Question

matrix A=[2 5 ] then det [A²⁰²⁰+2A²⁰¹⁹] is
[1 3]

Answer 1

Answer:

Answer

Correct option is

A

1−y

2

−log

∣

∣

∣

∣

∣

∣

2

1+

1−y

2

∣

∣

∣

∣

∣

∣

=x+c

B

1−y

2

−log

∣

∣

∣

∣

∣

∣

1−

1−y

2

1+

1−y

2

∣

∣

∣

∣

∣

∣

=−x+c

Since, the length of tangent

=

∣

∣

∣

∣

∣

∣

∣

y

1+(

dy

dx

)

2

∣

∣

∣

∣

∣

∣

∣

=1⇒y

2

(1+(

dy

dx

)

2

)=1

∴

dx

dy

=±

1−y

2

y

⇒∫

y

1−y

2

dy=±∫xdx

⇒∫

y

1−y

2

dy=±x+c

Substitute y=sinθ⇒dy=cosθdθ

∴∫

sinθ

cosθ

.cosθdθ=±x+c⇒∫

sin

2

θ

cos

2

θ

.sinθdθ=±x+c

Again substitute cosθ=t⇒−sinθdθ=dt

∴∫

1−t

2

t

2

dt=±x+c,⇒∫(1−

1−t

2

1

)dt=±x+c

⇒t−log

∣

∣

∣

∣

∣

1−t

1+t

∣

∣

∣

∣

∣

=±x+c⇒

1−y

2

−log

∣

∣

∣

∣

∣

∣

1−

1−y

2

1+

1−y

2

∣

∣

∣

∣

∣

∣

=±x+c