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JEE Mathematics Solved Question Paper 2020

Multiple Choice Questions

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $∠$ BAC = 90°, and ar( $∆$ ABC) = 55sq. units, then the abscissa of the vertex C is :

$1 + \sqrt{5}$
$2 + \sqrt{5}$
$1 + 2 \sqrt{5}$
$2 \sqrt{5} - 1$

$If 1 + (1 - 2^{2} . 1) + (1 - 4^{2} . 3) + (1 - 6^{2} . 5) + . . . + (1 - 20^{2} . 19) = α - 220 β then an ordered pair (α, β) = ?$

(11, 97)
(10, 103)
(11, 103)
(10, 97)

Let P(3, 3) be a point on the hyperbola, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If the normal to it at P intersects the x-axis at (9,0) and e is its eccentricity, then the ordered pair (a², e²) is equal to :

$(\frac{9}{2}, 3)$
$(\frac{3}{2}, 2)$
$(9, 3)$
$(\frac{9}{2}, 2)$

The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)$ be agiven ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function $φ (t) = \frac{5}{12} + t - t^{2}$ then a²+ b² is equal to

$Let α and β be the roots of x^{2} - 3 x + p = 0 and γ and δ be the roots of x^{2} - 6 x + q = 0 . If α, β, γ, δ from a geometric progression . Then ratio (2 q + p) : (2 q - p) is$

3 : 1
5 : 3
9 : 7
33 : 31

7.
$If A = [\begin{matrix} \cos (θ) & isin (θ) \\ isin (θ) & \cos (θ) \end{matrix}], (θ = \frac{π}{24}) and A^{5} = [\begin{matrix} a & b \\ c & d \end{matrix}], where i = \sqrt{- 1}, then which one of the following is not true ?$

$0 \leq a^{2} + b^{2} \leq 1$

$a^{2} - d^{2} = 0$

$a^{2} - b^{2} = \frac{1}{2}$

$a^{2} - c^{2} = 1$

$a^{2} - b^{2} = \frac{1}{2}$

$A^{2} = [\begin{matrix} \cos (θ) & isin (θ) \\ isin (θ) & \cos (θ) \end{matrix}] [\begin{matrix} \cos (θ) & isin (θ) \\ isin (θ) & \cos (θ) \end{matrix}] = [\begin{matrix} \cos^{2} (θ) - \sin^{2} (θ) & 2 isin (θ) \cos (θ) \\ 2 iisin (θ) \cos (θ) & \cos^{2} (θ) - \sin^{2} (θ) \end{matrix}] = [\begin{matrix} \cos (2 θ) & isin (2 θ) \\ isin (2 θ) & \cos (2 θ) \end{matrix}] A^{3} = [\begin{matrix} \cos (2 θ) & isin (2 θ) \\ isin (2 θ) & \cos (2 θ) \end{matrix}] [\begin{matrix} \cos (θ) & isin (θ) \\ isin (θ) & \cos (θ) \end{matrix}]$

$= [\begin{matrix} \cos (3 θ) & isin (3 θ) \\ isin (3 θ) & \cos (3 θ) \end{matrix}] A^{5} = [\begin{matrix} \cos (5 θ) & isin (5 θ) \\ isin (5 θ) & \cos (5 θ) \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] a = \cos (5 θ), b = isin (5 θ), c = isin (5 θ), d = \cos (5 θ) a = d, b = c (1) a^{2} + b^{2} = \cos^{2} (5 θ) - \sin^{2} (5 θ) = \cos (10 θ) \in (0, 1) as θ = \frac{π}{24} (2) a^{2} - d^{2} = 0 (3) a^{2} - b^{2} = \cos^{2} (5 θ) + \sin^{2} (5 θ) = 1 (4) a^{2} - c^{2} = a^{2} - b^{2} = 1$

$Let y = y (x) be the solution of the differential equation, xy' - y = x^{2} (xcosx + sinx), x > 0 . If y (π) = π, then y'' (\frac{π}{2}) + y (\frac{π}{2}) is equal to :$