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**Question 1:** A quadrilateral has distinct integer side lengths. If the second-largest side has length 10, then the maximum possible length of the largest side is

- a. 25
- b. 26
- c. 27
- d. 28

Solution:

**Answer: (B)**Let a, b, c, d are the four sides of a quadrilateral

Where a > b > c > d are distinct positive integers and b = 10 (given)

Now, we know that in a quadrilateral

Sum of any three sides > 4

^{th}side⇒ a < b + c + d

Since b = 10 and b > c > d, so the largest possible values of c and d are 9 and 8 respectively.

⇒ a < 10 + 9 + 8

⇒ a < 27

Therefore, the largest possible value of a is 26.

**Question 2:** The largest power of 2 that divides 200!/100!

- a. 98
- b. 99
- c. 100
- d. 101

Solution:

**Answer: (C)**The exponent of 2 in 200!

= (200/2) + (200/2

^{2}) + ……… + (200/2^{7})= 100 + 50 + 12 + 6 + 3 + 1

= 197.

The exponent of 2 in 100!

= (100/2)+(100/2

^{2})+…(100/2^{6})= 50+12+6+3+1

= 97

Therefore, the exponent of 2 in 200!/100! = 197 – 97 = 100.

**Question 3:** Let a_{1}, a_{2}, a_{3}, a_{4} be real numbers such that a_{1}+ a_{2} +a_{3 }+ a_{4 }= 0 and a_{1}^{2}+ a_{2}^{2} +a_{3}^{2} + a_{4}^{2} = 1. Then the smallest possible value of the expression (a_{1} – a_{2})^{2} + (a_{2 }– a_{3})^{2} + (a_{3 }– a_{4})^{2} + (a_{4 }– a_{1})^{2} lies in the interval

- a. (0, 1.5)
- b. (1.5, 2.5)
- c. (2.5, 3)
- d. (3, 3.5)

Solution:

**Answer: BONUS**Given, a

_{1}^{2}+ a_{2}^{2}+a_{3}^{2}+ a_{4}^{2}= 1.Therefore (a

_{1}– a_{2})^{2}+ (a_{2 }– a_{3})^{2}+ (a_{3 }– a_{4})^{2}+ (a_{4 }– a_{1})^{2 }= 0When a

_{1}= a_{2}= a_{3 }= a_{4 }= +1/2 or -1/2.

**Question 4:** Let S be the set of all ordered pairs (x, y) of positive integers satisfying the condition x^{2} – y^{2 }= 12345678. Then

- a. S is an infinite set
- b. S is the empty set
- c. S has exactly one element
- d. S is a finite set and has at least two elements

Solution:

**Answer: (B)**x

^{2}– y^{2}= 12345678OR

(x – y) (x + y) = 12345678 x, y belongs to z

^{+}Case I: x, y belongs to odd

Difference of squares of two odd integers is multiple of 8 (or divisible by 8)

⇒ x

^{2}– y^{2}= 8k, k belongs to IBut R.H.S is not divisible by ‘8’

Therefore, equality can’t hold.

Case II:x, y belongs to even

⇒ x – y belongs to even, x + y belongs to even

L.H.S = (x – y) (x + y)

= 2n×2m (n, m belongs to I)

= 4 mn (multiple of 4)

But R.H.S is not divisible by ‘4’

Again not possible.

Case III:x belongs to odd, y belongs to even

⇒ x – y belongs to odd, x + y belongs to odd

⇒ L.H.S = (x – y) (x + y)

= odd × odd

= odd

But R.H.S is even.

So not possible.

Therefore, S is an empty Set.

