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JEE Main 2020 Maths Paper With Solutions January 9 Shift 1

JEE Main 2020 solved shift 1 Maths paper with solutions are available on this page. Students are recommended to revise these questions and understand how the solutions are formed so that they can improve their problem-solving skills. These question paper given here will help candidates to understand the question types along with the paper pattern and the difficulty level of questions from each chapter. Students can easily download the solutions in PDF format for free.
January 9 Shift 1 – Maths
Question 1. If C be the centroid of the triangle having vertices (3, -1), (1, 3) and (2, 4). Let P be the point of intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0, then the line passing through the points C and P also passes through the point:
a) (-9, -7)
b) (-9, -6)
c) (7, 6)
d) (9, 7)

Coordinates of C are ((3+1+2)/3 , (-1+3+4)/3) = (2,2)

Point of intersection of two lines

x + 3y – 1 = 0 and 3x – y + 1 = 0 is P(-1/5, 2/5).

Equation of line CP is 8x-11y+6 = 0

Point (-9,-6) lies on CP.

Answer:(b)

Question 2. The product 21/4Γ—41/16Γ—81/48Γ—161/128… ..∞ to is equal to
a) 21/4
b) 2
c) 21/2
d) 1

21/4Γ—41/16Γ—81/48Γ—161/128… ∞= 21/4Γ—22/16Γ—23/48×….∞

21/4Γ—21/8Γ—21/16×… ∞= 21/4 + 1/8 + 1/16 β€¦βˆž

\(\begin{array}{l}2^{\left ( \frac{\frac{1}{4}}{1-\frac{1}{2}} \right )}\end{array} \)

= √2

Answer: (c)

Question 3. A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at the rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is:
a) 5/6Ο€
b) 1/54Ο€
c) 1/36Ο€
d) 1/18Ο€

Let thickness of ice be π‘₯ cm.

Therefore, net radius of sphere = (10+π‘₯) cm

Volume of sphere V = (4Ο€/3)(10+π‘₯)3

β‡’dV/dt = 4πœ‹(10+π‘₯)2 dx/dt

At x = 5, dV/dt = 50 cm3/min

β‡’ 50 = 4π×225×𝑑x/𝑑t

dx/dt = 1/18Ο€ cm/min

Answer: (d)

Question 4. Let f be any function continuous on [π‘Ž,𝑏] and twice differentiable on (π‘Ž,𝑏). If for all π‘₯∈(π‘Ž,𝑏),f’(x)> 0 and f’’(x) < 0, then for any c ∈ (a,b), (f(c) – f(a)])- (f(b) – f(c)]) is greater than:
a) (b-c)/(c-a)
b) 1
c) (c-a)/(b-c)
d) (b+a)/(b-a)

c∈(a,b) and f is twice differentiable and continuous function on (π‘Ž,𝑏)

LMVT is applicable.

For p ∈ (a,c) f’(𝑝) = (f(𝑐)βˆ’π‘“(π‘Ž))/(π‘βˆ’π‘Ž)

For q ∈ (c,b), f’(q) = (f(b)-f(c)/(b-c)

f’’(x) < 0 f’(x) is decreasing

f’(p) > f’(q)

(f(c) – f(a))/(c-a) > (f(b) – f(c))/(b-c)

(f(c) – f(a))/ (f(b) – f(c)) > (c-a)/(b-c) (as f’(x)>0 f(x) is increasing.

