KVPY-SX 2019 Maths Paper with Solutions

Question 1: The number of four-letter words that can be formed with letters a, b, c such that all three letters occur is:

(a) 30
(b) 36
(c) 81
(d) 256

Answer: (b)

The 4 letter code will have a, b, c and a repeat letter from either a, b or c

The possible selections are

{a, a, b, c}, {b, b, a, c} and {c, c, a, b}

First selection is {a, a, b, c} = (arrangement of 4 words)/(repetition of a’s two times)

= 4!/2!

= 4×3×2/2

= 12

Second selection is {b, b, a, c} = (arrangement of 4 words)/(repetition of b’s two times)

= 4!/2!

= 4×3×2/2

= 12

Third selection is {c, c, a, b} = (arrangement of 4 words)/(repetition of c’s two times)

= 4!/2!

= 4×3×2/2

= 12

Number of four letter words that can be formed with letters a, b, c such that all three letters occur is = 12 + 12 + 12 = 36

Question 2: Let A = {θ ∈ R:{(1/3)sin θ+(2/3) cos θ}²= (1/3)sin² θ +(2/3)cos² θ} . Then

(a) A ⋂ [0, π] is an empty set
(b) A ⋂ [0, π] has exactly one point
(c) A ⋂ [0, π] has exactly two points
(d) A ⋂ [0, π] has more than two points

Answer: (b)

Given A = {θ ∈ R: {(1/3)sin θ+(2/3) cos θ}²= (1/3)sin² θ +(2/3)cos² θ}

((1/3)sin θ+(2/3) cos θ)² = (1/3)sin² θ+(2/3)cos² θ { since (A+B)²= A²+2AB+B²}

((1/9)sin² θ+(4/9)sin θ cos θ+(4/9)+(4/9)cos² θ = (1/3) sin² θ+(2/3)cos² θ

(2/9)sin 2θ = ((1/3)-(1/9))sin² θ+((2/3)-(4/9)) cos² θ {Since 2 sin A cos A = sin 2A}

(2/9)sin 2θ = (2/9) sin² θ+(2/9) cos² θ

(2/9)sin 2θ = (2/9)[sin²θ+cos²θ]

sin 2 θ = 1

sin 2 θ = sin 2(π/4)

2 θ = n π+(-1)ⁿ(π/4), n ∈ I

θ = (π/4) , θ ∈ [0, π]

A ⋂ [0, π] has exactly one point.

Question 3: The area of the region bounded by the lines x = 1, x = 2 and the curves x(y – e^x) = sin x and 2xy = 2sin x + x³ is:

(a) e² – e – 1/6
(b) e² – e – 7/6
(c) e² – e + 1/6
(d) e² – e + 7/6

Answer: (b)

Given curve

x(y – e^x) = sin x

y = (sin x/x)+e^x …(1)

second curve

2xy = 2 sin x+x³ (given)

y = (sin x/x)+(x²/2) …(2)

= |[e²-(2)³/6 -e¹+(1)³/6] |

= e²-e-(8/6)+(1/6)

Area = e²-e-7/6

Question 4: Let AB be a line segment with midpoint C, and D be the midpoint of AC. Let C₁ be the circle with diameter AB, and C₂ be the circle with diameter AC. Let E be a point on C₁ such that EC is perpendicular to AB. Let F be a point on C₂ such that DF is perpendicular to AB, and E and F lie on opposite sides of AB. Then the value of sin ∠ FEC is

(a) 1/√10
(b) 2/√10
(c) 1/√13
(d) 2/√13

Answer: (a)

Given that

∠ FEC = 90° –θ…………….. (1)

We are going to find out slope of FE

tanθ = slope of FE = (r-(-r/2))/(r-r/2)

tan θ = 3 = perpendicular/base

⇒ cos θ = base/hypotenuse = 1/√10

⇒ sin (90° – θ) = 1/√10 {According to reduction formula}

Hence sin (∠ FEC) = 1//√10 {from equation (1)}

Question 5: The number of integers x satisfying -3x⁴ + det

(a) 1
(b) 2
(c) 5
(d) 8

Answer: (b)

Given -3x⁴ + det

⇒ x = 0 or x(1 – x)²(x² – 1) = 3x

⇒ x = 0 or (1 – x)²(x² – 1) = 3

⇒ x = 0 or x⁴–2x³+ 2x – 4 = 0

⇒ x = 0 or (x–2)(x³ + 2) = 0

Integer value are 0, 2.

