JEE Main 2022 June 26 – Shift 1 Maths Question Paper with Solutions (Download PDF)

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JEE Main 2022 June 26 – Shift 1 Maths Question Paper with 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

\begin{array}{l} f (x) = \frac{x - 1}{x + 1}, x ϵ R - {0, - 1, 1} \end{array}

If ƒⁿ⁺¹(x) = ƒ(ƒⁿ(x)) for all n∈N, then ƒ⁶(6) + ƒ⁷(7) is equal to :

(A)

\begin{array}{l} \frac{7}{6} \end{array}

(B)

\begin{array}{l} - \frac{3}{2} \end{array}

(C)

\begin{array}{l} \frac{7}{12} \end{array}

(D)

\begin{array}{l} - \frac{11}{12} \end{array}

Answer (B)

Sol.

\begin{array}{l} f (x) = \frac{x - 1}{x + 1} \Rightarrow f (f (x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = - \frac{1}{x} \end{array}

⇒

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

So, ƒ⁶(6) + ƒ⁷(7) = ƒ²(6) + ƒ³(7)

\begin{array}{l} = - \frac{1}{6} - \frac{7 + 1}{7 - 1} = - \frac{9}{6} = - \frac{3}{2} \end{array}

2. Let

\begin{array}{l} A = {z ϵ C : | \frac{z + 1}{z - 1} | < 1} \end{array}

and

\begin{array}{l} B = {z ϵ C : a r g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}} \end{array}

Then A∩Bis :

(A) A portion of a circle centred at

\begin{array}{l} (0, - \frac{1}{\sqrt{3}}) \end{array}

that lies in the second and third quadrants only

(B) A portion of a circle centred at

\begin{array}{l} (0, - \frac{1}{\sqrt{3}}) \end{array}

that lies in the second quadrant only

(C) An empty set

(D) A portion of a circle of radius

\begin{array}{l} \frac{2}{\sqrt{3}} \end{array}

that lies in the third quadrant only

Answer (B)

Sol.

\begin{array}{l} | \frac{z + 1}{z - 1} | < 1 \Rightarrow | z + 1 | < | z - 1 | \Rightarrow R e (z) < 0 \end{array}

and

\begin{array}{l} a r g (\frac{z - 1}{z + 1}) = \frac{2 π}{3} \end{array}

is a part of circle as shown.

JEE Main 2022 June 26 Shift 1 Maths A2

3. Let A be a 3 × 3 invertible matrix. If |adj (24A)| = |adj (3 adj (2A))|, then |A|² is equal to :

(A) 6⁶

(B) 2¹²

(C) 2⁶

(D) 1

Answer (C)

Sol.

\begin{array}{l} | a d j (24 A) | = | a d j (3 a d j (2 A)) | \end{array}

⇒

\begin{array}{l} {| 24 A |}^{2} = {| 3 a d j (2 A) |}^{2} \end{array}

⇒

\begin{array}{l} {(24^{3})}^{2} \cdot {| A |}^{2} = {(3^{3})}^{2} {| a d j (2 A) |}^{2} \end{array}

⇒

\begin{array}{l} 24^{6} \cdot {| A |}^{2} = 3^{6} {| 2 A |}^{4} \end{array}

⇒

\begin{array}{l} 24^{6} {| A |}^{2} = 3^{6} \cdot {(2^{3})}^{4} {| A |}^{4} \end{array}

⇒

\begin{array}{l} {| A |}^{2} = \frac{24^{6}}{3^{6} \cdot 2^{12}} = \frac{2^{18} . \cdot 3^{6}}{3^{6} \cdot 2^{12}} = 2^{6} \end{array}

4. The ordered pair (a, b), for which the system of linear equations

3x – 2y + z = b

5x – 8y + 9z = 3

2x + y + az = –1

has no solution, is :

(A)

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

(B)

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

(C)

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

(D)

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

Answer (C)

Sol.

\begin{array}{l} | \begin{array}{c} 3 & - 2 & 1 \\ 5 & - 8 & 9 \\ 2 & 1 & a \end{array} | = 0 \Rightarrow - 14 a - 42 = 0 \Rightarrow a = - 3 \end{array}