Answer: (c)

Question 5. The value of cos3 (Ο€/8) cos (3Ο€/8) + sin3 (Ο€/8) sin (3Ο€/8) is:
a) 1/4
b) 1/2√2
c) 1/2
d) 1/√2

cos3 (Ο€/8) cos (3Ο€/8) + sin3 (Ο€/8) sin (3Ο€/8) = cos3 (Ο€/8) [ 4 cos3 (Ο€/8) – 3 cos (Ο€/8)] + sin3 (Ο€/8)[ 3 sin (Ο€/8) – 4 sin3 (Ο€/8)]

= 4[cos6 (Ο€/8) – sin6 (Ο€/8)] + 3[sin4 (Ο€/8) – cos4(Ο€/8)]

= 4[cos2 (Ο€/8) – sin2 (Ο€/8)] [cos4 (Ο€/8) + sin4 (Ο€/8) + cos2 (Ο€/8) sin2 (Ο€/8)] -3[cos2 (Ο€/8) – sin2 (Ο€/8)]

= [cos2 (Ο€/8) – sin2 (Ο€/8)][4(1-cos2(Ο€/8) sin2 (Ο€/8) -3]

= cos (Ο€/4)[1- sin2 (Ο€/4)]

= 1/2√2

Answer: (c)

Question 6. The number of real roots of the equation, e4x + e3x – 4e2x + 1 = 0 is
a) 3
b) 4
c) 1
d) 2

e4x + e3x – 4e2x + ex + 1 = 0

e2x + ex – 4 + (1/ex) + (1/e2x) = 0

(e2x + (1/e2x)) + (ex + 1/ex) – 4 = 0

(ex + 1/ex)2 – 2 + (ex + 1/ex) – 4 = 0

Let (ex + 1/ex) = u

Then u2 + u – 6 = 0

u = 2, -3

u β‰  -3 as u > 0 ( since ex > 0)

(ex + 1/ex) = 2

(ex – 1)2 = 0

ex = 1

x = 0

Hence, only one real solution is possible.

Answer: (c)

Question 7. The value of
\(\begin{array}{l}\int_{0}^{2\pi }\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
is equal to

a) 2
b) 4
c) 22
d) Ο€2

Let I =

\(\begin{array}{l}\int_{0}^{2\pi }\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
…(i)

I =

\(\begin{array}{l}\int_{0}^{2\pi }\frac{(2\pi -x)\sin ^{8}(2\pi -x)}{\sin ^{8}(2\pi -x)+\cos ^{8}(2\pi -x)}dx\end{array} \)

=

\(\begin{array}{l}\int_{0}^{2\pi }\frac{(2\pi -x)\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
..(ii)

Adding (i) and (ii)

\(\begin{array}{l}2I = 2\pi \int_{0}^{2\pi }\frac{\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
\(\begin{array}{l}I = \pi \int_{0}^{2\pi }\frac{\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
\(\begin{array}{l}I = 4\pi \int_{0}^{\frac{\pi }{2} }\frac{\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
…(iii)

\(\begin{array}{l}I = 4\pi \int_{0}^{\frac{\pi }{2} }\frac{\sin ^{8}(\frac{\pi }{2}-x)}{\sin ^{8}(\frac{\pi }{2}-x)+\cos ^{8}(\frac{\pi }{2}-x)}dx\end{array} \)

=

\(\begin{array}{l}4\pi \int_{0}^{\frac{\pi }{2} }\frac{\cos ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx\end{array} \)
..(iv)

Adding (iii) and (iv), we get:

\(\begin{array}{l}I = 2\pi \int_{0}^{\frac{\pi }{2} }1dx = 2\pi \times \frac{\pi }{2}=\pi ^{2}\end{array} \)

Answer: (d)

Question 8. If for some Ξ± and Ξ² in 𝑅, the intersection of the following three planes

x + 4y βˆ’ 2z = 1

x + 7y βˆ’ 5z =Ξ²

x + 5y + Ξ±z = 5

is a line in 𝑅3, then Ξ±+Ξ² is equal to:


a) 0
b) 10
c) -10
d) 2

The given planes intersect in a line

D = Dx = Dy = Dz = 0

D = 0

\(\begin{array}{l}\begin{vmatrix} 1 &4 & -2\\ 1 &7 & -5\\ 1 & 5& \alpha \end{vmatrix} = 0\end{array} \)

7Ξ± + 25 – 4Ξ± -20 + 4 = 0

Ξ±= -3

Dz = 0

\(\begin{array}{l}\begin{vmatrix} 1 &4 & 1\\ 1 &7 & \beta \\ 1 & 5& 5 \end{vmatrix} = 0\end{array} \)

