SECTION – 1 (Maximum Marks = 18)
- This section contains SIX (06) questions.
- Each question has FOUR options. ONLY ONE of these four options is the correct answer.
- For each question, choose the option corresponding to the correct answer.
- Answer to each question will be evaluated according to the following marking scheme :
- Full Marks: +3 If ONLY the correct option is chosen;
- Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
- Negative Marks: –1 In all other cases.
Q.1 Suppose a, b denote the distinct real roots of the quadratic polynomial x² + 20x – 2020 and suppose c, d denote the distinct complex roots of the quadratic polynomial x² – 20x + 2020. Then the value of ac(a – c) + ad(a – d) + bc(b – c) + bd(b – d) is
(A) 0
(B) 8000
(C) 8080
(D) 16000
Solution:
x² + 20x – 2020 = 0 has two real roots a, b
x² – 20x + 2020 = 0 has two complex roots c, d
now ac(a – c) + ad(a – d) + bc(b – c) + bd(b – d)
⇒ a²c – ac² + a²d – ad² + b²c – bc² + b²d – bd²
⇒ a²(c + d) + b²(c + d) – c²(a + b) – d²(a + b)
⇒ (c + d)(a² + b²) – (a + b)(c² + d²)
⇒ (c + d) ((a + b)² – 2ab) – (a + b) ((c + d)² – 2cd)
⇒ 20 [(-20)² + 4040] + 20 [(20)² – 4040]
⇒ 20 [(20)² + 4040 + (20)² – 4040]
⇒ 20 × 800 ⇒ 16000
⇒ Option (D) is correct
Q.2 If the function f : R → R is defined by f(x) = |x|(x – sinx), then which of the following statements is TRUE ?
(A) f is one-one, but NOT onto
(B) f is onto, but NOT one-one
(C) f is BOTH one-one and onto
(D) f is NEITHER one-one NOR onto
Solution:
f(x) is a non-periodic, continuous and odd function
f(x) = -x² + xsin x, x < 0 and f(x) = x² - xsin x, x ≥ 0
⇒ f(-∞) → -∞ and f(∞) → ∞ ⇒ Range of f(x) = R ⇒ f(x) is an onto function
f'(x) = -2x + sin(x) + xcos(x), x < 0 and f'(x) = 2x - sin(x) - xcos(x), x ≥ 0
⇒ f'(x) ≥ 0, ∀ x ∈ R {equality holds at x = 0}
⇒ f is one-one ⇒ f is BOTH one-one and onto
⇒ Option (C) is correct
Q.3 Let the functions f : R → R and g : R → R be defined by f(x) = e⁽ˣ⁻¹⁾ – e⁻|ˣ⁻¹| and g(x) = (e⁽ˣ⁻¹⁾ + e⁽¹⁻ˣ⁾)/2. Then the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
(A) (2 – √3) + (e – e⁻¹)/2
(B) (2 + √3) + (e – e⁻¹)/2
(C) (2 – √3) + (e + e⁻¹)/2
(D) (2 + √3) + (e + e⁻¹)/2
Solution:
Here f(x) = 0, x ≤ 1 and f(x) = e⁽ˣ⁻¹⁾ + e⁽¹⁻ˣ⁾, x > 1
solving f(x) and g(x) we have x = 1 + ln√3
So the required area is [e – 1/e]/2 + [(-√3/2 – √3/2) + 2] = (2 – √3) + (e – e⁻¹)/2
⇒ Option (A) is correct
Q.4 Let a, b and c be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y² = 4cx and suppose the ellipse x²/a² + y²/b² = 1 passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
(A) 1/√2
(B) 1/2
(C) 1/3
(D) 2/5
Solution:
Let coordinates of the point P is (c, 2c)