Now 3(equation (1)) – (equation (2)) – 2(equation (3)) is

3(3x – 2y + z – b) – (5x – 8y + 9z – 3) – 2(2x + y + az + 1) = 0

⇒ –3b + 3 – 2 = 0

⇒

\begin{array}{l} b = \frac{1}{3} \end{array}

So for no solution a = –3 and

\begin{array}{l} b \neq \frac{1}{3} \end{array}

5. The remainder when (2021)²⁰²³ is divided by 7 is :

(A) 1

(B) 2

(C) 5

(D) 6

Answer (C)

Sol. 2021 ≡ –2 (mod 7)

⇒ (2021)²⁰²³ ≡–(2)²⁰²³ (mod 7)

≡ –2(8)⁶⁷⁴ (mod 7)

≡ –2(1)⁶⁷⁴ (mod 7)

≡ –2(mod 7)

≡ 5(mod 7)

So when (2021)²⁰²³ is divided by 7, remainder is 5.

6.

\begin{array}{l} lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin (\cos^{- 1} x) - x}{1 - \tan (\cos^{- 1} x)} \end{array}

is equal to :

(A)

\begin{array}{l} \sqrt{2} \end{array}

(B)

\begin{array}{l} - \sqrt{2} \end{array}

(C)

\begin{array}{l} \frac{1}{\sqrt{2}} \end{array}

(D)

\begin{array}{l} - \frac{1}{\sqrt{2}} \end{array}

Answer (D)

Sol.

\begin{array}{l} lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin (\cos^{- 1} x) - x}{1 - \tan (\cos^{- 1} x)} \end{array}

\begin{array}{l} let \cos^{- 1} x = \frac{π}{4} + θ \end{array}

\begin{array}{l} = lim_{θ \to 0} \frac{\sin (\frac{π}{4} + θ) - \cos (\frac{π}{4} + θ)}{1 - \tan (\frac{π}{4} + θ)} \end{array}

\begin{array}{l} = lim_{θ \to 0} \frac{\sqrt{2} \sin (\frac{π}{4} + θ - \frac{π}{4})}{1 - \frac{1 + \tan θ}{1 - \tan θ}} \end{array}

\begin{array}{l} = lim_{θ \to 0} \frac{\sqrt{2} \sin θ}{- 2 \tan θ} (1 - \tan θ) = - \frac{1}{\sqrt{2}} \end{array}

7. g :R→R be two real valued functions defined as

\begin{array}{l} f (x) = {\begin{array}{c} - | x + 3 |, & x < 0 \\ e^{x}, & x \geq 0 \end{array} \end{array}

and

\begin{array}{l} g (x) = {\begin{array}{c} x^{2} + k_{1} x, & x < 0 \\ 4 x + k_{2}, & x \geq 0 \end{array} \end{array}

where k₁ and k₂ are real constants. If (goƒ) is differentiable at x = 0, then (goƒ) (–4) + (goƒ) (4) is equal to :

(A) 4(e⁴ + 1)

(B) 2(2e⁴ + 1)

(C) 4e⁴

(D) 2(2e⁴ – 1)

Answer (D)

Sol. ∵ goƒ is differentiable at x = 0

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

\begin{array}{l} \frac{d}{d x} (4 e^{x} + k_{2}) = \frac{d}{d x} ({(- | x + 3 |)}^{2} - k_{1} | x + 3 |) \end{array}

⇒ 4 = 6 –k₁⇒k₁ = 2

Also g(ƒ(0⁺)) = g(ƒ(0^–))

⇒ 4 + k₂ = 9 – 3k₁⇒k₂ = –1

Now g(ƒ(–4)) + g(ƒ(4))

= g(–1) + g(e⁴) = (1 – k₁) + (4e⁴ + k₂)

= 4e⁴ – 2

= 2(2e⁴ – 1)

8. The sum of the absolute minimum and the absolute maximum values of the function ƒ(x) = |3x – x² + 2| – x in the interval [–1, 2] is :

(A)

\begin{array}{l} \frac{\sqrt{17} + 3}{2} \end{array}

(B)

\begin{array}{l} \frac{\sqrt{17} + 5}{2} \end{array}

(C) 5

(D)

\begin{array}{l} \frac{9 - \sqrt{17}}{2} \end{array}

Answer (A)

