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**Question 1: **Let BC be a fixed-line segment in the plane. The locus of a point A such that the triangle ABC is isosceles, is (with finitely many possible exceptional points)

- a. A line
- b. A circle
- c. The union of a circle and a line
- d. The union of two circles and a line

Solution:

**Answer: (d)**Let the coordinates of fixed base BC be B (0, 0), C (a, 0) and the variable point A (h, k).

Since △ABC is isosceles, three cases arises:

Case-I: AB = AC

√h

^{2}+ k^{2}= √(h - a)^{2}+ k^{2}h

^{2}+ k^{2}= h^{2}+ a^{2}– 2ah + k^{2}h = a / 2

x = a / 2

Locus of A is straight line,

Case-II: AB = BC

√h

^{2}+ k^{2}= ah

^{2}+ k^{2}= a^{2}x

^{2}+ y^{2}= a^{2}Circle

Case-III: AB = BC

√(h - a)

^{2}+ k^{2}= a(h – a)

^{2}+ k^{2}= a^{2}(x – a)

^{2}+ k^{2}= a^{2}Circle

So, the locus of A is union of two circles and a line.

**Question 2: **The number of solution pairs (x, y) of the simultaneous equations log_{1/3}(x + y) + log_{3}(x – y) = 2 and 2y^{2 }= 512^{x+1 }is

- a. 0
- b. 1
- c. 2
- d. 3

Solution:

**Answer: (b)**Given, log

_{1/3}(x + y) + log_{3}(x – y) = 2log

_{3^(-1)}(x + y) + log_{3 }(x – y) = 2–log

_{3 }(x + y) + log_{3 }(x – y) = 2 {log_{b^n}a = (1 / n) log_{b}a}log

_{3}[(x - y) / (x + y)] = 2 {log m – log n = log m / n}Taking antilog

(x - y) / (x + y) = 9

x – y = 9x + 9y

8x + 10y = 0

4x + 5y = 0 …(i)

Also, 2

^{y^2}= 512^{x+1}

----- (ii)From eq.(i) & (ii)

y

^{2}= 9 (-5y / 4) + 94y

^{2}+ 45y – 36 = 04y

^{2}+ 48y – 3y – 36 = 0(4y – 3) (y + 12) = 0

y = 3 / 4, –12

Using (i)

x = – 15 / 16 , 15

We have (–15 / 16, 3 / 4) and (15, –12)

Since, (–15 / 16, 3 / 4) do not satisfy the original equation, rejected.

So the only answer is (15, –12).

**Question 3: **The value of the limit lim_{x→∞} (√(4x^{2} - x) + 2x) is

- a. –∞
- b. − 1 / 4
- c. 0
- d. 1 / 4

Solution:

**Answer: (b)**lim

_{x→∞}(√(4x^{2}- x) + 2x)Put x = –y; x → –∞, y → +∞

Therefore lim

_{y→∞}(√(4y^{2}+ y) – 2y)Rationalizing

lim

_{y→∞}(y / √(4y^{2}+ y) + 2y)Dividing numerator and denominator by y

lim

_{y→∞}1 / [√(4 + [1 / y])] + 2= 1 / [√(4 + 0) + 2] {1 / ∞ → 0}

= 1 / 4

**Question 4: **Let R be a relation on the set of all-natural numbers given by a R b ⇔ a divides b^{2}. Which of the following properties does R satisfy?

I. Reflexivity

II. Symmetry

III. Transitivity

- a. I only
- b. III only
- c. I and III only
- d. I and II only

Solution:

**Answer: (a)**R: N → N, aRb a R a ⇒ a divides b

^{2}Reflexivity: a R a ⇒ a divides a

^{2}True, for every natural number.

‘R’ is reflexive.

Symmetry: a R b ⇒ a divides b

^{2}⇏ b divides a

^{2}For example: 1 R 2 i.e. 1 divides 2

^{2}, but 2~~R~~1 i.e. 2 does not divides 1^{2}.‘R’ is not symmetric.

Transitivity:

Again by taking example,

8 R 4 i.e. 8 divides 4

^{2 }= 164 R 2 i.e. 4 divides 2

^{2 }= 4But 8

~~R~~2 i.e. 8 do not divide 2^{2 }= 4R is not transitive.

