JEE Main 2017 Maths Paper With Solutions (April 2)

Question 1: The function f : R → [(-1/2), (1/2)] is defined as f(x) = [x]/[1 + x²], is:

1) injective but not surjective.
2) surjective but not injective.
3) neither injective nor surjective.
4) invertible.

Answer: (2)

f(x) = [x]/[1 + x²]

f’(x) changes sign in different intervals.

∴ Not injective.

y = (x / (1 + x²))

yx² – x + y = 0

For y ≠ 0

D = 1 – 4y² ≥ 0

y ∈ [(-1 / 2), (1 / 2)] – {0}

For, y = 0 ⇒ x = 0

∴Part of range

∴Range = [(-1 / 2), (1 / 2)]

∴Surjective but not injective.

Question 2: If, for a positive integer n, the quadratic equation,has two consecutive integral solutions, then n is equal to:

1) 9
2) 10
3) 11
4) 12

Answer: (3)

Rearranging equation, we get

nx² + {1 + 3 + 5 + ….. + (2n – 1)x + {1.2 + 2.3 + ….. + (n – 1)n} = 10n

nx² + n²x + [(n – 1) n (n + 1)] / 3 = 10n

x² + nx + [n² – 31] / 3 = 0

Given difference of roots = 1

|ɑ – β| = 1

D = 1

n² – (4 / 3) (n² – 31) = 1

n = 11

Question 3: Let ω be a complex number such that 2ω + 1 = z where z = √-3. If then k is equal to,

1) z
2) −1
3) 1
4) − z

Answer: (4)

2ω + 1 = z, z = √3i

ω = (-1 + √3i) / 2 [Cube root of unity]

= 3 (ω² – ω⁴)

= 3 [{(-1 – √3i) / 2} – {(-1 + √3i) / 2}]

= – 3√3i

= -3z

k = -z

Question 4: Ifthen adj (3A² + 12A) is equal to :

Answer: (1)

= (2 – 2λ – λ + λ²) – 12

f (λ) = λ² – 3λ – 10

∵ A satisfies f (λ).

A² – 3A – 10I = 0

A² – 3A = 10I

3A² – 9A = 30I

3A² + 12A = 30I + 21A

Question 5: If S is the set of distinct values of ‘b’ for which the following system of linear equations

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

has no solution, then S is :

1) an infinite set
2) a finite set containing two or more elements
3) a singleton
4) an empty set

Answer: (3)

⇒ – (1 – a²) = 0

⇒ a = 1

For a = 1

Eq. (1) & (2) are identical i.e.,x + y + z = 1.

To have no solution with x + by + z = 0.

b = 1

Question 6: A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is :

1) 468
2) 469
3) 484
4) 485

Answer: (4)

Required number of ways

= ⁴C₃ . ⁴C₃ + (⁴C₂ . ³C₁)² + (⁴C₁ . ³C₂)² + (³C₃)²

= 16 + 324 + 144 + 1

= 485

Question 7: The value of (²¹C₁ – ¹⁰C₁) + (²¹C₂ – ¹⁰C₂) + (²¹C₃ – ¹⁰C₃) + (²¹C₄ – ¹⁰C₄) + … (²¹C₁₀ – ¹⁰C₁₀) is

1) 2²¹ − 2¹⁰
2) 2²⁰ − 2⁹
3) 2²⁰ − 2¹⁰
4) 2²¹ − 2¹¹

Answer: (3)

²¹C₁ + ²¹C₂ + …… ²¹C₁₀ = (1 / 2) (²¹C₀ + ²¹C₁ + ….. ²¹C₂₁) – 1

= 2²⁰ – 1

¹⁰C₁ + ¹⁰C₂ + …. ¹⁰C₁₀ = 2¹⁰ – 1

Required sum = (2²⁰ – 1) – (2¹⁰ – 1)

= 2²⁰ – 2¹⁰

Question 8: For any three positive real numbers a, b and c, 9 (25a² + b²) + 25 (c² − 3ac) = 15b (3a + c). Then :

1) b, c and a are in A.P.
2) a, b and c are in A.P.
3) a, b and c are in G.P.
4) b, c and a are in G.P.

