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JEE Advanced 2020 Question Paper
January 9 Shift 2 – Maths
January 9 Shift 2 – Maths
1. If A = {x∈ R∶ |x| <2} and B = {x∈ R∶ |x-2| ≥3} then :
a. A – B = [-1,2]
b. B – A = R-(-2,5)
c. A⋃B = R-(2,5)
d. A∩B = (-2,-1)
A = {x: x ∈ (-2,2)}
B = {x: x ∈ (∞,-1] ⋃ [5, ∞)}
A∩B = {x: x ∈ (-2,-1] }
B – A = {x: x ∈ (∞,-2] ⋃ [5, ∞)}
A – B = {x: x ∈ (-1,2)}
A⋃B = {x: x ∈ (∞,2] ⋃ [5, ∞)}
Answer: (b)
2. If 10 different balls has to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :
a. 965/210
b. 945/210
c. 945/211
d. 965/211
Total ways to distribute 10 balls in 4 boxes is = 410
Total ways of placing exactly 2 and 3 balls in any two of these boxes is = 4C2×10C5×(5!/(2!3!))×2×25
P(E) = 945/210
Answer: (b)
3. If x = 2 sin θ – sin 2 θ and y = 2 cos θ – cos 2 θ, θ ∈ [0, 2π], then d2y/dx2 at θ =π is:
a. -3/8
b. 3/4
c. 3/2
d. -3/4
dx/d θ = 2 cos θ -2 cos2 θ
dy/d θ = -2 sin θ + 2 sin2 θ
dy/dx = cot (3 θ/2)
d2y/dx2 = -(3/2) cosec2(3 θ/2)(d θ/dx)
\(\begin{array}{l}\frac{d^{2}y}{dx^{2}}= \left ( -\frac{3}{2}cosec^{2} \: \frac{3\theta}{2} \right )\frac{1}{(2\cos \theta -2\cos 2\theta )}\end{array} \)
(
\(\begin{array}{l}\frac{d^{2}y}{dx^{2}})_{\theta =\pi }= \frac{3}{8}\end{array} \)
Answer : None of the above option satisfies the answer.
4. Let f and g be differentiable functions on R, such that fog is the identity function. If for some a, b ∈ R, g’(a) = 5 and g(a) = b, then f'(b) is equal to :
a. 2/5
b. 5
c. 1
d. 1/5
f(g(x) = x
f’(g(x))g’(x) = 1
put x = a
f’(g(a))g’(a) = 1
f’(b) ×5 = 1
f’(b) = 1/5
Answer: (d)
5. In the expansion of \(\begin{array}{l}\left ( \frac{x}{\cos \theta }+\frac{1}{x\sin \theta } \right )^{16}\end{array} \)
, if l1 is the least value of the term independent of x when (π/8) ≤ θ ≤ (π/4) and l2 is the least value of the term independent of x when (π/16) ≤ θ ≤ (π/8), then the ratio l2 : l1 is equal to :
a. 16:1
b. 8:1
c. 1:8
d. 1:16
\(\begin{array}{l}T_{r+1} =\: ^{16}C_{r}\left ( \frac{x}{\cos \theta } \right )^{16-r}\left ( \frac{1}{x\sin\theta } \right )^{r}\end{array} \)
For term independent of x, 16-2r = 0
⇒ r = 8
T9 =
\(\begin{array}{l}\: ^{16}C_{8}\left ( \frac{1}{\sin \theta \cos \theta } \right )^{8}=\: ^{16}C_{8}2^{8}\left ( \frac{1}{\sin 2\theta } \right )^{8}\end{array} \)
l1 = 16C8 28 at θ = π/4
l2 =
\(\begin{array}{l}^{16}C_{8}\frac{2^{8}}{\left ( \frac{1}{\sqrt{2}} \right )^{8}}= ^{16}C_{8}2^{12} at\: \theta =\frac{\pi }{8}\end{array} \)
l2/l1 = 16:1
Answer: (a)
6. Let a,b ∈R, a ≠ 0, such that the equation, ax2-2bx+5 = 0 has a repeated root α, which is also a root of the equation x2-2bx-10 = 0. If β is the root of this equation, then α2+β2 is equal to:
a. 24
b. 25
c. 26
d. 28
ax2-2bx+5 = 0 has both roots as α
2α = 2b/a
α= b/a
And α2 = 5/a
b2 = 5a (a≠0) ..(i)
α+β = 2b
And αβ = -10
α= b/a is also a root of x2 -2bx-10 = 0
b2 -2ab2-10a2 = 0
b2 = 5a
5a-10a2 -10a2 = 0
a = 1/4
b2 = 5/4
α2 = 20
β2 = 5
α2+β2 = 25
Answer: (b)
7. Let a function f:[0,5] R, be continuous, f(1) = 3 and F be defined as:
F(x) =
\(\begin{array}{l}\int_{1}^{x}t^{2}g(t)dt\end{array} \)
,where \(\begin{array}{l}g(t)=\int_{1}^{t}f(u)du\end{array} \)
Then for the function F, the point x = 1 is
a. a point of inflection.
