JEE Main 2022 July 28 – Shift 1 Maths Question Paper with Solutions

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JEE Main 2022 28th July Shift 1 Mathematics Question Paper and Solutions

SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. Let the solution curve of the differential equation

$\begin{array}{l} x d y = (\sqrt{x^{2} + y^{2}} + y) d x, x > 0 \end{array}$

intersect the line x = 1 at y = 0 and the line x = 2 at y = α. Then the value of α is

\begin{array}{l} (A) \frac{1}{2} \end{array}

\begin{array}{l} (B) \frac{3}{2} \end{array}

\begin{array}{l} (C) - \frac{3}{2} \end{array}

\begin{array}{l} (D) \frac{5}{2} \end{array}

Answer (B)

Sol.

\begin{array}{l} \frac{x d y - y d x}{\sqrt{x^{2} + y^{2}}} = d x \end{array}

\begin{array}{l} \Rightarrow \frac{d y}{d x} = \frac{\sqrt{x^{2} + y^{2}}}{x} + \frac{y}{x} \end{array}

\begin{array}{l} \Rightarrow \frac{d y}{d x} = \sqrt{1 + \frac{y^{2}}{x^{2}}} + \frac{y}{x} \end{array}

\begin{array}{l} Let \frac{y}{x} = v \end{array}

\begin{array}{l} \Rightarrow v + x \frac{d v}{d x} = \sqrt{1 + v^{2}} + v \end{array}

\begin{array}{l} \Rightarrow \frac{d v}{\sqrt{1 + v^{2}}} = \frac{d x}{x} \end{array}

\begin{array}{l} OR \ln (v + \sqrt{1 + v^{2}}) = \ln x + C \end{array}

at x = 1, y = 0

⇒ C = 0

\begin{array}{l} \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} = x \end{array}

At x = 2,

\begin{array}{l} \frac{y}{2} + \sqrt{1 + \frac{y^{2}}{4}} = 2 \end{array}

\begin{array}{l} \Rightarrow 1 + \frac{y^{2}}{4} = 4 + \frac{y^{2}}{4} - 2 y \end{array}

\begin{array}{l} OR y = \frac{3}{2} \end{array}

2. Considering only the principal values of the inverse trigonometric functions, the domain of the function

$\begin{array}{l} f (x) = \cos^{- 1} (\frac{x^{2} - 4 x + 2}{x^{2} + 3}) \end{array}$

is

\begin{array}{l} (A) (- \infty, \frac{1}{4}] \end{array}

\begin{array}{l} (B) [- \frac{1}{4}, \infty) \end{array}

\begin{array}{l} (C) (\frac{- 1}{3}, \infty) \end{array}

\begin{array}{l} (C) (- \infty, \frac{1}{3}] \end{array}

Answer (B)

Sol.

\begin{array}{l} - 1 \leq \frac{x^{2} - 4 x + 2}{x^{2} + 3} \leq 1 \end{array}

\begin{array}{l} \Rightarrow - x^{2} - 3 \leq x^{2} - 4 x + 2 \leq x^{2} + 3 \end{array}

⇒ 2x² – 4x + 5 ≥ 0 & -4x ≤ 1

x ∈ R & x ≥ -1/4 ]

\begin{array}{l} So domain is [- \frac{1}{4}, \infty) \end{array}

3. Let the vectors

$\begin{array}{l} \vec{a} = (1 + t) \hat{i} + (1 - t) \hat{j} + \hat{k}, \vec{b} = (1 - t) \hat{i} + (1 + t) \hat{j} + 2 \hat{k} \end{array}$

and
$\begin{array}{l} \vec{c} = t \hat{i} - t \hat{j} + \hat{k}, t \in R \end{array}$
be such that for
$\begin{array}{l} α, β, γ \in R, α \vec{a} + β \vec{b} + γ \vec{c} = \vec{0} \Rightarrow α = β = γ = 0. \end{array}$
Then, the set of all values of t is

(A) A non-empty finite set

(B) Equal to N

\begin{array}{l} (C) Equal to R - {0} \end{array}

\begin{array}{l} (C) Equal to R \end{array}

Answer (C)

Sol.

\begin{array}{l} Clearly \vec{a}, \vec{b}, \vec{c} are non-coplanar \end{array}

\begin{array}{l} | \begin{array}{c} 1 + t & 1 - t & 1 \\ 1 - t & 1 + t & 2 \\ t & - t & 1 \end{array} | \neq 0 \end{array}

\begin{array}{l} \Rightarrow (1 + t) (1 + t + 2 t) - (1 - t) (1 - t - 2 t) + 1 (t^{2} - t - t - t^{2}) \neq 0 \end{array}

\begin{array}{l} \Rightarrow (3 t^{2} + 4 t + 1) - (1 - t) (1 - 3 t) - 2 t \neq 0 \end{array}

\begin{array}{l} \Rightarrow (3 t^{2} + 4 t + 1) - (3 t^{2} - 4 t + 1) - 2 t \neq 0 \end{array}

\begin{array}{l} \Rightarrow t \neq 0 \end{array}

4. Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation

$\begin{array}{l} c o s^{- 1} (x) - 2 s i n^{- 1} (x) = c o s^{- 1} (2 x) \end{array}$

is equal to

\begin{array}{l} (A) 0 \end{array}

\begin{array}{l} (B) 1 \end{array}

\begin{array}{l} (C) \frac{1}{2} \end{array}

\begin{array}{l} (D) - \frac{1}{2} \end{array}

Answer (A)

Sol.