**Question 5:** Let A_{1}A_{2}A_{3}....A_{9} be a nine-sided regular polygon with side length 2 units. The difference between the lengths of the diagonals A_{1}A_{5} and A_{2}A_{4 }equals

- a. 2+ √12
- b. √12-2
- c. 6
- d. 2

Solution:

**Answer: (D)**In ΔA

_{1}OA_{2}∠ A

_{1}OA_{2}= 2 π/9 = 40^{0}⇒ cos 40° = (x

^{2}+x^{2}-2^{2})/2.x.x {using cosine rule}⇒ 2x

^{2}(cos 40^{0}–1) = –4⇒ x

^{2}= 2/(1-cos 40^{0}) = 2/2sin^{2}20^{0}⇒ x = 1/sin 20

^{0}……(1)In Δ A

_{1}OA_{5}Now, ∠ A

_{1}OA_{5}= 8π/9⇒ cos 8π/9 = (x

^{2}+x^{2}-(A_{1}A_{5})^{2})/2x^{2}{using cosine rule}⇒ A

_{1}A_{5}^{2}= 2x^{2}(1 – cos 160^{0})⇒ A

_{1}A_{5}= 2 x sin80^{0}……. (ii)Similarly, In Δ A

_{2}OA_{4}A

_{2}A_{4}= 2 x sin40^{0}Therefore A

_{1}A_{5 }– A_{2}A_{4}= 2 x (sin 80^{0}– sin40^{0})= 2(1/sin 20

^{0})2.sin 20^{0}cos 60^{0}(sin C – sin D = 2 cos(C+D)/2 sin (C-D)/2)= 2

**Question 6:** Let a_{1}, a_{2},.... an be n nonzero real numbers, of which p are positive and remaining are negative. The number of ordered pairs (j, k), j < k, for which a_{j }a_{k} is positive, is 55. Similarly, the number of ordered pairs (j, k), j < k, for which a_{j }a_{k} is negative is 50. Then the value of p^{2} + (n – p)^{2} is

- a. 629
- b. 325
- c. 125
- d. 221

Solution:

**Answer: (C)**Given, a

_{1}, a_{2}, .... a_{n}be n nonzero real numbers, among which p are positive and n – p are negative.The number of ordered pair (j,k), j < k such that the product a

_{j}. a_{k}is positive is 55.Which is possible when either both a

_{j}, a_{k}are positive or both are negative.⇒

^{p}C_{2}+^{n–p}C_{2}= 55 (No. of ways of selecting two positive and two negative numbers)p(p-1)/2 +(n-p)(n-p-1)/2 = 55

p

^{2}+ (n – p)^{2}– n = 55 ….(i)The number of ordered pair (j,k), j < k such that the product aj . ak is negative is 50.

Which is possible when one of aj, ak is positive and other negative.

Required number of ways = Total number of ways of selecting two numbers - No. of

ways of selecting two positive and two negative numbers

^{n}C_{2}– 55 = 50n(n-1)/2 = 105

n

^{2 }– n – 210 = 0n = 15, –14 (n can’t be negative therefore – 14 is rejected)

Now, put n = 15 in equation (i)

So p

^{2}+ (n – p)^{2}= 125

**Question 7:** If a, b, c, d are four distinct numbers chosen from the set {1,2,3 ....,9}, then the minimum value of is

- a. 3/8
- b. 1/3
- c. 13/36
- d. 25/72

Solution:

**Answer: (D)**a, b, c, d ∈ {1, 2, 3, ………9}

Now, (a/b)+(c/d) is minimum if (a/b) and (c/d) individually become minimum.

a/b -> min, when a -> min and b -> max.

c/d -> min, when c -> min and d -> max.

Also a, b, c, d are distinct

So minimum (a/b)+(c/d) = (1/9)+(2/8) Or (1/8)+(2/9)

= 26/72 or 25/72

Hence 25/72 is required answer.

**Question 8:** If 72^{x}. 48^{y} = 6^{xy}, where x and y are non zero rational numbers, then x + y equals

- a. 3
- b. 10/3
- c. –3
- d. –10/3

Solution:

**Answer: (D)**72

^{x}. 48^{y}= 6^{x. y}(2

^{3}.3^{2})^{x}.(2^{4}.3^{1})^{y}= (2.3)^{x.y}2

^{3x}^{+ 4y}. 3^{2x + y }= 2^{x. y}. 3^{xy}Equating the powers of 2 and 3, we get

3x + 4y = xy …………(i)

2x + y = xy ……………(ii)