35 – 5Ξ² -20 + 4Ξ² -2 = 0

Ξ²= 13

Ξ±+ Ξ²= 10

Answer: (b)

Question 9. If e1 and e2 are the eccentricities of the ellipse, (x2/18) + (y2/4) = 1 and the hyperbola, (x2/9) -(y2/4) =1 respectively and (e1 , e2) is a point on the ellipse, 15x2 + 3y2 = k. Then k is equal to:
a) 14
b) 15
c) 17
d) 16

e1 = √(1-(4/18))

= √7/3

e2 = √(1+(4/9))

= √13/3

Since (e1, e2) lies on the ellipse 15x2 + 3y2 = k

15Γ—(7/9) + 3Γ—(13/9) = k

k = 16

Answer: (d)

Question 10. If f(x) =
JEE Main 2020 Paper With Solutions Maths Shift 1 Jan 9
is continuous at x = 0 then a + 2b is equal to:

a) -2
b) 1
c) 0
d) 1

f(x) is continuous at x = 0

\(\begin{array}{l}\lim_{x\to0^{-}}f(x)= b= \lim_{x\to0^{+}}f(x)\end{array} \)
\(\begin{array}{l}b = \lim_{h\to0}f(0+h)= \lim_{h\to0}\frac{(h+3h^{2})^{\frac{1}{3}}-h^{\frac{1}{3}}}{h^{\frac{4}{3}}}\end{array} \)
\(\begin{array}{l}b = \lim_{h\to0} \frac{h^{\frac{1}{3}\left [ (1+3h)^{\frac{1}{3}}-1\right ]}}{h\frac{4}{3}}\end{array} \)
\(\begin{array}{l}b = \lim_{h\to0} \frac{(1+3h)^{\frac{1}{3}}-1}{h}\end{array} \)
\(\begin{array}{l}b = \lim_{h\to0} \frac{1}{3}(1+3h)^{\frac{-2}{3}}\times 3\end{array} \)

Or b = 1.

\(\begin{array}{l}\lim_{x\to0^{-}} f(x)=1\Rightarrow \lim_{h\to0}\frac{\sin ((a+2)(-h))+\sin (-h)}{-h}= 1\end{array} \)

a + 3 = 1 a = -2

a + 2b = 0

Answer: (c)

Question 11. If the matrices A =
\(\begin{array}{l}\begin{bmatrix} 1 & 1 &2 \\ 1& 3&4 \\ 1 & -1&3 \end{bmatrix}\end{array} \)
, B = adj A and C = 3A, then
\(\begin{array}{l}\frac{\left | adj \: B \right |}{\left | C \right |}\end{array} \)
is equal to

a) 16
b) 2
c) 8
d) 72

A =

\(\begin{array}{l}\begin{bmatrix} 1 & 1 &2 \\ 1& 3&4 \\ 1 & -1&3 \end{bmatrix}\end{array} \)
\(\begin{array}{l}\left | A \right |= \begin{vmatrix} 1 & 1 & 2\\ 1& 3 & 4\\ 1& -1 & 3 \end{vmatrix}\end{array} \)

= 13 + 1 – 8

= 6

B = adj(A) Η€adj BΗ€ = Η€adj (adj A)Η€ = Η€AΗ€4 = 64

Η€CΗ€ = Η€3AΗ€ = 33Η€AΗ€ = 33 Γ— 6

\(\begin{array}{l}\frac{\left | adj \: B \right |}{\left | C \right |}\end{array} \)
= 64/(33Γ—6)

= 8

Answer: (c)

Question 12. A circle touches the y-axis at the point (0,4) and passes through the point (2,0). Which of the following lines is not a tangent to the circle?
a) 4xβˆ’3y+17 = 0
b) 3x+4yβˆ’6 = 0
c) 4x+3yβˆ’8 = 0
d) 3xβˆ’4yβˆ’24 = 0

JEE Main 2020 Papers With Solutions Maths Shift 1 Jan 9

OD2 = OAΓ—OB

16 = 2Γ— OB

OB = 8

AB = 6

AM = 3

CM = 4

CA = 5

OM = 5

Centre will be (5,4) and radius is 5

Now checking all the options

Option (c) is not a tangent.