\begin{array}{l} c o s^{- 1} x - 2 s i n^{- 1} x = c o s^{- 1} 2 x \end{array}

\begin{array}{l} For Domain : x \in [\frac{- 1}{2}, \frac{1}{2}] \end{array}

\begin{array}{l} \cos^{- 1} x - 2 (\frac{π}{2} - \cos^{- 1} x) = \cos^{- 1} (2 x) \end{array}

\begin{array}{l} \Rightarrow c o s^{- 1} x + 2 c o s^{- 1} x = π + c o s^{- 1} 2 x \end{array}

\begin{array}{l} \Rightarrow c o s (3 c o s^{- 1} x) = - c o s (c o s^{- 1} 2 x) \end{array}

\begin{array}{l} \Rightarrow 4 x^{3} = x \end{array}

\begin{array}{l} \Rightarrow x = 0, \pm \frac{1}{2} \end{array}

5. Let the operations *, ◉ ∈ {∧, ∨}. If (p * q) ◉ (p ◉ ~q) is a tautology, then the ordered pair (*, ◉) is

(A) (∨, ∧)

(B) (∨, ∨)

(C) (∧, ∧)

(D) (∧, ∨)

Answer (B)

Sol. *, ◉ ∈ {∧, ∨}.

Now for (q * q)◉(p◉~q) is tautology

(A) (∨, ∧) : (p ∨ q) ∧ (p ∧ ~q) not a tautology

(B) (∨, ∨) : (p ∨ q) ∨ (p ∨ ~q)

= P ∨ T is tautology

(C) (∧ , ∧) : (p ∧ q) ∧ (p ∧ ~q)

= (p ∧ p) ∧ (q ∧ ~q) = p ∧ F not a tautology (Fallasy)

(D) (∧, ∨) : (p ∧ q) ∨ (p ∨ ~q) not a tautology

6. Let a vector

$\begin{array}{l} \vec{a} has magnitude 9. Let a vector \vec{b} \end{array}$

be such that for every
$\begin{array}{l} (x, y) \in R \times R - {(0, 0)}, the vector (x \vec{a} + y \vec{b}) \end{array}$

$\begin{array}{l} is perpendicular to the vector (6 y \vec{a} - 18 x \vec{b}) . \end{array}$

$\begin{array}{l} Then the value of | \vec{a} \times \vec{b} | \end{array}$
is equal to

\begin{array}{l} (A) 9 \sqrt{3} \end{array}

\begin{array}{l} (B) 27 \sqrt{3} \end{array}

\begin{array}{l} (C) 9 \end{array}

\begin{array}{l} (D) 81 \end{array}

Answer (B)

Sol.

\begin{array}{l} (x \vec{a} + y \vec{b}) \cdot (6 y \vec{a} - 18 x \vec{b}) = 0 \end{array}

\begin{array}{l} \Rightarrow (6 x y | \vec{a} |^{2} - 18 x y | \vec{b} |^{2}) + (6 y^{2} - 18 x^{2}) \vec{a} . \vec{b} = 0 \end{array}

As given equation is identity

Coefficient of x² = coefficient of y² = coefficient of
xy = 0

\begin{array}{l} \Rightarrow | \vec{a} |^{2} = 3 | \vec{b} |^{2} \Rightarrow | \vec{b} | = 3 \sqrt{3} \end{array}

\begin{array}{l} and \vec{a} . \vec{b} = 0 \end{array}

\begin{array}{l} | \vec{a} \times \vec{b} | = | \vec{a} | | \vec{b} | \sin θ \end{array}

\begin{array}{l} = 9.3 \sqrt{3} .1 = 27 \sqrt{3} \end{array}

7. For t ∈ (0, 2π), if ABC is an equilateral triangle with vertices A(sint, – cost), B(cost, sint) and C(a, b) such that its orthocentre lies on a circle with centre (1, 1/3), then (a² – b²) is equal to

\begin{array}{l} (A) \frac{8}{3} \end{array}

\begin{array}{l} (B) 8 \end{array}

\begin{array}{l} (C) \frac{77}{9} \end{array}

\begin{array}{l} (D) \frac{80}{9} \end{array}

Answer (B)

Sol. Let P(h, k) be the orthocentre of ΔABC.

Then

\begin{array}{l} h = \frac{\sin t + \cos t + a}{3}, k = \frac{- \cos t + \sin t + b}{3} \end{array}

(orthocentre coincide with centroid)

\begin{array}{l} ∴ {(3 h - a)}^{2} + {(3 k - b)}^{2} = 2 \end{array}

\begin{array}{l} ∴ {(h - \frac{a}{3})}^{2} + {(k - \frac{b}{3})}^{2} = \frac{2}{9} \end{array}

\begin{array}{l} ∵ \end{array}

orthocentre lies on circle with centre

\begin{array}{l} (1, \frac{1}{3}) \end{array}

∴ a = 3, b = 1

∴ a² – b² = 8

8. For α ∈ N, consider a relation R on N given by R = {(x, y) : 3x + αy is a multiple of 7}. The relation R is an equivalence relation if and only if

(A) α = 14

(B) α is a multiple of 4

(C) 4 is the remainder when α is divided by 10

(D) 4 is the remainder when α is divided by 7

Answer (D)

Sol.

