JEE Main 2021 March 17 – Shift 2 Maths Question Paper with Solutions

Section – A

Question 1: If the Boolean expression (p ∧ q) ⍟ (p ⊗ q) is a tautology, then ⍟ and ⊗ are respectively given by :

a. ∧, →
b. →, →
c. ∨, →
d. ∧, ∨

Answer: (b)

(p ∧ q) → (p → q)

(p ∧ q) → (~ p ∨ q)

(~ p ∨ ~ q) ∨ (~ p ∨ q)

~ p ∨ (~ q ∨ q) ⇒ Tautology

⍟ ⇒ →

⊗ ⇒ →

Question 2: Let the tangent to the circle x² + y² = 25 at the point R (3, 4) meet the x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having a centre at the incentre of the triangle OPQ, then r² is equal to:

a. 625 / 72
b. 585 / 66
c. 125 / 72
d. 529 / 64

Answer: (a)

Given equation of circle is x² + y² = 25.

Tangent equation at (3, 4) is given by T : 3x + 4y = 25

Incentre of ΔOPQ is I = {[(25 / 4) * (25 / 3)] / [(25 / 3) + (25 / 4) + (125 / 12)], [(25 / 4) * (25 / 3)] / [(25 / 3) + (25 / 4) + (125 / 12)]

I = {[625 / (75 + 100 + 125)], [625 / (75 + 100 + 125)]} = (25 / 12, 25 / 12)

Distance from the origin to incentre is r.

∴ r² = (25 / 12)² + (25 / 12)² = 625 / 72

Therefore, the correct answer is (1).

Question 3: Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with a probability of occurrence of 0 at even places be 1 / 2 and probability of occurrence of 0 at the odd place be 1 / 3. Then the probability that ‘10’ is followed by ‘01’ is equal to :

a. 1 / 6
b. 1 / 18
c. 1 / 9
d. 1 / 3

Answer: (c)

P (0 at even place) = 1 / 2, P(0 at odd place) = 1 / 3

P (1 at even place) = 1 / 2, P(1 at odd place) = 2 / 3

P (10 is followed by 01) = [(2 / 3) * (1 / 2) * (1 / 3) * (1 / 2)] + [(1 / 2) * (1 / 3) * (1 / 2) * (2 / 3)]

= (1 / 18) + (1 / 18)

= 1 / 9

Question 4: The number of solutions of the equation x + 2 tanx = π / 2 in the interval [0, 2π] is :

a. 5
b. 2
c. 4
d. 3

Answer: (d)

x + 2 tan x = π / 2 in [0, 2π]

2 tan x = π / 2 – x

2 tan x = π / 2 – x

tan x = π / 4 – x / 2

y = tan x and y = (- x / 2) + π / 4

3 intersection points

∴ 3 solutions

Question 5: If the equation of the plane passing through the mirror image of a point (2, 3, 1) with respect to line (x + 1) / 2 = (y – 3) / 1 = (z + 2) / -1 and containing the line (x – 2) / 3 = (1 – y) / 2 = (z + 1) / 1 is ɑx + βy + 𝜸z = 24, then, then ɑ + β + 𝜸 is equal to:

a. 21
b. 19
c. 18
d. 20

Answer: (b)

Let point M is (2λ – 1, λ + 3, – λ – 2)

D.R.’s of AM line are: 2λ – 1 – 2, λ + 3 – 3, – λ – 2 – 1

2λ – 3, λ, –λ – 3

AM ⊥ line L₁

∴ 2 (2λ – 3) + 1 (λ) – 1 (–λ – 3) = 0

6λ = 3, λ = 1 / 2

∴ M ≡ (0, 7 / 2, – 5 / 2)

M is mid-point of A & B

M = [A + B] / 2

B = 2 M – A

B ≡ (–2, 4, –6)

Now we have to find the equation of the plane passing through B (–2, 4, –6) & also containing the line