Sol. ƒ(x) = |x²– 3x – 2| – x

\begin{array}{l} \forall x ϵ [- 1, 2] \end{array}

⇒

\begin{array}{l} f (x) = {\begin{array}{c} x^{2} - 4 x - 2 if - 1 \leq x < \frac{3 - \sqrt{17}}{2} \\ - x^{2} + 2 x + 2 if \frac{3 - \sqrt{17}}{2} \leq x \leq 2 \end{array} \end{array}

JEE Main 2022 June 26 Shift 1 Maths A8

ƒ(x)_max = 3

\begin{array}{l} f {(x)}_{min} = f (\frac{3 - \sqrt{17}}{2}) \end{array}

\begin{array}{l} = \frac{\sqrt{17} - 3}{2} \end{array}

9. Let S be the set of all the natural numbers, for which the line

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

is a tangent to the curve

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

at the point (a, b), ab ≠ 0. Then :

(A) S = ɸ

(B) n(S) = 1

(C) S = {2k : k ∈ N }

(D) S = N

Answer (D)

Sol.

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

\begin{array}{l} \Rightarrow \frac{n}{a} {(\frac{x}{a})}^{n - 1} + \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = 0 \end{array}

\begin{array}{l} \Rightarrow \frac{d y}{d x} = - \frac{b}{a} {(\frac{b x}{a y})}^{n - 1} \end{array}

\begin{array}{l} \frac{d y}{d x_{(a, b)}} = - \frac{b}{a} \end{array}

So line always touches the given curve.

10. The area bounded by the curve

\begin{array}{l} y = | x^{2} - 9 | \end{array}

and the line y = 3 is

(A)

\begin{array}{l} 4 (2 \sqrt{3} + \sqrt{6} - 4) \end{array}

(B)

\begin{array}{l} 4 (4 \sqrt{3} + \sqrt{6} - 4) \end{array}

(C)

\begin{array}{l} 8 (4 \sqrt{3} + 3 \sqrt{6} - 9) \end{array}

(D)

\begin{array}{l} 8 (4 \sqrt{3} + \sqrt{6} - 9) \end{array}

Answer (*)

Sol. y = 3 and y = |x² – 9|

Intersect in first quadrant at

\begin{array}{l} x = \sqrt{6} and x = \sqrt{12} \end{array}

JEE Main 2022 June 26 Shift 1 Maths A10

Required area

\begin{array}{l} = 2 [\frac{2}{3} (6 \times \sqrt{6}) + \int_{\sqrt{6}}^{3} (3 - (9 - x^{2})) d x + \int_{3}^{\sqrt{12}} (3 - (x^{2} - 9)) d x] \end{array}

\begin{array}{l} = 2 [4 \sqrt{6} + (\frac{x^{3}}{3} - 6 x) {|_{\sqrt{6}}^{3} + (12 x - \frac{x^{3}}{3}) |}_{3}^{\sqrt{12}}] \end{array}

\begin{array}{l} = 2 [4 \sqrt{6} + (4 \sqrt{6} - 9) + (8 \sqrt{12} - 27)] \end{array}

\begin{array}{l} = 2 [8 \sqrt{6} + 16 \sqrt{3} - 36] = 8 [2 \sqrt{6} + 4 \sqrt{3} - 9] \end{array}

11. Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle, Then the area of ΔPQRis :

(A)

\begin{array}{l} \frac{25}{4 \sqrt{3}} \end{array}

(B)

\begin{array}{l} \frac{25 \sqrt{3}}{2} \end{array}

(C)

\begin{array}{l} \frac{25}{\sqrt{3}} \end{array}

(D)

\begin{array}{l} \frac{25}{2 \sqrt{3}} \end{array}

Answer (D)

Sol.