4x+3yβˆ’8=0

(20 + 12 – 8)/√(32 + 42) = 24/5

(p β‰  r)

Answer: (c)

Question 13. Let 𝑧 be a complex number such that
\(\begin{array}{l}{\left | \frac{z-i}{z+2i} \right |}= 1\end{array} \)
and
\(\begin{array}{l}{\left | z\right |}= \frac{5}{2}\end{array} \)
. Then the value of
\(\begin{array}{l}{\left | z+3i\right |}\end{array} \)
is:

a) √10
b) 7/2
c) 15/4
d) 2√3

If

\(\begin{array}{l}{\left | \frac{z-i}{z+2i} \right |}= 1\end{array} \)
and
\(\begin{array}{l}{\left | z\right |}= \frac{5}{2}\end{array} \)
\(\begin{array}{l}{\left | z-i\right |} = {\left | z+2i\right |}\end{array} \)

y-1 = Β± (y+2)

y-1 = -y-2

y = -1/2

Η€zΗ€ = 5/2

x2 + y2 = 25/4

x2 + y2 = 25/4

x2 + ΒΌ = 25/4

x = ±√6

\(\begin{array}{l}{\left | z+3i\right |} = \sqrt{(x^{2}+(y+3)^{2}}\end{array} \)
\(\begin{array}{l}{\left | z+3i\right |} = \frac{7}{2}\end{array} \)

Answer: (b)

Question 14. If f’(x) = tan -1 (sec x + tan x ), (-Ο€/2) < x < (Ο€/2), and f(0) = 0, then f(1) is equal to:
a) (Ο€+1)/4
b) (Ο€+2)/4
c) 1/4
d) (Ο€-1)/4

f’(x) = tan-1 (sec x + tan x) = tan-1 ((1+ sin x)/cos x )

JEE Main 2020 Paper With Solution Maths Shift 1 Jan 9

f’(x) = (Ο€/4) + (x/2)

f(x) = (Ο€/4) x+ (x2/4) +c

f(0) = 0

c = 0

f(1) = (Ο€/4) + (1/4) = (Ο€+1)/4

Answer: (a)

Question 15. Negation of the statement: β€²βˆš5 is an integer or 5 is irrationalβ€² is:
a) √5 is irrational or 5 is an integer.
b) √5 is not an integer or 5 is not irrational.
c) √5 is an integer and 5 is irrational.
d) √5 is not an integer and 5 is not irrational.

p:√5 is an integer

q:5 is an irrational number

Given statement : (p∨q)

Required negation statement: ~(p∨q) = ~p∧ ~ q

β€²βˆš5 is not an integer and 5 is not irrational’

Answer: (d)

Question 16. If for all real triplets (a,b,c), f(x) = a+ bx+ cx2; then
\(\begin{array}{l}\int_{0}^{1}f(x)dx\end{array} \)
is equal to

a) 2(3f(1)+2f(1/2))
b) (1/3)(f(0)+f(1/2))
c) (1/2) (f(1)+3f(1/2))
d) (1/6)(f(0)+f(1)+4f(1/2))

f(x) = a+ bx + cx2

f(0) = a, f(1) = a + b + c

f(1/2) = (c/4) + (b/2) + a

\(\begin{array}{l}\int_{0}^{1}f(x)dx = \int_{0}^{1}(a+bx+cx^{2})dx = a+\frac{b}{2}+\frac{c}{3}\end{array} \)

= (1/6) (6a + 3b + 2c)

= (1/6) (a + (a + b + c) + (4a + 2b +c))

= (1/6)(f(0) + f(1) + 4f(1/2))

Answer: (d)