Subtract (ii) from (i)

x = – 3y { put in (i) }

3(–3y)+ 4y = (–3y).y

– 5y = – 3y

^{2 }y = 5/3, x = – 5

Hence x + y = (5/3)-5

= -10/3

**Question 9:** Let AB be a line segment of length 2. Construct a semicircle S with AB as diameter. Let C be the midpoint of the arc AB. Construct another semicircle T external to the triangle ABC with chord AC as diameter. The area of the region inside the semicircle T but outside S is

- a. π/2
- b. 1/2
- c. π/√2
- d. 1/√2

Solution:

**Answer: (B)**C being midpoint of arc ACB ⇒ AX = BX ⇒ ∠CAB = ∠CBA

Also ∠ ACB = 90 (Angle in semicircle)

OC AB (Construction)

⇒ ∠OAC = ∠OCA = 45

OA =OX = 1

In ΔOAC

AC

^{2}= OA^{2}+ OC2⇒ AC = √2

Now, Required Area = Area of semicircle on AC as diameter – Area of segment A.M.C.A

= (1/2) ×π (√2/2)

^{2}-((π /4)-(1/2) )= ((π /4)-(π /4)+(1/2) )

= 1/2

**Question 10:** Let r(x) be the remainder when the polynomial x^{135} + x^{125} – x^{115} + x^{5} + 1 is divided by x^{3}– x. Then.

- a. r(x) is the zero polynomial
- b. r(x) is a nonzero constant
- c. degree of r(x) is one
- d. degree of r(x) is two

Solution:

**Answer: (C)**x

^{135}+ x^{125}– x^{115}+ x^{5 }+ 1 = q(x) (x^{3}– x) + Ax^{2}+Bx + CWhere q(x) is quotient and r(x) = Ax

^{2}+Bx + C is remainderNow, putting x = 0, 1, –1

We get

⇒ 1 = C

⇒ 3 = A + B + C

⇒ A+B = 2 ….(i)

⇒ –1 = A – B + C

⇒ -1 = A-B+C

⇒ A - B = -2

Adding (i) and (ii)

⇒ 2A = 0

⇒ A = 0 and B = 2

So r(x) = 2x + 1 a polynomial of degree 1.

**Question 11:** It is given that the number 43361 can be written as a product of two distinct prime numbers p_{1} , p_{2}. Further, assume that there are 42900 numbers which are less than 43361 and are co-prime to it. Then, p_{1}+p_{2} is

- a. 462
- b. 464
- c. 400
- d. 402

Solution:

**Answer: (A)**The prime factors given number are,

43361 = 131 × 331

p

_{1}+ p_{2 }= 131 + 331= 462

**Question 12:** Let ABC be a triangle with ∠ C = 90^{0}. Draw CD perpendicular to AB. Choose points M and N on sides AC and BC respectively such that DM is parallel to BC and DN is parallel to AC. If DM = 5, DN = 4, then AC and BC are respectively equal to

- a. 41/4, 41/5
- b. 39/4, 39/5
- c. 38/4, 38/5
- d. 37/4, 37/5

Solution:

**Answer: (A)**CM || ND and CN || MD

Therefore CMDN is a rectangle

⇒ CM = ND = 4 and CN = MD = 5

In Δ CMD

By Pythagoras theorem

CD

^{2}= CM^{2}+ MD^{2}CD = √41

Also, Δ AMD ~ Δ DNB

⇒ y/5 = 4/x = z/t …….(i) {Sides of similar triangles are in proportion}

In Δ ADC

By Pythagoras Theorem

(x + 4)

^{2}= 41 + t^{2}……..(ii)In Δ BDC

By Pythagoras Theorem

(y + 5)

^{2}= 41 +z^{2}………(iii)From (ii) & (iii)

((x+4)

^{2}-41)/((y+5)^{2}-41) = (t/z)^{2}((x+4)

^{2}-41)/((20/x)+5)^{2}-41) = (x/4)^{2}{using (i)}16 (x

^{2}+ 16 + 8x – 41) = 400 +25x^{2}+ 200x – 41x^{2}4x

^{2}– 9x – 100 = 0x = 25/4 , y = 16/5 {using (i) again }

Hence AC = x + 4 = 41/4 and BC = 5 + y = 41/5.

