

Question 12.Let f be a twice differentiable function on (1,6). If f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4, for all x ∈ (1,6), then:
a) f(5)+f’(5) ≥28
b) f’(5)+f’’(5) ≤20
c) f(5) ≤10
d) f(5)+f’(5) ≤26

f(2) = 8, f’(2) = 5, f’(x) ≥1 and f’’(x) ≥4,

x∈(1,6)

adding (1) and (2) we get,

f(5)+f’(5) ≥28

Answer: (a)

Question 13.Let α and β be the roots of x²-3x+p = 0 and 1/γ and 1/δ be the roots of x²-6x+q = 0. If α, β, γ, δ form a geometric progression. Then ratio (2q+p): (2q-p) is:
a) 33 :31
b) 9 : 7
c) 3 : 1
d) 5 : 3

Roots of x²-3x+p = 0 are α and β.

The roots of x²-6x+q = 0 are γ and δ.

α + β = 3

γ + δ = 6

α = a , β = ar, γ = ar², δ = ar³

a(1+r) = 3 …(i)

ar²(1+r) = 6 …(ii)

Divide (ii) by (i)

r²= 2

α.β = p = a²r

γ.δ = q = a²r⁵

(2q+p)/(2q-p) = (2r⁴+1)/( 2r⁴-1)

= (2×2²+1)/( 2×2²-1) = 9/7

Answer: (b)

Question 14. Let u = (2z+i)/(z-ki), z = x+iy and k>0. If the curve represented by Re(u) +Im(u) =1 intersects the y-axis at the points P and Q where PQ =5, then the value of k is:
a) 4
b) 1/2
c) 2
d) 3/2

u = (2z+i)/(z-ki)

z = x+iy

JEE Main Solution Sept 4 Shift 1 2020 Maths Papers

Re(u)+Img(u) = 1

2x²+(2y+1)(y -k )+x+2xk = x² + (y-k)²

at y – axis, x = 0

(2y +1)(y- k) = (y- k )²

2y²+y- 2yk- k = y²+k²-2yk

Roots of y²+y-(k +k²) = 0 are y₁ and y₂

Diff. of roots = 5

√(1+4k+4k²) = 5

4k²+4k = 24

k²+k-6 = 0

(k +3)(k-2) = 0

k = 2

Answer: (c)

Question 15. If A =

\begin{array}{l} [\begin{array}{c} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{array}] \end{array}

, (θ = π/24) and A⁵ =

\begin{array}{l} [\begin{array}{c} a & b \\ c & d \end{array}] \end{array}

, where i = √-1, then which one of the following is not true?
a) a²-d² = 0
b) a²-c² = 1
c) 0 ≤a²+b² ≤1
d) a²-b² = 1/2

A =

\begin{array}{l} [\begin{array}{c} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{array}] \end{array}

A² =

\begin{array}{l} [\begin{array}{c} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{array}] [\begin{array}{c} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{array}] \end{array}

JEE Main Sept 4 Shift 1 2020 Solved Maths Papers

a = d = cos (5θ)

b = c = i sin (5θ)

a²-b² = cos²5θ + sin²5θ

= 1

Answer: (d)

Question 16. The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is:
a) 3
b) 9
c) 7
d) 5

(5+7+10+12+14+15+x+y)/8 = 10

x+y = 17 ..(i)

Variance = (5²+7²+10²+12²+14²+15²+x²+y²)/8 -100 = 13.5

(739+x²+y²)/8 – 100 = 13.5

x²+y² = 169 ..(ii)

∴ x = 12, y = 5

|x-y| = 7

Answer: (c)

Question 17. A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:
a) 37
b) 29
c) 65
d) 55

Shift 1 JEE Main Sept 4 2020 Solved Maths Papers

n(B) ≤n(A∪B) ≤n(U)

76 ≤76+63-x ≤100

-63 ≤-x ≤-39

63 ≥x ≥ 39

Answer (d)

Question 18.Given the following two statements

(S₁): (q˅p)→(p ↔ ~q) is a tautology.

(S₂): ~q ˄ (~p ↔ q) is a fallacy. Then:

a) only (S₁) is correct.
b) both (S₁) and (S₂) are correct.
c) only (S₂) is correct
d) both (S₁) and (S₂) are not correct.

