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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 = ? and the straight line ? = ?, is

a) (1/3)(12? − 1)
b) (1/6)(12? − 1)
c) (1/3)(6? − 1)
d) (1/6)(24? − 1)

Solution:

Required area = area of the circle – area bounded by given line and parabola

Required area = πr² -
$\int_{0}^{1}(y-y^{2})dy$

Area =
$2\pi -(\frac{y^{2}}{2}-\frac{y^{3}}{3})_{0}^{1}$

= 2? - (1/6)

= (1/6)(12? - 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!

Solution:

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 =
$\frac{6!}{2!}$

Total numbers = ⁵C₅×⁵C₁×(6!/2!)

= (5/2) 6!

Answer: (d)

Question 3. An unbiased coin is tossed 5 times. Suppose that a variable ? is assigned the value ? when ? consecutive heads are obtained for ? = 3, 4, 5, otherwise ? takes the value -1. The expected value of ?, is

a) 1/8
b) 3/16
c) -1/8
d) -3/16

Solution:

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}

$P(\bar{3}\cap \bar{4}\cap \bar{5})= 1-(\frac{5}{32}+\frac{2}{32}+\frac{1}{32}) = \frac{24}{32}$

ƩXP(X) =
$(-1\times \frac{24}{32})+(3\times \frac{5}{32})+(4\times \frac{2}{32})+(5\times \frac{1}{32})$

= 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.

Solution:

z = x + iy

$\frac{x+iy-1}{2x+2iy+i} = \frac{(x-1)+iy}{2x+i(2y+1)}\left ( \frac{2x-i(2y+1)}{2x-i(2y+1)}\right )$

$\frac{2x(x-1)+y(2y+1)}{4x^{2}+(2y+1)^{2}}= 1$

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

$\frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx$

is equal to

a)
$\int_{a-1}^{b-1}f(x)dx$
b)
$\int_{a+1}^{b+1}f(x+1)dx$
c)
$\int_{a-1}^{b-1}f(x+1)dx$
d)
$\int_{a+1}^{b+1}f(x)dx$

Solution:

f(a + b + 1 - x) = f(x) …(i)

f(a + b - x) = f(x+1) …(ii)

$I = \frac{1}{(a+b)}\int_{a}^{b}x(f(x)+f(x+1))dx$
…(iii)

From (i) and (ii)

$I = \frac{1}{(a+b)}\int_{a}^{b}(a+b-x)(f(x+1)+f(x))dx$
…(iv)

Adding (iii) and (iv)

$2I = \int_{a}^{b}(f(x)+f(x+1))dx$

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

Solution:

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

Solution:

?	?	? ⇒ ?	∼ ?	? ⇒∼ ?	(? ⇒ ?) ∧ (? ⇒∼ ?)
T	T	T	F	F	F
T	F	F	F	T	F
F	T	T	T	T	T
F	F	T	T	T	T

Clearly (? ⇒ ?) ∧ (? ⇒∼ ?) is equivalent to ∼ ?.

Answer: (a)

Question 8. The greatest positive integer ?, for which 49^? + 1 is a factor of the sum 49¹²⁵ + 49¹²⁴ + ⋯ + 49² + 49 + 1, is

a) 32
b) 60
c) 65
d) 63

Solution:

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 ?⃗ = ??̂ + 2?̂ + ??̂ (?, ? ∈ ?) lies in the plane of the vectors, ⃗? = ?̂ + ?̂ and ?⃗ = ?̂ − ?̂ + 4?̂. If ?⃗ bisects the angle between ?⃗ and ?⃗, then

a) ?⃗. ?̂ + 3 = 0
b) ?⃗. ?̂ + 4 = 0
c) ?⃗. ?̂ + 1 = 0
d) ?⃗. ?̂ + 2 = 0

Solution:

Question 10. If y(α) =

$\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}$

where α ∈ (3π/4, π) then dy/dα at α = 5π/6 is

a) -1/4
b) 4/3
c) 4
d) -4

Solution:

y(α) =
$\sqrt{2(\frac{\tan \alpha + \cot \alpha }{1+\tan ^{2}\alpha })+\frac{1}{\sin ^{2}\alpha }}$

$y(\alpha )= \sqrt{2\frac{1}{\sin \alpha \cos \alpha \times \frac{1}{\cos ^{2}\alpha }}+\frac{1}{\sin ^{2}\alpha }}$

$y(\alpha )= \sqrt{2\cot \alpha +cosec^{2} \alpha }$

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

Solution:

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 ? be a root of the equation x²+ x+ 1= 0 and the matrix A =

$\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}$

then the matrix A³¹ is equal to

a) A
b) A²
c) A³
d) I₃

Solution:

The roots of equation ?² + ? + 1=0 are complex cube roots of unity.

= ω or ω²

$A= \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \alpha & \alpha ^{2}\\ 1& \alpha ^{2} & \alpha ^{4} \end{bmatrix}$

=
$\frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}$

A² =
$\frac{1}{{3}}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\ 1& \omega & \omega ^{2}\\ 1& \omega ^{2} & \omega ^{4} \end{bmatrix}$

$A^{2}=\frac{1}{3}\begin{bmatrix} 3 & 0 & 0\\ 0 & 0 & 3\\ 0 & 3 & 0 \end{bmatrix}$

$A^{2}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$

$A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$

$A^{4}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}= I$

A⁴ = I

A²⁸ = ?

