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JEE Advanced 2020 Question Paper
January 7 Shift 1 – Maths
January 7 Shift 1 – Maths
Question 1. The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = 𝑥 and the straight line 𝑦 = 𝑥, is
a) (1/3)(12𝜋 − 1)
b) (1/6)(12𝜋 − 1)
c) (1/3)(6𝜋 − 1)
d) (1/6)(24𝜋 − 1)
Required area = area of the circle – area bounded by given line and parabola
Required area = πr2 –
\(\begin{array}{l}\int_{0}^{1}(y-y^{2})dy\end{array} \)
Area =
\(\begin{array}{l}2\pi -(\frac{y^{2}}{2}-\frac{y^{3}}{3})_{0}^{1}\end{array} \)
= 2𝜋 – (1/6)
= (1/6)(12𝜋 – 1)
Answer:(b)
Question 2. Total number of six-digit numbers in which only and all the five digits 1,3,5,7 and 9 appear, is
a) 56
b) (½)(6!)
c) 6!
d) (5/2) 6!
Selecting all 5 digits = 5C5 = 1 way
Now, we need to select one more digit to make it a 6 digit number = 5C1 = 5 ways
Total number of permutations =
\(\begin{array}{l}\frac{6!}{2!}\end{array} \)
Total numbers = 5C5×5C1×(6!/2!)
= (5/2) 6!
Answer: (d)
Question 3. An unbiased coin is tossed 5 times. Suppose that a variable 𝑋 is assigned the value 𝑘 when 𝑘 consecutive heads are obtained for 𝑘 = 3, 4, 5, otherwise 𝑋 takes the value -1. The expected value of 𝑋, is
a) 1/8
b) 3/16
c) -1/8
d) -3/16
k = no. of consecutive heads
P(k = 3) = 5/32 {HHHTH, HHHTT, THHHT, HTHHH, TTHHH}
P(k = 4) = 2/32 {HHHHT, THHHH}
P(k = 5) = 1/32 {HHHHH}
\(\begin{array}{l}P(\bar{3}\cap \bar{4}\cap \bar{5})= 1-(\frac{5}{32}+\frac{2}{32}+\frac{1}{32}) = \frac{24}{32}\end{array} \)
ƩXP(X) =
\(\begin{array}{l}(-1\times \frac{24}{32})+(3\times \frac{5}{32})+(4\times \frac{2}{32})+(5\times \frac{1}{32})\end{array} \)
= 1/8
Answer: (a)
Question 4. If Re (z-1)/(2z + i) = 1, where z = x+iy, then the point (x,y) lies on a
a) circle whose centre is at (-1/2, -3/2)
b) straight line whose slope is 3/2.
c) circle whose diameter is √5/2
d) straight line whose slope is -2/3.
z = x + iy
\(\begin{array}{l}\frac{x+iy-1}{2x+2iy+i} = \frac{(x-1)+iy}{2x+i(2y+1)}\left ( \frac{2x-i(2y+1)}{2x-i(2y+1)}\right )\end{array} \)
\(\begin{array}{l}\frac{2x(x-1)+y(2y+1)}{4x^{2}+(2y+1)^{2}}= 1\end{array} \)
2x2 + 2y2 – 2x + y = 4x2 + 4y2 + 4y + 1
x2 + y2 + x + (3/2)y + (1/2) = 0
Circle’s centre will be (-1/2, -3/4)
Radius = √[(1/4) + (9/16) – (1/2)] = √5/4
Diameter = √5/2
Answer: (c)
Question 5. If f(a + b + 1 – x) = f(x) ∀x, where a and b are fixed positive real numbers, then \(\begin{array}{l}\frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx\end{array} \)
is equal to
a) \(\begin{array}{l}\int_{a-1}^{b-1}f(x)dx\end{array} \)
b) \(\begin{array}{l}\int_{a+1}^{b+1}f(x+1)dx\end{array} \)
c) \(\begin{array}{l}\int_{a-1}^{b-1}f(x+1)dx\end{array} \)
d) \(\begin{array}{l}\int_{a+1}^{b+1}f(x)dx\end{array} \)
f(a + b + 1 – x) = f(x) …(i)
f(a + b – x) = f(x+1) …(ii)
\(\begin{array}{l}I = \frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx\end{array} \)
…(iii)
From (i) and (ii)
\(\begin{array}{l}I = \frac{1}{(a+b)}\int_{a}^{b}(a+b-x)(f(x+1)+f(x))dx\end{array} \)
…(iv)
Adding (iii) and (iv)
\(\begin{array}{l}2I = \int_{a}^{b}(f(x)+f(x+1))dx\end{array} \)
Answer: (c)
Question 6. If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is
a) 2√3
b) √3
c) 3/√2
d) 3√2
Let the equation of ellipse be
x2/a2 + y2/b2 = 1
a>b
Now 2ae = 6
2a/e = 12
ae = 3 and a/e = 6
a2 = 18
a2e2 = c2 = a2-b2 = 9
b2 = 9
length of latus rectum = 2b2/a = 2×9/√18 = 3√2
Answer: (d)