Answer: (1)

9 (25a² + b²) + 25 (c² − 3ac) = 15b (3a + c)

⇒ (15a)² + (3b)² + (5c)² – 45b – 15b – 75ac = 0

(15a – 3b)² + (3b – 5c)² + (15a – 5c)² = 0

It is possible when

15a – 3b = 0 and 3b – 5c = 0 and 15a – 5c = 0

15a = 3b = 5c

(a / 1) = (b / 5) = (c / 3)

∴ b, c, a are in A.P.

Question 9: Let a, b, c ∈ R. If f (x) = ax² + bx + c is such that a + b + c = 3 and f (x + y) = f (x) + f (y) + xy, ∀ x, y ∈ R, then

1) 165
2) 190
3) 255
4) 330

Answer: (4)

As f (x + y) = f (x) + f (y) + xy

Given, f (1) = 3

Putting, x = y = 1

⇒ f (2) = 2 f (1) + 1 = 7

Similarly, x = 1, y = 2,

⇒ f (3) = f (1) + f (2) + 2 = 12

Now,

= 3 + 7 + 12 + 18 + …

= S (let)

Now, S_n = 3 + 7 + 12 + 18 … + t_n

Again, S_n = 3 + 7 + 12 + 18 … + t_n-1 + t_n

We get, t_n = 3 + 4 + 5 + …. n terms

= [n (n + 5)] / 2

Question 10:

1) 1 / 16
2) 1 / 8
3) 1 / 4
4) 1 / 24

Answer: (1)

Question 11: If for x ∈ (0, (1 / 4)), the derivative of tan^-1 (6x√x / (1 – 9x³)) is √x . g(x), then g(x) equals:

1) (3x√x / (1 – 9x³))
2) (3x/(1 – 9x³))
3) (3/(1 + 9x³))
4) (9/(1 + 9x³))

Answer: (4)

f (x) = 2 tan^-1 (3x√x) for x ∈ (0, (1/4))

f ‘ (x) = (9√x/(1 + 9x³))

g (x) = (9/(1 + 9x³))

Question 12: The normal to the curve y (x − 2) (x − 3) = x + 6 at the point where the curve intersects the y-axis passes through the point :

1) (1/2, 1/2)
2) (1/2, -1/3)
3) (1/2, 1/3)
4) (-1/2, -1/2)

Answer: (1)

y (x – 2) (x – 3) = x + 6

At y-axis, x = 0, y = 1

Now, on differentiation,

(dy / dx) (x – 2) (x – 3) + y (2x – 5) = 1

(dy / dx) (6) + 1 * (-5) = 1

(dy / dx) = 6 / 6 = 1

Now slope of normal = –1

Equation of normal y – 1 = –1 (x – 0)

y + x – 1 = 0 … (i)

Line (i) passes through (1 / 2, 1 / 2).

Question 13: Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :

1) 10
2) 25
3) 30
4) 12.5

Answer: (2)

2r + θr = 20 —- (i)

A = Area = (θ / 2π) * (πr²) = (θr² / 2) —- (ii)

A = (r² / 2) ([20 – 2r] / r)

A = [(20r – 2r²) / 2] = 10r – r²

A to be maximum

dA / dr = 10 – 2r = 0 ⇒ r = 5

d²A / dr² = -2 < 0

Hence for r = 5, A is maximum

Now, 10 + (θ * 5) = 20 ⇒ θ = 2 (radian)

Area = (2 / 2π) * (π) (5)² = 25 sq m

Question 14: Let I_n = ∫tanⁿ x dx, (n > 1). If I₄ + I₆ = a tan⁵ x + bx⁵ + C, where C is a constant of integration, then the ordered pair (a, b) is equal to :

1) (1/5, 0)
2) (1/5, -1)
3) (-1/5, 0)
4) (-1/5, 1)

Answer: (1)

I_n = ∫tanⁿ x dx, (n > 1)

I₄ + I₆ = ∫(tan⁴ x + tan⁶ x) dx

= ∫tan⁴ x sec² x dx

Let tan x = t

sec² x dx = dt

= ∫t⁴ dt

= (t⁵/5) + C

= (1/5) tan⁵ x + C

a = (1/5), b = 0

Question 15: The integral

1) 2
2) 4
3) −1
4) −2

Answer: (1)

Question 16: The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x² ≤ 4y and y ≤ 1 + x} is :

1) 3 / 2
2) 7 / 3
3) 5 / 2
4) 59 / 12

Answer: (3)

Area of the shaded region

=

Question 17: If (2 + sinx) (dy/dx) + (y + 1) cosx = 0 and y(0) = 1, then y(π/2) is equal to:

1) -2/3
2) -1/3
3) 4/3
4) 1/3

Answer: (4)

(2 + sinx) (dy/dx) + (y + 1) cos x = 0

y(0) = 1, y(π / 2) = ?

