JEE Main 2020 Maths Paper With Solutions Shift 2 September 6
1. If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:
1) e^{4}+2e^{2}-1 = 0
2) e^{2}+2e-1 = 0
3) e^{4}+e^{2}-1 = 0
4) e^{2}+e-1 = 0
Equation of normal at (ae, b^{2}/a)
(a^{2}x/ae)-b^{2}y/(b^{2}/a) = a^{2}e^{2}
(ax/e)-ay = a^{2}e^{2}
⇒ (x/e)-y = ae^{2}
It passes through (0,-b)
-b = ae^{2} ⇒b^{2} = a^{2}e^{4}
a^{2 }(1- e^{2} ) = a^{2}e^{4}
⇒ e^{4}+e^{2}-1 = 0
Answer: (3)
2. The set of all real values of λ for which the function f(x) = (1-cos^{2} x)(λ+sinx) , x∈(-π/2, π/2), has exactly one maxima and exactly one minima, is:
1) (-3/2, 3/2)-{0}
2) (-1/2, ½)-{0}
3) (-3/2, 3/2)
4) (-1/2, 1/2)
f(x) = (1-cos^{2}x) (λ+sinx)
f(x) = sin^{2}x (λ+sinx)
f’(x) = 2sinx cosx (λ +sinx) + sin^{2}x (cosx)
= sin 2x(λ+sin x+ (sin x)/2)
= sin 2x(2λ+3sin x)
For extreme value f’(x) = 0
sin2x = 0 ⇒ sinx = 0
⇒ x = 0 → One point
2λ+3sinx = 0
⇒ sinx = -2λ/3
sinx ∈(-1,1)-{0}
-1<-2λ/3 <1
⇒ -3/2 <λ<3/2
λ∈(-3/2, 3/2)-{0}
Answer: (1)
3. The probabilities of three events A, B and C are given by P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P (A∪B) = 0.8, P (A∩C) = 0.3, P (A∩B∩C) = 0.2, P (B∩C) = β and P(A∪B ∪C) = α , where 0.85 ≤α≤ 0.95 , then β lies in the interval:
1) [0.36,0.40]
2) [0.25,0.35]
3) [0.35,0.36]
4) [0.20,0.25]
P(A∪B∪C) = P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(C∩A) + P(A∩B∩C)
α = 0.6+0.4+0.5-P(A∩B)-β-0.3+0.2
α = 1.4-P(A∩B)-β
⇒ α+β = 1.4-P(A∩B) ……..(i)
again
P(A∪B) = P(A)+P(B)-P(A∩B)
0.8 = 0.6+0.4-P(A∩B)
P(A∩B) = 0.2 ……..(ii)
Put the value P(A∩B) in equation (i)
α+β = 1.2
α = 1.2-β
0.85 ≤α≤ 0.95
⇒ 0.8 ≤1.2-β≤ 0.95
β∈ [0.25, 0.35]
Answer: (2)
4. The common difference of the A.P. b_{1}, b_{2},….. b_{m} is 2 more than the common difference >of A.P. a_{1}, a_{2}, …a_{n}. If a_{40} = -159, a_{100} = -399 and b_{100} = a_{70}, then b_{1} is equal to:
1) -127
2) 81
3) 127
4) -81
A.P (a_{1}, a_{2}, a_{3} ……….a_{n}) (CD = D_{a})
(b_{1}, b_{2}, b_{3} ………. b_{m}) (CD = D_{b})
D_{b} = D_{a}+2
a_{40} = -159
a_{1}+39D_{a} = -159 …(i)
a_{100} = -399
a_{1}+99D_{a} = -399 …(ii)
Eqn (i)-(ii)
-60D_{a} = 240
⇒ D_{a} = -4
D_{b} = -4+2 = -2
a_{1}+39(-4) = -159
⇒ a_{1} = -3
b_{100 }= a_{70}
b_{1}+99D_{b} = a_{1}+69D_{a}
b_{1}+99 (-2) = (-3)+69(-4)
b_{1} = -81
Answer: (4)
5. The integral
1) e(4e-1)
2) e(4e+1)
3) 4e^{2}-1
4) e(2e-1)
Put e^{x}.x^{x} = t
Since upper limit = e^{2}.2^{2}, Lower Limit = e
(e^{x}.x^{x} + e^{x} x^{x}(1+lnx)) dx = dt
e^{x}.x^{x}(2+lnx) dx = dt
= 4e^{2}-e
= e(4e-1)
Answer: (1)
