^{nd}Maths

^{nd}– Maths

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

1) injective but not surjective.

2) surjective but not injective.

3) neither injective nor surjective.

4) invertible.

**Answer: (2)**

f(x) = [x]/[1 + x^{2}]

f’(x) changes sign in different intervals.

∴ Not injective.

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

yx^{2} – x + y = 0

For y ≠ 0

D = 1 – 4y^{2} ≥ 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

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

nx^{2} + n^{2}x + [(n – 1) n (n + 1)] / 3 = 10n

x^{2} + nx + [n^{2} – 31] / 3 = 0

Given difference of roots = 1

|ɑ – β| = 1

D = 1

n^{2} – (4 / 3) (n^{2} – 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 (ω^{2} – ω^{4})

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

= – 3√3i

= -3z

k = -z

**Question 4: **Ifthen adj (3A^{2 }+ 12A) is equal to :

**Answer: (1)**

= (2 – 2λ – λ + λ^{2}) – 12

f (λ) = λ^{2} – 3λ – 10

∵ A satisfies f (λ).

A^{2} – 3A – 10I = 0

A^{2} – 3A = 10I

3A^{2}_{ }– 9A = 30I

3A^{2} + 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 – a^{2}) = 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

= ^{4}C_{3} . ^{4}C_{3} + (^{4}C_{2} . ^{3}C_{1})^{2} + (^{4}C_{1} . ^{3}C_{2})^{2} + (^{3}C_{3})^{2}

= 16 + 324 + 144 + 1

= 485

**Question 7: **The value of (^{21}C_{1} – ^{10}C_{1}) + (^{21}C_{2} – ^{10}C_{2}) + (^{21}C_{3} – ^{10}C_{3}) + (^{21}C_{4} – ^{10}C_{4}) + … (^{21}C_{10} – ^{10}C_{10}) is

1) 2^{21 }− 2^{10}

2) 2^{20 }− 2^{9}

3) 2^{20 }− 2^{10}

4) 2^{21 }− 2^{11}

**Answer: (3)**

^{21}C_{1} + ^{21}C_{2 }+ …… ^{21}C_{10} = (1 / 2) (^{21}C_{0} + ^{21}C_{1} + ….. ^{21}C_{21}) – 1

= 2^{20} – 1

^{10}C_{1} + ^{10}C_{2} + …. ^{10}C_{10} = 2^{10} – 1

Required sum = (2^{20} – 1) – (2^{10} – 1)

= 2^{20} – 2^{10}

**Question 8: **For any three positive real numbers a, b and c, 9 (25a^{2 }+ b^{2}) + 25 (c^{2} − 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 (25a^{2 }+ b^{2}) + 25 (c^{2} − 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) = ax^{2 }+ 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 – 9x^{3})) is √x . g(x), then g(x) equals:

1) (3x√x / (1 – 9x^{3}))

2) (3x/(1 – 9x^{3}))

3) (3/(1 + 9x^{3}))

4) (9/(1 + 9x^{3}))

**Answer: (4)**

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

f ‘ (x) = (9√x/(1 + 9x^{3}))

g (x) = (9/(1 + 9x^{3}))

**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) * (πr^{2}) = (θr^{2} / 2) —- (ii)

A = (r^{2} / 2) ([20 – 2r] / r)

A = [(20r – 2r^{2}) / 2] = 10r – r^{2}

A to be maximum

dA / dr = 10 – 2r = 0 ⇒ r = 5

d^{2}A / dr^{2} = -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} = ∫tan^{n} x dx, (n > 1). If I_{4 }+ I_{6 }= a tan^{5} x + bx^{5 }+ 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} = ∫tan^{n} x dx, (n > 1)

I_{4 }+ I_{6 }= ∫(tan^{4} x + tan^{6} x) dx

= ∫tan^{4} x sec^{2} x dx

Let tan x = t

sec^{2} x dx = dt

= ∫t^{4} dt

= (t^{5}/5) + C

= (1/5) tan^{5} 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, x^{2 }≤ 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)**

(k^{2} – 7k + 10) + 4k^{2} + 20k = ± 56

5k^{2} + 13k – 46 = 0

5k^{2} + 13k + 66 = 0

5k^{2} + 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 − x^{2} and the lines, y = |x| is :

1) 2 (√2 – 1)

2) 4 (√2 – 1)

3) 4 (√2 + 1)

4) 2 (√2 + 1)

**Answer: (2)**

x^{2} = – (y – 4)

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

Equation of normal at P is

y + (t^{2} / 4) – 4 = (1 / t) (x – [t / 2])

x – ty – (t^{3} / 4) + (7 / 2)t = 0

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

-tk – (t^{3} / 4) + (7 / 2)t = 0 —- (i)

t = 0, t^{2} = 14 – 4k

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

r = k / √2

Equation of circle is x^{2} + (y – k)^{2} = k^{2} / 2

It passes through point P

(t^{2} / 4) + [4 – (t^{2} / 4) – k]^{2} = k^{2} / 2

t^{4} + t^{2} [8k – 28] + 8k^{2} – 128k + 256 = 0 —- (ii)

For t = 0 ⇒ k^{2} – 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) + 8k^{2} – 128k + 256 = 0

2k^{2} + 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, b^{2} = a^{2} (1 – e^{2}) = 3

Equation to ellipse

(x^{2}/4) + (y^{2}/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) **

^{2}/ a

^{2}] – [y

^{2}/ b

^{2}] = 1

a^{2} + b^{2} = 4 and [2 / a^{2}] – [3 / b^{2}] = 1

(2 / [4 – b^{2}]) – (3 / [b^{2}]) = 1

b^{2} = 3

a^{2} = 1

x^{2} – (y^{2} / 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 √1^{2} + 4^{2} + 5^{2} = 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 = ^{11}C_{2} = 55

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

Probability = 6 / 55

**Question 28: **If 5(tan^{2} x − cos^{2} x) = 2cos^{2}x + 9, then the value of cos 4x is :

1) 1 / 3

2) 2 / 9

3) −7 / 9

4) −3 / 5

**Answer: (3)**

5(tan^{2} x − cos^{2} x) = 2cos^{2}x + 9

5 sec^{2} x – 5 = 9 cos^{2} x + 7

Let cos^{2} x = t

(5 / t) = 9t + 12

9t^{2} + 12t – 5 = 0

t = 1 / 3 as t ≠ – 5 / 3

cos^{2} x = 1 / 3, cos 2x = 2cos^{2} x – 1 = -1 / 3

cos 4x = 2cos^{2} 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)**

## Video Lessons – April 2^{nd} – Maths

## JEE Main 2017 Maths Paper With Solutions 2nd April

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