Class 12th

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New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

 f (x)=tan−1 (sinx−cosx)

f' (x)=cosx+sinx (sinx−cosx)2+1 = 0

∴x=3π4

Sum = tan-1 2−π4

= cos−113−π4

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

 f (x)=xex (1−x)

f' (x)=−ex (1−x) (2x+1) (x−1)

f (x)  is  ↑  in   (−12, −1)

New answer posted

a year ago

0 Follower 42 Views

V
Vishal Baghel

Contributor-Level 10

Note : n should be given as a natural number:

f (x)= {−sin (x−1)x−1, x<−1− (sin2+1), x=−1cos2πx, −1<x<1          1                    ,             x=1

−sin (x−1)x−1, x>1

f (x) is discontinuous at x = 1 and x = 1

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

 f (0)+3+λ+4=14

∴f (0)=7−λ=c

f (1)=a+b+c=3 ……. (i)

f (3)=9a+3b+c=4 …… (ii)

f (−2)=4a−2b+c=λ …. (iii)

(ii) – (iii)

a+b=4−λ5,  put in equation (i)

6λ=24λ=4

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Given that AT=A, BT=−B

(A) C=A4−B4

=A4−B4=C

(B)C = AB – BA

=BTAT−ATBT=−BA+AB=C

(C) C=B5−A5

CT= (B5−A5)T= (B5)T− (A5)T=−B5−A5

(D)C = AB + BA

= BA – AB = C

∴ option is true

New question posted

a year ago

0 Follower 3 Views

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 a=−1α2−1β2−2, b=1α2+1β2+1+1α2β2

6x2+17x+7=0, x=−73, x=−12 are roots

Both roots, real and negative

New answer posted

a year ago

0 Follower 38 Views

V
Vishal Baghel

Contributor-Level 10

S ∩ T = { − 5 , − 4 , 3 }

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

The following statement is true.

New answer posted

a year ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

The following statement is false.

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