JEE Main 2022 June 26 Shift 1 Maths A11

Altitude of equilateral triangle,

\begin{array}{l} \frac{\sqrt{3} l}{2} = \frac{5}{\sqrt{2}} \end{array}

\begin{array}{l} l = \frac{5 \sqrt{2}}{\sqrt{3}} \end{array}

Area of triangle

\begin{array}{l} = \frac{\sqrt{3}}{4} l^{2} = \frac{\sqrt{3}}{4} \cdot \frac{50}{3} = \frac{25}{2 \sqrt{3}} \end{array}

12. Let C be a circle passing through the points
A(2, –1) and B (3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle

\begin{array}{l} {(x - 5)}^{2} + {(y - 1)}^{2} = \frac{13}{2} \end{array}

then r² is equal to :

(A) 32

(B)

\begin{array}{l} \frac{65}{2} \end{array}

(C)

\begin{array}{l} \frac{61}{2} \end{array}

(D) 30

Answer (B)

Sol. Equation of perpendicular bisector of AB is

\begin{array}{l} y - \frac{3}{2} = - \frac{1}{5} (x - \frac{5}{2}) \Rightarrow x + 5 y = 10 \end{array}

Solving it with equation of given circle,

\begin{array}{l} {(x - 5)}^{2} + {(\frac{10 - x}{5} - 1)}^{2} = \frac{13}{2} \end{array}

⇒

\begin{array}{l} {(x - 5)}^{2} (1 + \frac{1}{25}) = \frac{13}{2} \end{array}

⇒

\begin{array}{l} x - 5 = \pm \frac{5}{2} \Rightarrow x = \frac{5}{2} or \frac{15}{2} \end{array}

But

\begin{array}{l} x \neq \frac{5}{2} \end{array}

because AB is not the diameter.

So, centre will be

\begin{array}{l} (\frac{15}{2}, \frac{1}{2}) \end{array}

Now

\begin{array}{l} r^{2} = {(\frac{15}{2} - 2)}^{2} + {(\frac{1}{2} + 1)}^{2} \end{array}

\begin{array}{l} = \frac{65}{2} \end{array}

13. Let the normal at the point P on the parabola
y² = 6x pass through the point (5, –8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

(A) –3

(B)

\begin{array}{l} - \frac{9}{4} \end{array}

(C)

\begin{array}{l} - \frac{5}{2} \end{array}

(D) –2

Answer (B)

Sol. Let P(at², 2at) where

\begin{array}{l} a = \frac{3}{2} \end{array}

T :yt = x + at² So point Q is

\begin{array}{l} (- a, a t - \frac{a}{t}) \end{array}

N :y = –tx + 2at + at³ passes through (5, –8)

\begin{array}{l} - 8 = - 5 t + 3 t + \frac{3}{2} t^{3} \end{array}

⇒ 3t³ – 4t + 16 = 0

⇒ (t + 2)(3t² – 6t + 8) = 0

⇒ t = –2

So ordinate of point Q is

\begin{array}{l} - \frac{9}{4} \end{array}

14. If the two lines

\begin{array}{l} l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 \end{array}

and

\begin{array}{l} l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2} \end{array}

are perpendicular, then an angle between the lines l₂ and

\begin{array}{l} l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} \end{array}

is :

(A)

\begin{array}{l} \cos^{- 1} (\frac{29}{4}) \end{array}

(B)

\begin{array}{l} \sec^{- 1} (\frac{29}{4}) \end{array}

(C)

\begin{array}{l} \cos^{- 1} (\frac{2}{29}) \end{array}

(D)

\begin{array}{l} \cos^{- 1} (\frac{2}{\sqrt{29}}) \end{array}

Answer (B)

Sol. ∵ l₁ and l₂ are perpendicular, so

\begin{array}{l} 3 \times 1 + (- 2) (\frac{α}{2}) + 0 \times 2 = 0 \end{array}

⇒ α = 3

Now angle between l₂ and l₃,

\begin{array}{l} \cos θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt{1 + \frac{α^{2}}{4}} + 4 \sqrt{9 + 4 + 16}} \end{array}

⇒

\begin{array}{l} \cos θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = \cos^{- 1} (\frac{4}{29}) = \sec^{- 1} (\frac{29}{4}) \end{array}

15. Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point

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

in the rotated plane is B( a, b, c),then :

