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  • 1 answers

Shaina Rajput 6 years, 10 months ago

I think question is wrong coz the possibile terms in the expansion is 3. So, how can we find 6th term
  • 1 answers

Gaurav Seth 6 years, 10 months ago

The component statements of given statement are:
q : x is an integer and x2  is even
r : x is an even integer.
In method of contrapositive,
We assume that ~r is true and we have to prove that ~q is also true.
Now, ~r is true,
Therefore, x is not an even integer or x is an odd integer.

or x = 2n + 1 for some integer n.
or 
or 
or 
or  is an odd integer.
or q is false.
or ~q is true.
Thus, 
Hence, the given statement p is true.

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Ansh Kukreja 6 years, 10 months ago

1
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2#2
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Shaina Rajput 6 years, 10 months ago

?
  • 2 answers

Mohit Shrivastav 6 years, 10 months ago

Divide numerator and denominator by x , Then ==> limit x->0 tanx/x/sin3x/x ==> lim x->0 1/sin3x/x (lim y-->0 tany/y=1) Now multiply and divide denominator by 3 ==> lim x->0 1/sin3x*3/x*3 ==> lim x->0 1/1*3 (lim y-->0 siny/y=1) ==> 1/3 I hope, this will help you.

Lokesh Chayal 6 years, 10 months ago

1/3
  • 2 answers

Aman Kumar 6 years, 10 months ago

Wel explained answer.. thanks

Shaina Rajput 6 years, 10 months ago

First, its important to find the value of tan x – cot x. So, tan x – cot x = tan x - 1/tan x = tan² x-1/tan x = -2 (1-tan² x/ 2tan x) = -2 ( 1/2tan x ÷ 1- tan² x) = -2× 1/tan² x = 2cot 2x ....(1) . Now— tan x + 2 tan 2x + 4 tan 4x + 8 cot 8x = tan x – cot x + 2 tan 2x + 4 tan 4x + 8 cot 8x + cot x = –2 cot 2x + 2 tan 2x + 4 tan 4x + 8 cot 8x + cot x = 2 (tan 2x – cot 2x) + 4 tan 4x + 8 cot 8x + cot x = 2 (– 2 cot 4x) + 4 tan 4x + 8 cot 8x + cot x = 4 (tan 4x – cot 4x) + 8 cot 8x + cot x = 4 (– 2 cot 8x) + 8 cot 8x + cot x = – 8 cot 8x + 8 cot 8x + cot x = cot x . Hence, tan x + 2 tan 2x + 4 tan 4x + 8 cot 8x = cot x
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Mohit Shrivastav 6 years, 10 months ago

Let the sq. root of √ -15-8i=+-(x+iy) On Sq. both sides we get, -15-8i=x^2+(yi)^2+2xyi -15-8i=^2-y^2+2xyi (i^2=-1) on comparing, we get, =x^2-y^2=-15 ........(1) 2xy=8 We know that (x^2+y^2)^2=(x^2-y^2)^2+(2xy) ^2 On putting the values, we get (x^2+y^2)^2=(-15)^2+(8)^2=279 (x^2+y^2)=17 .........(2) On solving (1)and(2),we get X=+-(1),y=+-(4) Hence, √-15-8i=+-(1+4i)
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Ayush Pandey 6 years, 10 months ago

Sb me thoda pr trigo 7, 8 and 14 is most important

Ashish Kumar 6 years, 10 months ago

1 to 27
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Anuj Singh 6 years, 10 months ago

The total no. Of event can happen.
  • 1 answers

Nikhita V 6 years, 10 months ago

Let A = 18. Then Sin 5A=1 which means that sin (2A+3A)=1. Put the formula and solve
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Shaina Rajput 6 years, 10 months ago

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚ ⇒ 2θ = 90˚ - 3A Taking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A ⇒ 2 sin A cos A = 4 cos^3 A - 3 cos A ⇒ 2 sin A cos A - 4 cos^3A + 3 cos A = 0 ⇒ cos A (2 sin A - 4 cos^2 A + 3) = 0 Dividing both sides by cos A = cos 18˚ ≠ 0, we get ⇒ 2 sin θ - 4 (1 - sin^2 A) + 3 = 0 ⇒ 4 sin^2 A + 2 sin A - 1 = 0, which is a quadratic in sin A Therefore, sin θ = −2±−4(4)(−1)√2(4) ⇒ sin θ = −2±4+16√8 ⇒ sin θ = −2±25√8 ⇒ sin θ = −1±√5/4 Now sin 18° is positive, as 18° lies in first quadrant. Therefore, sin 18° = sin A = −1±√5/4
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T S 6 years, 10 months ago

f'(x) =d/dx[cos(x-pi/16)] =-sin(x-pi/16){d/dx(x-pi/16)} =-sin(x-pi/16){1-0} =-sin(x-pi/16).......ANS
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Sb Junaid 6 years, 10 months ago

Domain can only be 3 or more than 3

Nikhita V 6 years, 10 months ago

Use wavy curve method

Anisha Sharma 6 years, 10 months ago

koi answer krdo...??
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Anamika Gupta 6 years, 10 months ago

Total needed area=l*b =29*6=174m sq. Change it into cm=174*100=17400cm sq. Now find the area of square tile=40*40=1600cm sq. Total square tiles needed=17400/1600 =10.87ans.
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Ragraj Singh Chauhan 6 years, 10 months ago

??

Ragraj Singh Chauhan 6 years, 10 months ago

This was my brother's query and he is in 1+1 class

Rajat Singh 6 years, 10 months ago

62+4???
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Nikhita V 6 years, 10 months ago

Domain is all real numbers. Range is R+ U {0}

Ananya Gupta 6 years, 10 months ago

x€R, y€R
  • 1 answers

Amod Kumar 6 years, 10 months ago

Range=[0,4]

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