\begin{array}{l} R = {(x, y) : 3 x + α y is a multiple of 7} \end{array}

, Now R to be an equivalence relation

\begin{array}{l} (1) R should be reflexive : (a, a) \in R \forall a \in N \\ ∴ 3 a + a α = 7 k \end{array}

\begin{array}{l} ∴ (3 + α) a = 7 k \end{array}

\begin{array}{l} ∴ 3 + α = 7 k_{1} \Rightarrow α = 7 k_{1} - 3 \\ = 7 k_{1} + 4 \end{array}

\begin{array}{l} (2) R should be symmetric : a R b \Leftrightarrow b R a \end{array}

\begin{array}{l} a R b : 3 a + (7 k - 3) b = 7 m \end{array}

\begin{array}{l} \Rightarrow 3 (a - b) + 7 k b = 7 m \\ \Rightarrow 3 (b - a) + 7 k a = 7 m \end{array}

\begin{array}{l} S o, a R b \Rightarrow b R a \end{array}

\begin{array}{l} ∴ R will be symmetric for a = 7 k_{1} - 3 \end{array}

\begin{array}{l} (3) Transitive : Let (a, b) \in R, (b, c) \in R \end{array}

\begin{array}{l} \Rightarrow 3 a + (7 k - 3) b = 7 k_{1} and \\ 3 b + (7 k_{2} - 3) c = 7 k_{3} \end{array}

Adding 3a + 7kb + (7k₂ – 3) c = 7 (k₁ + k₃)

3a + (7k₂ – 3) c = 7 m

\begin{array}{l} ∴ (a, c) \in R \end{array}

∴ R is transitive

\begin{array}{l} ∴ a = 7 k - 3 = 7 k + 4 \end{array}

9. Out of 60% female and 40% male candidates appearing in an exam, 60% of candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability that the chosen candidate is a female, is

\begin{array}{l} (A) \frac{3}{4} \end{array}

\begin{array}{l} (B) \frac{11}{16} \end{array}

\begin{array}{l} (C) \frac{23}{32} \end{array}

\begin{array}{l} (D) \frac{13}{16} \end{array}

Answer (*) None of the given option is correct

Sol.

\begin{array}{l} P (Female) = \frac{60}{100} = \frac{3}{5} \end{array}

\begin{array}{l} P (Male) = \frac{2}{5} \end{array}

\begin{array}{l} P (Female/Qualified) = \frac{40}{60} = \frac{2}{3} \end{array}

\begin{array}{l} P (Male/Qualified) = \frac{20}{60} = \frac{1}{3} \end{array}

10. If y = y(x), x ∈ (0, π/2) be the solution curve of the differential equation

$\begin{array}{l} (\sin^{2} 2 x) \frac{d y}{d x} + (8 \sin^{2} 2 x + 2 \sin 4 x) y = 2 e^{- 4 x} (2 \sin 2 x + \cos 2 x), \end{array}$

$\begin{array}{l} with y (\frac{π}{4}) = e^{- π}, then y (\frac{π}{6}) \end{array}$
is equal to

\begin{array}{l} (A) \frac{2}{\sqrt{3}} e^{- 2 π / 3} \end{array}

\begin{array}{l} (B) \frac{2}{\sqrt{3}} e^{2 π / 3} \end{array}

\begin{array}{l} (C) \frac{1}{\sqrt{3}} e^{- 2 π / 3} \end{array}

\begin{array}{l} (D) \frac{1}{\sqrt{3}} e^{2 π / 3} \end{array}

Answer (A)

Sol.

\begin{array}{l} (\sin^{2} 2 x) \frac{d y}{d x} + (8 \sin^{2} 2 x + 2 \sin 4 x) y = 2 e^{- 4 x} (2 \sin 2 x + \cos 2 x) \end{array}

\begin{array}{l} \frac{d y}{d x} + (8 + 4 \cot 2 x) y = 2 e^{- 4 x} (\frac{2 \sin 2 x + \cos 2 x}{\sin^{2} 2 x}) \end{array}

\begin{array}{l} Integrating factor (I . F .) = e^{\int (8 + 4 \cot 2 x) d x} \end{array}

\begin{array}{l} = e^{8 x + 2 \ln \sin 2 x} \end{array}

Solution of differential equation

\begin{array}{l} y . e^{8 x + 2 \ln \sin 2 x} = \int 2 e^{(4 x + 2 \ln \sin 2 x)} \frac{(2 \sin 2 x + \cos 2 x)}{\sin^{2} 2 x} d x \end{array}

\begin{array}{l} = 2 \int e^{4 x} (2 \sin 2 x + \cos 2 x) d x \end{array}

\begin{array}{l} y . e^{8 x + 2 \ln \sin 2 x} = e^{4 x} \sin 2 x + c \end{array}

\begin{array}{l} y (\frac{π}{4}) = e^{- π} \end{array}

\begin{array}{l} e^{- π} . e^{2 π} = e^{π} + c \Rightarrow c = 0 \end{array}

\begin{array}{l} y (\frac{π}{6}) = \frac{e^{\frac{2 π}{3}} \frac{\sqrt{3}}{2}}{e^{(\frac{4 π}{3} + 2 \ln \frac{\sqrt{3}}{2})}} \end{array}

\begin{array}{l} = e^{\frac{- 2 π}{3}} . \frac{2}{\sqrt{3}} \end{array}

11. If the tangents drawn at the points P and Q on the parabola y² = 2x – 3 intersect at the point R(0, 1), then the orthocentre of the triangle PQR is :

(A) (0, 1)

(B) (2, –1)

(C) (6, 3)

(D) (2, 1)

Answer (B)

Sol.

Equation of chord PQ

⇒ y × 1 = x – 3

⇒ x – y = 3

For point P & Q

Intersection of PQ with parabola P : (6, 3) Q : (2, –1)

Slope of RQ = –1 & Slope of PQ = 1

\begin{array}{l} Therefore ∠ P Q R = 90^{\circ} \Rightarrow Orthocentre is at Q : (2, - 1) \end{array}

12. Let C be the centre of the circle

$\begin{array}{l} x^{2} + y^{2} - x + 2 y = \frac{11}{4} \end{array}$

and P be a point on the circle. A line passes through the point C, makes an angle of π/4with the line CP and intersects the circle at the Q and R. Then the area of the triangle PQR (in unit²) is :

\begin{array}{l} (A) 2 \end{array}

\begin{array}{l} (B) 2 \sqrt{2} \end{array}

\begin{array}{l} (C) 8 \sin (\frac{π}{8}) \end{array}

\begin{array}{l} (D) 8 \cos (\frac{π}{8}) \end{array}

Answer (B)

Sol.