⇒ Slope of the tangent to the parabola at point P is 1
⇒ Slope of the tangent to the ellipse at point P is -1
⇒ b² = 2a² ⇒ eccentricity = 1/√2
⇒ Option (A) is correct
Q.5 Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are 2/3 and 1/3 respectively. Suppose ‘a’ is the number of heads that appear when C1 is tossed twice, independently, and suppose b is the number of heads that appear when C2 is tossed twice,
independently, Then probability that the roots of the quadratic polynomial x² – ax + b are real and
equal, is
(A) 40/81
(B) 20/81
(C) 1/2
(D) 1/4
Solution:
For coin C1: (No. of Heads a, Probability) ≡ (0, 1/9), (1, 4/9), (2, 4/9)
For coin C2: (No. of Heads b, Probability) ≡ (0, 4/9), (1, 4/9), (2, 1/9)
Since the roots of the quadratic polynomial x² – ax + b are real and equal ⇒ Discriminant of the quadratic polynomial x² – ax + b will be zero
⇒ a² = 4b ⇒ (a, b) ≡ (0, 0) and (2, 1)
⇒ Then the required probability is = (1/9)(4/9) + (4/9)(4/9) ⇒ (4/81) + (16/81) ⇒ 20/81
⇒ Option (B) is correct
Q.6 Consider all rectangles lying in the region {(x, y) ∈ R x R : x ∈ [0, π/2] and y ∈ [0, 2Sin(2x)]} and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is
(A) 3π/2
(B) π
(C) π/(2√3)
(D) (π√3)/2
Solution:
Let the length of the rectangle is ‘2a’ ⇒ width of the rectangle is 2Sin(2((π/4) + a)) ⇒ 2Cos(2a)
⇒ Perimeter = 2(2a + 2Cos(2a)) ⇒ P = 4(a + Cos(2a))
⇒ dP/da = 4(1 – 2Sin(2a)) ⇒ for max/min dP/da = 0 ⇒ 4(1 – 2Sin(2a)) = 0 ⇒ Sin(2a) = 1/2
⇒ 2a = π/6, 5π/6 ; d²P/da² = -4Cos(2a)
For maxima a = π/12 ⇒ Area = (2a)(2Cos(2a)) = (π/6)(2(√3/2)) = π/(2√3)
⇒ Option (C) is correct
SECTION-2 – (Maximum Marks = 24)
- This section contains SIX (06) questions.
- Each question has FOUR options. ONE OR MORE THAN ONE of these four options is(are) correct answer(s).
- For each question, choose the option(s) corresponding to (all) the correct answer(s).
- Answer to each question will be evaluated according to the following marking scheme :
- Full Marks : +4 If only (all) the correct option(s) is(are) chosen;
- Partial Marks: +3 If all the four options are correct but ONLY three options are chosen;
- Partial Marks: +2 If three or more options are correct but ONLY two options are chosen, both of
- which are correct;
- Partial Marks: +1 If two or more options are correct but ONLY one option is chosen and it is a
- correct option;
- Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
- Negative Marks: –2 In all other cases
Q.1 Let the function f : R → R be defined by f(x) = x³ – x² + (x – 1)Sin(x) and let g : R → R be an arbitrary function. Let fg : R → R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following statements is/are TRUE ?
