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1

### JEE Main 2021 (Online) 27th August Evening Shift

Let A(a, 0), B(b, 2b + 1) and C(0, b), b $$\ne$$ 0, |b| $$\ne$$ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :
A
$${{ - 2b} \over {b + 1}}$$
B
$${{2b} \over {b + 1}}$$
C
$${{2{b^2}} \over {b + 1}}$$
D
$${{ - 2{b^2}} \over {b + 1}}$$

## Explanation

$$\left| {{1 \over 2}\left| {\matrix{ a & 0 & 1 \cr b & {2b + 1} & 1 \cr 0 & b & 1 \cr } } \right|} \right| = 1$$

$$\Rightarrow \left| {\matrix{ a & 0 & 1 \cr b & {2b + 1} & 1 \cr 0 & b & 1 \cr } } \right| = \pm \,2$$

$$\Rightarrow a(2b + 1 - b) - 0 + 1({b^2} - 0) = \pm \,2$$

$$\Rightarrow a = {{ \pm \,2 - {b^2}} \over {b + 1}}$$

$$\therefore$$ $$a = {{2 - {b^2}} \over {b + 1}}$$ and $$a = {{ - 2 - {b^2}} \over {b + 1}}$$

Sum of possible values of 'a' is

$$= {{ - 2{b^2}} \over {a + 1}}$$
2

### JEE Main 2021 (Online) 27th August Evening Shift

Let $$A = \left( {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right)$$, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval :
A
[68, 69)
B
[62, 63)
C
[65, 66)
D
[60, 61)

## Explanation

$$\left| {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right| = 192$$

R1 $$\to$$ R1 $$-$$ R3 & R2 $$\to$$ R2 $$-$$ R3

$$\left[ {\matrix{ 1 & 0 & { - 1} \cr 0 & 1 & { - 1} \cr {[x]} & {[x] + 2} & {[x] + 4} \cr } } \right] = 192$$

$$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$
3

### JEE Main 2021 (Online) 27th August Morning Shift

If the matrix $$A = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is :
A
$${1 \over 2}$$
B
$$-$$$${1 \over 2}$$
C
$$-$$1
D
1

## Explanation

Given matrix $$A = \left[ {\matrix{ 0 & 2 \cr k & { - 1} \cr } } \right]$$

$${A^4} + 3IA = 2I$$

$$\Rightarrow {A^4} = 2I - 3A$$

Also characteristic equation of A is $$|A - \lambda I|\, = 0$$

$$\Rightarrow \left| {\matrix{ {0 - \lambda } & 2 \cr k & { - 1 - \lambda } \cr } } \right| = 0$$

$$\Rightarrow \lambda + {\lambda ^2} - 2k = 0$$

$$\Rightarrow A + {A^2} = 2K.I$$

$$\Rightarrow {A^2} = 2KI - A$$

$$\Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$

Put $${A^2} = 2KI - A$$

and $${A^4} = 2I - 3A$$

$$2I - 3A = 4{K^2}I + 2KI - A - 4AK$$

$$\Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)$$

$$\Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)$$

$$\Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)$$

$$\Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0$$

$$\Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0$$

$$\Rightarrow K = {1 \over 2}$$
4

### JEE Main 2021 (Online) 26th August Evening Shift

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right)$$. Then A2025 $$-$$ A2020 is equal to :
A
A6 $$-$$ A
B
A5
C
A5 $$-$$ A
D
A6

## Explanation

$$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^n} = \left[ {\matrix{ 1 & 0 & 0 \cr {n - 1} & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$

$${A^{2025}} - {A^{2020}} = \left[ {\matrix{ 0 & 0 & 0 \cr 5 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$

$${A^6} - A = \left[ {\matrix{ 0 & 0 & 0 \cr 5 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]$$

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