QR = 2r = 4[

\begin{array}{l} P = (\frac{1}{2} + 2 \cos \frac{π}{4}, - 1 + 2 \sin \frac{π}{4}) \end{array}

\begin{array}{l} = (\frac{1}{2} + \sqrt{2}, - 1 + \sqrt{2}) \end{array}

\begin{array}{l} Area of Δ P Q R = \frac{1}{2} \times 4 \times \sqrt{2} \end{array}

\begin{array}{l} = 2 \sqrt{2} sq. units \end{array}

13. The remainder 7²⁰²² + 3²⁰²² is divided by 5 is:

(A) 0

(B) 2

(C) 3

(D) 4

Answer (C)

Sol. Let E = 7²⁰²² + 3²⁰²²

= (15 – 1)¹⁰¹¹ + (10 – 1)¹⁰¹¹

= –1 + (multiple of 15) –1 + multiple of 10

= –2 + (multiple of 5)

Hence remainder on dividing E by 5 is 3.

14. Let the matrix

$\begin{array}{l} A = | \begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} | \end{array}$

and the matrix
$\begin{array}{l} B_{0} = A^{49} + 2 A^{98} . If B_{n} = A d j (B_{n - 1}) for all n \geq 1, \end{array}$
then det(B₄) is equal to :

(A) 3²⁸

(B) 3³⁰

(C) 3³²

(D) 3³⁶

Answer (C)

Sol.

\begin{array}{l} A = [\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}] \end{array}

\begin{array}{l} \Rightarrow A^{2} = [\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}] \times [\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}] = [\begin{array}{c} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}] \end{array}

\begin{array}{l} \Rightarrow A^{3} = [\begin{array}{c} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}] [\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}] = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = I \end{array}

\begin{array}{l} Now B_{0} = A^{49} + 2 A^{98} = {(A^{3})}^{16} . A + 2 {(A^{3})}^{32} . A^{2} \end{array}

\begin{array}{l} B_{0} = A + 2 A^{2} = [\begin{array}{c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \end{array}] = [\begin{array}{c} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{array}] \end{array}

\begin{array}{l} | B_{0} | = 9 \end{array}

\begin{array}{l} Since, B_{n} = A d j | B_{n - 1} | \Rightarrow | B_{n} | = {| B_{n - 1} |}^{2} \end{array}

\begin{array}{l} Hence | B_{4} | = {| B_{3} |}^{2} = {| B_{2} |}^{4} = {| B_{1} |}^{8} = {| B_{0} |}^{16} = {| 3^{2} |}^{16} = 3^{32} \end{array}

15. Let

$\begin{array}{l} S_{1} = {z_{1} \in C : | z_{1} - 3 | = \frac{1}{2}} and S_{2} = {z_{2} \in C : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |} . \end{array}$

Then, for z₁ ∈ S₁ and z₂ ∈ S₂, the least value of |z₂ – z₁| is :

\begin{array}{l} (A) 0 \end{array}

\begin{array}{l} (B) \frac{1}{2} \end{array}

\begin{array}{l} (C) \frac{3}{2} \end{array}

\begin{array}{l} (D) \frac{5}{2} \end{array}

Answer (C)

Sol.

\begin{array}{l} ∵ | Z_{2} + | Z_{2} - 1 | |^{2} = | Z_{2} - | Z_{2} + 1 | |^{2} \end{array}

\begin{array}{l} \Rightarrow (Z_{2} + | Z_{2} - 1 |) ({\bar{Z}}_{2} + | Z_{2} - 1 |) = (Z_{2} - | Z_{2} - 1 |) ({\bar{Z}}_{2} - | Z_{2} + 1 |) \end{array}

\begin{array}{l} \Rightarrow Z_{2} (| Z_{2} - 1 | + | Z_{2} + 1 |) + {\bar{Z}}_{2} (| Z_{2} - 1 | + | Z_{2} + 1 |) = | Z_{2} + 1 |^{2} - | Z_{2} - 1 |^{2} \end{array}

\begin{array}{l} \Rightarrow (Z_{2} + {\bar{Z}}_{2}) (| Z_{2} + 1 | + | Z_{2} - 1 |) = 2 (Z_{2} + {\bar{Z}}_{2}) \end{array}

\begin{array}{l} \Rightarrow Either Z_{2} + {\bar{Z}}_{2} = 0 or | Z_{2} + 1 | + | Z_{2} + 1 | = 2 \end{array}

So, Z₂ lies on imaginary axis or on real axis within [–1, 1]

\begin{array}{l} Also | Z_{1} - 3 | = \frac{1}{2} \Rightarrow Z_{1} \end{array}

lies on the circle having center 3 and radius _½.

\begin{array}{l} Clearly | Z_{1} - Z_{2} |_{min} = \frac{3}{2} \end{array}

16. The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

\begin{array}{l} (A) {(6 x + 5 y - 12)}^{2} + 4 {(3 x + 7 y - 8)}^{2} = 1, z = 6 - 2 x - 3 y \end{array}

\begin{array}{l} (B) {(5 x + 6 y - 12)}^{2} + 4 {(3 x + 5 y - 9)}^{2} = 1, z = 6 - 2 x - 3 y \end{array}

\begin{array}{l} (C) {(6 x + 5 y - 14)}^{2} + 9 {(3 x + 5 y - 7)}^{2} = 1, z = 6 - 2 x - 3 y \end{array}

\begin{array}{l} (D) {(5 x + 6 y - 14)}^{2} + 9 {(3 x + 7 y - 8)}^{2} = 1, z = 6 - 2 x - 3 y \end{array}