Question 6: Let P be a non-zero polynomial such that P(1 + x) = P(1 – x) for all real x, and P(1) = 0. Let m be the largest integer such that (x – 1)^m divides P(x) for all such P(x). Then m equals

(a) 1
(b) 2
(c) 3
(d) 4

Answer: (b)

P(x) is non-zero polynomial and P(1 + x) = P(1 – x) for all x

Differentiate with respect to x

P’(1 + x) = (–1)P’(1 – x)

⇒ P’(1 + x) = –P’(1 – x)

Put x = 0

⇒ P’(1) = –P’(1)

⇒ P’(1) + P’(1) = 0

⇒ 2P’(1) = 0

⇒ P’(1) = 0

and P(1) = 0

⇒ P(x) touches the x-axis at x = 1

⇒ P(x) = (x – 1)² Q(x)

⇒ m = 2 such that (x – 1)^m divides P(x) for all such P(x).

Question 7: Let

Then A has

(a) exactly one element
(b) exactly two element
(c) exactly three element
(d) infinitely many elements

Answer: (a)

Given

A = {x ∈ R: f(x) = 1}

f(x) = 1 for x = 0

for x ≠ 0,

f(x) = 1

⇒ x sin 1/x = 1

⇒ sin 1/x = 1/x

⇒ sin θ = θ which is true only when θ = 0

As θ ≠ 0 so it is not possible.

Question 8: Let S be a subset of the plane defined by S = {(x, y); |x| + 2|y| = 1}. Then the radius of the smallest circle with centre at the origin and having non-empty intersection with S is

(a) 1/5
(b) 1/√5
(c) 1/2
(d) 2/√5

Answer: (b)

Given

S = {(x, y); |x| + 2|y| = 1}

So |x| + 2|y| = 1

Equations are

x + 2y = 1 ……………. (1)

x – 2y = 1 ……………. (2)

–x + 2y = 1 ……………. (3)

–x – 2y = 1 ……………. (4)

Find intersecting point equation (1) & (2)

x + 2y = 1

x – 2y = 1

Adding above 2 equations

2x = 2

x = 1

So y = 0 {from equation (1)}

Intersecting point of equations (1) & (2) is (1, 0).

Similarly,

Find intersecting point equation (1) & (3)

x + 2y = 1

–x + 2y = 1

Adding above 2 equations

4y = 2

y = 1/2

x = 0 {from equation (1)}

Intersecting point of equations (1) & (3) is (0, 1/2).

Find intersecting point equation (2) & (4)

x – 2y = 1

–x – 2y = 1

Adding above 2 equations

–4y = 2

y = –1/2

x = 0 {from equation (2)}

Intersecting point of equations (2) & (4) is (0, –1/2).

Find intersecting point equation (3) & (4)

–x + 2y = 1

–x – 2y = 1

Adding above 2 equations

–2x = 2

x = –1

So y = 0 {from equation (3)}

Intersecting point of equations (3) & (4) is (–1, 0)

Length of the perpendicular from a point (x₁, y₁) to a line ax + by = c is

According to question minimum radius is the distance between point (0, 0) and line x + 2y = 1.

So, Minimum radius = |(0+2×0-1)/ √(1²+2²)|

= |-1/ √(1+4)|

= 1/√5

Question 9: The number of solutions of the equation sin (9x) + sin (3x) = 0 in the closed interval [0, 2 π] is

(a) 7
(b) 13
(c) 19
(d) 25

Answer: (b)

Given

sin(9x) + sin(3x) = 0

Question 10: Among all the parallelograms whose diagonals are 10 and 4, the one having maximum area has its perimeter lying in the interval

(a) (19, 20]
(b) (20, 21]
(c) (21, 22]
(d) (22, 23]

Answer: (c)

Area of parallelogram = (1/2)d₁d₂sin φ

Where d₁ and d₂ is the diagonal of the parallelogram.