(x – 2) / 3 = (1 – y) / 2 = (z + 1) / 1 … (1)

(x – 2) / 3 = (y – 1) / – 2 = (z + 1) / 1

Point P on line is (2, 1, –1)

b₂ of line L₂ is 3, –2, 1

n || (b₂ x PB)

b₂ = 3i – 2j + k

PB = – 4i + 3j – 5k

n = 7i + 11j + k

∴ equation of plane is r . n = a . n

r . (7i + 11j + k) = (–2i + 4j – 6k) . (7i + 11j + k)

7x + 11y + z = – 14 + 44 – 6

7x + 11 y + z = 24

∴ α = 7

β = 11

γ = 1

∴ α + β + γ = 19

Question 6: Consider the function f : R → R defined by f(x) =

a. monotonic on (0, ∞) only
b. Not monotonic on (–∞, 0) and (0, ∞)
c. monotonic on (–∞, 0) only
d. monotonic on (–∞, 0) ⋃ (0, ∞)

Answer: (b)

Then f is not monotonic on (–∞, 0) and (0, ∞).

Question 7: Let O be the origin. Let OP = xi + yj – k and OQ = – i + 2j + 3xk, x, y ∈ R, x > 0, be such that |PQ| = √20 and the vector OP is perpendicular to OQ. If OR = 3i + zk – 7k, z ∈ R, is coplanar with OP and OQ, then the value of x² + y² + z² is equal to:

a. 2
b. 9
c. 1
d. 7

Answer: (b)

OP = xi + yj – k

OP is perpendicular to OQ

OQ = – i + 2j + 3xk

PQ = (-1 – x) i + (2 – y) j + (3x + 1) k

|PQ| = √(- 1 – x)² + (2 – y)² + (3x + 1)²

OP . OQ = 0

– x + 2y – 3x = 0

4x – 2y=0

√20 = √(- 1 – x)² + (2 – y)² + (3x + 1)²

20 = 1 + x² + 2x + 4 + y² – 4y + 9x² + 1 + 6x

20 = 10x² + 4x² + 8x + 6 – 8x

14 = 14x²

x² = 1

y² = 4x²

y² = 4

x = 1 as x > 0 and y = 2

1 (- 14 – 3z) – 2 (7 – 9) – 1 (- z – 6)=0

– 14 – 3z + 4 + z + 6 = 0

2z = – 4

z = – 2

x² + y² + z² = 9

Question 8: Let L be a tangent line to the parabola y² = 4x – 20 at (6, 2). If L is also a tangent to the ellipse (x² / 2) + (y² / b) = 1, then the value of b is equal to:

a. 20
b. 14
c. 16
d. 11

Answer: (b)

Parabola y² = 4x – 20

Tangent at P (6, 2) will be

2y = 4 [(x + 6) / 2] – 20

2y = 2x + 12 – 20

2y = 2x – 8

y = x – 4

x – y – 4 = 0 ….(1)

This is also tangent to ellipse (x² / 2) + (y² / b) = 1

Apply c² = a²m² + b²

(–4)² = (2)(1) + b

b = 14

Question 9: Let f : R → R be defined as f (x) = e^–x sinx. If F : [0,1] R→ is a differentiable function such that F (x) = ∫₀^x f (t) dt, then the value of ∫₀¹ [F’ (x) + f (x)] e^x dx lies in the interval

a. [330 / 360, 331 / 360]
b. [327 / 360, 329 / 360]
c. [331 / 360, 334 / 360]
d. [335 / 360, 336 / 360]

Answer: (a)

F’ (x) = f (x) by Leibnitz theorem

∫ ₀¹ [F’ (x) + f (x)] e^x dx = ∫₀¹ 2 f (x) e^x dx

I = ∫₀¹ 2 sinx dx

I = 2 (1 – cos 1)

= {1 – [1 – (1² / 2!) + (1⁴ / 4!) – (1⁶ / 6!) + ……]}

= 2 {1 – [1 – (1 / 2) + (1 / 24)]} < 2 (1 – cos 1) < 2 {1 – [1 – (1 / 2) + (1 / 24) – (1 / 720)]}

330 / 360 < 2(1 – cos 1) < 331 / 360

330 / 360 < I < 331 / 360

Choice (a) is correct.