Answer (B)

Sol. Any point on x² + y² = 1, z = 0 is p(cosθ, sinθ, 0)

If foot of perpendicular of p on the plane 2x + 3y + z = 6 is (h, k, l) then

\begin{array}{l} \frac{h - \cos θ}{2} = \frac{k - \sin θ}{3} = \frac{l - 0}{1} \end{array}

\begin{array}{l} = - (\frac{2 \cos θ + 3 \sin θ + 0 - 6}{2^{2} + 3^{2} + 1^{2}}) = r (let) \end{array}

h = 2r + cosθ, k = 3r + sinθ, l = r

Hence, h – 2l = cosθ and k – 3l = sinθ

Hence (h – 2l)² + (k – 3l)² = 1

When l = 6 – 2h – 3k

Hence required locus is

\begin{array}{l} {(x - 2 (6 - 2 x - 3 y))}^{2} + {(y - 3 (6 - 2 x - 3 y))}^{2} = 1 \end{array}

\begin{array}{l} \Rightarrow {(5 x + 6 y - 12)}^{2} + 4 {(3 x + 5 y - 9)}^{2} = 1, z = 6 - 2 x - 3 y \end{array}

17. If the minimum value of

$\begin{array}{l} f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0 \end{array}$

is 14, then the value of α is equal to

(A) 32

(B) 64

(C) 128

(D) 256

Answer (C)

Sol.

\begin{array}{l} f (x) = \frac{5 x^{2}}{2} + \frac{α}{x^{5}} {x > 0} \end{array}

\begin{array}{l} f^{'} (x) = 5 x - \frac{5 α}{x^{6}} = 0 \end{array}

\begin{array}{l} \Rightarrow x = (α)^{1 / 7} \end{array}

\begin{array}{l} f (x)_{min} = \frac{5 (α)^{2 / 7}}{2} + \frac{α}{a^{5 / 7}} = 14 \end{array}

\begin{array}{l} \frac{5}{2} α^{2 / 7} + α^{2 / 7} = 14 \end{array}

\begin{array}{l} \frac{7}{2} α^{2 / 7} = 14 \\ α = 128 \end{array}

18. Let α, β and γ be three positive real numbers.

$\begin{array}{l} L e t f (x) = α x^{5} + β x^{3} + γ x, x \in R and g : R \to R be such that g (f (x)) = x for all x \in R . \end{array}$

If a₁, a₂, a₃, …, a_n be in arithmetic progression with mean zero, then the value of
$\begin{array}{l} f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i}))) \end{array}$
is equal to

(A) 0

(B) 3

(C) 9

(D) 27

Answer (A)

Sol.

\begin{array}{l} f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i}))) \end{array}

\begin{array}{l} \frac{a_{1} + a_{2} + a_{3} + \dots + a_{n}}{n} = 0 \end{array}

∴ First and last term, second and second last and so on are equal in magnitude but opposite in sign.

\begin{array}{l} f (x) = α x^{5} + β x^{3} + γ x \end{array}

\begin{array}{l} \sum_{i = 1}^{n} f (a_{i}) = α (a_{1}^{5} + a_{2}^{5} + a_{3}^{5} + \dots + a_{n}^{5}) + β (a_{1}^{3} + a_{2}^{3} + \dots + a_{n}^{3}) + γ (a_{1} + a_{2} + \dots + a_{n}) \end{array}

\begin{array}{l} = 0 α + 0 β + 0 γ = 0 \end{array}

\begin{array}{l} ∴ f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i}))) = \frac{1}{n} \sum_{i = 1}^{n} f (a_{i}) = 0 \end{array}

19. Consider the sequence a₁, a₂, a₃, … such that a₁ = 1, a₂ = 2 and

$\begin{array}{l} a_{n + 2} = \frac{2}{a_{n + 1}} + a_{n} for n = 1, 2, 3, \dots \end{array}$

. If
$\begin{array}{l} (\frac{a_{1} + \frac{1}{a_{2}}}{a_{3}}) (\frac{a_{2} + \frac{1}{a_{3}}}{a_{4}}) (\frac{a_{3} + \frac{1}{a_{4}}}{a_{5}}) \dots (\frac{a_{30} + \frac{1}{a_{31}}}{a_{32}}) = 2^{α} (^{61} C_{31}), \end{array}$
then α is equal to

(A) –30

(B) –31

(C) –60

(D) –61

Answer (C)

Sol.

\begin{array}{l} a_{n + 2} = \frac{2}{a_{n + 1}} + a_{n} \end{array}

\begin{array}{l} \Rightarrow a_{n} a_{n + 1} + 1 = a_{n + 1} a_{n + 2} - 1 \end{array}

\begin{array}{l} \Rightarrow a_{n + 2} a_{n + 1} + 1 - a_{n} . a_{n + 1} = 2 \end{array}