b. a point of local maxima
c. a point of local minima
d. not a critical point
F'(x) = x2g(x)
Put x = 1
F'(1) = g(1) = 0 ..(i)
Now F’’(x) = 2xg(x) + g’(x)x2
F’’(1) = 2g(1) + g’(1) {∵g’(x) = f(x)}
F’’(1) = f(1) = 3 ..(ii)
From (1) and (2), F(x) has local minimum at x = 1.
Answer: (c)
8. Let [t] denotes the greatest integer ≤ t and \(\begin{array}{l}\lim_{x\to0} x\left [ \frac{4}{x} \right ]= A\end{array} \)
. Then the function, f(x)=[x2] sin πx is discontinuous, when x is equal to
a. √(A+1)
b. √A
c. √(A+5)
d. √(A+21)
f(x)=[x2]sin π x
It is continuous ∀x ∈ Z as sinπx 🡪 as 🡪 Z.
f(x) is discontinuous at points where [x2] is discontinuous i.e. x2∈Z with an exception that f(x) is continuous as x is an integer.
∴ Points of discontinuity for f(x) would be at
x = ±√2, ±√3, ±√5,……
Also, it is given that
\(\begin{array}{l}\lim_{x\to0} x\left [ \frac{4}{x} \right ]= A\end{array} \)
(indeterminate form (0×∞))
\(\begin{array}{l}4-\lim_{x\to0} (\frac{4}{x}) = A\end{array} \)
A = 4
√(A+5) = 3
√(A+1) = √5
√(A+21) =5
√A = 2
Points of discontinuity for f(x) is x = √5
Answer: (a)
9. Let a-2b+c = 1. If f(x) = \(\begin{array}{l}\begin{vmatrix} x+a & x+2&x+1 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}\end{array} \)
, then
a. f(-50) = 501
b. f(-50) = -1
c. f(50) = 1
d. f(-50) = -501
Given f(x) =
\(\begin{array}{l}\begin{vmatrix} x+a & x+2&x+1 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}\end{array} \)
a -2b+c = 1
Applying R1 🡪 R1 -2R2+ R3
f(x) =
\(\begin{array}{l}\begin{vmatrix} a -2b+c& 0&0 \\ x+b& x+3 & x+2\\ x+c &x+4 & x+3 \end{vmatrix}\end{array} \)
Using a-2b+c = 1
f(x) = (x+3)2 -(x+2)(x+4)
f(x) = 1
f(50) = 1
f(-50) = 1
Answer: (c)
10. Given:
and g(x)= (x− 1/2)2, x ∈ R. Then the area (in sq. units) of the region bounded by the curves y = f(x) and y = g(x) between the lines 2x = 1 to 2x = √3 is :
a. (√3/4) – (1/3)
b. (1/3) + (√3/4)
c. (1/2) + (√3/4)
d. (1/2) – (√ 3/4)
The area between f(x) and g(x) from x = 1/2 to x = √3/2 :
Points of intersection of f(x) and g(x):
1-x = (x- 1/2)2
x = √3/2 ,-√3/2
Required area =
\(\begin{array}{l}\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}(f(x)-g(x))dx\end{array} \)
=
\(\begin{array}{l}\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}(1-x-(x-\frac{1}{2})^{2})dx\end{array} \)
=
\(\begin{array}{l}[x-\frac{x^{2}}{2}-\frac{1}{3}(x-\frac{1}{2})^{3}]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\end{array} \)
= (√3/4) – (1/3)
Answer: (a)
11. The following system of linear equations
7x+6y-2z = 0
3x+4y+2z = 0
x-2y-6z = 0, has
a. infinitely many solutions, (x,y,z) satisfying y = 2z
b. infinitely many solutions (x,y,z) satisfying x = 2z
c. no solution
d. only the trivial solution
7x+6y−2z = 0
3x+4y+2z = 0
x−2y−6z = 0
As the system of equations are homogeneous the system is consistent.
