Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

1

Active Users

0

Followers

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

( i ) W e h a v e , A = { x : x R , 2 x + 1 1 = 1 5 } 2 x + 1 1 = 1 5 2 x = 1 5 1 1 2 x = 4 x = 2 A = { 2 } ( i i ) W e h a v e , B = { x | x 2 = x , x R } x 2 = x x 2 x = 0 x ( x 1 ) = 0 x = 0 , 1 B = { 0 , 1 } ( i i i ) W e h a v e , C = { x | x i s a p o s i t i v e f a c t o r o f p r i m e n u m b e r p . } Since,positivefactorsofaprimenumberare1andthenumberitself. C = { 1 , p }

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Answer: (a) A catalyst favours the formation of products

New question posted

a year ago

0 Follower 2 Views

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Answer: (d) decreases

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Answer: (a) increases

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

(i) f(x)=sin x cos x

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?0sin(x+h)cos(x+h)?sinxcosxh

=limh?012h*[2sin(x+h)cos(x+h)?2sinxcosx]

=limh?012h[sin2(x+h)?sin2x]

=limh?012h[2cos2(x+h)+2x2sin2(x+h)?2x2]

=limh?01h[cos(2x+h)sinh]

=limh?0cos(2x+h)*limh?0sinhh

=cos(2x+0)

=cos2x

(ii) f(x)=secx

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[sec(x+h)?secx]

=limh?01h[1cos(x+h)?1cosx]

=limh?01h[cosx?cos(x+h)cos(x+h)cosx]

=limh?01h[?2sin(x+x+h2)sin(x?(x+h)2)cos(x+h)cosx]

=limh?01h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]

=limh?0(?1sin(2x+h2)cos(x+h)cosx*limh?0(?1)sinh/2h/2

=sinxcosx?cosx*1

=tanx?secx.

(iii) Given f(x)=5 sec x+4 cosx.

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?05sec(x+h)+4cos(x+h)?[5secx+4cosx]h.

=limh?05h[sec(x+h)?secx]+limh?04h[cos(x+h)?cosx]

=limh?05h[1cos(x+h)?1cosx]+limh?04h[?2sin(x+h+x2)sin(x+h?x2)]

=limh?05h[cosx?cos(x+h)cos(x+h)(cosx)]+limh?04h[?2sin(2x+h2)sinh2]

=limh?05h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]?4limh?0sin(2x+h2)limh?0sinh/2h/2

=sin(2x+02)cos(x+0)cosx*1?4sin(2x2)

=5sinxcosx?1cosx?4sinx

=5tanx?secx?4sinx

(iv) Given f(x)=cosecx

f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[cosec(x+h)?cosecx]

=limh?01h[1sin(x+h)?1sinx]

=limh?01h[sinx?sin(x+h)sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)]sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx]

=limh?01h[2cos(2x+h2)sin(?h/2)]sin(x+h)sinx]

=limh?0cos(2x+h2)sin(x+h)sinx*(?1)sin(2)h/2)

=cos(2x+02)sin(x+0)sinx*(?1)

=?cosxsinx*1sinx

=?cotx?cosecx

(v) Given,f(x)=3 cot x+5cosecx.

So, f?(x)=limh?0f(x+h)?f(x)h =2cos(x+0)cosx*1+7sin(2x+02)*(?1)

=limh?03cot(x+h)+5cosec(x+h)?[3cotx+5cosx)

h?03h[cot(x+h)?cotx]+limh?0

=?5h[cosec(x+h)?cosecx]

=limh?03h[cos(x+h)sin(x+h)?cosxsinx]+limh?05h[1sin(x+h)?1sinx]

=limh?03h[cos(x+h)sinx?cosxsin(x+h)sin(x+h)sinx]+limh?05h[sinx?sin(x+h)sin(x+h)sinx]

=limh?03h[sin(x?(x+h))sin(x+h)sinx+limh?05h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx

=limh?03sin(x+h)sinx*limh?0(?1)sinhh+limh?05h[2cos(2x+h2)sin(?h/2)sin(x+h)sinx

=3sinx?sinx*(?1)+5limh

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

29.

Given, f (x) = (x-a1) (x-a2)… (x-an)

So,  limxa1f (x)=limxa1 (xa1)limxa1 (xa2)? limxa1 (xan)

= (a1-a1) (a1-a2) … (a1-an)

= 0 (a1-a2) … (a1-an)

 = 0

And limxaf (x)=limxa (xa1)limxa (xa2)limxa (xan)

= (a-a1) (a-a2) … (a-an)

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

17. Kindly go through the solution

==limx02sin2x2sin2x2

=limx0 (sinxx)2*x2limx0 (sinx2x2)2x22

= (1)2*x2*4 (1)2*x2

= 4

New question posted

a year ago

0 Follower 2 Views

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

44. Secx = 2.                      .

We have, Secx = 2, | Secx is (+)ve  the principal solution lies in Ist and IVthquadrent.

π3 and 6ππ3

π3 and 3 .

As secx = sec π3 . cosx = cos π3  [secx=1cosx]

The general solution is

x = 2nπ ±π3, n ∈ z.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 703k Reviews
  • 1850k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.