(A)

\begin{array}{l} \frac{a}{8} = \frac{b}{5} = \frac{c}{- 4} \end{array}

(B)

\begin{array}{l} \frac{a}{4} = \frac{b}{5} = \frac{c}{- 2} \end{array}

(C)

\begin{array}{l} \frac{a}{8} = \frac{b}{- 5} = \frac{c}{4} \end{array}

(D)

\begin{array}{l} \frac{a}{4} = \frac{b}{5} = \frac{c}{2} \end{array}

Answer (A)

Sol. Consider the equation of plane,

P : (2x + 3y + z + 20) + λ(x – 3y + 5z – 8) = 0

P : (2 + λ)x + (3 – 3λ)y + (1 + 5λ)z + (20 – 8λ) = 0

∵ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So, 4 + 2λ + 9 – 9λ + 1 + 5λ = 0

λ = 7

P :9x – 18y + 36z – 36 = 0

Or P :x – 2y + 4z = 4

If image of

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

in plane P is (a, b, c) then

\begin{array}{l} \frac{a - 2}{1} = \frac{b + \frac{1}{2}}{- 2} = \frac{c - 2}{4} \end{array}

and

\begin{array}{l} (\frac{a + 2}{2}) - 2 (\frac{b - \frac{1}{2}}{2}) + 4 (\frac{c + 2}{2}) = 4 \end{array}

Clearly

\begin{array}{l} a = \frac{4}{3}, b = \frac{5}{6} and c = - \frac{2}{3} \end{array}

So, a :b : c = 8 : 5 : – 4

16. If

\begin{array}{l} \vec{a} \cdot \vec{b} = 1, \vec{b} \cdot \vec{c} = 2 and \vec{c} \cdot \vec{a} = 3 \end{array}

, then the value of

\begin{array}{l} [\vec{a} \times (\vec{b} \times \vec{c}), \vec{b} \times (\vec{c} \times \vec{a}), \vec{c} \times (\vec{b} \times \vec{a})] \end{array}

is :

(A) 0

(B)

\begin{array}{l} - 6 \vec{a} \cdot (\vec{b} \times \vec{c}) \end{array}

(C)

\begin{array}{l} 12 \vec{c} \cdot (\vec{a} \times \vec{b}) \end{array}

(D)

\begin{array}{l} - 12 \vec{b} \cdot (\vec{c} \times \vec{a}) \end{array}

Answer (A)

Sol. ∵

\begin{array}{l} \vec{a} \times (\vec{b} \times \vec{c}) = 3 \vec{b} - \vec{c} = \vec{u} \end{array}

\begin{array}{l} \vec{b} \times (\vec{c} \times \vec{a}) = \vec{c} - 2 \vec{a} = \vec{v} \end{array}

\begin{array}{l} \vec{c} \times (\vec{b} \times \vec{a}) = 3 \vec{b} - 2 \vec{a} = \vec{w} \end{array}

∴

\begin{array}{l} \vec{u} + \vec{v} = \vec{w} \end{array}

So vectors

\begin{array}{l} \vec{u}, \vec{v} and \vec{w} \end{array}

are coplanar, hence their Scalar triple product will be zero.

17. Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:

(A)

\begin{array}{l} \frac{275}{6^{5}} \end{array}

(B)

\begin{array}{l} \frac{36}{5^{4}} \end{array}

(C)

\begin{array}{l} \frac{181}{5^{5}} \end{array}

(D)

\begin{array}{l} \frac{46}{6^{4}} \end{array}

Answer (D)

Sol. Let probability of getting head = p

So,

\begin{array}{l} ^{5} C_{4} p^{4} (1 - p) =^{5} C_{5} p^{5} \end{array}

⇒

\begin{array}{l} p = 5 (1 - p) \Rightarrow p = \frac{5}{6} \end{array}

Probability of getting atmost two heads =

\begin{array}{l} ^{5} C_{0} {(1 - p)}^{5} +^{5} C_{1} p {(1 - p)}^{4} +^{5} C_{2} p^{2} {(1 - p)}^{3} \end{array}

\begin{array}{l} = \frac{1 + 25 + 250}{6^{5}} \end{array}

\begin{array}{l} = \frac{276}{6^{5}} = \frac{46}{6^{4}} \end{array}

18. The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to:

(A) 60

(B) 55

(C) 50

(D) 45

Answer (A)