\(\begin{array}{l}\begin{vmatrix} 7 & 6&-2 \\ 3&4 & 2\\ 1 & -2 & -6 \end{vmatrix}=0\end{array} \)
⇒ Infinite solutions exist (both trivial and non-trivial solutions)
When y = 2z
Let’s take y = 2, z = 1
When (x,2,1)is substituted in the system of equations
⇒7x+10 = 0 ,3x+10 = 0 , x-10 = 0 (which is not possible)
∴ y = 2z ⇒ Infinitely many solutions does not exist.
For x = 2z, lets take x = 2, z = 1, y = y
Substitute (2,y,1)in system of equations ⇒ y = -2
∴For each pair of (x,z), we get a value of y.
Therefore, for x = 2z infinitely many solutions exists.
Answer: (b)
12. If p -> (p ∧~ q) is false. Then the truth values of p and q are respectively
a. F, T
b. T, F
c. F, F
d. T, T
Given p ->(p ∧~ q)
Truth table:
|
p
|
q
|
~q
|
(p ∧~ q)
|
p->(p ∧~ q)
|
|
T
|
T
|
F
|
F
|
F
|
|
T
|
F
|
T
|
T
|
T
|
|
F
|
T
|
F
|
F
|
T
|
|
F
|
F
|
T
|
F
|
T
|
p→(p ∧~ q) is false when p is true and q is true.
Answer: (d)
13. The length of minor axis (along y-axis) of an ellipse of the standard form is 4/√3. If this ellipse touches the line x + 6y = 8, then its eccentricity is:
a. (1/2)(√5/3)
b. (1/2)√(11/3)
c. √(5/6)
d. (1/3)√(11/3)
If 2b = 4/√3
b = 2/√3
Comparing y = (-x/6)+(8/6) with y = mx ±√(a2m2+b2)
m = -1/6 and a2m2 + b2 = 16 / 9
(a2/36) + (4/3) = 16/9
(a2/36) = (16/9) – (4/3)
a2 = 16
e = √(1-b2/a2)
e = √(11/12)
Answer: (b)
14. If z be a complex number satisfying |Re(z)|+|Im(z)| = 4, then |𝑧| cannot be:
a. √7
b. √(17/2)
c. √10
d. √8
|Re(z)|+|Im(z)| = 4
Let z = x+iy
⇒|x|+|y| = 4
z lies on the rhombus.
Maximum value of |z| = 4 when z = 4, -4, 4i, -4i
Minimum value of |z| = 2√2 when z = 2±2i, ±2+2i
|z|∈[2√2 ,4]
|z|∈[√8 ,√16]
|z| ≠ √7
Answer: (a)
15. If \(\begin{array}{l}x = \sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta\end{array} \)
and \(\begin{array}{l}y = \sum_{n=0}^{\infty }\cos ^{2n}\theta\end{array} \)
where 0 < θ < π/4, then:
a. y(1+x) = 1
b. x(1-y) = 1
c. y(1-x) = 1
d. x(1+y) = 1
y = 1+cos2 θ +cos4 θ + …
\(\begin{array}{l}y =\frac{1}{1-\cos ^{2}\theta }\end{array} \)
1/y = sin2 θ
x = 1-tan2 θ + tan4 θ-…
\(\begin{array}{l}x =\frac{1}{1-(-\tan ^{2}\theta) } = \cos ^{2}\theta\end{array} \)
x +(1/y) = 1
y(1-x) = 1
Answer: (c)
16. If \(\begin{array}{l}\frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}\end{array} \)
, y(1) = 1, then a value of x satisfying y(x)= e is:
a. √3e
b. (1/2)√3e
c. √2e
d. e/√2
Let y = vx
dy/dx = v+x(dv/dx)
v+x(dv/dx) = vx2/(x2(1+v2) = v/(1+v2)
x(dv/dx) = -v3/(1+v2)
(1/x)dx = (-1/v3)-(1/v)dv
log x = (1/2v2)-log v + log c
log x = (x2/2y2 )-log y + log x + log c
log c + (x2/2y2) – log y = 0
y(1) = 1
log C + (1/2)-0 = 0
log c = −1/2
y(x) = e
-(1/2) + (x2/2e2)-1 = 0
x2/e2 = 3
x = ±√3e
Answer: (a)
17. If one end of focal chord AB of the parabola y2 = 8x is at A(1/2,-2) , then the equation of tangent to it at B is
a. x+2y+8 = 0
b. 2x-y-24 = 0
c. x-2y+8 = 0
d. 2x+y-24 = 0
Let PQ be the focal chord of the parabola y2 = 8x
P(t1) = (2t12, 4t1) and Q(t2) = (2t22, 4t2)
t1t2 = -1
(1/2, -2) is one of the ends of the focal chord of the parabola
Let (1/2, -2) = (2t22, 4t2)
Other end of focal chord will have parameter t1= 2
The co-ordinate of the other end of the focal chord will be (8,8).