**Question 5: **The fractional part of a real number x is x – [x], where [x] is the greatest integer less than or equal to x. Let F_{1 }and F_{2} be the fractional parts of (44 – √2017)^{2017} and (44 + √2017)^{2017} respectively. Then F_{1 }+ F_{2} lies between the numbers

- a. 0 and 0.45
- b. 0.45 and 0.9
- c. 0.9 and 1.35
- d. 1.35 and 1.8

Solution:

**Answer: (c)**Let x = (44 – √2017)

^{2017}x = – (√2017 – 44)

^{2017}≅ – (0.911)^{2017}–x = (√2017 – 44)

^{2017}{Pure fractional number}{–x} = {(√2017 – 44)

^{2017}}1 – {x} = (√2017 – 44)

^{2017}1 – F

_{1}= (√2017 – 44)^{2017}…(1)Again, let y = (√2017 + 44)

^{2017}= I + F_{2}…(2)Where [y] = I & {y} = F

_{2}Now subtracting eq.(2) – eq.(1)

I + F

_{2}– 1 + F_{1 }= (√2017 + 44)^{2017}– (√2017 – 44)^{2017}= 2{

^{2017}C_{1}(√2017)^{2016}(44)^{1}+ ……}= 2 (Integer)

F

_{1}+ F_{2}= 2I_{1}– I + 1 ∈ Integer= 1 (only possible value)

0.9 < F

_{1}+ F_{2 }< 1.35

**Question 6: **The number of real solutions of the equation 2sin 3x + sin 7x – 3 = 0 which lie in the interval [–2𝛑, 2𝛑] is

- a. 1
- b. 2
- c. 3
- d. 4

Solution:

**Answer: (b)**2 sin3x + sin 7x – 3 = 0 ………(1)

Since maximum value of sin θ is 1, therefore only possible solution of (1) is,

sin3x = 1 and sin7x = 1

3x = (4n + 1) 𝛑 / 2, n ∈ I and 7x = (4m + 1) 𝛑 / 2, m ∈ I

x = (4n + 1) 𝛑 / 6, n ∈ I

x = (4m + 1) 𝛑 / 14, m ∈ I

x = …., – 7𝛑 / 6, – 𝛑 / 2, 𝛑 / 6, 5𝛑 / 6, 3𝛑 / 2 …..

x = …, – 3𝛑 / 14, – 𝛑 / 2, 𝛑 / 14, 5𝛑 / 14, 13𝛑 / 14, 3𝛑 / 2 …

Therefore, two solutions, x = – 𝛑 / 2 and 3𝛑 / 2.

**Question 7: **Suppose p, q, r are real numbers such that q = p (4 – p), r = q (4 – q), p = r (4 – r). The maximum possible value of p + q + r is

- a. 0
- b. 3
- c. 9
- d. 27

Solution:

**Answer: (c)**Given, q = p (4 – p) ⇒ q / p = 4 – p ..……(i)

r = q (4 – q) ⇒ r / q = 4 – q ……. (ii)

p = r (4 – r) ⇒ p / r = 4 – r ……..(iii)

Multiplying (i),(ii) and (iii)

(q / p) . (r / q) . (p / r) = (4 – p) (4 – q) (4 – r) = 1 ……..(iv)

Let x = 4 – p = 4 – 4r + r

^{2}= (r – 2)^{2}≥ 0y = 4 – q = 4 – 4p + p

^{2}= (r – 2)^{2}≥ 0z = 4 – r = 4 – 4q + q

^{2}= (q – 2)^{2 }≥ 0Now,

Using A.M. ≥ G.M.

(x + y + z) / 3 ≥ (x . y . z)

^{⅓}[4 – p + 4 – q + 4 – r] / 3 ≥ {(4 – p) (4 – q) (4 – r)}

^{⅓}4 – p + 4 – q + 4 – r ≥ 3 {Using eq.(iv)}

p + q + r ≤ 9 (Equality holds for p = q = r = 3)

**Question 8: **The parabola y^{2 }= 4x + 1 divides the disc x^{2 }+ y^{2} ≤ 1 into two regions with areas A_{1 }and A_{2}. Then |A_{1 }– A_{2}| equals

- a. 1 / 3
- b. 2 / 3
- c. 𝛑 / 4
- d. 𝛑 / 3

Solution:

**Answer: (b)**C

_{1}: y^{2}= 4x + 1; C_{2}: x^{2}+ y^{2}= 1A

_{1}= 2 ∫_{-1/4}^{0}√4x + 1 dx + 2 ∫_{0}^{1}√1 - x^{2}dx= 2 [(1 / 6) – 0] + 2 [0 + (1 / 2) * (𝛑 / 2) – 0]