**Question 13:** Let A, G and H be the arithmetic mean, geometric mean and harmonic mean, respectively of two distinct positive real numbers. If α is the smallest of the two roots of the equation A(G – H)x^{2} + G (H – A) x + H(A –G) = 0, then

- a. –2 < α < –1
- b. 0 < α < 1
- c. –1 < α < 0
- d. 1 < α < 2

Solution:

**Answer: (B)**Given, A(G–H) x

^{2}+ G(H–A) x + H(A–G) = 0Put x = 1 in above equation

A(G–H) + G(H–A) + H(A–G) = 0

0 = 0

⇒ x = 1 is one root (let it be ‘α’ and other root be ‘β’)

Now using product of roots,

α. β = A(A-G)/A(G-H)

1. β = (H.A-H.G)/A(G-H)

β = (G

^{2}-H.G)/A(G-H) {A.H = G^{2}}β = G(G-H)/A(G-H)

β = G/A < 1{ as A > G}

**Question 14:** In the figure, ABCD is a unit square. A circle is drawn with centre O on the extended line CD and passing through A. If the diagonal AC is tangent to the circle, then the area of the shaded region is

- a. (9-π )/6
- b. (8-π )/6
- c. (7-π )/4
- d. (6-π )/4

Solution:

**Answer: (D)**AC is tangent to circle with radius OA

Therefore ∠ OAC = 90

^{0}Also, AC is diagonal of square

Therefore ∠ CAD = 45

^{0}⇒ ∠ OAD = 45

^{0}In Δ OAD

⇒ OA = AD cosec(45

^{0}) = √2Now, Required Area = Area of square – (area of sector OAX – area of Δ OAD)

= 1-( π(√2)

^{2}. ((π/4)/2 π)-(1/2)(1).(1))= 1-((π/4)-(1/2))

= (3/2)-( π/4)

= (6-π )/4

**Question 15:** The sum of all non-integer roots of the equation x^{5 }– 6x^{4} + 11x^{3} – 5x^{2} – 3x + 2 = 0 is

- a. 6
- b. –11
- c. –5
- d. 3

Solution:

**Answer: (D)**x

^{5}– 6x^{4}+ 11x^{3 }– 5x^{2}– 3x + 2 = 0(x – 1) (x – 2) (x

^{3}– 3x + 1) = 0 {factorizing}Now, x

^{3}– 3x + 1 has non integral roots whose sum is 3.

**Question 16:** Let S be the circle in xy-plane which touches the x-axis at point A, the y-axis at point B and the unit circle x^{2} + y^{2 }= 1 at point C externally. If O denotes the origin, then the angle OCA equals

- a. 5π/8
- b. π/2
- c. 3π/4
- d. 3π/5

Solution:

**Answer: (A)**Circle S touches both the axes, therefore its center lies on y = x, let it be O’(a,a).

So A(a,0) and B(0,a)

⇒ ∠ O’OA = 45

‘C’ lies on x

^{2 }+ y^{2}= 1 & on OO’Therefore C (1/√2, 1/√2)

OO’ = OC + CO’

√2a = 1+a

a = 1/√2-1 = √2+1

Now, slope of AC = (0-1/√2)/(1+√2-(1/√2)

= – 1/(√2+1)

= – (√2-1)

= – tan π/8

= tan (π –π /8)

= tan (7π /8)

So ∠ CAX = 7π/8

Also ∠ CAX = ∠ OCA + ∠ COA {Exterior angle property}

⇒ 7 π /8 = ∠ OCA + π /4

⇒ ∠ OCA = 5 π /8.

**Question 17:** In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to 30. In order to maximize the area of the trapezium, the smallest angle should be

- a. π/6
- b. π/4
- c. π/3
- d. π/2

Solution:

**Answer: (C)**In ΔADN

x = 30 cos θ; y = 30 sin θ

Now, Area of trapezium

A = ½ (30+30+2x)×(y) { Area =1/2 (sum of parallel sides)height }

= (30+30cos θ) 30 sin θ

= 900 sin θ (1+cos θ)

Differentiating both sides with respect to θ

dA/d θ = 900 {sin θ. (– sin θ) +(1 + cos θ). cos θ } = 0 {for critical points}

⇒ cos

^{2}θ – sin^{2 }θ + cos θ = 0⇒ 2cos

^{2}θ + cos θ – 1 =0⇒ (2cos θ – 1) (cos θ +1) = 0

⇒ cos θ = ½ or –1

θ = π /3 OR π (not possible).