Sol. ∵

\begin{array}{l} \overset{―}{x} = 6 = \frac{a + b + 8 + 5 + 10}{5} \Rightarrow a + b = 7 \dots (i) \end{array}

And

\begin{array}{l} σ^{2} = \frac{a^{2} + b^{2} + 8^{2} + 5^{2} + 10^{2}}{5} - 6^{2} = 6.8 \end{array}

⇒ a² + b² = 25 …(ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

\begin{array}{l} M = \frac{1}{5} (3 + 2 + 2 + 1 + 4) = \frac{12}{5} \end{array}

⇒ 25M = 60

19. Let

\begin{array}{l} f (x) = 2 \cos^{- 1} x + 4 \cot^{- 1} x - 3 x^{2} - 2 x + 10, χ ϵ [- 1, 1] \end{array}

If [a, b] is the range of the function,f then 4a – b is equal to :

(A) 11

(B) 11 – π

(C) 11 + π

(D) 15 – π

Answer (B)

Sol.

\begin{array}{l} f (x) = 2 \cos^{- 1} x + 4 \cot^{- 1} x - 3 x^{2} - 2 x + 10 \forall x ϵ [- 1, 1] \end{array}

⇒

\begin{array}{l} f^{‘} (x) = - \frac{2}{\sqrt{1 - x^{2}}} - \frac{4}{1 + x^{2}} - 6 x - 2 < 0 \forall x ϵ [- 1, 1] \end{array}

Sof(x) is decreasing function and range of f(x) is

[f(1), f(-1)], which is [π + 5, 5π + 9]

Now 4a – b = 4(π + 5) – (5π + 9)

= 11 – π

20. Let

\begin{array}{l} Δ, ▽ ϵ {\land, \lor} \end{array}

be such that

\begin{array}{l} p ▽ q \Rightarrow ((p Δ q) ▽ r) \end{array}

is a tautology. Then

\begin{array}{l} (p ▽ q) Δ r \end{array}

is logically equivalent to :

(A)

\begin{array}{l} (p Δ r) \lor q \end{array}

(B)

\begin{array}{l} (p Δ r) \land q \end{array}

(C)

\begin{array}{l} (p \land r) Δ q \end{array}

(D)

\begin{array}{l} (p ▽ r) \land q \end{array}

Answer (A)

Sol. Case-I

If ∇ is same as ∧

Then (p∧q) ⇒ ((pΔq) ∧r) is equivalent to ~ (p∧q) ∨ ((pΔq) ∧r) is equivalent to (~ (p∧q) ∨ (pΔq))∧ (~ (p∧q) ∨r)

Which cannot be a tautology

For both Δ (i.e.∨ or ∧)

Case-II

If ∇ is same as ∨

Then (p∨q) ⇒ ((pΔq) ∨r) is equivalent to

~(p∨q) ∨ (pΔq) ∨r which can be a tautology if Δ is also same as ∨.

Hence both Δ and ∇ are same as ∨.

Now (p∇q) Δr is equivalent to (p∨q∨r).

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. The sum of the cubes of all the roots of the equation x⁴ – 3x³ –2x² + 3x +1 = 0 is _______.

Answer (36)

Sol. x⁴ – 3x³ – x² – x² + 3x + 1 = 0

(x² – 1) (x² – 3x – 1) = 0

Let the root of x² – 3x – 1 = 0 be α and β and other two roots of given equation are 1 and –1

So sum of cubes of roots = 1³ + (–1)³ + α³ + β³

= (α + β)³ – 3αβ(α + β)

= (3)³ – 3(–1)(3)

= 36

2. There are ten boys B₁, B₂, …, B₁₀ and five girls G₁, G₂,…, G₅ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B₁ and B₂ together should not be the members of a group, is ________.

Answer (1120)

Sol. Required number of ways = Total ways of selection – ways in which B₁ and B₂ are present together.

\begin{array}{l} =^{10} C_{3} \cdot^{5} C_{3} -^{8} C_{1} \cdot^{5} C_{3} = 10 (120 - 8) \end{array}

= 1120

3. Let the common tangents to the curves
4(x² + y²) = 9 and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then

\begin{array}{l} \frac{l}{e^{2}} \end{array}

is equal to ________.