For n = 1 a₃a₂ – a₁a₂ = 2

n = 2 a₄a₃ – a₃a₂ = 2

n = 3 a₅a₄ – a₄a₃ = 2

_:

n = na_n_{+ 2}a_n_{+ 1} – a_na_n_{+ 1} = 2

a_n_{+ 2}a_n_{+ 1} = 2n + a₁a₂

Now,

\begin{array}{l} \frac{(a_{1} a_{2} + 1)}{a_{2} a_{3}} . \frac{(a_{2} a_{3} + 1)}{a_{3} a_{4}} . \frac{(a_{3} a_{4} + 1)}{a_{4} a_{5}} . \dots . \frac{(a_{30} a_{31} + 1)}{a_{31} a_{32}} \end{array}

\begin{array}{l} = \frac{3}{4} . \frac{5}{6} . \frac{7}{8} . \dots . \frac{61}{62} \end{array}

\begin{array}{l} = 2^{- 60} (^{61} C_{31}) \end{array}

20. The minimum value of the twice differentiable function

$\begin{array}{l} f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R \end{array}$

is

\begin{array}{l} (A) - \frac{2}{\sqrt{e}} \end{array}

\begin{array}{l} (B) - 2 \sqrt{e} \end{array}

\begin{array}{l} (C) - \sqrt{e} \end{array}

\begin{array}{l} (D) \frac{2}{\sqrt{e}} \end{array}

Answer (A)

Sol.

\begin{array}{l} f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x} \end{array}

\begin{array}{l} f (x) = e^{x} \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1) e^{x} \end{array}

\begin{array}{l} e^{- x} f (x) = \int_{0}^{x} e^{- t} f^{'} (t) d t - (x^{2} - x + 1) \end{array}

Differentiate on both sides

\begin{array}{l} e^{- x} f^{'} (x) + (- f (x) e^{- x}) = e^{- x} f^{'} (x) - 2 x + 1 \end{array}

\begin{array}{l} f (x) = e^{x} (2 x - 1) \end{array}

\begin{array}{l} f^{'} (x) = e^{x} (2) + e^{x} (2 x - 1) \end{array}

\begin{array}{l} = e^{x} (2 x + 1) \end{array}

\begin{array}{l} x = - \frac{1}{2} \end{array}

\begin{array}{l} f ” (x) = e^{x} (2) + (2 x + 1) e^{x} \end{array}

\begin{array}{l} = e^{x} (2 x + 3) \end{array}

\begin{array}{l} For x = - \frac{1}{2} f ” (x) > 0 \end{array}

\begin{array}{l} \Rightarrow M a x i m a ∴ Max = e^{- \frac{1}{2}} (- 1 - 1) \end{array}

\begin{array}{l} ∴ M a x . = - \frac{2}{\sqrt{e}} \end{array}

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. Let S be the set of all passwords which are six to eight characters long, where each character is either an alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the number of passwords in S whose at least one character is a number from {1, 2, 3, 4, 5} is α × 5⁶, then α is equal to _______.

Answer (7073)

Sol. If password is 6 character long, tehn

Total number of ways having atleast one number = 10⁶ – 5⁶

Similarly, if 7 character long = 10⁷ – 5⁷

and if 8-character long = 10⁸ – 5⁸

\begin{array}{l} Number of password = (10^{6} + 10^{7} + 10^{8}) - (5^{6} + 5^{7} + 5^{8}) \end{array}

\begin{array}{l} = 5^{6} (2^{6} + {5.2}^{7} + {25.2}^{8} - 1 - 5 - 25) \end{array}

\begin{array}{l} = 56 (64 + 640 + 6400 - 31) \end{array}

\begin{array}{l} = 7073 \times 5^{6} \\ ∴ α = 7073 \end{array}

.

2. Let P(–2, –1, 1) and Q(56/17, 43/17, 111/17) be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are α, –1, β, where both α and β are integers of minimum absolute values, then α² + β² is equal to ___________.

Answer (450)

Sol.

d.r’s of RS = < α, –1, β >

\begin{array}{l} d.r’s of P Q =< \frac{90}{17}, \frac{60}{17}, \frac{94}{17} >=< 45, 30, 47 > \end{array}

as PQ and RS are diagonals of rhombus

\begin{array}{l} α (45) + 30 (- 1) + 47 (β) = 0 \\ \Rightarrow 45 α + 47 β = 30 \end{array}

\begin{array}{l} i . e ., α = \frac{30 - 47 β}{45} \end{array}

for minimum integral value α = – 15 and β = 15

\begin{array}{l} \Rightarrow α^{2} + β^{2} = 450. \end{array}

3. Let f : [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line
y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1) then the least number of points x ∈ (0, 1) at which f”(x) = 0, is ___________.

Answer (2)

Sol.

If a graph cuts y = 2x + 5 in (0, 1) twice then its concavity changes twice

∴ f”(x) = 0 at atleast two points.

4. If

$\begin{array}{l} \int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1 + x^{2} + \sqrt{(1 + x^{2})^{3}}}} d x = α \sqrt{2} + β \sqrt{3}, \end{array}$

where α, β are integers, then α + β is equal to

Answer (10)

Sol. Put

\begin{array}{l} x = t a n θ \Rightarrow d x = s e c^{2} θ d θ \end{array}

\begin{array}{l} \Rightarrow I = \int_{0}^{\frac{π}{3}} \frac{15 \tan^{3} θ . \sec^{2} θ d θ}{\sqrt{1 + \tan^{2} θ + \sqrt{\sec^{6} θ}}} \end{array}

\begin{array}{l} \Rightarrow I = \int_{0}^{\frac{π}{3}} \frac{15 \tan^{2} θ \sec^{2} θ d θ}{\sec θ \sqrt{1 + \sec θ}} \end{array}

\begin{array}{l} \Rightarrow I = \int_{0}^{\frac{π}{3}} \frac{15 (\sec^{2} θ - 1) \sec θ \tan θ d θ}{(\sqrt{1 + \sec θ})} \end{array}