Maximum area of parallelogram = ½ d₁d₂ {since where φ =π/2 }

It is a rhombus

ΔAOB is a right a triangle so apply Pythagoras theorem

So, a² = (d₁/2)²

Side length (a) = √((d₁/2)²+ (d₂/2)²)

⇒ (a) = √((10/2)²+ (4/2)²)

⇒ (a) = √((5)²+ (2)²)

⇒ (a) = √(25+4)

⇒ (a) = √29

Perimeter = 4a

= 4(√29 ) ∈ [21, 22)

Question 11: The number of ordered pairs (a, b) of positive integers such that (2a-1)/b and (2b-1)/a are both integers is

(a) 1
(b) 2
(c) 3
(d) more than 3

Answer: (c)

Given (2a-1)/b and (2b-1)/a are both integers.

From equation (1)

⇒ 4a – 2 = 2b λ

⇒ 4a – 2 = λ (µa + 1) {From equation (2)}

⇒ (4 – λ µ)a = λ + 2

Since multiplication of λ and µ lie between only 1 & 3 because a and b are positive integer.

So 1 ≤µ ≤ 3

In equation (2) put λ = 1, µ = 1

(4 – 1) a = 1 + 2 ⇒ a = 1, b = 1

Similarly,

λ = 1, µ = 3 ⇒ a = 3, b = 5

λ = 3, µ = 1 ⇒ a = 5, b = 3

So, total set = 3

(1, 1), (3, 5), (5, 3)

Question 12: Let z = x + iy and w = u + iv be complex numbers on the unit circle such that z² + w² = 1. Then the number of ordered pairs (z, w) is

(a) 0
(b) 4
(c) 8
(d) infinite

Answer: (c)

Let

z = e^iα = cos α + i sin α

w = e^iβ = cos β + i sin β

Since, z² + w² = 1

⇒ (cos α + i sin α)² + (cos β + i sin β)² = 1 {(A + iB)² = A²–B² + 2iAB}

⇒ cos²α – sin²α + 2i sinαcosα + cos²β – sin²β + 2i sinβcosβ = 1

⇒ (cos2 α + cos2 β) + i(sin2 α + sin2 β ) = 1{cos2A = cos²A – sin²A, sin2A = 2sinA cosA}

Comparing real and imaginary part

So cos2 α + cos2 β = 1 and sin2 α + sin2 β = 0

2cos(α + β)cos(α -β) = 1

And sin2 α = –sin2 β

Squaring both side

sin²2 α = sin²2 β

1 – cos²2 α = 1 – cos²2 β

cos²2 α = cos²2 β

Taking square root both side

cos2 α = ±cos2 β

cos2 α = –cos2 β

or cos2 α + cos2 β = 0 (cancelled)

If cos2 α = cos2 β

So 2cos2 α = 1

cos 2α = ½

Question 13: Let E denote the set of letters of the English alphabet, V = {a, e, i, o, u}, and C be the complement of V in E. Then, the number of four-letter words (where repetitions of letters are allowed) having at least one letter from V and at least one letter from C is

(a) 261870
(b) 314160
(c) 425880
(d) 851760

Answer: (a)

‘V’ denotes vowels and ‘C’ denotes consonants.

Total alphabet words = 26

Total vowel in alphabet = 5

Total consonant in alphabet = 21

Total 4 letter words = (26)⁴

Number of 4 letter words which contains only vowels = (5)⁴

Number of 4 letter words which contains only consonant = (21)⁴

Number of 4 letter words which contains atleast one vowel and atleast one consonants:

(26)⁴ – (21)⁴ – (5)⁴ = 261870

Question 14: Let σ₁, σ₂, σ₃ be planes passing through the origin. Assume that σ₁ is perpendicular to the vector (1, 1, 1), σ₂ is perpendicular to a vector (a, b, c), and σ₃ is perpendicular to the vector (a², b², c²). What are all the positive values of a, b, and c so that σ₁⋂ σ₂ ⋂σ₃ is a single point?