(1/(y + 1)) dy + (cos x/[2 + sinx]) dx = 0

ln |y + 1| + ln (2 + sinx) = ln C

(y + 1) (2 + sinx) = C

Put x = 0, y = 1

(1 + 1) . 2 = C ⇒ C = 4

Now, (y + 1) (2 + sinx) = 4

(y + 1) = 4 / 3

y = (4 / 3) – 1 = 1 / 3

Question 18: Let k be an integer such that the triangle with vertices (k, −3k), (5, k) and (−k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :

1) (1, 3/4)
2) (1, -3/4)
3) (2, 1/2)
4) (2, -1/2)

Answer: (3)

(k² – 7k + 10) + 4k² + 20k = ± 56

5k² + 13k – 46 = 0

5k² + 13k + 66 = 0

5k² + 13k – 46 = 0

k = [-13 ± √169 + 920]/10

= 2, – 4.6 [reject]

For k = 2

Equation of AD,

x = 2 …(i)

Also, equation of BE,

(y – 2) = (1/2) (x – 5)

2y – 4 = x – 5

x – 2y – 1 = 0 …(ii)

Solving (i) & (ii), 2y = 1

y = 1 / 2

Orthocentre is (2, 1/2)

Question 19: The radius of a circle, having minimum area, which touches the curve y = 4 − x² and the lines, y = |x| is :

1) 2 (√2 – 1)
2) 4 (√2 – 1)
3) 4 (√2 + 1)
4) 2 (√2 + 1)

Answer: (2)

x² = – (y – 4)

Let a point on the parabola P [(t/2) , (4 – (t²/4))]

Equation of normal at P is

y + (t² / 4) – 4 = (1 / t) (x – [t / 2])

x – ty – (t³ / 4) + (7 / 2)t = 0

It passes through centre of circle, say (0, k)

-tk – (t³ / 4) + (7 / 2)t = 0 —- (i)

t = 0, t² = 14 – 4k

Radius = r = |(0 – k) / √2| (Length of perpendicular from (0, k) to y = x)

r = k / √2

Equation of circle is x² + (y – k)² = k² / 2

It passes through point P

(t² / 4) + [4 – (t² / 4) – k]² = k² / 2

t⁴ + t² [8k – 28] + 8k² – 128k + 256 = 0 —- (ii)

For t = 0 ⇒ k² – 16k + 32 = 0

k = 8 ± 4√2

r = k / √2 = 4 (√2 – 1) (discarding 4 (√2 + 1)) —– (iii)

For t = ± √(14 – 4k)

(14 – 4k)² + (14 – 4k) (8k – 28) + 8k² – 128k + 256 = 0

2k² + 4k – 15 = 0

k = [-2 ± √34] /[2]

r = k / √2 = [√17 – √2] / 2 (Ignoring negative value of r) ..(iv)

From (iii) & (iv),

r_min = [√17 – √2] / 2

But from options, 4 (√2 – 1)

Question 20: The eccentricity of an ellipse whose centre is at the origin is 1 / 2. If one of its directrices is x = −4, then the equation of the normal to it at (1, (3 / 2)) is:

1) 4x − 2y = 1
2) 4x + 2y = 7
3) x + 2y = 4
4) 2y − x = 2

Answer: (1)

x = –4

e = 1 / 2

(-a / e) = -4

-a = -4 * e

a = 2

Now, b² = a² (1 – e²) = 3

Equation to ellipse

(x²/4) + (y²/3) = 1

Equation of normal is

Question 21: A hyperbola passes through the point P (√2, √3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point :

1) (2√2 , 3√3)
2) (√3, √2)
3) (− √2 , –√3)
4) (3√2 , 2√3)

Answer: (1)

a² + b² = 4 and [2 / a²] – [3 / b²] = 1

(2 / [4 – b²]) – (3 / [b²]) = 1

b² = 3

a² = 1

x² – (y² / 3) = 1

∴ Tangent at P (√2 , √3) is √(2)x – (y / √3)) = 1

Clearly it passes through (2√2 , 3√3).