(A) If g is continuous at x = 1, then fg is differentiable at x = 1
(B) If fg is differentiable at x = 1, then g is continuous at x = 1
(C) If g is differentiable at x = 1, then fg is differentiable at x = 1
(D) If fg is differentiable at x = 1, then g is differentiable at x = 1
Solution:
Since f : R → R ; f(x) = x³ – x² + (x – 1)Sin(x) ⇒ f(1⁺) = f(1⁻) = f(1) = 0
And fg : R → R such that (fg)(x) = f(x)g(x) ; Let F(x) = f(x)g(x)
Now Taking Option (A) g is continuous at x = 1 ⇒ g(1⁺) = g(1⁻) = g(1)
(fg)(x) = f(x)g(x) = F(x) will be differentiable at x = 1 if F'(1⁺) = F'(1⁻)
Since F'(1⁺) = F'(1⁻) = f'(1)g(1) ⇒ fg is differentiable at x = 1
Now Taking Option (B) and (D) fg is differentiable at x = 1
Since F'(1⁺) = f'(1)g(1⁺) and F'(1⁻) = f'(1)g(1⁻) ⇒ g(1⁺) = g(1⁻)
So it cannot be concluded the continuity and differentiability of the function ‘g’
Now Taking Option (C) g is differentiable at x = 1 ⇒ g'(x) exists at x = 1
Since F(x) = f(x)g(x) ⇒ F'(x) = f'(x)g(x) + f(x)g'(x) ⇒ F'(1) = f'(1)g(1) + f(1)g'(1)
⇒ F'(1) = f'(1)g(1) {f(1) = 0 as g'(1) exists}
⇒ If g is differentiable at x = 1, then fg is differentiable at x = 1
⇒ Option (A) and (C) are correct
Q.2 Let M be a 3 × 3 invertible matrix with real entries and let I denote the 3 × 3 identity matrix. If M⁻¹ = adj(adj M), then which of the following statement is/are ALWAYS TRUE ?
(A) M = I
(B) det M = 1
(C) M² = I
(D) (adj M)² = I
Solution:
Since M be an invertible matrix ⇒ M ≠ 0
Given that M⁻¹ = adj(adj M) ⇒ M⁻¹ = (det(M)).M ⇒ M⁻¹.M = (det M).M² ⇒ I = (det M).M²
⇒ det(I) = det((det M).M²) ⇒ det(I) = (det(M))⁵ ⇒ det(M) = 1 {det(I) = 1}
SInce I = (det M).M² and det(M) = 1 ⇒ M² = I
Further (adj M)² = adj(M)² = adj(I) = I ⇒ (adj M)² = I
⇒ Option (B), (C), and (D) are correct
Q.3 Let S be the set of all complex numbers z satisfying |z² + z + 1| = 1. Then which of the following statements is/are correct
(A) |z + 1/2| ≤ 1/2 for all z ∈ S
(B) |z| ≤ 2 for all z ∈ S
(C) |z + 1/2| ≥ 1/2 for all z ∈ S
(D) The set S has exactly four elements
Solution:
Given that |z² + z + 1| = 1 ⇒ |(z + 1/2)² + 3/4| = 1 ⇒ |(z + 1/2)² + 3/4| ≤ |(z + 1/2)²| + 3/4
1 ≤ |(z + 1/2)²| + 3/4 ⇒ |z + 1/2| ≥ 1/2 for all z ∈ S
Also |z² + z + 1| = 1 ≥ ||z² + z| – 1| ⇒ |z² + z| – 1 ≤ 1 ⇒ |z² + z| ≤ 2
⇒ ||z²| – |z|| ≤ |z² + z| ≤ 2 ⇒ |z| ≤ 2 for all z ∈ S
Also the roots of the equation |z² + z + 1| = 1 exist infinitely
⇒ Option (B) and (C) are correct
Q.4 Let x, y and z be positive real numbers. Suppose x, y and z are lengths of the sides of a triangle opposite to its angles X, Y and Z, respectively. If Tan(X/2) + Tan(Z/2) = (2y)/(x + y + z), then which of the following statements is/are CORRECT?