(a) Any positive value of a, b, and c other than 1
(b) Any positive value of a, b, and c where either a ≠ b, b ≠ c or a ≠ c
(c) Any three distinct positive values of a, b, and c
(d) There exist no such positive real numbers a, b, and c

Answer: (c)

σ₁ is perpendicular to

σ₂ is perpendicular to

σ₃ is perpendicular to

since σ₁ , σ₂ and σ₃ planes are passing through the origin

Hence the planes are

σ₁: x + y + z = 0

σ₂: ax + by + cz = 0

σ₃: a²x + b²y + c²z = 0

Δ = (a – b)(b – c)(c – a)

For unique solution, Δ ≠ 0

⇒ (a – b)(b – c)(c – a) ≠ 0

Hence a ≠ b, b ≠ c, c ≠ a

Question 15: Ravi and Rashmi are each holding 2 red cards and 2 black cards (all four red and all four black cards are identical). Ravi picks a card at random from Rashmi, and then Rashmi picks a card at random from Ravi. This process is repeated a second time. Let p be the probability that both have all 4 cards of the same colour. Then p satisfies

(a) p≤ 5%
(b) 5% < p ≤10%
(c) 10% < p ≤ 15%
(d) 15% < p

Answer: (a)

If Ravi withdraw red cards from Rashmi, then Rashmi withdraw black card from Ravi and this process repeat again, vice-versa. If Ravi withdraw black card from Rashmi

P = 2(2/4)(2/5)(1/4)(1/5)

P = 2/100

P = 2%

Question 16: Let A₁, A₂ and A₃ be the regions on R² defined by

A₁ = {(x, y); x ≥ 0, y ≥0, 2x + 2y – x² – y²> 1 > x + y},

A₂ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x² + y²},

A₃ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x³ + y³},

Denote by |A₁|, |A₂| and |A₃| the areas of the regions A₁, A₂, and A₃ respectively. Then

(a) |A₁| > |A₂| > |A₃|
(b) |A₁| > |A₃| > |A₂|
(c) |A₁| = |A₂| < |A₃|
(d) |A₁| = |A₃| > |A₂|

Answer: (c)

Given

A₁ = {(x, y); x ≥ 0, y ≥0, 2x + 2y – x² – y²> 1 > x + y}

A₂ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x² + y²}

A₃ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x³ + y³}

For A₁:

Find intersecting point of circle & line

⇒ (x – 1)² + (1 – x – 1)2 = 1 {Line x + y = 1 put in circle equation.}

⇒ x² – 2x + 1 + x² = 1

⇒ 2x² – 2x = 0

⇒ x (x –1) = 0

⇒ x = 0, 1

Hence y = 1, 0

(0, 1) and (1, 0)

Area of A₁ = (1/4)of area of circle- area of triangle

= π(1)²/4 -(1/2)(1)(1)

= (π/4)-(1/2)

For A₂:

A₂ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x² + y²},

⇒ 1 > x² + y², x + y > 1, x ≥ 0, y ≥ 0

Find intersecting point of circle & line

⇒ (x)² + (1 – x)² = 1 {Line x + y = 1 put in circle equation.}

⇒ x² + 1– 2x + x² = 1

⇒ 2x² – 2x = 0

⇒ x (x –1) = 0

⇒ x = 0, 1

So y = 1, 0

(0, 1) and (1, 0)

Area of A₂= 1/4 of area of circle – Area of triangle

= π(1)²/4 – 1/2(1)(1)

A₂ = π/4 – 1/2

For A₃:

A₃ = {(x, y); x ≥ 0, y ≥0, x + y > 1 > x³ + y³}

⇒ x + y > 1, x³ + y³< 1, x ≥ 0, y ≥ 0

Hence |A₃| > |A₂| = |A₁|

Question 17: Let f : R-> R be a continuous function such that f (x²) = f (x³) for all x ∈ R. Consider the following statements.