6. If the tangent to the curve, y = f(x) = xlog_{e}x, (x>0) at a point (c,f(c)) is parallel to the line-segment joining the points (1,0) and (e,e), then c is equal to:
1) e^{(1/1-e)}
2) (e-1)/e
3) 1/(e-1)
4) e^{(1/e-1)}
y = f(x) = x lnx
slope of the line joining (1, 0), (e, e)
m_{2} = (e/e-1)
m_{2} = m_{1} ⇒ ln c+1 = e/(e-1)
ln c = (e/e-1)-1
= 1/(e-1)
c = e^{(1/e-1)}
Answer: (4)
7. If y = ((2/π)x-1) cosec x is the solution of the differential equation, (dy/dx)+p(x)y = ((2/π)-1)cosec x, 0<x,π/2, then the function p(x) is equal to:
1) cosec x
2) cot x
3) tan x
4) sec x
y = ((2/π)x-1)cosec x
Differentiate w.r.t.x
(dy/dx) = (2/π)cosec x-((2x/π)-1)cosec x.cot x
(dy/dx) +(2x/π-1)cosec x.cot x = (2/π) cosec x
(dy/dx)+y cot x = (2/π) cosec x
Compare this differential equation with given differential equation
p(x) = cot x
Answer: (2)
8. If α and β are the roots of the equation 2x(2x+1) = 1, then β is equal to:
1) 2α(α-1)
2) -2α(α+1)
3) 2α^{2}
4) 2α(α+1)
2x(2x+1) = 1
If α and β are the roots i.e α and β satisfy this equation
2α(2α+1) = 1
⇒ α(2α+1) = ½
4x^{2}+2x-1 = 0
α + β = -1/2
= -α(2α+1)
β = -α(2α+1)-α
= -α(2α+2)
= -2α(α+1)
Answer: (2)
9. For all twice differentiable functions f: R→ R, with f(0) = f(1) = f’(0) = 0,
1) f”(x) = 0, at every point x∈(0,1)
2) f”(x) ≠ 0, at every point x∈(0,1)
3) f”(x) = 0, for some x∈(0,1)
4) f”(0) = 0
Applying rolle’s theorem in [0,1] for function f(x)
f’(c) = 0, c ∈(0,1)
again applying rolles theorem in [0,c] for function f'(x) s
f’’(c_{1}) = 0, c_{1}∈(0,c)
Answer: (3)
10. The area (in sq.units) of the region enclosed by the curves y = x^{2}-1 and y = 1-x^{2} is equal to:
1) 4/3
2) 7/2
3) 16/3
4) 8/3
Total area =
= 4(1-1/3)
= 8/3 sq. unit
Answer: (4)
11. For a suitably chosen real constant a, let a function, f:R-{-a}→R be defined by f(x) = (a-x)/(a+x). further suppose that for any real number x ≠ -a and f(x) ≠ -a, (fof)(x) = x. Then f(-1/2) is equal to:
1) -3
2) 3
3) 1/3
4) -1/3
f(x) = (a-x)/(a+x)
f(f(x)) = (a-f(x))/(a+f(x)) = x
(a-ax)/(1+x) = f(x) = (a-x)/(a+x)
a(1-x)/(1+x) = (a-x)/(a+x)
⇒ a = 1
So f(x) = (1-x)/(1+x)
f(-1/2) = 3
Answer: (2)
12. Let θ = π/5 and A =
1) is one
2) lies in (1,2)
3) lies in (2,3)
4) is zero
B = A+A^{4}
A^{2} =
=
Similarly
A^{4} =
B = A^{4}+A
=
=
det B = (cos4θ+cos θ)^{2} +(sin 4θ+sin θ)^{2}
= cos^{2}4θ + cos^{2}θ+2cos 4θ cos θ+sin^{2}4 θ + sin^{2} θ+2sin4 θ sin θ
= 2+2(cos4θ cos θ+sin4 θ sin θ)
= 2+2 cos3 θ
At θ = π/5,
det B = 2+2 cos3×π/5
= 2(1-sin 18)
= 2(1-(√5-1)/4)
= 2(5-√5)/4)
= (5-√5)/2
lies in (1,2)
Answer: (2)