Question 10: If x, y, z are in arithmetic progression with common difference d, x ≠ 3d, and the determinant of the matrix

a. 6
b. 36
c. 72
d. 12

Answer: (c)

Question 11: If the integral ∫₀¹⁰ [sin 2πx] / e^x-[x] dx = ɑe^-1 + βe^-½ + 𝛾, where ɑ, β, 𝞬 are integers and [x] denotes the greatest integer less than or equal to x, then the value of ɑ + β + 𝞬 is equal to:

a. 20
b. 0
c. 25
d. 10

Answer: (b)

Given integral ∫₀¹⁰ [sin 2πx] / e^x-[x] dx = 10 ∫₀¹ [sin 2πx] / e^[x] dx (using the property of definite integral]

= 10 [∫₀^1/2 0 . dx + ∫_1/2¹ – 1 / e^x dx]

= – 10 [e^-x / – 1]_1/2¹

= 10 [e^-1 – e^-½]

= 10 e^-1 – 10e^-1/2

Comparing with the given relation,

α = 10, β = –10, γ = 0

α + β + γ = 0.

Therefore, the correct answer is (2).

Question 12: Let y = y (x) be the solution of the differential equation cosx (3 sinx + cosx + 3) dy = (1 + y sinx (3 sinx + cosx + 3)) dx, 0 ≤ x ≤ π / 2, y(0) = 0. Then, y (π / 3) is equal to :

a. 2 log_e [(2√3 + 10) / 11]
b. 2 log_e [(√3 + 7) / 2]
c. 2 log_e [(3√3 – 8)] / 4
d. 2 log_e [(2√3 + 9) / 6]

Answer: (a)

cosx (3 sin x + cos x + 3) dy = (1 + y sin x (3 sin x + cos x + 3)) dx …(1)

(3 sin x + cos x + 3) (cos x dy – y sin x dx) = dx

∫ d (y . cosx) = ∫ dx / [3 sinx + cosx + 3]

y cos x = ln |(tan (x / 2) + 1) / [tan (x / 2) + 2]| + c

Put x = 0 & y = 0

C = – ln (1 / 2) = ln (2)

y (π / 3) = 2 ln |(1 + √3) / (1 + 2√3)| + ln 2

= 2 ln |(5 + √3) / 11| + ln 2

= 2 ln |(2 √3 + 10) / 11|

Question 13: The value of the limit lim_θ→0 [tan (π cos² θ)] / [sin (2π sin² θ)] is equal to:

a. – 1 / 2
b. – 1 / 4
c. 0
d. 1 / 4

Answer: (a)

Given,

lim_θ→0 [tan (π cos² θ)] / [sin (2π sin² θ)]

= lim_θ→0 [tan (π – π sin² θ)] / [sin (2π sin² θ)]

= lim_θ→0 [- tan (π sin² θ)] / [sin (2π sin² θ)]

= lim_θ→0 (-1 / 2) {[tan (π sin² θ)] / [π sin² θ]} * {[2π sin² θ] / [sin (2π sin² θ)]} = – 1 / 2

Therefore, the correct answer is (1).