**Question 18:** Let A_{1}, A_{2}, A_{3} be regions in the xy-plane defined by

A_{1 }= {(x, y) : x^{2} + 2y^{2} ≤1},

A_{2 }= {(x, y) : |x|^{3} + 2√2 |y|^{3} ≤ 1},

A_{3} = {(x, y) : max (|x|, √2 |y|) ≤1. Then

- a. A
_{1 }⊃A_{2}⊃A_{3} - b. A
_{3 }⊃A_{1}⊃A_{2} - c. A
_{2}⊃A_{3 }⊃A_{1} - d. A
_{3}⊃A_{2}⊃A_{1}

Solution:

**Answer: (D)**$A_{1}=4\int_{0}^{1}\frac{{\sqrt{1-x^{2}}}}{\sqrt{2}}dx$ {ellipse}=

$\frac{4}{\sqrt{2}}\left \{ (\frac{x^{2}}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin ^{-1}x)_{0}^{1} \right \}$ = (4/√2)(0+π/4 -0)

= π/√2 ..(1)

Now for x(0,1)

x

^{2}> x^{3}1-x

^{2}<1-x^{3}$\sqrt[3]{1-x^{3}}> \sqrt{1-x^{2}}$ Therefore A

_{2}=$4\int_{0}^{1}\frac{\sqrt[3]{1-x^{3}}}{\sqrt{2}}dx> 4\int_{0}^{1}\frac{\sqrt{1-x^{2}}}{\sqrt{2}}dx$ …(2)= 2√2 ……..(3)

So from (1), (2) and (3)

A

_{3}⊃A_{2}⊃A_{1}

**Question 19:** Let ABCD be a square and E be a point outside ABCD such that E, A, C are collinear in that order. Suppose EB = ED = √130 and the areas of triangle EAB and square ABCD are equal. Then the area of square ABCD is

- a. 8
- b. 10
- c. √120
- d. √125

Solution:

**Answer: (B)**Let side of square be ‘a’

⇒ Diagonal BD = a ⇒ OB = BD/2 = a/2

Now, In Δ OBE

BE

^{2}= OE^{2}+OB^{2}(using Pythagoras theorem)⇒ OE = √130-a

^{2}/2Now, ar (Δ EAB) = ar (Δ EOB) – ar (ΔAOB)

= (½)(a/√2) √(130-a

^{2}/2)-(1/2)(a/√2)^{2}= (a/4) √(130-a

^{2}/2)-a^{2}/4ar (square ABCD) = a

^{2}Given, ar (Δ EAB) = ar (square ABCD)

⇒ (a

^{2}/4)√(260-a^{2})-a^{2}/4 = a^{2}a

^{2}(260 – a^{2}) = 25a^{4}260 – a

^{2}= 25a^{2}a

^{2 }= 10 (area of square).

**Question 20:** Consider the set A = {1, 2, 3, ..., 30}. The number of ways in which one can choose three distinct numbers from A so that the product of the chosen numbers is divisible by 9 is

- a. 1590
- b. 1505
- c. 1110
- d. 1025

Solution:

**Answer: (A)**A = {1, 2, 3 …..….30}

Let X, Y , Z be subsets of A :

X = {9, 18, 27}

Y = {3, 6, 12, 15, 21, 24, 30}

Z = {1, 2, 4, 5, …….}

n(X) = 3, n(Y) = 7, n(Z) = 20

Now, the product of three selected numbers is divisible by 9, if

1) One from X, two from remaining

2) Two from X, one from remaining

3) All three from X

4) Two from Y, one from Z

5) All three from Y

So required no. of cases

=

^{3}C_{1}×^{27}C_{2 }+_{3C2 ×27C1 +3C3 + 7C2×20C1 +7C3 }= 1590

KVPY-SA 2017 Maths Paper with Solutions