13. The centre of the circle passing through the point (0,1) and touching the parabola y = x^{2} at the point (2,4) is :
1) (3/10, 16/5)
2) (6/5, 53/10)
3) (-16/5, 53/10)
4) (-53/10, 16/5)
Circle passing through point (0,1) and touching curve y = x^{2} at (2,4)
tangent at (2,4) is (y+4)/2 = x(2)
⇒ y-4x+4 = 0
Equation of circle
(x-2)^{2} +(y-4)^{2}+λ(4x-y-4) = 0
Passing through (0,1) 4+9 +λ(-5) = 0
λ = 13/5
Circle is
x^{2}-4x+4+y^{2}-8y+16+(13/5)(4x-y-4) = 0
x^{2}+y^{2}+((52/5)-4)x-(8+13/5)y+20-(52/5) = 0
x^{2}+y^{2}+(32/5)x-(53/5)y+(48/5) = 0
centre is (-16/5, 53/10)
Answer: (3)
14. A plane P meets the coordinate axes at A, B and C respectively. The centroid of a triangle ABC is given to be (1,1,2). Then the equation of the line through this centroid and perpendicular to the plane P is:
1) (x-1)/2 = (y-1)/1 = (z-2)/1
2) (x-1)/2 = (y-1)/2 = (z-2)/1
3) (x-1)/1 = (y-1)/2 = (z-2)/2
4) (x-1)/1 = (y-1)/1 = (z-2)/2
G = (α/3, β/3,
α = 3, β = 3,
Equation of plane is
(x/α)+(y/β)+(z/
(x/3)+(y/3)+(z/6) = 1
2x+2y+z = 6
Required line perpendicular to plane is (x-1)/2 = (y-1)/2 = (z-2)/1
Answer: (2)
15. Let f : R→ R be a function defined by f(x) = max {x,x^{2}}. Let S denote the set of all points in R, where f is not differentiable. Then
1) {0,1}
2) an empty set
3) {1}
4) {0}
Function is not differentiable at two point {0,1}.
Answer: (1)
16. The angle of elevation of the summit of a mountain from a point on the ground is 45^{0}. After climbing up one km towards the summit at an inclination of 30^{0} from the ground, the angle of elevation of the summit is found to be 60^{0}. Then the height (in km) of the summit from the ground is:
If ΔCDF
Sin 30^{0} = z/1
Z = ½ Km
Cos 30^{0} = y/1
y = √3/2 Km
Now in ABC
tan 45^{0} = h/(x+y)
⇒ h = x+y
Or x = h -√3/2
Now, in ΔBDE
tan 60^{0} = (h-z)/x
After substituting the value of x, we get
h = 1/ (√3-1) Km
Answer: (1)
17. If the constant term in the binomial expansion of (√x – k/x^{2})^{10} is 405, then |k| equals:
1) 1
2) 9
3) 2
4) 3
^{10}C_{r} (-k/x^{2})^{r} (√x)^{10-r}
^{10}C_{r} (-k)^{r} (x)^{5-5r/2}
For constant term
5-5r/2 = 0
Or r = 2
Now, T_{3} = ^{10}C_{2} k^{2} = 405
Or k^{2} = 405/45 = 9
Or |k| = 3
Answer: (4)
18. Let z = x+iy be a non-zero complex number such that z^{2} = i|z|^{2}, where i = √-1 , then z lies on the
1) line, y = x
2) real axis
3) imaginary axis
4) line, y = -x
z^{2} = i |z|^{2}
x^{2}-y^{2}+2i xy = i(x^{2}+y^{2})
equating real terms
x^{2}-y^{2} = 0
⇒ x^{2} = y^{2}
equating imaginary terms
2xy = x^{2}+y^{2}
(x-y)^{2} = 0
⇒ x = y
Answer: (1)
19. Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (-1, -4) in this line is:
1) (11/5, 28/5)
2) (8/5, 29/5)
3) (29/5, 11/5)
4) (29/5, 8/5)
Let (h, k) is the image of the point (-1, -4) in the line
x/3 + y/1 = 1
x+3y = 3
L2: 3x-y + λ = 0
-3+4+λ = 0
λ = -1
3x-y = 1
(h,k) satisfy the equation of line L2
3h-k = 1 …(1)
16 = |h+3k-3|
h +3k = 19 …(2)
h +3k = -13 …(3)
From equation (2) and (3), put value of h in (1)
h = 19-3k,
3(19 -3k) -k=1
-10k = -56 = 28/5
k = 28/5
h = 19-3(28/5) = 11/5
h = -13-3k
3(-13-3k) -k=1
-10k = 40
k = -4
Image = (11/5, 28/5)
Answer: (1)
20. Consider the statement : “For an integer n, if n^{3}-1 is even, then n is odd.” The contrapositive statement of this statement is:
1) For an integer n, if n is even, then n^{3}-1 is even
2) For an integer n, if n is odd, then n^{3}-1 is even
3) For an integer n, if n^{3}-1 is not even, then n is not odd.
4) For an integer n, if n is even, then n^{3}-1 is odd
P:n^{3}-1 is even, q : n is odd
Contrapositive of p →q = ∼q → ∼p
⇒ If n is not odd then n^{3}-1 is not even
⇒ For an integer n, if n is even, then n^{3}-1 is odd
Answer: (4)
21. The number of words (with or without meaning) that can be formed from all the letters of the word “LETTER” in which vowels never come together is:
Consonants → LTTR
Vowels → EE
Total No of words = 6!/(2!2!)
= 180 (T and E repeated)
Total no of words if vowels are together
= 5!/2!
= 60 (E is repeated)
Required = 180-60 = 120
Answer: 120
22. If
⇒
And
⇒
by comparing (i) and (ii)
we get λ = 1
Answer: 1
23. Consider the data on x taking the values 0, 2, 4, 8, …..,2n with frequencies ^{n}C_{0}, ^{n}C_{1}, ^{n}C_{2…}^{ n}C_{n}, respectively. If the mean of this data is 728/2^{n}, then n is equal to:
X_{i} |
0 |
2 |
2^{2} |
… |
2^{n} |
f_{i} |
^{n}C_{0} |
^{n}C_{1} |
^{n}C_{2} |
… |
^{n}C_{n} |
3^{n} = 729 = 3^{6}
n = 6
Answer: 6
24. Suppose that function f : R→R satisfies f(x+y) = f(x)f(y) for all x, y∈R and f(1) = 3. If
f(x+y) = f(x) f(y)
f(x) = a^{x}
f(1) = a = 3
So f(x) = 3^{x}
⇒ 3+3^{2}+3^{3}+….+3^{n} = 363
⇒ 3(3^{n}-1)/2 = 363
⇒ n = 5
Answer: 5
25. The sum of distinct values of λ for which the system of equations
(λ-1)x+(3λ+1)y+2λz = 0
(λ-1)x+(4λ-2)y+(λ+3)z = 0
2x+(3λ+1)y+3(λ-1)z = 0 has non-zero solutions, is:
R_{2} → R_{2}-R_{1}
R_{3} → R_{3}-R_{1}
C_{1} → C_{1 }+ C_{3}
(λ-3)[(3λ-1)(λ-3)-3(3-λ)(3λ+1)] = 0
(λ-3)[3λ^{2}-10λ+3-(8λ-3λ^{2}+3)] = 0
(λ-3)(6λ^{2}-18λ) = 0
(6λ)(λ-3)^{2} = 0
λ = 0, 3
Sum of the values of λ = 0+3 = 3
Answer: 3
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