1
Numerical

JEE Main 2021 (Online) 1st September Evening Shift

A man starts walking from the point P($$-$$3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $$5\left( {{{(PR)}^2} + {{(RQ)}^2}} \right)$$ is equal to ____________.
Your Input ________

Answer

Correct Answer is 1250

Explanation


50(PR2 + RQ2)

= 50(20 + 5)

= 50(25)

= 1250
2
Numerical

JEE Main 2021 (Online) 1st September Evening Shift

Let the points of intersections of the lines x $$-$$ y + 1 = 0, x $$-$$ 2y + 3 = 0 and 2x $$-$$ 5y + 11 = 0 are the mid points of the sides of a triangle ABC is _____________.
Your Input ________

Answer

Correct Answer is 6

Explanation

Intersection point of given lines are (1, 2), (7, 5), (2, 3)


$$\Delta = {1 \over 2}\left| {\matrix{ 1 & 2 & 1 \cr 7 & 5 & 1 \cr 2 & 3 & 1 \cr } } \right|$$

$$ = {1 \over 2}[1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$$

$$ = {1 \over 2}[2 - 10 + 11]$$

$$\Delta$$DEF $$ = {1 \over 2}(3) = {3 \over 2}$$

$$\Delta$$ABC = 4$$\Delta$$DEF $$ = 4\left( {{3 \over 2}} \right) = 6$$
3
Numerical

JEE Main 2021 (Online) 20th July Evening Shift

Consider a triangle having vertices A($$-$$2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point $$\left( {0,{\alpha \over 2}} \right)$$, then the value of real number $$\alpha$$ is ________________.
Your Input ________

Answer

Correct Answer is 9

Explanation


$${\left( {\sqrt {50} } \right)^2} = {\left( {\sqrt {45} } \right)^2} + {\left( {\sqrt 5 } \right)^2}$$

$$\angle B = 90^\circ $$

Circum-center $$ = \left( {{1 \over 2},{{11} \over 2}} \right)$$

Mid point of BC $$ = \left( {2,{{17} \over 2}} \right)$$

Line : $$\left( {y - {{11} \over 2}} \right) = 2\left( {x - {1 \over 2}} \right) \Rightarrow y = 2x + {9 \over 2}$$

Passing through $$\left( {0,{\alpha \over 2}} \right)$$

$${\alpha \over 2} = {9 \over 2} \Rightarrow \alpha = 9$$
4
Numerical

JEE Main 2021 (Online) 18th March Morning Shift

A square ABCD has all its vertices on the curve x2y2 = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is
Your Input ________

Answer

Correct Answer is 80

Explanation

x2y2 = 1

$$ \Rightarrow $$ y2 = $${1 \over {{x^2}}}$$

$$ \Rightarrow $$ y = $$ \pm {1 \over x}$$

Graph of this equation,

$$OA \bot OB$$

$$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$$

$$ \Rightarrow {p^2}{q^2} = 1$$

$$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$$ midpoint of AB lies

On $${x^2}{y^2} = 1$$

$$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$$

$$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$$

$$ \Rightarrow {({p^2} - {q^2})^2} = 16$$

$$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$$

$$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$$

$$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $$

$$ \Rightarrow {p^2} = 2 + \sqrt 5 $$ or $$ - 2 + \sqrt 5 $$

$$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$$ or $$ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $$

Area $$ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $$

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