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

Solution:

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 ? and ? are two real roots of the equation (? + 1) tan² ? − √2 ? tan ? = 1 − ?, where (? ≠ −1) and are real numbers. If tan²(α + β) = 50, then value of ? is

a) 5√2
b) 10√2
c) 10
d) 5

Solution:

(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)

$\left ( \frac{\frac{\sqrt{2 }\lambda}{k+1}}{1-\frac{k-1}{k+1}} \right )^{2} = 50$

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 ? be any point (2, 1, 6). Then the image of ? in the plane ? is:

a) (6, 5, 2)
b) (6, 5, −2)
c) (4,3, 2)
d) (3,4, −2)

Solution:

Points A(2, 1, 0), B(4, 1, 1) C(5, 0, 1)

$\vec{AB } = (2,0,1)$

$\vec{AC } = (3,-1,1)$

$\vec{n } = \vec{AB}\times \vec{AC }$

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 x^k + y^k = a^k, (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

Solution:

x^k + y^k = a^k

kx^k-1 + ky^k-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 ?^′(?) ≤ 2, for all ? ∈ (−7, 0), then for all such functions ?, ?(−1) + ?(0) lies in the interval:

a) [-6, 20]
b) (-∞, 20]
c) (-∞, 11]
d) [-3, 11]

Solution:

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 e^y(dy/dx)-1 = e^x such that y(0) = 0, then y(1) is equal to

a) log_e2
b) 2e
c) 2 + log_e 2
d) 1+log_e 2

Solution:

?^y(?^′ − 1) = ?^?

⇒ dy/dx = ?^?−? + 1

Let x-y = t

1 - dy/dx = dt/dx

So, we can write

⇒ 1 − dt/dx = ?^? + 1

⇒ −?^−? ?? = ??

⇒ ?^−? = ? + ?

⇒ ?^?−? = ? + ?

1 = 0 + ?

⇒ ?^?−? = ? + 1

at ? = 1

⇒ ?^?−1 = 2

⇒ ? = 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

Solution:

Let 5 numbers be ? − 2?, ? − ?, ?, ? + ?, ? + 2?

5? = 25

? = 5

(? − 2?)(? − ?) a (? + ?)(? + 2?) = 2520

(25 − 4?²)(25 − ?²) = 504

4?⁴ − 125?² + 121 = 0

4?⁴− 4?² − 121?²+ 121 = 0

?²= 1 ?? ?² = 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

Solution:

$\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c & c \end{vmatrix}=0$

R₂ R₂ - R₁

R₃ R₃ - R₁

$\begin{vmatrix} 2 & 2a &a \\ 0& 3b-2a &b-a \\ 0& 4c-2a & c-a \end{vmatrix}=0$

⇒ (3b − 2?)(? − ?) − (4? − 2?)(? − ?) = 0

⇒ 3?? − 2?? − 3?? + 2?² − [4?? − 4?? − 2?? + 2?²] = 0

⇒ −?? + 2?? − ?? = 0

⇒ ?? + ?? = 2??

1/c + 1/a = 2/b

Question 21.

$\lim_{x\to 2}\frac{3^{x}+3^{3-x}-12}{3^{-\frac{x}{2}}-3^{1-x}}$

is equal to

Solution:

Put 3^x/2= t

Answer: (36)

Question 22.If variance of first ? natural numbers is 10 and variance of first ? even natural numbers is 16, ? + ? is equal to

Solution:

For ? natural number variance is given by

$\sigma ^{2} = \frac{\sum x_{i}^{2}}{n}- (\frac{\sum x_{i}}{n})^{2}$

$\frac{\sum x_{i}^{2}}{n}= \frac{1^{2}+2^{2}+..n^{2}}{n} = \frac{n(n+1)(2n+1)}{6n}$

$\frac{\sum x_{i}}{n}= \frac{1+2+..n}{n} = \frac{n(n+1)}{2n}$

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 ? in the product

(1 + x + x² + x³ … . +x^2?)(1 − x + x² − x³ … . + x^2?) is 61, then ? is equal to

Solution:

Let (1 + x + x² + ⋯ + x^2?)(1 − x + x² − ⋯ + x^2?) = a? + a1? + a2?² + ⋯

Put x = 1

2n + 1 = a_o + a₁ + a₂ + a₃ + . . . . . . . . . . . . (1)

Put x = −1

2n + 1 = a_o - a₁ + a₂ - a₃ +. . . . . . . . . . . . . (2)

Add (1) and (2)

2(2n + 1) = 2(a_? + a₂ + a₄ + . . . . . . . . . . .

2n + 1 = 61

n = 30

Answer: (30)

Question 24. Let S be the set of points where the function, (?) = |2 − |? − 3||, ? ∈ ?, is not differentiable. Then, the value of ∑?∈S f(f(x)) is equal to

Solution:

Solution:

There will be three points ? = 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

Solution:

P is the centroid which is =
$\left ( \frac{1+6+\frac{3}{2}}{3} , \frac{0+2+6}{3}\right )$

P = (17/6, 8/3)

Q = (-7/6, -1/3)

PQ =
$\sqrt{4^{2}+3^{2}} = 5$

Answer (5)

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Video Lessons - January 7 Shift 1 Maths

JEE Main 2020 Maths Paper January 7 Shift 1

JEE Main 2020 Maths Paper With Solutions Jan 7 Shift 1