Answer (4)

Sol. Let y = mx + c is the common tangent

So

\begin{array}{l} c = \frac{1}{m} = \pm \frac{3}{2} \sqrt{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3} \end{array}

So equation of common tangents will be

\begin{array}{l} y = \pm \frac{1}{\sqrt{3}} x \pm \sqrt{3} \end{array}

which intersects at Q(–3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

\begin{array}{l} e^{2} = 1 - \frac{1}{4} = \frac{3}{4} \end{array}

and length of latus rectum

\begin{array}{l} = \frac{2 b^{2}}{a} = 3 \end{array}

Hence

\begin{array}{l} \frac{ℓ}{e^{2}} = \frac{3}{3 / 4} = 4 \end{array}

4. Let

\begin{array}{l} f (x) = max {| x + 1 |, | x + 2 |, \dots \dots, | x + 5 |} \end{array}

Then

\begin{array}{l} \int_{- 6}^{0} f (x) d x \end{array}

is equal to _______.

Answer (21)

Sol.

\begin{array}{l} \int_{- 6}^{0} f (x) d x = 2 [\frac{1}{2} (2 + 5) 3] = 21 \end{array}

5. Let the solution curve y = y(x) of the differential equation (4 + x²)dy – 2x(x² + 3y + 4)dx = 0 pass through the origin. Then y(2) is equal to _______.

Answer (12)

Sol. (4 + x²) dy – 2x(x² + 3y + 4)dx = 0

⇒

\begin{array}{l} \frac{d y}{d x} = (\frac{6 x}{x^{2} + 4}) y + 2 x \end{array}

⇒

\begin{array}{l} \frac{d y}{d x} - (\frac{6 x}{x^{2} + 4}) y = 2 x \end{array}

I.F. =

\begin{array}{l} e^{- 3 In (x^{2} + 4)} = \frac{1}{{(x^{2} + 4)}^{3}} \end{array}

So

\begin{array}{l} \frac{y}{{(x^{2} + 4)}^{3}} = \int \frac{2 x}{{(x^{2} + 4)}^{3}} d x + c \end{array}

⇒

\begin{array}{l} y = - \frac{1}{2} (x^{2} + 4) + c {(x^{2} + 4)}^{3} \end{array}

When x = 0, y = 0 gives

\begin{array}{l} c = \frac{1}{32} \end{array}

So, for x = 2,y = 12

6. If sin²(10°)sin(20°)sin(40°)sin(50°)sin(70°)

\begin{array}{l} = α - \frac{1}{16} \sin (10^{\circ}) \end{array}

, then 16 + α^–1 is equal to _______.

Answer (80)

Sol: (sin10° ⋅ sin50° ⋅ sin70°) ⋅ (sin10° ⋅ sin20° ⋅ sin40°)

=

\begin{array}{l} (\frac{1}{4} \sin 30^{\circ}) \cdot [\frac{1}{2} \sin 10^{\circ} (\cos 20^{\circ} - \cos 60^{\circ})] \end{array}

=

\begin{array}{l} \frac{1}{16} [\sin 10^{\circ} (\cos 20^{\circ} - \frac{1}{2})] \end{array}

=

\begin{array}{l} \frac{1}{32} [2 \sin 10^{\circ} \cdot \cos 20^{\circ} - \sin 10^{\circ}] \end{array}

=

\begin{array}{l} \frac{1}{32} [\sin 30^{\circ} - \sin 10^{\circ} - \sin 10^{\circ}] \end{array}

=

\begin{array}{l} \frac{1}{64} - \frac{1}{16} \sin 10^{\circ} \end{array}

Clearly

\begin{array}{l} α = \frac{1}{64} \end{array}

Hence 16 + α^–1 = 80

7. Let A = {n∈N : H.C.F. (n, 45) = 1} and

Let B = {2k :k∈ {1, 2, …,100}}. Then the sum of all the elements of A∩B is ___________.