(I) f is an odd function

(II) f is an even function

(III) f is differentiable everywhere.

Then

(a) I is true and III is false
(b) II is true and III is false
(c)both I and III are true
(d) both II and III are true

Answer: (d)

f: R -> R be continuous function such that

f(x²) = f(x³) …(1)for all x belongs to R

Put x = –x

f(x²) = f(–x³)

From equation (1) we have f(x³) = f(–x³)

Put x³ = t we have f(t) = f(–t)

So f(x) is an even function.

(ii) Now take x³ = t

Then f(t^2/3) = f(t)

Put t = t^2/3

Then f((t ^2/3)²) = f(t)

f(t) = f(t ^2/3) = f((t ^2/3)²) = f((t ^2/3)³) = ……….. = f((t ^2/3) ⁿ)

This is true for all t belongs to R and any n belongs to I

Hence if we take n -> ∞, (2/3)ⁿ -> 0

then f(t) = f(t⁰) = 1

So f(x) is a constant function, hence it is differentiable everywhere.

Question 18: Suppose a continuous function f : [0, ∞) -> R satisfies

(a) e
(b) e²

(c) e⁴

(d) e⁶

Answer: (a)

Given

f: [0, ∞)-> R

f(x) =

Differentiate with respect to x

So f ‘(x) = 2.x.f(x)

f'(x)/f(x) = 2x

Integrate both sides

Ln f(x) = x² + c

f(x) =

Since f(0) = 1

put x = 0, y = 1

f(0) = e¹ + c

c = 0

So f(x) =

f(x) =

f(1) = e

Question 19: Let a > 0, a ≠ 1. Then the set S of all positive real numbers b satisfying (1 + a²) (1+ b²) = 4ab is

(a) an empty set
(b) a singleton set
(c) a finite set containing more than one element
(d) (0, ∞)

Answer: (a)

a> 0, a ≠ 1

(1 + a²)(1 + b²) = 4ab

Equality holds true when a = b = 1

But it is given in the question that a ≠ 1

Hence, there are no values of b.

Question 20: Let f : R -> R be a function defined by

Then, at x = 0, f is:

(a) Not continuous
(b) Continuous but not differentiable
(c) Differentiable and the derivative is not continuous
(d) Differentiable and the derivative is continuous

Answer: (d)

f : R -> R

Check continuity:

Hence, f(x) is continuous at x = 0

Check differentiability:

R.H.D. = L.H.D.

f(x) is differentiable at x = 0

For continuity

= 2 – 1 = 1

Derivative is continuous at x = 0.

Question 21: The points C and D on a semicircle with AB as diameter are such that AC = 1, CD = 2, and DB = 3. Then the length of AB lies in the interval

(a) [4, 4.1)
(b) [4.1, 4.2)
(c) [4.2, 4.3)
(d) [4.3, ∞ )

Answer: (b)

Let AB = x

ΔABC,

(AB)² = (AC)² + (BC)² { since ΔABC is a right angle triangle}

BC = √[x²-1]

ΔABD,

(AB)² = (AD)² + (BD)² {since ΔABD is a right angle triangle}

AD = √[x²-9]

Apply Ptolemy theorem, we get

AB × CD + AC × BD = AD × BC

2x + 3 = √[x²-1] √[x²-9]

On squaring we get

(2x + 3)² = [√[x²-1] √[x²-9]]²

4x² + 12x + 9 = (x² – 1)(x² – 9) {(A + B)² = A² + B² + 2AB}

4x² + 12x + 9 = x⁴ – 10x² + 9

x⁴ – 14x² – 12x = 0

x³ – 14x – 12 = 0

f(x) = x³ – 14x – 12

f(4.1) = (4.1)³ – 14(4.1) – 12 = –0.479

f(4.2) = (4.2)³ – 14(4.2) – 12 = 3.288

f(4.1).f(4.2) < 0

As f(x) is a continuous function therefore one root of f(x) lies in [4.1, 4.2) i.e. length of AB lies in this interval.