Question 7. The logical statement (p⇒q) ∧ (q ⇒∼ p) is equivalent to
a) ∼ p
b) p
c) q
d) ∼ q
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𝒑
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𝒒
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𝒑 ⇒ 𝒒
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∼ 𝒑
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𝒒 ⇒∼ 𝒑
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(𝒑 ⇒ 𝒒) ∧ (𝒒 ⇒∼ 𝒑)
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T
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T
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T
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F
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F
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F
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T
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F
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F
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F
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T
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F
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F
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T
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T
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T
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T
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T
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F
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F
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T
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T
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T
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T
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Clearly (𝑝 ⇒ 𝑞) ∧ (𝑞 ⇒∼ 𝑝) is equivalent to ∼ 𝑝.
Answer: (a)
Question 8. The greatest positive integer 𝑘, for which 49𝑘 + 1 is a factor of the sum 49125 + 49124 + ⋯ + 492 + 49 + 1, is
a) 32
b) 60
c) 65
d) 63
1 + 49 + 492 + …. + 49125 = (49126 – 1)/(49 – 1)
= (4963 + 1)( 4963 – 1)/48
= (4963 + 1)((1 + 48)63-1)/48
= (4963 + 1)(1 +48)63 -1)/48
= (4963 + 1)(1 +48 I-1)/48: where I is an integer
= (4963 + 1)I
Greatest positive integer is k = 63
Answer: (d)
Question 9. A vector 𝑎⃗ = 𝛼𝑖̂ + 2𝑗̂ + 𝛽𝑘̂ (𝛼, 𝛽 ∈ 𝑹) lies in the plane of the vectors, ⃗𝑏 = 𝑖̂ + 𝑗̂ and 𝑐⃗ = 𝑖̂ − 𝑗̂ + 4𝑘̂. If 𝑎⃗ bisects the angle between 𝑏⃗ and 𝑐⃗, then
a) 𝑎⃗. 𝑖̂ + 3 = 0
b) 𝑎⃗. 𝑘̂ + 4 = 0
c) 𝑎⃗. 𝑖̂ + 1 = 0
d) 𝑎⃗. 𝑘̂ + 2 = 0
Question 10. If y(α) = \(\begin{array}{l}\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}\end{array} \)
where α ∈ (3π/4, π) then dy/dα at α = 5π/6 is
a) -1/4
b) 4/3
c) 4
d) -4
y(α) =
\(\begin{array}{l}\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}\end{array} \)
\(\begin{array}{l}y(\alpha )= \sqrt{2\frac{1}{\sin \alpha \cos \alpha \times \frac{1}{\cos ^{2}\alpha }}+\frac{1}{\sin ^{2}\alpha }}\end{array} \)
\(\begin{array}{l}y(\alpha )= \sqrt{2\cot \alpha +cosec^{2} \alpha }\end{array} \)
y(α) = √(1+cotα )2
y(α) = -1 – cotα
dy/dα = 0 + cosec2α
dy/dα = cosec2 5π/6
dy/dα = 4
Answer: (c)
Question 11. If y = mx + 4 is a tangent to both the parabolas, y2 = 4x and x2 = 2by, then b is equal to
a) -64
b) 128
c) -128
d) -32
Any tangent to the parabola y2 = 4x is y = mx + a/m
Comparing it with y = mx + 4, we get 1/m = 4
So m = ¼.