Question 14: If the curve y = y(x) is the solution of the differential equation 2 (x² + x^5/4) dy – y (x + x^1/4) dx = 2x^9/4 dx, x > 0 which passes through the point [1, 1 – (4 / 3) log_e 2] , then the value of y (16) is equal to:

a. [(31 / 3) – (8 / 3) log_e 3]
b. 4 [(31 / 3) + (8 / 3) log_e 3]
c. [(31 / 3) + (8 / 3) log_e 3]
d. 4 [(31 / 3) – (8 / 3) log_e 3]

Answer: (d)

(dy / dx) – (y / 2x) = x^9/4 / [x^5/4 (x^3/4 + 1)]

If = e^{-∫ dx / 2x} = e^{(- 1 / 2) lnx} = 1 / x^1/2

y . x^–1/2 = ∫ [x^9/4 . x^-½] / [x^5/4 (x^3/4 + 1)] dx

∫ [x^1/2] / [x^3/4 + 1] dx

x = t⁴ ⇒ dx = 4t³dt

4 ∫ {[t² (t³ + 1 – 1)] / [t³ + 1]} dt

4 ∫ t² dt – 4 ∫ {t² / (t³ + 1)} dt

(4t³ / 3) – (4 / 3) ln (t³ + 1) + c

yx^–1/2 = (4x^3/4) / 3 – (4 / 3) ln(x^3/4 + 1) + C

When x = 1, y = (1 – 4/3)log_e2

1 – (4 / 3) log_e 2=(4 / 3) – (4 / 3) log_e 2 + C

⇒ C = – (1 / 3)

y = (4 / 3) x^5/4 – (4 / 3) √x ln (x^3/4 + 1) – (√x / 3)

y (16) = (4 / 3) × 32 – (4 / 3) × 4 ln 9 – (4 / 3)

= (124 / 3) – (32 / 3) ln 3

= 4 [(31 / 3) – (8 / 3) ln 3]

Question 15: Let S₁, S₂ and S₃ be three sets defined as S₁ = {z ∈ C : |z – 1| ≤ √2}, S₂ = {z ∈ C : Re ((1 – i) z) ≥ 1}, S₃ = {z ∈ C : Im (z) ≤ 1}. Then the set S₁ ⋂ S₂ ⋂ S₃

a. has infinitely many elements
b. has exactly two elements
c. has exactly three elements
d. is a singleton

Answer: (a)

Let, z = x + iy

S₁ ≡ (x – 1)² + y² ≤ 2 …(1)

S₂ ≡ x + y ≥ 1 …(2)

S₃ ≡ y ≤ 1 …(3)

⇒ S₁ ∩ S₂ ∩ S₃ has infinitely many elements.

Question 16: If the sides AB, BC, and CA of a triangle ABC have, 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices is equal to:

a. 360
b. 240
c. 333
d. 364

Answer: (c)

Total number of triangles

= ³C₁ × ⁵C₁ × ⁶C₁

+ ³C₁ × ⁵C₂ + ⁵C₁ × ³C₂

+ ³C₁ × ⁶C₂ + ⁶C₁ × ³C₂

+ ⁵C₁ × ⁶C₂ + ⁶C₁ × ⁵C₂

= 90 + 30 + 15 + 45 + 18 + 75 + 60 = 333

Question 17: The value of lim_x→∞ [r] + [2r] + ……. [nr] / [n²], where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to:

a. 0
b. r
c. r / 2
d. 2r

Answer: (c)

We know that

r ≤ [r] < r + 1 and

2r ≤ [2r] < 2r + 1

3r ≤ [3r] < 3r + 1

nr ≤ [nr] < nr + 1

^{______________________}

r + 2r + ….. + nr ≤ [r] + [2r] + …..+ [nr] < (r + 2r + …..+ nr) + n

{[n (n + 1) / 2] * r / (n²)} ≤ [r] + [2r] + …… [nr] / n² < {[n (n + 1) / 2] * r + (n) / (n²)}

Now,

lim_n→∞ [n (n + 1) . r / (2 . n²)] = r / 2

And lim_n→∞ {[n (n + 1) / 2] * r + (n) / (n²)} = r / 2

So, by Sandwich Theorem, we can conclude that lim_n→∞ [r] + [2r] + ……. [nr] / [n²] = r / 2