Answer (5264)

Sol: Sum of all elements of A∩B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]

=

\begin{array}{l} 2 [\frac{100 \times 101}{2} - 3 (\frac{33 \times 34}{2}) - 5 (\frac{20 \times 21}{2}) + 15 (\frac{6 \times 7}{2})] \end{array}

= 10100 – 3366 – 2100 + 630

= 5264

8. The value of the integral

\begin{array}{l} \frac{48}{π^{4}} \int_{0}^{π} (\frac{3 π x^{2}}{2} - x^{3}) \frac{\sin x}{1 + \cos^{2} x} d x \end{array}

is equal to ________.

Answer (6)

Sol:

\begin{array}{l} I = \frac{48}{π^{4}} \int_{0}^{π} [{(\frac{π}{2} - x)}^{3} - \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{\sin x d x}{1 + \cos^{2} x} \end{array}

Using

\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{array}

we get

\begin{array}{l} I = \frac{48}{π^{4}} \int_{0}^{π} [- {(\frac{π}{2} - x)}^{3} + \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{\sin x d x}{1 + \cos^{2} x} \end{array}

Adding these two equations, we get

\begin{array}{l} 2 I = \frac{48}{π^{4}} \int_{0}^{π} \frac{π^{3}}{2} \cdot \frac{\sin x d x}{1 + \cos^{2} x} \end{array}

⇒

\begin{array}{l} I = \frac{12}{π} {[- \tan^{- 1} (\cos x)]}_{0}^{π} = \frac{12}{π} \cdot \frac{π}{2} = 6 \end{array}

9. Let

\begin{array}{l} A = \sum_{i = 1}^{10} \sum_{j = 1}^{10} min {i, j} \end{array}

and

\begin{array}{l} B = \sum_{i = 1}^{10} \sum_{j = 1}^{10} max {i, j} \end{array}

. Then A + B is equal to ____________.

Answer (1100)

Sol: Each element of ordered pair {i, j} is either present in A or in B.

So,A + B =Sum of all elements of all ordered pairs {i, j} for 1 ≤i≤ 10 and 1 ≤j≤ 10

= 20 (1 + 2 + 3+…. + 10)

= 1100

10. Let

\begin{array}{l} S = (0, 2 π) - {\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}} \end{array}

. Let y = y(x), x∈S, be the solution curve of the differential equation

\begin{array}{l} \frac{d y}{d x} = \frac{1}{1 + \sin 2 x}, y (\frac{π}{4}) = \frac{1}{2} \end{array}

. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve

\begin{array}{l} y = \sqrt{2} \sin x \end{array}

is

\begin{array}{l} \frac{k π}{12} \end{array}

, then k is equal to _________.

Answer (42)

Sol:

\begin{array}{l} \frac{d y}{d x} = \frac{1}{1 + \sin 2 x} \end{array}

⇒

\begin{array}{l} d y = \frac{\sec^{2} x d x}{{(1 + \tan x)}^{2}} \end{array}

⇒

\begin{array}{l} y = - \frac{1}{1 + \tan x} + c \end{array}

When

\begin{array}{l} x = \frac{π}{4}, y = \frac{1}{2} gives c = 1 \end{array}

So

\begin{array}{l} y = \frac{\tan x}{1 + \tan x} \Rightarrow y = \frac{\sin x}{\sin x + \cos x} \end{array}

Now,

\begin{array}{l} y = \sqrt{2} \sin x \Rightarrow \sin x = 0 \end{array}

or

\begin{array}{l} \sin x + \cos x = \frac{1}{\sqrt{2}} \end{array}

sinx = 0 gives x = π only.

and

\begin{array}{l} \sin x + \cos x = \frac{1}{\sqrt{2}} \Rightarrow \sin (x + \frac{π}{4}) = \frac{1}{2} \end{array}

So

\begin{array}{l} x + \frac{π}{4} = \frac{5 π}{6} or \frac{13 π}{6} \Rightarrow x = \frac{7 π}{12} or \frac{23 π}{12} \end{array}

Sum of all solutions =

\begin{array}{l} π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{42 π}{12} \end{array}

Hence k = 42.

❑ ❑ ❑

2,477

3,767

4,411

$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$
$JEE Main 2022 Maths Question Paper with Solutions June 26 – Shift 1$

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