NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### JEE Advanced 2021 Paper 2 Online

Let $$M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\}$$, where r > 0. Consider the geometric progression $${a_n} = {1 \over {{2^{n - 1}}}}$$, n = 1, 2, 3, ...... . Let S0 = 0 and for n $$\ge$$ 1, let Sn denote the sum of the first n terms of this progression. For n $$\ge$$ 1, let Cn denote the circle with center (Sn$$-$$1, 0) and radius an, and Dn denote the circle with center (Sn$$-$$1, Sn$$-$$1) and radius an.
Consider M with $$r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}$$. The number of all those circles Dn that are inside M is
A
198
B
199
C
200
D
201

## Explanation

$$\because$$ $$r = {{({2^{199}} - 1)\sqrt 2 } \over {{2^{198}}}}$$

Now, $$\sqrt 2 {s_{n - 1}} + {a_n} < \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2$$

$$2\sqrt 2 \left( {1 - {1 \over {{2^{n - 1}}}}} \right) + {1 \over {{2^{n - 1}}}} < \left( {{{{2^{199}} - 1} \over {{2^{198}}}}} \right)\sqrt 2$$

$$\therefore$$ $$2\sqrt 2 - {{\sqrt 2 } \over {{2^{n - 2}}}} + {1 \over {{2^{n - 1}}}} < 2\sqrt 2 - {{\sqrt 2 } \over {{2^{198}}}}$$

$${1 \over {{2^{n - 2}}}}\left( {{1 \over 2} - \sqrt 2 } \right) < - {{\sqrt 2 } \over {{2^{198}}}} \Rightarrow {{2\sqrt 2 - 1} \over {2\,.\,{2^{n - 2}}}} > {{\sqrt 2 } \over {{2^{198}}}}$$

$$\Rightarrow {2^{n - 2}} < \left( {2 - {1 \over {\sqrt 2 }}} \right)\,.\,{2^{197}}$$

$$\therefore$$ n $$\le$$ 199

So, number of circles = 199
2

### JEE Advanced 2021 Paper 2 Online

Let $$M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\}$$, where r > 0. Consider the geometric progression $${a_n} = {1 \over {{2^{n - 1}}}}$$, n = 1, 2, 3, ...... . Let S0 = 0 and for n $$\ge$$ 1, let Sn denote the sum of the first n terms of this progression. For n $$\ge$$ 1, let Cn denote the circle with center (Sn$$-$$1, 0) and radius an, and Dn denote the circle with center (Sn$$-$$1, Sn$$-$$1) and radius an.
Consider M with $$r = {{1025} \over {513}}$$. Let k be the number of all those circles Cn that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then
A
k + 2l = 22
B
2k + l = 26
C
2k + 3l = 34
D
3k + 2l = 40

## Explanation

$$\because$$ $${a_n} = {1 \over {{2^{n - 1}}}}$$ and $${S_n} = 2\left( {1 - {1 \over {{2^n}}}} \right)$$

For circle Cn to be inside M.

$${S_{n - 1}} + {a_n} < {{1025} \over {513}} \Rightarrow {S_n} < {{1025} \over {513}}$$

$$\Rightarrow 2\left( {1 - {1 \over {{2^n}}}} \right) < {{1025} \over {513}} \Rightarrow 1 - {1 \over {{2^n}}} < {{1025} \over {1026}}$$

$$\Rightarrow 1 - {1 \over {{2^n}}} < 1 - {1 \over {1026}} \Rightarrow {{ - 1} \over {{2^n}}} < {{ - 1} \over {1026}}$$

$$\Rightarrow {1 \over {{2^n}}} > {1 \over {1026}} \Rightarrow {2^n} < 1026 \Rightarrow n \le 10$$

$$\therefore$$ Number of circles inside be 10 = k. Clearly, alternate circle do not intersect each other i.e. C1, C3, C5, C7, C9 do not intersect each other as well as C2, C4, C6, C8 and C10 do not intersect each other.

Hence, maximum 5 set of circles do not intersect each other.

$$\therefore$$ l = 5

So, 3k + 2l = 40
3

### JEE Advanced 2021 Paper 1 Online

Consider a triangle $$\Delta$$ whose two sides lie on the x-axis and the line x + y + 1 = 0. If the orthocenter of $$\Delta$$ is (1, 1), then the equation of the circle passing through the vertices of the triangle $$\Delta$$ is
A
x2 + y2 $$-$$ 3x + y = 0
B
x2 + y2 + x + 3y = 0
C
x2 + y2 + 2y $$-$$ 1 = 0
D
x2 + y2 + x + y = 0

## Explanation

Equation of circle passing through C(0, 0) is

x2 + y2 + 2gx + 2fy = 0 ..... (i)

Since Eq. (i), also passes through ($$-$$1, 0) and (1, $$-$$2).

Then, 1 $$-$$ 2g = 0 $$\Rightarrow$$ g = 1 / 2

and 5 + 1 $$-$$ 4f = 0 $$\Rightarrow$$ f = 3 / 2

$$\therefore$$ Equation of circumcircle is x2 + y2 + 2 $$\times$$ $${1 \over 2}$$x + 2 $$\times$$ $${3 \over 2}$$y = 0

i.e. x2 + y2 + x + 3y = 0
4

### JEE Advanced 2019 Paper 1 Offline

A line y = mx + 1 intersects the circle $${(x - 3)^2} + {(y + 2)^2}$$ = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate $$- {3 \over 5}$$, then which one of the following options is correct?
A
6 $$\le$$ m < 8
B
$$-$$3 $$\le$$ m < $$-$$1
C
4 $$\le$$ m < 6
D
2 $$\le$$ m < 4

## Explanation

It is given that points P and Q are intersecting points of circle $${(x - 3)^2} + {(y + 2)^2}$$ = 25 .....(i)

Line y = mx + 1 .....(ii)

And, the mid-point of PQ is A having x-coordinate $$- {3 \over 5}$$

so y-coordinate is $$1 - {3 \over 5}$$ m.

So, $$A\left( { - {3 \over 5},1 - {3 \over 5}m} \right)$$

From the figure,

$$\because$$ AC $$\bot$$ PQ

$$\Rightarrow$$ (slope of AC) $$\times$$ (slope of PQ) = $$-$$1

$$\Rightarrow$$ $$\left( {{{ - 2 - 1 + {3 \over 5}m} \over {3 + {3 \over 5}}}} \right) \times m = - 1$$

$$\Rightarrow {{(3/5)m - 3} \over {18/5}}m = - 1$$

$$\Rightarrow \left( {{{3m - 15} \over {18}}} \right)m = - 1$$

$$\Rightarrow 3{m^2} - 15m + 18 = 0$$

$$\Rightarrow {m^2} - 5m + 6 = 0$$

$$\Rightarrow$$ m = 2 or 3

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12