Shift 1 2020 JEE Main Sept 4 Solved Maths Papers

Answer (d)

Question 19. Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is:
a) 5
b) 20/3
c) 10/3
d) 6

Shift 1 Sept 4 JEE Main 2020 Solved Maths Papers

Using similar triangle concept

tan θ₁ = 10/x = H/x₁

x₁ = Hx/10

tan θ₂ = 15/x = H/x₂

x₂ = Hx/15

Since x₁+x₂ = x

(Hx/10)+( Hx/15) = x

15H+10H = 150

H = 150/25 = 6 m

Answer: (d)

Question 20. If (a+√2b cos x)( a-√2b cos y) = a²-b², where a>b>0, then dy/dx at (π/4, π/4) is:
a) (a+b)/(a-b)
b) (a-2b)/(a+2b)
c) (a-b)/(a+b)
d) (2a+b)/(2a-b)

(a+√2b cos x)( a-√2b cos y) = a²-b²

Differentiating both sides w.r.t.y

-√2b sin x (dx/dy)(a-√2b cos y)+ (a+√2b cos x)(√2b sin y) = 0

x = y = π/4

-b(dx/dy) (a-b)+(a+b)b = 0

dx/dy = (a+b)/(a-b)

Answer: (a)

Question 21. Suppose a differentiable function f(x) satisfies the identity f(x+y) = f(x)+f(y)+xy²+x²y,for all real x and y. If

\begin{array}{l} lim_{x \to 0} \frac{f (x)}{x} = 1 \end{array}

then f’(3) is equal to:

f(x+y) = f(x)+f(y)+xy²+x²y

x = y = 0

f(0) = 2f(0)

f(0) = 0

Now,

Maths Shift 1 JEE Main Sept 4 2020 Solved Papers

f’(x) = 1+0+x²

f’(x) = 1+x²

f’(3) = 10

Answer: (10)

Question 22.If the equation of a plane P, passing through the intersection of the planes, x+4y-z+7 = 0 and 3x+y+5z = 8 is ax+by+6z = 15 for some a, b∈R, then the distance of the point (3,2,-1) from the plane P is ….. units.

p₁+λp₂ = 0

(x +4y-z+7)+λ(3x+y+5z-8) = ax+by+6z-15

(1+3λ)/a = (4+λ)/b = (-1+5λ)/6 = (7-8λ)/-15

∴ 15-75λ = 42-48λ

-27 = 27λ

= -1

∴ plane is -2x+3y-6z+15 = 0

d = |-6+6+6+15)/√(4+9+36)|

= 3 units

Answer: (3)

Question 23. If the system of equations

x-2y+3z = 9

2x+y+z = b

x-7y+az = 24, has infinitely many solutions, then a-b is equal to:

D = 0

\begin{array}{l} | \begin{array}{c} 1 & - 2 & 3 \\ 2 & 1 & 1 \\ 1 & - 7 & a \end{array} | \end{array}

= 0

1(a +7)+2(2a -1)+3(-14-1) = 0

a+7+4a-2-45 = 0

5a = 40

a = 8

D₁ =

\begin{array}{l} | \begin{array}{c} 9 & - 2 & 3 \\ b & 1 & 1 \\ 24 & - 7 & 8 \end{array} | = 0 \end{array}

9(8+7)+2(8b-24)+3(-7b-24) = 0

135+16b-48-21b-72 = 0

15 = 5b

b = 3

Hence a – b = 8-3 = 5

Answer: (5)

Question 24.Let (2x²+3x+4)¹⁰ =

\begin{array}{l} \sum_{r = 0}^{20} a_{r} x^{r} \end{array}

. Then a₇/a₁₃ is equal to:

Given (2x²+3x+4)¹⁰ =

\begin{array}{l} \sum_{r = 0}^{20} a_{r} x^{r} \end{array}

Replace x by 2/x in above identity:

Maths Shift 1 Sept 4 2020 JEE Main Solved Papers

Now, comparing coefficient of x⁷ from both sides

(take r = 7 in L.H.S. and r = 13 in R.H.S)

2¹⁰a₇ = a₁₃2¹³

a₇/a₁₃ = 2³ = 8

Answer: (8)

Question 25.The probability of a man hitting a target is 1/10. The least number of shots required, so that the probability of his hitting the target at least once is greater than 1/4 is:

Probability of hitting, P(H) = 1/10

probability of missing, P(M) = 9/10

We have, 1-(probability of all shots result in failure ) >1/4

= 1-P(M)ⁿ > 1/4

= 1-(9/10)ⁿ > 1/4

(9/10)ⁿ<3/4

n≥3

Answer (3)

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