Question 22: Let ABC be a triangle and let D be the midpoint of BC. Suppose cot(∠CAD) : cot(∠BAD) = 2 : 1. If G is the centroid of triangle ABC, then the measure of ∠ BGA is

(a) 90⁰

(b) 105⁰

(c) 120⁰

(d) 135⁰

Answer: (a)

AD² + b² –a²/4 = 2AD² + 2c² – a²/2

(Using the formula of length of median, AD = 1/2 √(2b²+2c²-a²)

(a²/4)+b² = 2c²+(1/4)(2b²+2c²-a²)

a²/2+ b²/2= 5c²/2

a²+ b²= 5c²

Question 23: Let f(x) = x⁶ – 2x⁵ + x³ + x² – x – 1 and g(x) = x⁴ – x³ – x² – 1 be two polynomials. Let a, b, c and d be the roots of g(x) = 0. Then the value of f(a) + f(b) + f(c) + f(d) is

(a) -5
(b) 0
(c) 4
(d) 5

Answer: (b)

Given

g(x) = x⁴ – x³ – x² – 1 = 0

f(x) = x⁶ – 2x⁵ + x³ + x² – x – 1

since a, b, c, d are roots of x⁴ – x³ – x² – 1 = 0

Hence ∑a = 1

∑ab = -1

Now,

f(x) = x⁶ – 2x⁵ + x³ + x² – x – 1

f(x) = x²(x⁴ – x³ – x² – 1) – x(x⁴ – x³ – x² – 1) + (2x² – 2x – 1)

f(x) = (x² – x) (x⁴ – x³ – x² – 1) + (2x² – 2x – 1)

f(x) = 2x² – 2x – 1 {since x⁴ – x³ – x² – 1 = 0}

Now,

⇒f(a) + f(b) + f(c) + f(d)

⇒2a² – 2a – 1 + 2b² – 2b – 1 + 2c² – 2c – 1 + 2d² – 2d – 1

⇒2[a² + b² + c² + d²] – 2[a + b + c + d] – 4

⇒2[(a + b + c + d)² – 2(ab + bc + cd + ac + bd)] – 2[a + b + c + d] – 4

⇒2[1 – 2(–1)] – 2(1) – 4 = 0

Question 24: Let

(a) 4
(b) 9
(c) 14
(d) 19

Answer: (a)

Given

⇒ x+y+z = 5

Sum of distances of a point

=

=

=

With

Area in the plane constitutes an ellipse

Distance between

=

2ae =

2a = 4

So 4e =

e =

Eccentricity (e) = √(1-b² /a²)

(√14/4)² = 1-b²/a²

b² = ½

area of ellipse = πab

= π(2)1/√2

= √2 π

Question 25: The number of solutions to sin(π sin²(θ)) + sin(π cos²(θ)) = 2cos (π/2 cos θ ) satisfying 0 ≤ θ ≤ 2 π is

(a) 1
(b) 2
(c) 4
(d) 7

Answer: (d)

Given sin(π sin²(θ)) + sin(π cos²(θ)) = 2cos (π/2 cos θ )

{sin C+sin D = 2 sin ((C+D)/2 )cos((C-D)/2)}

θ = 2n π, 2n π ±2n π/3

θ belongs to {0, 2 π, 2 π/3, 4 π/3}

Case-II: {Take – ve}

cos2 θ + cos θ = 4n

Only possible cos2 θ + cos θ = 0 {cos2 θ = 2cos² θ – 1}

2cos² θ + cos θ – 1 = 0

(cos θ + 1)(2cos θ – 1) = 0

cos θ = –1, 1/2

θ = (2n + 1) π, 2n π ± π/3

θ belongs to { π, π/3, 5 π/3}

Total solutions is 7.

Question 26: Let J =

Consider the following assertions:

I. J > 1/4

II. J < π/8

Then

(a) Only I is true
(b) only II is true
(c) both I and II are true
(d) neither I nor II is true

Answer: (a)

Given J =

Now,1 + x⁸< 2 {Limit 0 to 1}

1/(1+x⁸)>1/2 ⇒ x/(1+x⁸)>x/2

Apply integration both sides

J>1/4

Statement I is true.