(A) 2Y = X + Z
(B) Y = X + Z
(C) Tan(X/2) = (x)/(y + z)
(D) x² + z² – y² = xz
Solution:
Given that Tan(X/2) + Tan(Z/2) = (2y)/(x + y + z) ⇒ Δ/(s(s-x)) + Δ/(s(s-z)) = 2y/2s
⇒ (Δ/s)(2s-(x+z))/(s-x)(s-z) = y/s ⇒ (Δy)/(s(s-x)(s-z)) = y/s
⇒ Δ = (s-x)(s-z) ⇒ Δ² = (s-x)²(s-z)² ⇒ s(s-y) = (s-x)(s-z)
⇒ (x + y + z) (x + z – y) = (y + z – x) (x + y – z)
⇒ (x + z)² – y² = y² – (z – x)²
⇒ (x + z)² + (x – z)² = 2y²
⇒ x² + z² = y² ⇒ Angle Y is π/2 ⇒ Y = X + Z
Also Tan(X/2) = Δ/(s(s-x))
⇒ Tan(X/2) = (xz/2)/((y+z)² – x²)/4
⇒ Tan(X/2) = (xz/2)/((y+z)² – x²)/4
⇒ Tan(X/2) = 2xz/(y² + z² + 2yz- x²)
⇒ Tan(X/2) = 2xz/(2z² + 2yz) {x² + z² = y²}
⇒ Tan(X/2) = (x)/(y + z)
⇒ Option (B) and (C) are correct
Q.5 Let L1 and L2 be the following straight lines. L1 : (x – 1)/1 = -y/1 = (z – 1)/3 and L2 : -(x – 1)/3 = -y/1 = (z – 1)/1. Suppose the straight line L : (x – a)/l = (y – 1)/m = -(z – b)/2 lies in the plane containing L1 and L2, and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following
statements is/are TRUE?
(A) a – b = 3
(B) l + m = 2
(C) a – b = 1
(D) l + m = 0br>
Solution:
Point of intersection of L1 & L2 is (1, 0, 1)
⇒ Line L passes through (1, 0, 1)
⇒ (1 – a)/l = -1/m = -(1 – b)/2
⇒ Vector equation of the acute angle bisector of L1 & L2 is given by
r = i + k + t(i + j – 2k) {where t is any scaler quantity, i, j, k are standard unit vectors}
⇒ l/1 = m/1 = 2/2 ⇒ l = m = 1
Since (1 – a)/l = -1/m = -(1 – b)/2
⇒ (1 – a)/l = -1 ⇒ a = 2 AND
-(1 – b)/2 = -1 ⇒ b = -1
⇒ a – b = 3 and l + m = 2
⇒ Option (A) and (B) are correct
Q.6 Which of the following inequalities is/are TRUE?
(A) \int ^{1}_{0}xCosxdx\geq \dfrac{3}{8}
(B) \int ^{1}_{0}xSinxdx\geq \dfrac{3}{10}
(C) \int ^{1}_{0}x^{2}Cosxdx\geq \dfrac{1}{2}
(D) \int ^{1}_{0}x^{2}Sinxdx\geq \dfrac{2}{9}
Solution:
Since Cosx = 1 – (x²/2!) + …
and Sinx = x – (x³/3!) + …
Taking Option (A)
\begin{aligned}\int ^{1}_{0}x\cos xdx\geq \int ^{1}_{0}x\left( 1-\dfrac{x^{2}}{2}\right) dx\\ =\int ^{1}_{0}\left( x-\dfrac{x^{3}}{2}\right) dx=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\end{aligned}
Taking Option (B)
\begin{aligned}\int ^{1}_{0}x\sin xdx\geq \int ^{1}_{0}x\left( x-\dfrac{x^{3}}{6}\right) dx\\ =\int ^{1}_{0}\left( x^{2}-\dfrac{x^{4}}{6}\right) dx=\dfrac{1}{3}-\dfrac{1}{30}=\dfrac{9}{30}\\ =\dfrac{3}{10}\end{aligned}
Similarly \int ^{1}_{0}x^{2}Sinxdx\geq \dfrac{2}{9}
But Option C is not correct
⇒ Option (A), (B) and (D) are correct
SECTION – 3 : (Maximum Marks = 24)
- This section contains SIX (06) questions.
- The answer to each question is a NUMERICAL VALUE.
- For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.
- If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
- Answer to each question will be evaluated according to the following marking scheme :
- Full Marks: +4 If ONLY the correct numerical value is entered;
- Zero Marks: 0 In all other cases.