Question 17. If the number of five digit numbers with distinct digits and 2 at the 10th place is 336k, then k is equal to:
a) 8
b) 7
c) 4
d) 6

Shift 1 Jan 9 JEE Main 2020 Paper With Solutions Maths

Total numbers that can be formed are = 8Γ—8Γ—7Γ—6

= 8Γ—336

k = 8

Answer (a)

Question 18. Let the observations xi(1 ≀ i ≀ 10) satisfy the equations,
\(\begin{array}{l}\sum_{i=1}^{10}(x_{i}-5)=10\end{array} \)
and
\(\begin{array}{l}\sum_{i=1}^{10}(x_{i}-5)^{2}=40\end{array} \)
. If ΞΌ and Ξ» are the mean and the variance of observations, (x1-3),(x2-3)…..(x10-3) , then the ordered pair (ΞΌ,Ξ») is equal to:

a) (6,3)
b) (3,6)
c) (3,3)
d) (6,6)

Solution:

\(\begin{array}{l}\sum_{i=1}^{10}(x_{i}-5)=10\end{array} \)
\(\begin{array}{l}\sum_{i=1}^{10}x_{i}=60\end{array} \)
\(\begin{array}{l}\mu = \frac{\sum_{i=1}^{10}(x_{i}-3)}{10}=\frac{\sum_{i=1}^{10}x_{i}-30}{10} = 3\end{array} \)

Variance is unchanged, if a constant is added or subtracted from each observation.

Ξ»= Var (xi-3)

= Var (xi-5)

=

\(\begin{array}{l}\frac{\sum_{i=1}^{10}(x_{i}-5)^{2}}{10}-\left ( \frac{\sum_{i=1}^{10}(x_{i}-5)}{10} \right )^{2}= \frac{40}{10}-(\frac{10}{10})^{2} = 3\end{array} \)

Answer (c)

Question 19. The integral
\(\begin{array}{l}\int \frac{dx}{\left (x+4\right )^{\frac{8}{7}}\left (x-3 \right )^{\frac{6}{7}}}\end{array} \)
is equal to: (where C is a constant of integration)

a)
\(\begin{array}{l}-\left ( \frac{x-3}{x+4} \right )^{-\frac{1}{7}}+ C\end{array} \)

b)
\(\begin{array}{l}\frac{1}{2}\left ( \frac{x-3}{x+4} \right )^{\frac{3}{7}}+ C\end{array} \)

c)
\(\begin{array}{l}\left ( \frac{x-3}{x+4} \right )^{\frac{1}{7}}+ C\end{array} \)

d)
\(\begin{array}{l}-\frac{1}{13}\left ( \frac{x-3}{x+4} \right )^{-\frac{13}{7}}+ C\end{array} \)

Shift 1 Jan 9 JEE Main 2020 Paper With Solutions Maths

Answer: (c)

Question 20. In a box, there are 20 cards out of which 10 are labelled as A and remaining 10 are labelled as B. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third 𝐡-card is:


a) 15/16
b) 9/16
c) 13/16
d) 11/16

Here P(A) = P(B) = 12

Then, these following cases are possible AA, BAA, ABA, ABBA, BBAA, BABA

So, the required probability is = (1/4)+(1/8)+(1/8) +(1/16)+(1/16)+(1/16) = 11/16

Answer: (d)

Question 21. If the vectors
\(\begin{array}{l}\vec{p} = (a+1)\hat{i}+a\hat{j}+a\hat{k}\end{array} \)
,
\(\begin{array}{l}\vec{q} = a\hat{i}+(a+1)\hat{j}+a\hat{k}\end{array} \)
and
\(\begin{array}{l}\vec{r} = a\hat{i}+a\hat{j}+(a+1)\hat{k}\end{array} \)
(a∈R) are coplanar and
\(\begin{array}{l}3(\vec{p} .\vec{q})^{2}-\lambda \left |\vec{r}\times \vec{q} \right |^{2}=0\end{array} \)
then value of Ξ» is