= 1 / 3 + π / 2

A

_{2}= π(1)^{2}– A_{1}Now,

|A

_{1}- A_{2}| = |2A_{1}- π| = | 2 * [(1 / 3) + (π / 2)] - π | = 2 / 3

**Question 9: **A shooter can hit a given target with probability 1 / 4. She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval

- a. (0.5272, 0.5274)
- b. (0.2636, 0.2638)
- c. (0.1317, 0.1319)
- d. (0.0658, 0.0660)

Solution:

**Answer: (d)**Since she hits the target successfully three times in exactly six attempts.

So 3

^{rd}hit must be the sixth time and in the first five attempts, there will be two hits. Now, Required Probability == 10 × [27 / 4

^{6}] = 270 / 4096 = 0.06591

**Question 10: **Consider the following events:

E_{1}: Six fair dice are rolled and at least one dice shows six.

E_{2}: Twelve fair dice are rolled and at least two dice show six.

Let p_{1} be the probability of E_{1 }and p_{2} be the probability of E_{2}.

Which of the following is true?

- a. p
_{1}> p_{2} - b. p
_{1}= p_{2}= 0.6651 - c. p
_{1}< p_{2} - d. p
_{1}= p_{2}= 0.3349

Solution:

**Answer: (a)**E

_{1}: Six fair dice are rolled and at least one dice shows six.E

_{2}: Twelve fair dice are rolled and at least two dice show six.P(E

_{1}) = p_{1}= 1 – {no dice shows six}= 1 – (5 / 6)

^{6}(Using Binomial Theorem)= 0.6651

P(E

_{2}) = p_{2}= 1 – {no dice shows six + one dice shows six}= 1 – {(5 / 6)

^{12}+^{12}C_{1}(1 / 6) (5 / 6)^{11}} (Using Binomial Theorem)= 0.61866

p

_{1}> p_{2}

**Question 11: **For how many different values of a does the following system have at least two distinct solutions? ax + y = 0

x + (a + 10) y = 0

- a. 0
- b. 1
- c. 2
- d. Infinitely many

Solution:

**Answer: (c)**ax + y = 0; x + (a + 10) y = 0

The given system is homogeneous system of linear equations in two variables Therefore, it will either have unique solution i.e. x = y = 0 or infinite number of solutions (when lines are coincident).

Atleast two distinct solutions → Infinite solutions

a / 1 = 1 / [a + 10] (Co-incident lines)

a

^{2}+ 10a – 1 = 0 (Discriminant > 0)Therefore, two values of ‘a’ are possible.

**Question 12: **Let R be the set of real numbers and f : R →R be defined by f (x) = {x} / [1 + [x]^{2}], where [x] is the greatest integer less than or equal to x, and {x} = x – [x]. Which of the following statements are true?

I. The range of f is a closed interval

II. f is continuous on R.

III. f is one-one on R.

- a. I only
- b. II only
- c. III only
- d. None of I, II and III

Solution:

**Answer: (d)**f (x) = {x} / [1 + [x]

^{2}]Since, 0 ≤ {x} < 1 & 1 ≤ 1 + [x]

^{2 }< ∞0 < {x} / [1 + [x]

^{2}] < 1Range of f (x) = (0, 1) (I is FALSE)

Also, f (x) is discontinuous ∀ x ∈ Z. (II is TRUE)

f(x) = 0 ∀ x ∈ Z

Not one-one (III is FALSE)

**Question 13: **Let x_{n }= (2^{n} + 3^{n})^{1/2n }for all natural numbers n. Then

- a. lim
_{x→∞}x_{n}= ∞ - b. lim
_{x→∞}x_{n}= √3 - c. lim
_{x→∞}x_{n}= √3 + √2 - d. lim
_{x→∞}x_{n}= √5

Solution:

**Answer: (b)**

**Question 14: **One of the solutions of equation 8 sin^{3 }θ – 7 sinθ + √3 cos θ = 0 lies in the interval

- a. (0°, 10°]
- b. (10°, 20°]
- c. (20°, 30°]
- d. (30°, 40°]

Solution:

**Answer: (b)**Given, 8 sin

^{3 }θ – 7 sin θ + √3 cos θ = 02 . 4 sin

^{3}θ – 7 sin θ + √3 cos θ = 02 (3 sinθ – sin 3θ) – 7 sin θ + √3 cos θ = 0

{sin 3θ = 3 sin θ – 4 sin

^{3}θ ⇒ 4 sin^{3}θ = 3 sin θ – sin 3θ}√3 cos θ – sin θ – 2 sin 3θ = 0

[√3 / ]2 cos θ – [1 / 2] sin θ = sin 3θ

sin (π / 3) cos θ – cos (π / 3) sin θ = sin 3θ (sin (A – B) = sin A cos B – cos A sin B)

sin [(π / 3) - θ] = 3θ

(π / 3) - θ = nπ + (–1)

^{n }3θfor n = 0

(π / 3) – θ = 3θ

θ = π / 12 = 15° ∈ (10°, 20°]

**Question 15: **Let a, b, c, d, e, be real numbers such that a + b < c + d, b + c < d + e, c + d < e + a, d + e < a + b. Then

- a. The largest is a and the smallest is b
- b. The largest is a and the smallest is c
- c. The largest is c and the smallest is e
- d. The largest is c and the smallest is b

Solution:

**Answer: (a)**a + b < c + d …(i)

b + c < d + e …(ii)

c + d < e + a …(iii)

d + e < a + b …(iv)

From (i) & (iii)

a + b < e + a (x < y & y < z ⇒ x < z)

b < e

From (i) & (iv)

d + e < c + d (x < y & y < z ⇒ x < z)

e < c

From (ii) & (iv)

b + c < a + b (x < y & y < z ⇒ x < z)

c < a

b < e < c < a … (1)

Again from (i)

a + b < c + d

(a – c) + b < d

Positive real number + b < d {as a > c}

d > b … (2)

From (iii)

c + d < e + a

(c – e) + d < a

Positive real number + d < a {as c > e}

d < a … (3)

From (1), (2) & (3), we can say that a is largest and b is smallest.

**Question 16: **If a fair coin is tossed 5 times, the probability that heads does not occur two or more times in a row is:

- a. 12 / 2
^{5} - b. 13 / 2
^{5} - c. 14 / 2
^{5} - d. 15 / 2
^{5}

Solution:

**Answer: (b)**The following are the cases for no two heads to come together

Case - I: There is no Head

{TTTTT}

Probability = (1 / 2)

^{5}Case - II: Exactly one Head

{HTTTT, THTTT, TTHTT, TTTHT, TTTTH}

Probability = 5 * (1 / 2)

^{4}(1 / 2) = 5 / 32Case - III: Exactly Two Heads

{HTHTT, HTTHT, HTTTH, THTTH, THTHT, TTHTH}

Probability = 6 × (1 / 2)

^{3}(1 / 2)^{2}= 6 / 32Case - IV: Exactly Three Heads

{HTHTH}

Probability = 1 × (1 / 2)

^{2}* (1 / 2)^{3}= 1 / 32Total probability = 13 / 2

^{5}

**Question 17: **Consider the following parametric equation of a curve:

x (θ) = | cos 4 θ | cos θ

y (θ) = | cos 4 θ | sin θ

for 0 ≤ θ ≤ 2π

Which one of the following graphs represents the curve?

Solution:

**Answer: (a)**x (θ) = | cos 4 θ | cos θ

y (θ) = | cos 4 θ | sin θ

where, 0 ≤ θ ≤ 2π

For θ = 0°

x = 1 and y = 0

(1, 0) lies on the curve

Option (c) and (d) are rejected.

For θ = π / 4

x = y = 1 / √2

(1 / √2, 1 / √2) lies on the curve.

Option (b) is rejected.

So, the correct option is (a).

**Question 18: **Let A = (a_{1}, a_{2}) and B = (b_{1}, b_{2}) be two points in the plane with integer coordinates. Which one of the following is not a possible value of the distance between A and B?