Now put 1 + secθ = t²

\begin{array}{l} \Rightarrow s e c θ t a n θ d θ = 2 t d t \end{array}

\begin{array}{l} \Rightarrow I = \int_{\sqrt{2}}^{\sqrt{3}} \frac{15 ((t^{2} - 1)^{2} - 1) 2 t d t}{t} \end{array}

\begin{array}{l} \Rightarrow I = 30 \int_{\sqrt{2}}^{\sqrt{3}} (t^{4} - 2 t^{2} + 1 - 1) d t \end{array}

\begin{array}{l} \Rightarrow I = 30 \int_{\sqrt{2}}^{\sqrt{3}} (t^{4} - 2 t^{2}) d t \end{array}

\begin{array}{l} \Rightarrow I = 30 {(\frac{t^{5}}{5} - \frac{2 t^{3}}{3})}_{\sqrt{2}}^{\sqrt{3}} \end{array}

\begin{array}{l} = 30 [(\frac{9}{5} \sqrt{3} - 2 \sqrt{3}) - (\frac{4 \sqrt{2}}{5} - \frac{4 \sqrt{2}}{3})] \end{array}

\begin{array}{l} = (54 \sqrt{3} - 60 \sqrt{3}) - (24 \sqrt{2} - 40 \sqrt{2}) \end{array}

\begin{array}{l} = 16 \sqrt{2} - 6 \sqrt{3} \end{array}

\begin{array}{l} ∴ α = 16 and β = - 6 \\ α + β = 10. \end{array}

5. Let

$\begin{array}{l} A = [\begin{array}{c} 1 & - 1 \\ 2 & α \end{array}] and B = [\begin{array}{c} β & 1 \\ 1 & 0 \end{array}], a l p h a, β \in R . \end{array}$

Let α₁ be the value of α which satisfies
$\begin{array}{l} (A + B)^{2} = A^{2} + [\begin{array}{c} 2 & 2 \\ 2 & 2 \end{array}] \end{array}$
and α₂ be the value of α which satisfies
$\begin{array}{l} {(A + B)}^{2} = B^{2} . \end{array}$
Then |α₁ – α₂| is equal to _________.

Answer (2)

Sol.

\begin{array}{l} {(A + B)}^{2} = A^{2} + B^{2} + A B + B A \end{array}

\begin{array}{l} = A^{2} + [\begin{array}{c} 2 & 2 \\ 2 & 2 \end{array}] \end{array}

\begin{array}{l} ∴ B^{2} + A B + B A = [\begin{array}{c} 2 & 2 \\ 2 & 2 \end{array}] \dots (1) \end{array}

\begin{array}{l} A B = [\begin{array}{c} 1 & - 1 \\ 2 & α \end{array}] [\begin{array}{c} β & 1 \\ 1 & 0 \end{array}] = [\begin{array}{c} β - 1 & 1 \\ α + 2 β & 2 \end{array}] \end{array}

\begin{array}{l} B A = [\begin{array}{c} β & 1 \\ 1 & 0 \end{array}] [\begin{array}{c} 1 & - 1 \\ 2 & α \end{array}] = [\begin{array}{c} β + 2 & α - β \\ 1 & - 1 \end{array}] \end{array}

\begin{array}{l} B^{2} = [\begin{array}{c} β & 1 \\ 1 & 0 \end{array}] [\begin{array}{c} β & 1 \\ 1 & 0 \end{array}] = [\begin{array}{c} β^{2} + 1 & β \\ β & 1 \end{array}] \end{array}

By (1) we get

\begin{array}{l} [\begin{array}{c} β^{2} + 2 β + 2 & α + 1 \\ α + 3 β + 1 & 2 \end{array}] = [\begin{array}{c} 2 & 2 \\ 2 & 2 \end{array}] \end{array}

\begin{array}{l} ∴ α = 1, β = 0, \Rightarrow α_{1} = 1 \end{array}

Similarly If A² + AB + BA = 0 then

\begin{array}{l} (A^{2} = [\begin{array}{c} 1 & - 1 \\ 2 & α \end{array}] [\begin{array}{c} 1 & - 1 \\ 2 & α \end{array}] = [\begin{array}{c} - 1 & - 1 - α \\ 2 + 2 α & α^{2} - 2 \end{array}]) \end{array}

\begin{array}{l} [\begin{array}{c} 2 β & α - β + 1 - 1 - α \\ α + 2 β + 1 + 2 + 2 α & α^{2} - 2 + 1 \end{array}] = [\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] \end{array}

\begin{array}{l} \Rightarrow β = 0 and α = - 1 \Rightarrow α_{2} = - 1 \end{array}

\begin{array}{l} ∴ | α_{1} - α_{2} | = | 2 | = 2 \end{array}

6. For p, q, ∈ R, consider the real valued function f(x) = (x – p)² – q, x ∈ R and q > 0, Let a₁, a₂, a₃ and a₄ be in an arithmetic progression with mean p and positive common difference. If |f(a_i)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is

Answer (50)

Sol. ∵ a₁, a₂, a₃, a₄ are in A.P and its mean is p.

∴ a₁ = p – 3d, a₂ = p –d, a₃ = p + d and a₄ = p + 3d

Where d > 0

\begin{array}{l} ∵ | f (a_{i}) | = 500 \end{array}

\begin{array}{l} \Rightarrow | 9 d^{2} - q | = 500 \end{array}

\begin{array}{l} and | d^{2} - q | = 500 \dots (i) \end{array}

either 9d² – q = d² – q

\begin{array}{l} \Rightarrow d = 0 not acceptable \\ ∴ 9 d^{2} - q = q - d^{2} \end{array}

\begin{array}{l} ∴ 5 d^{2} - q = 0 \dots \dots . (i i) \end{array}

\begin{array}{l} Roots of f (x) = 0 a r e p + \sqrt{q} and p - \sqrt{q} \end{array}

\begin{array}{l} ∴ absolute difference between roots = | 2 \sqrt{q} | = 50 \end{array}

7. For the hyperbola H: x² – y² = 1 and the ellipse

$\begin{array}{l} E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b > 0, let the \end{array}$

(1) eccentricity of E be reciprocal of the eccentricity of H, and

(2) the line y = √(5/2) x + K be a common tangent of E and H.