The equation of the tangent will be given as 8y = 4(x+8)
⇒2y−x = 8
Answer:(c)
18. Let an be the nth term of a G.P. of positive terms. If \(\begin{array}{l}\sum_{n=1}^{100}a_{2n+1}= 200\end{array} \)
and \(\begin{array}{l}\sum_{n=1}^{100}a_{2n}= 100\end{array} \)
a. 300
b. 175
c. 225
d. 150
an is a positive term of GP.
Let GP be a, ar, ar2,…..
\(\begin{array}{l}\sum_{n=1}^{100}a_{2n+1}= 200\end{array} \)
= a3+a5+a7+…+a201
200 = ar2 + ar4 + …+ar201
200 =
\(\begin{array}{l}\frac{ar^{2}(r^{200}-1)}{r^{2}-1}\end{array} \)
…(i)
Also,
\(\begin{array}{l}\sum_{n=1}^{100}a_{2n}= 100\end{array} \)
100 = a2 + a4+ …+a200
100 = ar + ar3 + …ar199
100 =
\(\begin{array}{l}\frac{ar(r^{200}-1)}{r^{2}-1}\end{array} \)
..(ii)
From (i) and (ii), r = 2
And
\(\begin{array}{l}\sum_{n=1}^{100}a_{2n+1} + \sum_{n=1}^{100}a_{2n}= 300\end{array} \)
a2 + a3+a4 ….+a200+a201 = 300
ar + ar2+ar3+…ar200 = 300
r(a+ar+ar2+…+ar199) = 300
2(a1+a2+a3+…+a200) = 300
\(\begin{array}{l}\sum_{n=1}^{200}a_{n} = 150\end{array} \)
Answer: (d)
19. A random variable X has the following probability distribution:
|
X
|
1
|
2
|
3
|
4
|
5
|
|
P(X)
|
K2
|
2K
|
K
|
2K
|
5K2
|
Then P(X>2) is equal to:
a. 7/12
b. 23/36
c. 1/36
d. 1/6
We know that
\(\begin{array}{l}\sum_{X=1}^{5}P(X) = 1\end{array} \)
K2 +2K+K+2K+5K2 = 1
6K2+5K-1 = 0
6K (K+1)-(K+1) = 0
(K+1)(6K-1) = 0
K = -1 or K = 1/6
K cannot be negative.