- a. √65
- b. √74
- c. √83
- d. √97

Solution:

**Answer: (c)**Distance between A (a

_{1}, b_{1}) and B (a_{2}, b_{2})AB = √(a

_{1}- b_{1})^{2}+ (a_{2}- b_{2})^{2}{Using Distance Formula}Given a

_{1}, a_{2}, b_{1}, b_{2}∈ I ⇒ (a_{1}– b_{1}), (a_{2}– b_{2}) ∈ INow, √65 = √8

^{2}+ 1^{2}→ Possible√74 = √7

^{2}+ 5^{2}→ Possible√97 = √9

^{2}+ 4^{2}→ Possible√83 → Not Possible

**Question 19: **Let f (x) = max {3, x^{2}, (1 / x^{2})} for (1 / 2) ≤ x ≤ 2. Then the value of the integral ∫_{1/2}^{2} f(x)dx is:

- a. 11 / 3
- b. 13 / 3
- c. 14 / 3
- d. 16 / 3

Solution:

**Answer: (c)**f (x) = max {3, x

^{2}, (1 / x^{2})} for (1 / 2) ≤ x ≤ 2

**Question 20: **Let a_{i} = I + (1 / i) for I = 1, 2,………., 20. Put p = (1 / 20) (a_{1} + a_{2} + …….. a_{20}) and q = (1 / 20) [(1 / a_{1}) + (1 / a_{2}) + …… (1 / a_{20}). Then

Solution:

**Answer: (a)**Clearly, q > 0

We will try to contra prove that q < [22 – p] / 21 OR q + (p / 21) < (22 / 21)

**Question 21: **Let x, y, z be positive integers such that HCF (x, y, z) = 1 and x^{2 }+ y^{2 }= 2z^{2}. Which of the following statements are true?

I. 4 divides x or 4 divides y.

II. 3 divides x + y or 3 divides x – y.

III. 5 divides z (x^{2 }– y^{2})

- a. I and II only
- b. II and III only
- c. II only
- d. III only

Solution:

**Answer: (b)**Given x, y, z ∈ Z

^{+}HCF (x, y, z) = 1 ⇒ x, y and z are co-prime numbers

x

^{2}+ y^{2}= 2z^{2 }= z^{2}+ z^{2}⇒ Sum of the squares of 2 co-prime numbers.Thus, x = 7, y = 1, z = 5 or x = 1, y = 7, z = 5 are possible combinations.

i.e. 2z

^{2}= 2 . 5^{2}= 50 and x^{2}+ y^{2}= 7^{2}+ 1^{2}= 49 + 1 = 50Now, x – y is divisible by 3 and z (x

^{2 }– y^{2}) is divisible by 5.Statement II and III are true.

**Question 22: **How many different (mutually non-congruent) trapeziums can be constructed using four distinct side lengths from the set {1, 3, 4, 5, 6}?

- a. 5
- b. 11
- c. 15
- d. 30

Solution:

**Answer: (b)**Here, the sides a, b, c, d ∈ {1, 3, 4, 5, 6}

Condition of trapezium,

|a – c| < |b – d| < (a + c)

Cases:

(a, c)

(b, d)

(1, 3)

(4, 5), (5, 6)

= 2

(1, 4)

Not possible

(1, 5)

Not possible

(1, 6)

Not possible

(3, 4)

(1, 5), (1, 6)

= 2

(3, 5)

(1, 4), (1, 6)

= 2

(3, 6)

(1, 5)

= 1

(4, 5)

(1, 3) , (1, 6), (3, 6)

Repeated

= 2

(4, 6)

(1, 5)

= 1

(5, 6)

(1, 3) , (1, 4)

Repeated

= 1

Total number of trapezium = 11

**Question 23: **A solid hemisphere is mounted on a solid cylinder, both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume, then the ratio of the height of the cylinder to the common radius is

- a. 1 : 1
- b. 1 : 2
- c. 2 : 1
- d. √2 : 1

Solution:

**Answer: (a)**Let Radius of hemisphere and cylinder is R height of cylinder be h.

The whole surface area, S = 2πR

^{2}+ 2πRh + πR^{2}S = 2πRh + 3πR

^{2}…(1)S – 3πR

^{2}= 2πRhh = S – 3πR

^{2}/ 2πR …(2)Now, V = (2 / 3) πR

^{3}+ πR^{2}hV = (2 / 3) πR

^{3}+ πR^{2}[S – 3πR^{2}/ 2πR] {from eq.(2)}Differentiate with respect to R

dV / dR = 2πR

^{2}+ (S / 2) - (9 / 2) πR^{2}For maximum and minimum dV / dR = 0

0 = 2πR

^{2}+ (S / 2) - (9 / 2) πR^{2}0 = –5πR

^{2}+ S5πR

^{2}= S5πR

^{2}= 3πR^{2}+ 2πRh {from eq.(1)}5R = 3R + 2h

2R = 2h

h : R = 1 : 1

**Question 24: **Let ABC be an acute scalene triangle, and O and H be its circumcentre and orthocenter respectively. Further, let N be the midpoint of OH. The value of the vector sum NA + NB + NC is:

Solution:

**Answer: (c)**Let circumcenter O of triangle ABC be at the origin.

Position vector of Centroid, G = [a + b + c] / 3

Now, we know that centroid(G) divides the distance from the orthocenter (H) to the circumcenter(O) in the ratio 2:1.

