**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 2

^{1/4}×4

^{1/16}×8

^{1/48}×16

^{1/128}… ..∞ to is equal to

a) 2

^{1/4}

b) 2

c) 2

^{1/2}

d) 1

2^{1/4}×4^{1/16}×8^{1/48}×16^{1/128}… ∞= 2^{1/4}×2^{2/16}×2^{3/48}×….∞

2^{1/4}×2^{1/8}×2^{1/16}×… ∞= 2^{1/4 + 1/8 + 1/16 …∞}

= √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 cm

^{3}/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is:

a) 5/6π

b) 1/54π

c) 1/36π

d) 1/18π

Let thickness of ice be 𝑥 cm.

Therefore, net radius of sphere = (10+𝑥) cm

Volume of sphere V = (4π/3)(10+𝑥)^{3}

⇒dV/dt = 4𝜋(10+𝑥)^{2} dx/dt

At x = 5, dV/dt = 50 cm^{3}/min

⇒ 50 = 4π×225×𝑑x/𝑑t

dx/dt = 1/18π cm/min

**Answer:** (d)

**Question 4.**Let f be any function continuous on [𝑎,𝑏] and twice differentiable on (𝑎,𝑏). If for all 𝑥∈(𝑎,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 (𝑎,𝑏)

LMVT is applicable.

For p ∈ (a,c) f’(𝑝) = (f(𝑐)−𝑓(𝑎))/(𝑐−𝑎)

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 cos

^{3}(π/8) cos (3π/8) + sin

^{3}(π/8) sin (3π/8) is:

a) 1/4

b) 1/2√2

c) 1/2

d) 1/√2

cos^{3} (π/8) cos (3π/8) + sin^{3} (π/8) sin (3π/8) = cos^{3} (π/8) [ 4 cos^{3 }(π/8) – 3 cos (π/8)] + sin^{3} (π/8)[ 3 sin (π/8) – 4 sin^{3} (π/8)]

= 4[cos^{6} (π/8) – sin^{6} (π/8)] + 3[sin^{4} (π/8) – cos^{4}(π/8)]

= 4[cos^{2} (π/8) – sin^{2} (π/8)] [cos^{4} (π/8) + sin^{4} (π/8) + cos^{2} (π/8) sin^{2} (π/8)] -3[cos^{2} (π/8) – sin^{2} (π/8)]

= [cos^{2} (π/8) – sin^{2} (π/8)][4(1-cos^{2}(π/8) sin^{2} (π/8) -3]

= cos (π/4)[1- sin^{2} (π/4)]

= 1/2√2

**Answer:** (c)

**Question 6.**The number of real roots of the equation, e

^{4x}+ e

^{3x}– 4e

^{2x}+ 1 = 0 is

a) 3

b) 4

c) 1

d) 2

e^{4x} + e^{3x} – 4e^{2x} + e^{x} + 1 = 0

e^{2x} + e^{x} – 4 + (1/e^{x}) + (1/e^{2x}) = 0

(e^{2x} + (1/e^{2x})) + (e^{x }+ 1/e^{x}) – 4 = 0

(e^{x }+ 1/e^{x})^{2 }– 2 + (e^{x }+ 1/e^{x}) – 4 = 0

Let (e^{x }+ 1/e^{x}) = u

Then u^{2} + u – 6 = 0

u = 2, -3

u ≠ -3 as u > 0 ( since e^{x} > 0)

(e^{x }+ 1/e^{x}) = 2

(e^{x }– 1)^{2} = 0

e^{x} = 1

x = 0

Hence, only one real solution is possible.

**Answer:** (c)

**Question 7.**The value of

a) 2

b) 4

c) 2

^{2}

d) π

^{2}

Let I =

I =

=

Adding (i) and (ii)

=

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

**Answer: **(d)

**Question 8.**If for some α and β in 𝑅, the intersection of the following three planes

x + 4y − 2z = 1

x + 7y − 5z =β

x + 5y + αz = 5

is a line in 𝑅^{3}, then α+β is equal to:

a) 0

b) 10

c) -10

d) 2

The given planes intersect in a line

D = D_{x} = D_{y} = D_{z} = 0

D = 0

7α + 25 – 4α -20 + 4 = 0

α= -3

D_{z} = 0

35 – 5β -20 + 4β -2 = 0

β= 13

α+ β= 10

**Answer: **(b)

**Question 9.**If e

_{1}and e

_{2}are the eccentricities of the ellipse, (x

^{2}/18) + (y

^{2}/4) = 1 and the hyperbola, (x

^{2}/9) -(y

^{2}/4) =1 respectively and (e

_{1 }, e

_{2}) is a point on the ellipse, 15x

^{2}+ 3y

^{2}= k. Then k is equal to:

a) 14

b) 15

c) 17

d) 16

e_{1 }= √(1-(4/18))