P(X>2) = P(X = 3) + P(X = 4)+ P(X = 5)
= K+2K + 5K2
= (1/6) + 2(1/6) +5(1/6)2
= 23/36
Answer: (b)
20. If \(\begin{array}{l}\int \frac{d\theta }{\cos ^{2}\theta (\tan 2\theta +\sec 2\theta )}= \lambda \tan \theta +2\: log_{e}\left | f(\theta ) \right |+C\end{array} \)
where C is constant if integration, then the ordered pair (λ, f(θ)) is equal to:
a. (-1, 1-tan θ)
b. (-1, 1+tan θ)
c. (1, 1+tan θ)
d. (1, 1-tan θ)
Let
\(\begin{array}{l}I=\int \frac{d \theta}{\cos ^{2} \theta(\sec 2 \theta+\tan 2 \theta)}\end{array} \)
\(\begin{array}{l}I=\int \frac{\sec ^{2} \theta d \theta}{\left(\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}\right)+\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)}\end{array} \)
\(\begin{array}{l}I=\int \frac{\left(1-\tan ^{2} \theta\right)\left(\sec ^{2} \theta\right) d \theta}{(1+\tan \theta)^{2}}\end{array} \)
Let tan θ= k
sec2 θ = dk
\(\begin{array}{l}I=\int \frac{\left(1-k^{2}\right)}{(1+k)^{2}} d k=\int \frac{(1-k)}{(1+k)} d k\end{array} \)
I = (2/(1+k))-1)dk
I = 2 lnǀ1+kǀ -k+c
I = 2 lnǀ1+tan θǀ -tan θ +c
Given I = tan θ+ 2 log f(θ) + c
λ= -1, f(θ) = ǀ1+ tan θǀ
Answer: (b)
21. Let \(\begin{array}{l}\vec{a}\end{array} \)
, \(\begin{array}{l}\vec{b}\end{array} \)
, and \(\begin{array}{l}\vec{c}\end{array} \)
be three vectors such that \(\begin{array}{l}\vec{a}=\sqrt{3}\end{array} \)
, \(\begin{array}{l}\vec{b}=5\end{array} \)
, \(\begin{array}{l}\vec{b}.\vec{c}=10\end{array} \)
and the angle between \(\begin{array}{l}\vec{b}\end{array} \)
and \(\begin{array}{l}\vec{c}\end{array} \)
is π/3. If \(\begin{array}{l}\vec{a}\end{array} \)
is perpendicular to \(\begin{array}{l}\vec{b}\times \vec{c}\end{array} \)
, then \(\begin{array}{l}\left | \vec{a}\times (\vec{b}\times \vec{c} )\right |\end{array} \)
is equal to
\(\begin{array}{l}\left | \vec{a}\times (\vec{b}\times \vec{c} )\right | = \left | \vec{a} \right |\left | \vec{b} \times \vec{c}\right |\sin \theta\end{array} \)
where θ is the angle between \(\begin{array}{l}\vec{a}\end{array} \)
and \(\begin{array}{l}\vec{b}\times \vec{c}\end{array} \)
.
\(\begin{array}{l}\left | \vec{c} \right | = 4\end{array} \)
\(\begin{array}{l}\left |\vec{a}\times(\vec{b}\times \vec{c}) \right | = 30\end{array} \)
Answer : (30)
22. If Cr = 25Cr and C0+5⋅C1+9⋅C2 +⋯+101. C25 =225 .k then k is equal to
S = 25C0 + 525C1 + 925C2 + ….+9725C24 +10125C25 = 225k ..(i)
Reverse and apply property nCr = nCn-r in all coefficients
S = 10125C0 + 9725C1+…+525C24+25C25 …(ii)
Adding (i) and (ii)
2S = 102[25C0+25C1+…+25C5]
S = 51×225
k = 51
Answer: (51)
23. If the curves x2 -6x +y2 +8 = 0 and x2-8y+y2+16-k = 0, (k>0) touch each other at a point, then the largest value of k is
Two circles touch each other if C1C2 = ǀr1± r2ǀ
√k +1 = 5 or ǀ√k-1ǀ = 5
k = 16 or 36
Maximum value of k is 36.
Answer: (36)
24. The number of terms common to the A.P.’s 3,7,11,…407 and 2,9,16,…709 is .
First common term is 23
Common difference = LCM(7,4) = 28
23+(n−1)28 ≤ 407
n−1 ≤ 13.71
n = 14
Answer: (14)
25. If the distance between the plane, 23x-10y-2z+48 = 0 and the plane containing the lines \(\begin{array}{l}\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}\end{array} \)
and \(\begin{array}{l}\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda }\end{array} \)
, λ∈ R is equal to \(\begin{array}{l}\frac{k}{\sqrt{633}}\end{array} \)
, then k is equal to:
We find the point of intersection of the two lines, and the distance of given plane from the two lines is the distance of plane from the point of intersection.
(2p-1, 4p+3, 3p-1) = (2q-3, 6q-2, λq+1)
p = -1/2 and q = 1/2
λ= -7
Point of intersection is (-2,1,-5/2)
\(\begin{array}{l}\frac{k}{\sqrt{633}}=\left|\frac{-46-10+5+48}{\sqrt{633}}\right|\end{array} \)
k = 3
Answer: (3)
Video Lessons – January 9 Shift 2 Maths
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