**Question 25: **The quotient when 1 + x^{2} + x^{4} + x^{6} + ………..x^{34} is divided by 1 + x + x^{2} + x^{3} +......+ x^{17} is

- a. x
^{17}– x^{15}+ x^{13}– x^{11}...... + x - b. x
^{17}+ x^{15 }+ x^{13 }+ x^{11}...... + x - c. x
^{17}+ x^{16 }+ x^{15 }+ x^{14}...... + x - d. x
^{17}– x^{16}+ x^{15}– x^{14}...... – x

Solution:

**Answer: (d)**1 + x

^{2}+ x^{4}+ x^{6}+ ………..x^{34}(Forms a G.P with common ratio x^{2})

**Question 26: **Let R be the region of the disc x^{2} + y^{2} ≤ 1 in the first quadrant. Then the area of the largest possible circle contained in R is

- a. π (3 − 2√2)
- b. π (4 − 3√2)
- c. π / 6
- d. π (2√2 - 2)

Solution:

**Answer: (a)**Let the radius of inner circle = r

This inner circle touches both the axes,

Centre of circle = (r, r)

Now, OP = OC + CP

1 = √(r

^{2}+ r^{2}) + r {Using Distance Formula}1 = √2r + r

r = 1 / √(2) + 1

r = [1 / √(2) + 1] * [√(2) - 1 / √(2) - 1]

r = √(2) - 1 / 1 = √(2) - 1

Area of Circle = πr

^{2}= π (√(2) - 1)

^{2}= π (2 + 1 - 2√2)

= π (3 - 2√2)

**Question 27: **Let R be the set of real numbers and f : R → R be given by f (x) = √|x| - log (1 + |x|). We now make the following assertions:

I. There exists a real number A such that f (x) ≤ A for all x.

II. There exists a real number B such that f (x) ≥ B for all x.

- a. I is true and II is false
- b. I is false and II is true
- c. I and II both are true
- d. I and II both are false

Solution:

**Answer: (b)**f (x) = √|x| - log (1 + |x|)

For x > 0

f (x) = x – log (1 + x)

On differentiating both sides with respect to x

f (x) is increasing when x > 0 …(i)

Also, f (0) = √|0| – log (1 + |0|) = 0 ...(ii) &

f (–x) = √|–x| – log (1 + |–x|) = f (x)

f(x) is Even Function ...(iii)

Therefore, using (i), (ii) and (iii) graph of f (x) can be drawn as

Clearly, f (x) is continuous everywhere but non-differentiable at x = 0.

**Question 28: **Define g(x) = ∫_{-3}^{3} f (x - y) f (y) dy, for all real x, where

Then

- a. g (x) is not continuous everywhere
- b. g (x) is continuous everywhere but differentiable nowhere
- c. g (x) is continuous everywhere and differentiable everywhere except at x = 0, 1
- d. g (x) is continuous everywhere and differentiable everywhere except at x = 0, 1, 2

Solution:

**Answer: (d)**Given, ∫

_{-3}^{3}f (x - y) f (y) dy and f (y) =g (x) = ∫

_{0}^{1}f (x – y) dyNow, put x – y = t

dy = –dt

Lower limit, y = 0 ; t = x

Upper limit, y = 1 ; t = x – 1

g (x) = – ∫

_{x}^{x-1}f (t) dt {Using property: – ∫_{a}^{b }f (x) dx = ∫_{b}^{a}f (x) dx}g (x) = ∫

_{x-1}^{x}f (t) dtg (x) is continuous everywhere, but non-diff. at x = 0, 1, 2 (see figure)

**Question 29: **The integer part of the number

is

- a. 50
- b. 52
- c. 57
- d. 59

Solution:

**Answer: (c)**

**Question 30: **The number of continuous functions f : [0, 1] → R that satisfy

- a. 0
- b. 1
- c. 2
- d. infinity

Solution:

**Answer: (b)**Therefore, the number of such continuous functions is 1.

KVPY SX 2017 Maths Paper with Solutions