Question 18: The value of ∑_r=0⁶ (⁶C_r . ⁶C_6-r) is equal to:

a. 1124
b. 924
c. 1324
d. 1024

Answer: (b)

∑_r=0⁶ (⁶C_r . ⁶C_6-r) = ⁶C₀.⁶C₆ + ⁶C₁.⁶C₅ + ……. + ⁶C₆.⁶C₀

Now, (1 + x)⁶ (1 + x)⁶ = (⁶C₀ + ⁶C₁x + ⁶C₂x² + …. + ⁶C₆x⁶) (⁶C₀ + ⁶C₁x + ⁶C₂x² + …. + ⁶C₆x⁶)

Comparing coefficient of x⁶ both sides

⁶C₀.⁶C₆ + ⁶C₁ +⁶C₅ + ….. + ⁶C₆.⁶C₀= ¹²C₆

= 924

Therefore, the correct answer is (2).

Question 19: Two tangents are drawn from a point P to the circle x² + y² – 2x – 4y + 4 = 0, such that the angle between these tangents is tan^-1 (12/5), where tan^-1 (12/5) ∈ (0, π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of △PAB and △CAB is :

a. 11 : 4
b. 9 : 4
c. 2 : 1
d. 3 : 1

Answer: (b)

Let θ = tan^-1 (12 / 5)

⇒ tan θ = 12 / 5

⇒ [2 tan (θ / 2)] / [1 – tan² (θ / 2)] = 12 / 5

⇒ tan (θ / 2) = 2 / 3 ⇒ sin (θ / 2) = 2 / √3 and cos (θ / 2) = 3 / √13

In ΔCAP, tan (θ / 2) = 1 / AP

⇒ AP = 3 / 2

In ΔAPM, sin (θ / 2) = AM / AP, cos (θ / 2) = AM / AP

⇒ AM = 3 / √13

⇒ PM = 9 / 2√13

∴ AB = 6 / √13

∴ Area of ΔPAB = (1 / 2) * AB * PM

(1 / 2) * (6 / √13) * (9 / 2√13) = 27 / 26

Now, φ = 90^o – (θ / 2)

In ΔCAM,

cos φ = CM / CA

⇒ CM = 1 . cos [(π / 2) – (θ / 2)]

= 1 . sin (θ / 2) = 2 / √13

∴ Area of ΔCAB =(1 / 2) × AB × CM

= (1 / 2) * (6 / √13) * (2 / √13) = 6 / 13

∴ [Area of ΔPAB] / [Area of ΔCAB] = (27 / 26) / (6 / 13) = 9 / 4

Therefore, the correct answer is (2).

Aliter

Area of ΔPAB = RL³ / [R² + L²] and Area of ΔCAB = LR³ / [R² + L]

Therefore, required ratio = {RL³ / [R² + L²]} / {LR³ / [R² + L]} = (L / R)²

= cot² (θ / 2) = [2 cos² (θ / 2)] / [2 sin² (θ / 2)]

= [1 + cos θ] / [1 – cos θ]

= [1 + (5 / 13)] / [1 – (5 / 13)]

= 9 / 4

Question 20: The number of solutions of the equation sin^-1 [x² + (1 / 3)] + cos^-1 [x² – (2 / 3)] = x², for x ∈ [- 1, 1], and [x] denotes the greatest integer less than or equal to x, is:

a. 0
b. 2
c. 4
d. Infinite

Answer: (a)

There are three cases possible for x ∈ [– 1, 1]

Case I : x ∈ [- 1, – √(2/3) )

sin^-1 (1) + cos^-1 0 = x²

x² = (π / 2) + (π / 2) = π ⇒ x = ± √π → (Reject)

Case II : x ∈ (- √(2/3), √(2/3) )

sin^-1 (0) + cos^-1 (- 1) = x²

0 + π = x² ⇒ x = ± √π → (Reject)

Case III: x ∈ (- √(2/3), 1)

sin^-1 (0) + cos^-1 (0) = x²

π = x² ⇒ x = ± √π → (Reject)

No solution exists.