Equation of tangent becomes y = (x/4) + 4
y = (x/4) + 4 is a tangent to x2 = 2by
x2 = 2b{(x/4)+4}
Or 2x2 – bx – 16b = 0
D = 0
b2 + 128b = 0
b = 0 (not possible)
b = -128
Answer: (c)
Question 12. Let 𝛼 be a root of the equation x2 + x+ 1= 0 and the matrix A = \(\begin{array}{l}\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}\end{array} \)
then the matrix A31 is equal to
a) A
b) A2
c) A3
d) I3
The roots of equation 𝑥2 + 𝑥 + 1=0 are complex cube roots of unity.
= ω or ω2
\(\begin{array}{l}A= \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}\end{array} \)
=
\(\begin{array}{l}\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}\end{array} \)
A2 =
\(\begin{array}{l}\frac{1}{{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}\end{array} \)
\(\begin{array}{l}A^{2}=\frac{1}{3}\begin{bmatrix} 3 & 0 & 0\\ 0 & 0 & 3\\ 0 & 3 & 0 \end{bmatrix}\end{array} \)
\(\begin{array}{l}A^{2}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\end{array} \)
\(\begin{array}{l}A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\end{array} \)
\(\begin{array}{l}A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}= I\end{array} \)
A4 = I
A28 = 𝐼
Therefore, we get
A31 = A28 A3
A31 = IA3
A31 = A3
Answer: (c)
Question 13. If g(x) = x2 + x – 1 and (gof)(x) = 4x2 – 10x + 5, then f(5/4) is equal to
a) -3/2
b) -1/2
c) 1/2
d) 3/2
g(x) = x2 + x – 1
gof(x) = 4x2 – 10x + 5
g(f(x) = 4x2 – 10x + 5
f2(x) + f(x) – 1 = 4x2 – 10x + 5
Putting x = 5/4 and f(5/4) = t
t2 + t + ¼ = 0
t = -1/2 or f(5/4) = -1/2
Answer: (b)
Question 14. Let 𝛼 and 𝛽 are two real roots of the equation (𝑘 + 1) tan2 𝑥 − √2 𝜆 tan 𝑥 = 1 − 𝑘, where (𝑘 ≠ −1) and are real numbers. If tan2(α + β) = 50, then value of 𝜆 is
a) 5√2
b) 10√2
c) 10
d) 5
(k+1) tan2x – √2 tan x = 1 – k
tan2(α + β) = 50
∵ tan α and tan β are the roots of the given equation
Now
tan α + tanβ = √2/(k+1),
tanα tan β= (k -1)/(k+1)
\(\begin{array}{l}\left ( \frac{\frac{\sqrt{2 }\lambda}{k+1}}{1-\frac{k-1}{k+1}} \right )^{2} = 50\end{array} \)
2λ2/4 = 50
λ2 = 100
λ = 10, λ = -10
Answer: (c)
Question 15. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and 𝑅 be any point (2, 1, 6). Then the image of 𝑅 in the plane 𝑃 is:
a) (6, 5, 2)
b) (6, 5, −2)
c) (4,3, 2)
d) (3,4, −2)
Points A(2, 1, 0), B(4, 1, 1) C(5, 0, 1)
\(\begin{array}{l}\vec{AB } = (2,0,1)\end{array} \)
\(\begin{array}{l}\vec{AC } = (3,-1,1)\end{array} \)
\(\begin{array}{l}\vec{n } = \vec{AB}\times \vec{AC }\end{array} \)
Equation of the plane is x + y -2z = 3….(1)
Let the image of point (2, 1, 6) is (l, m, n)
(l-2)/1 = (m-1)/1 = (n-6)/-2 = -2(-12)/6 = 4
⇒ l = 6, m = 5, n = −2
Hence the image of R in the plane P is (6, 5, −2)
Answer: (b)
Question 16. Let xk + yk = ak, (a,k>0) and (dy/dx) + (y/x)1/3 = 0, then k is
a) 1/3
b) 3/2
c) 2/3
d) 4/3
xk + yk = ak
kxk-1 + kyk-1 (dy/dx) = 0
dy/dx = -(x/y)k-1
dy/dx + (y/x)1-k = 0
1-k = 1/3
k = 2/3
Answer: (c)
Question 17. Let the function, f:[-7, 0] →R be continuous on [−7, 0] and differentiable on (−7, 0). If f(-7)= −3 and 𝑓′(𝑥) ≤ 2, for all 𝑥 ∈ (−7, 0), then for all such functions 𝑓, 𝑓(−1) + 𝑓(0) lies in the interval:
a) [-6, 20]
b) (-∞, 20]
c) (-∞, 11]
d) [-3, 11]
f(-7) = -3 and f’(x) ≤ 2
Applying LMVT in [-7,0], we get
(f(-7) – f(0))/-7 = f’(c) ≤ 2
(-3-f(0))/-7 ≤ 2
f(0) + 3 ≤ 14
f(0) ≤ 11
Applying LMVT in [-7, -1], we get
(f(-7) – f(-1))/(-7 +1) = f’(c) ≤ 2
-3 – f(-1))/-6 = f’(c) ≤ 2
f(-1) + 3 = ≤ 12
f(-1) ≤ 9
Therefore f(-1)+ f(0) ≤ 20
Answer (b)
Question 18. If y = y(x) is the solution of the differential equation ey(dy/dx)-1 = ex such that y(0) = 0, then y(1) is equal to
a) loge 2
b) 2e
c) 2 + loge 2
d) 1+loge 2
𝑒y(𝑦′ − 1) = 𝑒𝑥
⇒ dy/dx = 𝑒𝑥−𝑦 + 1
Let x-y = t
1 – dy/dx = dt/dx
So, we can write
⇒ 1 − dt/dx = 𝑒𝑡 + 1
⇒ −𝑒−𝑡 𝑑𝑡 = 𝑑𝑥
⇒ 𝑒−𝑡 = 𝑥 + 𝑐
⇒ 𝑒𝑦−𝑥 = 𝑥 + 𝑐
1 = 0 + 𝑐
⇒ 𝑒𝑦−𝑥 = 𝑥 + 1
at 𝑥 = 1
⇒ 𝑒𝑦−1 = 2
⇒ 𝑦 = 1 + log22
Question 19. Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is -1/2, then the greatest number amongst them is
a) 16
b) 27
c) 7
d) 21/2
Let 5 numbers be 𝑎 − 2𝑑, 𝑎 − 𝑑, 𝑎, 𝑎 + 𝑑, 𝑎 + 2𝑑
5𝑎 = 25
𝑎 = 5
(𝑎 − 2𝑑)(𝑎 − 𝑑) a (𝑎 + 𝑑)(𝑎 + 2𝑑) = 2520
(25 − 4𝑑2)(25 − 𝑑2) = 504
4𝑑4 − 125𝑑2 + 121 = 0
4𝑑4 − 4𝑑2 − 121𝑑2 + 121 = 0
𝑑2 = 1 𝑜𝑟 𝑑2 = 121/4
d = 11/2
For d = 11/2, a + 2d is the greatest term, a + 2d = 5 + 11 = 16.