Q.1 Let m be the minimum possible value of log₃(3ʸ¹ + 3ʸ² + 3ʸ³), where y1, y2, y3 are real numbers for which y1 + y2 + y3 = 9. Let M be the maximum possible value of (log₃(x1) + log₃(x2) + log₃(x3)), where x1, x2 , x3 are positive real numbers for which x1 + x2 + x3 = 9. Then the value of log₂(m³) + log₃(M²) is…
Solution:
(3ʸ¹ + 3ʸ² + 3ʸ³)/3 ≥ ((3)ʸ¹⁺ʸ²⁺ʸ³)¹/³
⇒ 3ʸ¹ + 3ʸ² + 3ʸ³ ≥ 3⁴
⇒ log₃(3ʸ¹ + 3ʸ² + 3ʸ³) ≥ log₃(3⁴)
⇒ m ≥ 4
Also (x1 + x2 + x3)/3 ≥ (x1x2x3)¹/³
⇒ x1x2x3 ≤ 27
⇒ log₃(x1x2x3) ≤ log₃(27)
⇒ (log₃(x1) + log₃(x2) + log₃(x3)) ≤ 3
⇒ M ≤ 3
⇒ log₂(m³) + log₃(M²) = log₂(4³) + log₃(3²) = 6 + 2 = 8
⇒ Answer is 8.00
Q.2 Let a1, a2, a3, … be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1, b2, b3, … be a sequence of positive integers in geometric progression with common ratio 2. If a1 = b1 = c, then the number of all possible values of c, for which the equality 2(a1 + a2 + … + an ) = b1 + b2 + … + bn holds for some positive integer n, is…
Solution:
Given that 2(a1 + a2 + … + an ) = b1 + b2 + … + bn
⇒ 2(n/2){2c + (n – 1)(2)} = c(2ⁿ – 1)
⇒ n{2c + (n – 1)(2)} = c(2ⁿ – 1)
⇒ 2cn + 2n² – 2n = c(2ⁿ – 1)
⇒ 2n² – 2n = c(2ⁿ – 1 – 2n)
Since c is natural number ⇒ 2n² – 2n ≥ 2ⁿ – 1 – 2n
⇒ 2n² ≥ 2ⁿ – 1
⇒ n < 7 ⇒ n = {1, 2, 3, 4, 5, 6}
Checking c against these values of n, we get c = 12 (when n = 3), Hence number of such c = 1
⇒ Answer is 1.00
Q.3 Let f : [0, 2] → R be the function defined by f(x) = (3 – Sin(2πx)).Sin(πx – π/4) – Sin(3πx + π/4). If a, b ∈ [0, 2] are such that {x ∈ [0, 2] : f (x) ≥ 0} = [a, b], then the value of (b – a) is…
Solution:
Let πx – π/4 = θ ∈ [-π/4, 7π/4]
Now for f (x) ≥ 0, (3 – Sin(2πx)).Sin(πx – π/4) – Sin(3πx + π/4) ≥ 0
⇒ (3 – Sin(π/2 + 2θ)).Sin(θ) – Sin(π + 3θ) ≥ 0
⇒ (3 – Cos2θ)).Sin(θ) + Sin(3θ) ≥ 0
⇒ (3 – (1 – 2Sin²(θ)))).Sin(θ) + 3Sin(θ) – 4Sin³(θ) ≥ 0
⇒ (2 + 2Sin²(θ)).Sin(θ) + 3Sin(θ) – 4Sin³(θ) ≥ 0
⇒ Sin(θ){5 – 2Sin²(θ)} ≥ 0
⇒ Sin(θ){4 + Cos2θ} ≥ 0
⇒ Sin(θ) ≥ 0
⇒ θ ∈ [0, π]
⇒ πx – π/4 ∈ [0, π]
⇒ x ∈ [1/4, 5/4]
⇒ (b – a) = 1
⇒ Answer is 1.00