Therefore, the correct answer is (1).

SECTION – B

Question 21: Let the coefficients of third, fourth and fifth terms in the expansion of [x + (a / x²)]ⁿ, x ≠ 0 be in the ratio 12:8:3. Then the term independent of x in the expansion is equal to ………

Answer: (4)

T_r+1 = ⁿc_r x^n-r (a / x²)^r = ⁿc_r a^r x^n-3r

T₃ = ⁿc₂ a² x^n-6, T₄ = ⁿc₃ a³ x^n-9

T₅ = ⁿc₄ a⁴ x^n-12

Now, [coefficient of T₃] / [coefficient of T₄] = [ⁿc₂ a²] / [ⁿc₃ a³] = 3 / [a (n – 2)] = 3 / 2

a(n – 2) = 2 ……… (i) and [coefficient of T₄] / [coefficient of T₅] = [ⁿc₃ a³] / [ⁿc₄ a⁴] = 4 / [a (n – 3)] = 8 / 3

a(n – 3) = 3 / 2 ……… (ii)

By (i) and (ii) n = 6, a = 1 / 2

For term independent of ‘x’,

n – 3r = 0

r = n / 3

r = 6 / 3 = 2

T₃ = ⁶c₂ (1 / 2)² x⁰ = 15 / 4 = 3.75 ≅ 4

Question 22: Let A =

Answer: (2020)

ɑ (a – 1) = – bβ and cɑ = β (1 – d)

(ɑ / β) = – b / (a – 1) and (ɑ / β) = (1 – d) / c

– b / (a – 1) = (1 – d) / c

–bc = (a – 1) (1 – d)

–bc = a – ad – 1 + d

ad – bc = a + d – 1

= 2021–1

= 2020

Question 23: Let f : [–1,1] → R be defined as f (x) = ax² + bx + c for all x ∈ [–1, 1], where a, b, c ∈ R such that f (–1) = 2, f’ (–1) = 1 and for x ∈ [–1, 1] the maximum value of f’’ (x) is 1 / 2. If f (x) ≤ ɑ, x ∈ [–1, 1], then the least value of ɑ is equal to …………… .

Answer: (5)

f (x) = ax² + bx + c

f’ (x) = 2ax + b,

f” (x) = 2a

Given f” (–1) = 1 / 2 ⇒ a =1 / 4

f’ (–1) = 1 ⇒ b – 2a = 1 ⇒ b = 3 /2

f (–1) = a – b + c = 2 ⇒ c = 13 / 4

Now f (x) = (1/4) (x² + 6x + 13), x ∈ [–1, 1]

f’ (x) = (1/4) (2x + 6) = 0 ⇒ x = –3 ∉ [–1, 1]

f (1) = 5, f (–1) = 2

f (x) ≤ 5

So α_minimum = 5

Question 24: Let I_n = ∫₁^e x¹⁹ (log |x|)ⁿ dx, where n ∈ N. If 20 (I₁₀) = αI₉ + βI₈, for natural numbers α and β, then α – β equal to ………………… .

Answer: (1)

Question 25: Let f : [–3, 1]→ R be given as

Answer: (41)

Area is ∫_-3^-2 (x + 6) dx + ∫_-2⁰ x² dx + ∫₀¹ √x dx = A

= (7 / 2) + (x³ / 3)_-2⁰ + [(2 / 3) x^3/2]₀¹

= (7 / 2) + (8 / 3) + (2 / 3)

= 41 / 6

So, 6A = 41

Question 26: Let x be a vector in the plane containing vectors a = 2i – j + k and b = i + 2j – k. If the vector x is perpendicular to (3i + 2j – k) and its projection on a is 17√6 / 2 , then the value of |x|² is equal to ……………….. .