= √7/3

e_{2 }= √(1+(4/9))

= √13/3

Since (e_{1}, e_{2}) lies on the ellipse 15x^{2} + 3y^{2} = k

15×(7/9) + 3×(13/9) = k

k = 16

**Answer: **(d)

**Question 10.**If f(x) =

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

Or b = 1.

a + 3 = 1 a = -2

a + 2b = 0

**Answer:** (c)

**Question 11.**If the matrices A =

a) 16

b) 2

c) 8

d) 72

A =

= 13 + 1 – 8

= 6

B = adj(A) ǀadj Bǀ = ǀadj (adj A)ǀ = ǀAǀ^{4} = 6^{4}

ǀCǀ = ǀ3Aǀ = 3^{3}ǀAǀ = 3^{3} × 6

^{4}/(3

^{3}×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) 4x−3y+17 = 0

b) 3x+4y−6 = 0

c) 4x+3y−8 = 0

d) 3x−4y−24 = 0

OD^{2} = OA×OB

16 = 2× 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+3y−8=0

(20 + 12 – 8)/√(3^{2} + 4^{2}) = 24/5

(p ≠ r)

**Answer: **(c)

**Question 13.**Let 𝑧 be a complex number such that

a) √10

b) 7/2

c) 15/4

d) 2√3

If

y-1 = ± (y+2)

y-1 = -y-2

y = -1/2

ǀzǀ = 5/2

x^{2} + y^{2} = 25/4

x^{2} + y^{2} = 25/4

x^{2} + ¼ = 25/4

x = ±√6

**Answer:** (b)

**Question 14.**If f’(x) = tan

^{-1}(sec x + tan x ), (-π/2) < x < (π/2), and f(0) = 0, then f(1) is equal to:

a) (π+1)/4

b) (π+2)/4

c) 1/4

d) (π-1)/4

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

f’(x) = (π/4) + (x/2)

f(x) = (π/4) x+ (x^{2}/4) +c

f(0) = 0

c = 0

f(1) = (π/4) + (1/4) = (π+1)/4

**Answer: **(a)

**Question 15.**Negation of the statement: ′√5 is an integer or 5 is irrational′ 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

′√5 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+ cx

^{2}; then

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 + cx^{2}

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

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

= (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

Total numbers that can be formed are = 8×8×7×6

= 8×336

k = 8

**Answer** (a)

**Question 18.**Let the observations x

_{i}(1 ≤ i ≤ 10) satisfy the equations,

_{1}-3),(x

_{2}-3)…..(x

_{10}-3) , then the ordered pair (μ,λ) is equal to:

a) (6,3)

b) (3,6)

c) (3,3)

d) (6,6)

**Solution**:

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

λ= Var (x_{i}-3)

= Var (x_{i}-5)

=

**Answer** (c)

**Question 19.**The integral

a)

b)

c)

d)

**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 𝐵-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

As vectors p, q and r are coplanar,

R_{1} 🡪 R_{1 }+ R_{2} + R_{3}

C_{2} 🡪 C_{2} – C_{1} and C_{3} 🡪 C_{3} – C_{1}

3a + 1 = 0

a = -1/3

**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 __.

Projection of vector AB on vector CD =

**Answer: **(8)

**Question 23.**The number of distinct solutions of the equation

sin 2x = ±1/2

∴ We have 8 solutions for x ∈ [0,2]

**Answer:** (8)

**Question 24.**If for x ≥ 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 =

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 = x^{2} -2x

= 9-6

= 3

y(3) = 3

**Answer:** (3)

**Question 25.**The coefficient of x

^{4}in the expansion of (1+x+x)

^{10}is

General term of the given expression is given by

Here, q+2r =4

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

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

For p = 8, q = 0, r = 2, coefficient = 10!/(8!×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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