Question 22: The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular to both the lines

1) 10 / √83
2) 5 / √83
3) 10 / √74
4) 20 / √74

Answer: (1)

Let the plane be a (x – 1) + b (y + 1) + c (z + 1) = 0.

It is perpendicular to the given lines

a – 2b + 3c = 0

2a – b – c = 0

Solving, a : b : c = 5 : 7 : 3

The plane is 5x + 7y + 3z + 5 = 0

Distance of (1, 3, –7) from this plane = 10 / √83

Question 23: If the image of the point P (1, −2, 3) in the plane, 2x + 3y − 4z + 22 = 0 measured parallel to the line, (x / 1) = (y / 4) = (z / 5) is Q, then PQ is equal to :

1) 2√42
2) √42
3) 6√5
4) 3√5

Answer: (1)

Equation of PQ,

Let M be (λ + 1, 4λ – 2, 5λ + 3)

As it lies on 2x + 3y – 4z + 22 = 0

λ = 1

For Q, λ = 2

Distance PQ = 2 √1² + 4² + 5² = 2√42

Question 24: Let a = 2i + j – 2k and b = i + j. Let c be a vector such that |c – a| = 3, |(a x b) x c| = 3 and the angle between c and a × b be 30^o. Then a ⋅ c is equal to :

1) 2
2) 5
3) 1 / 8
4) 25 / 8

Answer: (1)

Question 25: A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is :

1) 6
2) 4
3) 6 / 25
4) 12 / 5

Answer: (4)

n = 10

p (Probability of drawing a green ball) = 15 / 25

p = 3 / 5, q = 2 / 5

var (X) = n.p.q

= 10 * (3 / 5) * (2 / 5)

= 10 * (6 / 25)

= 12 / 5

Question 26: For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P (Exactly one of C or A occurs) = 1 / 4 and P (All the three events occur simultaneously) = 1 / 16. Then the probability that at least one of the events occurs is :

1) 7 / 16
2) 7 / 64
3) 3 / 16
4) 7 / 32

Answer: (1)

P (A) + P (B) – P (A ⋂ B) = 1 / 4

P (B) + P (C) – P (B ⋂ C) = 1 / 4

P (C) + P (A) – P (A ⋂ C) = 1 / 4

P (A) + P (B) + P (C) – P (A ⋂ B) – P (B ⋂ C) – P (A ⋂ C = 3 / 8

∵ P (A ⋂ B ⋂ C) = 1 / 16

∴ P (A ⋃ B ⋃ C) = (3 / 8) + (1 / 16) = 7 / 16

Question 27: If two different numbers are taken from the set {0, 1, 2, 3, ……, 10}; then the probability that their sum, as well as absolute difference, are both multiples of 4, is :

1) 12 / 55
2) 14 / 45
3) 7 / 55
4) 6 / 55

Answer: (4)

Total number of ways = ¹¹C₂ = 55

Favourable ways are (0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10)

Probability = 6 / 55

Question 28: If 5(tan² x − cos² x) = 2cos²x + 9, then the value of cos 4x is :

1) 1 / 3
2) 2 / 9
3) −7 / 9
4) −3 / 5

Answer: (3)

5(tan² x − cos² x) = 2cos²x + 9

5 sec² x – 5 = 9 cos² x + 7

Let cos² x = t

(5 / t) = 9t + 12

9t² + 12t – 5 = 0

t = 1 / 3 as t ≠ – 5 / 3

cos² x = 1 / 3, cos 2x = 2cos² x – 1 = -1 / 3

cos 4x = 2cos² 2x – 1

= (2 / 9) – 1

= – 7/ 9

Question 29: Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tan β is equal to :

1) 1/4
2) 2/9
3) 4/9
4) 6/7

Answer: (2)

tan θ = 1/4

tan (θ + β) = 1/2

Question 30: The following statement (p → q) → [(~p → q) → q] is :

1) equivalent to ~p → q
2) equivalent to p → ~q
3) a fallacy
4) a tautology

Answer: (4)