As vectors p, q and r are coplanar,

\(\begin{array}{l}\begin{vmatrix} (a+1) &a & a\\ a & (a+1)& a\\ a & a& (a+1) \end{vmatrix}=0\end{array} \)

R1 πŸ‘ͺ R1 + R2 + R3

\(\begin{array}{l}\begin{vmatrix} 3a+1 &3a+1 & 3a+1\\ a & a+1& a\\ a & a& a+1 \end{vmatrix}=0\end{array} \)
\(\begin{array}{l}(3a+1)\begin{vmatrix} 1 &1 & 1\\ a & a+1& a\\ a & a& a+1 \end{vmatrix}=0\end{array} \)

C2 πŸ‘ͺ C2 – C1 and C3 πŸ‘ͺ C3 – C1

\(\begin{array}{l}(3a+1)\begin{vmatrix} 1 &0 & 0\\ a & 1& 0\\ a & 0& 1 \end{vmatrix}=0\end{array} \)

3a + 1 = 0

a = -1/3

Shift 1 Jan 9 Maths JEE Main 2020 Paper With Solutions

Answer: (1)

Question 22.The projection of the line segment joining the points (1,-1,3) and (2,-4,11) on the line joining the points (-1,2,3) and (3,-2,10) is __.

\(\begin{array}{l}\vec{AB} = \hat{i}-3\hat{j}+8\hat{k}\end{array} \)
\(\begin{array}{l}\vec{CD} = 4\hat{i}-4\hat{j}+7\hat{k}\end{array} \)

Projection of vector AB on vector CD =

\(\begin{array}{l}\frac{\vec{AB}.\vec{CD}}{\left | \vec{CD} \right |}=\frac{4+12+56}{\sqrt{4^{2}+4^{2}+7^{2}}}=\frac{72}{9}=8\end{array} \)

Answer: (8)

Question 23. The number of distinct solutions of the equation
\(\begin{array}{l}log_{\frac{1}{2}}\left | \sin x \right |= 2-log_{\frac{1}{2}} \left | \cos x \right |\end{array} \)
in the interval [0, 2] is:

Shift 1 Jan 9 JEE Main 2020 Maths Paper With Solutions

sin 2x = Β±1/2

Shift 1 Jan 9 JEE Main 2020 Maths Paper With Solution

∴ We have 8 solutions for x ∈ [0,2]

Answer: (8)

Question 24. If for x β‰₯ 0,y = y(x) is the solution of the differential equation (1+x)dy = [(1+x)2+y-3]dx, y(2) = 0, then y(3) is equal to:

(1+x) dy/dx = [(1+x)2 + (y-3)]

(1+x) (dy/dx) -y = (1+x)2 -3

(dy/dx)-(1/1+x)y = 1+x – 3/(1+x)

I.F =

\(\begin{array}{l}e^{-\int \frac{1}{1+x}dx} = \frac{1}{1+x}\end{array} \)
\(\begin{array}{l}y\times \frac{1}{1+x} = \int 1-\frac{3}{(1+x)^{2}}dx\end{array} \)

y = x(1+x) + 3+c(1+x)

At x = 2, y = 0 we get

0 = 6+3+3c

c = -3

At x = 3,

y = x2 -2x

= 9-6

= 3

y(3) = 3

Answer: (3)

Question 25. The coefficient of x4 in the expansion of (1+x+x)10 is

General term of the given expression is given by

\(\begin{array}{l}\frac{10!}{p!q!r!}x^{q+2r}\end{array} \)

Here, q+2r =4

For p = 6, q = 4, r = 0, coefficient = 10!/(6!Γ—4!) = 210

For p = 7, q = 2, r = 1, coefficient = 10!/(7!Γ—2!Γ—1!) = 360

For p = 8, q = 0, r = 2, coefficient = 10!/(8!Γ—2!) = 45

Therefore, sum = 615

Answer (615)

Video Lessons – January 9 Shift 1 Maths

JEE Main 2020 Maths Paper With Solutions Jan 9 Shift 1

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