Answer: (486)

Let x = k (a + λb)

I. The vector x is perpendicular to 3i + 2j – k.

k{(2 + λ)3 + (2λ – 1)2 + (1 – λ)(–1) = 0

⇒ 8λ + 3 = 0

λ = – 3 / 8

II. Also projection of x on a is therefore

⇒ k [(a + λb) . a] / √6 = 17√6 / 2

⇒ k [6 + (3 / 8)] = [17 * 6] / 2

⇒ k = (51 / 51) * 8

k = 8

x = 8 [(13 / 8) i – (14 / 8) j + (11 / 8) k]

= 13i – 14j + 11k

|x|² = 169 + 196 + 121

= 486

Question 27: Consider a set of 3n numbers having variance 4. In this set, the mean of the first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of the first 2n numbers and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to ………………..

Answer: (68)

Let first 2n observations are x₁, x₂ ………… , x_2n and last n observations are y₁, y₂ ………… , y_n

= 29 + 6 + 1 – (16 / 3)²

= [324 – 256] / 9

= 68 / 9 = k

9k = 68

Therefore, the correct answer is 68.

Question 28: If 1, log₁₀ (4^x – 2) and log₁₀ [4^x + (18 / 5)] are in arithmetic progression for a real number x, then the value of the determinant

Answer: (2)

1, log₁₀ (4^x – 2) and log₁₀ [4^x + (18 / 5)] are in AP.

2. log₁₀ (4^x – 2) = 1 + log₁₀ [4^x + (18 / 5)]

log₁₀ [4^x – 2]² = log₁₀ [10 . [4^x + (18 / 5)] [4^x – 2]² = 10 . [4^x + (18 / 5)]

(4^x)² + 4 – 4 . 4^x = 10 . 4^x + 36

(4^x)² – 14 . 4^x – 32 = 0

(4^x)² + 2 . 4^x – 16 . 4^x – 32 = 0

4^x [4^x + 2] – 16 [4^x + 2] = 0

4^x = – 2 or 4^x = 16

Rejected the value of 4^x = – 2.

Therefore

=

= 3 (–2) – 1 (0 – 4) + 4 (1 – 0)

= – 6 + 4 + 4

= 2

Question 29: Let P be an arbitrary point having the sum of the squares of the distances from the planes x + y + z = 0, lx – nz = 0 and x – 2y + z = 0, equal to 9. If the locus of the point P is x² + y² +z² = 9, then the value of l – n is equal to ………………

Answer: (0)

Let point P is (α, β, γ)

{[α + β + γ] / √3}² + {[lα – nγ] / (√l² + n²)]² + {[α – 2β + γ)] / [√6]}² = 9

Locus is [(x + y + z)² / 3] + [(ln – nz)² / (l² + n²)] + [(x – 2y + z)² / 6] = 9

Since its given that x² + y² + z² = 9

After solving l = n, then l – n = 0

Question 30: Let tan α, tan β and tan γ; α, β, γ ≠ [2n – 1] π / 2, n ∈ N be the slopes of three-line segment OA, OB and OC, respectively, where O is the origin. If the circumcentre of triangle ABC coincides with the origin and its orthocentre lies on the y-axis, then the value of {[cos 3α + cos 3β + cos 3γ] / [cos α * cos β * cos γ]}² is equal to

Answer: (144)

Since orthocentre and circumcentre both lies on y-axis

Centroid also lies on y-axis

Σcos α = 0

cosα + cos β + cosγ = 0

cos³α + cos³β + cos³γ = 3 cos α * cos β * cos γ

{[cos 3α + cos 3β + cos 3γ] / [cos α * cos β * cos γ]}

= {4 ([cos³α + cos³β + cos³γ] – 3 [cos α + cos β + cos γ]} / [cos α * cos β * cos γ] = 12

Then, {[cos 3α + cos 3β + cos 3γ] / [cos α * cos β * cos γ]}² = 144