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New answer posted

10 months ago

0 Follower 6 Views

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Payal Gupta

Contributor-Level 10

(2x3+3xk)12

gen term =

=12Cr212r.3r.x363rrk

For constant term

36 – 3r – rk = 0

k=363rr

for r = 1, 2, 4

12Cr212-r>28

Possible values of k = 3, 1

New answer posted

10 months ago

0 Follower 15 Views

P
Payal Gupta

Contributor-Level 10

1 < a1 < a2 ……18 < 77

77 = 1 + (20 – 1) . d If n numbers are in A.P.

76 = 19 * d ⇒ d = 4

⇒ a1 = 5

a 1 + a 2 + . . . . + a 1 8 = 1 8 2 [ 2 * 5 + 1 7 * 4 ] = 7 0 2          

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

y=2x2+x+2.... (i)

dydx=4x+1

Slope of normal

dxdy=14x+1

Equation of PQ y - β = 14α+1 (xα)

It passes (6, 4)

(4β) (4α+1)= (6α)

4α3+3α23α3=0

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

R (10+α3, 83)

|mAQ|=|mAP|

|45α|=|32α|

= 7 not possible α=237.7α+3β=23+8=31

New answer posted

10 months ago

0 Follower 102 Views

P
Payal Gupta

Contributor-Level 10

 a=2i^+j^+3k^

b=3i^+3j^+k^c=c1i^+c2j^+c3k^

Coplnanar|213331c1c2c3|=0

8c1+7c2+12c3=0........(i)a.c=52c1+c2+3c3=5........(ii)b.c=03c1+3c2+c3=0........(iii)

Solving (i), (ii), (iii)

C1=10122,c2=85122,c3=225122

122(c1+c2+c3)=150

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

 Meani=115xi15=8

i=115xi=120....... (i)

S.D = 3

i=115xi2=15 (9+64)=15*73.......... (ii)

Given 20 has misread as 5

 in new case

q=115xi2=15*7325+400=1470

mean in new case

x¯=1205+2015

variance in new case

σnew2=115 (1470)81=17

New answer posted

10 months ago

0 Follower 29 Views

P
Payal Gupta

Contributor-Level 10

e4x+4e3x58e2x+4ex+1=0

(e2x+1e2x)+4 (ex+1ex)58=0

(ex+1ex+2)2=64

ex=6±322=3±2

New answer posted

10 months ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

A = {1, 2, 3, ….50}

R1 = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R2 = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

R 1 R 2 contains only 9 elements.

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Consider the following image

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

tanπ8=2h80......... (i)

tanθ=h80.......... (ii)

tanπ8=2tanθ

tan2θ=3224

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