Answer: (a)
Question 20. If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
where a, b, c ∈ R are non-zero and distinct; has non-zero solution, then
a) a + b + c = 0
b) a, b, c are in A.P
c) 1/a, 1/b, 1/c are in A.P
d) a, b, c are in G.P
\(\begin{array}{l}\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c & c \end{vmatrix}=0\end{array} \)
R2 R2 – R1
R3 R3 – R1
\(\begin{array}{l}\begin{vmatrix} 2 & 2a &a \\ 0& 3b-2a &b-a \\ 0& 4c-2a & c-a \end{vmatrix}=0\end{array} \)
⇒ (3b − 2𝑎)(𝑐 − 𝑎) − (4𝑐 − 2𝑎)(𝑏 − 𝑎) = 0
⇒ 3𝑏𝑐 − 2𝑎𝑐 − 3𝑎𝑏 + 2𝑎2 − [4𝑏𝑐 − 4𝑎𝑐 − 2𝑎𝑏 + 2𝑎2] = 0
⇒ −𝑏𝑐 + 2𝑎𝑐 − 𝑎𝑏 = 0
⇒ 𝑎𝑏 + 𝑏𝑐 = 2𝑎𝑐
1/c + 1/a = 2/b
Question 21. \(\begin{array}{l}\lim_{x\to 2}\frac{3^{x}+3^{3-x}-12}{3^{-\frac{x}{2}}-3^{1-x}}\end{array} \)
is equal to
Put 3x/2 = t
Answer: (36)
Question 22.If variance of first 𝑛 natural numbers is 10 and variance of first 𝑚 even natural numbers is 16, 𝑚 + 𝑛 is equal to
For 𝑛 natural number variance is given by
\(\begin{array}{l}\sigma ^{2} = \frac{\sum x_{i}^{2}}{n}- (\frac{\sum x_{i}}{n})^{2}\end{array} \)
\(\begin{array}{l}\frac{\sum x_{i}^{2}}{n}= \frac{1^{2}+2^{2}+..n^{2}}{n} = \frac{n(n+1)(2n+1)}{6n}\end{array} \)
\(\begin{array}{l}\frac{\sum x_{i}}{n}= \frac{1+2+..n}{n} = \frac{n(n+1)}{2n}\end{array} \)
σ2 = (n2-1)/12 = 10
n = 11
Variance of (2, 4, 6…) = 4×variance of (1,2,3,4…) = 4×(m2-1)/12
= (m2-1)/3
= 16
m = 7
Therefore, n + m = 11 + 7 = 18
Answer: (18)
Question 23. If the sum of the coefficients of all even powers of 𝑥 in the product
(1 + x + x2 + x3 … . +x2𝑛)(1 − x + x2 − x3 … . + x2𝑛) is 61, then 𝑛 is equal to
Let (1 + x + x2 + ⋯ + x2𝑛)(1 − x + x2 − ⋯ + x2𝑛) = a𝜊 + a1𝑥 + a2𝑥2 + ⋯
Put x = 1
2n + 1 = ao + a1 + a2 + a3 + . . . . . . . . . . . . (1)
Put x = −1
2n + 1 = ao – a1 + a2 – a3 +. . . . . . . . . . . . . (2)
Add (1) and (2)
2(2n + 1) = 2(a𝜊 + a2 + a4 + . . . . . . . . . . .
2n + 1 = 61
n = 30
Answer: (30)
Question 24. Let S be the set of points where the function, (𝑥) = |2 − |𝑥 − 3||, 𝑥 ∈ 𝐑, is not differentiable. Then, the value of ∑𝑥∈S f(f(x)) is equal to
Solution:
There will be three points 𝑥 = 1, 3, 5 at which f(x) is non-differentiable.
So f(f(1)) + f(f(3)) + f(f(5))
= f(0) + f(2) + f(0)
= 1+1+1
= 3
Answer: (3)
Question 25. Let A(1,0), B(6,2), C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangle APC, APB and BPC have equal areas, then the length of the line the segment PQ, where Q is the point ( -7/6, -1/3), is
P is the centroid which is =
\(\begin{array}{l}\left ( \frac{1+6+\frac{3}{2}}{3} , \frac{0+2+6}{3}\right )\end{array} \)
P = (17/6, 8/3)
Q = (-7/6, -1/3)
PQ =
\(\begin{array}{l}\sqrt{4^{2}+3^{2}} = 5\end{array} \)
Answer (5)
Video Lessons – January 7 Shift 1 Maths
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