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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

f : R defined as

f ( x ) = x − 1     a n d       g ( x ) : R − { 1 , − 1 } → R

g ( x ) = x 2 x 2 − 1

∴ f o g ( x ) = x 2 x 2 − 1 − 1 = 1 x 2 − 1

domain of fog (x) R – {-1, 1} and range   ( − ∞ , − 1 ] ∪ ( 0 , ∞ ) ∴

 fog (x) is neither one-one nor onto

New answer posted

a year ago

0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

(xa)n+ (yb)n=2

⇒na (xa)n−1+nb (yb)n−1dydx=0

⇒dydx=−ba (bxay)n−1

dydx (a, b)=−ba

So line always touches the given curve.

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

 f (x)=|x2−3x−2|−x

=| (x−3−172) (x−3+172)|−x

⇒f (x)= [x2−4x−2−1≤x≤3−172−x2+2x+23−172<x≤2]

absolute minimum f (3−172)=−3+172

absolute maximum = 3

∴sum  3+−3+172=3+172

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

GOF is differentiable at x = 0

So R.H.D = L.H.D.

d d x ( 4 e x + k 2 ) = d d x ( ( − | x + 3 | ) 2 − k 1 | x + 3 | )                

⇒ 4 = 6 – k1 Þ k1 = 2

Now g (f (-4) + g (f (4)

= 2 (2e4 – 1)

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

limx→12sin (cos−1x)−x1−tan (cos−1x)

let cos−1x=π4+θ

= limθ→∞2sinθ−2tanθ (1−tanθ)=−1

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

2021≡−2 mod (7)

⇒ (2021)2023≡ (−2)2023mod (7) …… (i)

Now,   (−2)3≡−1mod (7)

⇒ (−2)2023≡ (−2)mod (7)≡5mod (7) ……. (ii)

(i) & (ii)

⇒ (2021)2023≡5mod (7)

∴ Remainder = 5

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

System of equation can be written as

(3−215−8921a) (xyz)= (b3−1)

(3−2115−2427633a) (xyz)= (b9−3)

R3−2R1, R2−5R1

for no solution

3a + 9 = 0 but 32−9b2≠0

⇒a=−3            ⇒b≠13

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

|adj (24A)|=|adj (3adj (2A))|

⇒|24A|2=|3adj (2A)|2

246|A|2=36. (23)4|A|4

|A|2=24636.212=218.3636.212=26

New question posted

a year ago

0 Follower 4 Views

New question posted

a year ago

0 Follower 2 Views

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