Then 4(a² + b²) is equal to _______.

Answer (03.00)

Sol. The equation of tangent to hyperbola x² – y² = 1 within slope m is equal to

\begin{array}{l} y = m x \pm \sqrt{m^{2} + 1} \dots (i) \end{array}

And for same slope m, equation of tangent to ellipse

\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{array}

is

\begin{array}{l} y = m x \pm \sqrt{a^{2} m^{2} + b^{2}} \dots (i i) \end{array}

∵ Equation (i) and (ii) are identical

∴ a²m² + b² = m² – 1

\begin{array}{l} ∴ m^{2} = \frac{1 + b^{2}}{1 - a^{2}} \end{array}

But equation of common tangent is

\begin{array}{l} y = \sqrt{\frac{5}{2}} x + K \end{array}

\begin{array}{l} ∴ m = \sqrt{\frac{5}{2}} \Rightarrow \frac{5}{2} = \frac{1 + b^{2}}{1 - a^{2}} \end{array}

∴ 5a² + 2b² = 3 …(i)

\begin{array}{l} Eccentricity of ellipse = \frac{1}{\sqrt{2}} \end{array}

\begin{array}{l} ∴ 1 - \frac{b^{2}}{a^{2}} = \frac{1}{2} \end{array}

⇒a² = 2b² …(ii)

From equation (i) and (ii):

\begin{array}{l} a^{2} = \frac{1}{2}, b^{2} = \frac{1}{4} \end{array}

∴ 4(a² + b²) = 3

8. Let x₁, x₂, x₃, …, x₂₀ be in geometric progression with x₁ = 3 and the common ratio 1/2. A new data is constructed replacing each x_i by (x_i – i)². If x̄ is the mean of new data, then the greatest integer less than or equal to x̄ is _________.

Answer (142)

Sol. x₁, x₂, x₃, …, x₂₀ are in G.P.

\begin{array}{l} x_{1} = 3, r = \frac{1}{2} \end{array}

\begin{array}{l} \bar{x} = \frac{\sum x_{1}^{2} - 2 x_{i} i + i^{2}}{20} \end{array}

\begin{array}{l} = \frac{1}{20} [12 (1 - \frac{1}{2^{40}}) - 6 (4 - \frac{11}{2^{18}}) + 70 \times 41] \end{array}

\begin{array}{l} {\begin{array}{c} S = 1 + 2. \frac{1}{2} + 3. \frac{1}{2^{2}} + \dots \\ \frac{S}{2} = \frac{1}{2} + \frac{2}{2^{2}} + \dots \end{array} \end{array}

\begin{array}{l} \frac{S}{2} = 2 (1 - \frac{1}{2^{20}}) - \frac{20}{2^{20}} = 4 - \frac{11}{2^{18}}} \end{array}

\begin{array}{l} ∴ [\bar{x}] = [\frac{2858}{20} - (\frac{12}{240} - \frac{66}{2^{18}}) . \frac{1}{20}] = 142 \end{array}

9.

$\begin{array}{l} lim_{x \to 0} {(\frac{(x + 2 \cos x)^{3} + 2 (x + 2 \cos x)^{2} + 3 \sin (x + 2 \cos x)}{(x + 2)^{3} + 2 (x + 2)^{2} + 3 \sin (x + 2)})}^{\frac{100}{x}} \end{array}$

is equal to _________.

Answer (01)

Sol. Let x + 2cosx = a

x + 2 = b

as x → 0, a → 2 and b → 2

\begin{array}{l} lim_{x \to 0} {(\frac{a^{3} + 2 a^{2} + 3 \sin a}{b^{3} + 2 b^{2} + 3 \sin b})}^{\frac{100}{x}} \end{array}

\begin{array}{l} = e^{lim_{x \to 0} \frac{100}{x} . \frac{(a^{3} - b^{3}) + 2 (a^{2} - b^{2}) + 3 (\sin a - \sin b)}{b^{3} + 2 b^{2} + 3 \sin b}} \end{array}

\begin{array}{l} ∵ lim_{x \to 0} \frac{a - b}{x} = lim_{x \to 0} \frac{2 (\cos x - 1)}{x} = 0 \end{array}

= e⁰

= 1

10. The sum of all real value of x for which

$\begin{array}{l} \frac{3 x^{2} - 9 x + 17}{x^{2} + 3 x + 10} = \frac{5 x^{2} - 7 x + 19}{3 x^{2} + 5 x + 12} \end{array}$

is equal to ________.

Answer (06)

Sol.

\begin{array}{l} \frac{3 x^{2} - 9 x + 17}{x^{2} + 3 x + 10} = \frac{5 x^{2} - 7 x + 19}{3 x^{2} + 5 x + 12} \end{array}

\begin{array}{l} \Rightarrow \frac{3 x^{2} - 9 x + 17}{5 x^{2} - 7 x + 19} = \frac{x^{2} + 3 x + 10}{3 x^{2} + 5 x + 12} \end{array}

\begin{array}{l} \frac{- 2 x^{2} - 2 x - 2}{5 x^{2} - 7 x + 19} = \frac{- 2 x^{2} - 2 x - 2}{3 x^{2} + 5 x + 12} \end{array}

Either x² + x + 1 = 0 or 5x² – 7x + 19 = 3x² + 5x + 12 ⇒ 2x² – 12x + 7 = 0 ⇒ sum of roots = 6

For x² + x + 1 = 0, no real roots.

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