Now,

1 + x⁴> 1 + x⁸

1/(1+x⁴)<1/(1+x⁸)

x/1+x⁸>x/(1+x⁴)

Apply integration on both sides

Put x² = t

2xdx = dt

J>(1/2) π/4

J> π/8

Question 27: Let f: (–1, 1) -> R be a differentiable function satisfying (f ‘(x))⁴ = 16(f(x))² for all x belongs to (–1, 1), f(0) = 0. The number of such function is

(a) 2
(b) 3
(c) 4
(d) more than 4

Answer: (d)

Given

(f ‘(x))⁴ = 16(f(x))²

Taking square root both side.

(f ‘(x))² = ±4(f(x))

Case-I:

If f(x) > 0

(f ‘(x))² = 4(f(x))

Taking square root both side

f ‘(x) = ±2√f(x)

f ‘(x) = 2√f(x)

f ‘(x)/ √f(x) = 2

Integrate both side

Integrate both side

Case-II:

If f(x) < 0

(f ‘(x))² = –4(f(x))

f ‘(x) = ±2 √-f(x)

f ‘(x) = 2√-f(x) or f ‘(x) = –2√-f(x)

Similarly f(x) = –x², –1 < x < 0 or f(x) = –x², 1 > x > 0

Case-III:

Also, one singular solution of given differential equation is

f(x) = 0, –1 < x < 1

Hence, functions can be

f(x) = x², –1 < x < 1

f(x) = –x², –1 < x < 1

More functions are also possible.

Question 28: For x ∈ R, let f(x) = |sin x| and g(x) =

(a) p(x + π) = p(x) for all x
(b) p(x + π) ≠ p(x) for at least one but finitely many x
(c) p(x + π) ≠ (x) for infinitely many x
(d) p is a one-one function

Answer: (a)

Given

f(x) = |sin x| …(1)

g(x) =

p(x) = g(x) – (2/π) x …(3)

From equation (2)

g(x) =

Put x-> x+π

From equation (3)

p(x) = g(x) – (2/π)x

Put x -> x + π

p(x + π) = g(x + π) – 2/ π (x + π)

p(x + π) = g(x) + 2 – 2x/ π – 2 {From equation (4)}

p(x + π) = g(x) – 2x/ π …(5)

eq.(5) – eq.(3)

p(x + π) – p(x) = g(x + π) – g(x) – 2

p(x + π) – p(x) = 2 – 2

p(x + π) = p(x) for all x.

Question 29: Let A be the set of vectors

Then

(a) A is empty
(b) A contains exactly one element
(c) A has 6 elements
(d) A has infinitely many elements

Answer: (b)

Given

a₁²+(3a₂²/4)+(7a₃²/16) = a₁a₂+ a₂a₃ /4+ a₃a₁/2

(16 a₁²+12a₂²+7a₃²)/16 = (4a₁a₂+a₂a₃+2a₃a₁)/4

16 a₁²+12a₂²+7a₃² = 4(4a₁a₂+a₂a₃+2a₃a₁)

16 a₁²+12a₂²+7a₃² – 16a₁a₂-4a₂a₃-8a₃a₁ = 0

8 a₁²– 16a₁a₂ +8a₂² +8a₁²-8a₁a₃+2a₃²+4a₂ ²-4a₂a₁+a₃²+4a₃² = 0

(2√2a₁-2√2a₂)²+(2√2a₁-√2a₃)²+(2a₂-a₃)²+4a₃² = 0

Only possible when a₁ = a₂ = a₃ = 0

So only one element in the set.

Question 30: Let f : [0, 1] -> [0, 1] be a continuous function such that x² + (f(x))² ≤ 1 for all

x ∈ [0, 1] and

(a) π/12
(b) π/15
(c) (√2-1) π /2
(d) π/10

Answer: (a)

Given

x² + (f(x))² ≤ 1

Let y = f(x)

x² + y² ≤ 1 for all x belongs to [0, 1]

y² ≤ 1 –x²

y = λ√(1-x²)

Apply integration

=

=

= π/4- π/6

= π/12