Maths NCERT Exemplar Solutions Class 11th Chapter Eleven
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New answer posted
a month agoContributor-Level 10
Co-ordinate of Q (b+2, a)
⇒ 1/√2 + 7i/√2 = (b+2+ai)e^ (iπ/4)
= (b+2+ai) (cos (π/4)+isin (π/4)
⇒ b−a+2=−1
b+2+a=7
⇒a=4
b=1
⇒2a+b=9
New answer posted
a month agoContributor-Level 10
mean = Σx? f? /Σf? = (32+8α+9β)/ (8+α+β)=6
⇒2α+3β=16 . (i)
d? =x? −x? =−4,0,2,3
f? d? ²=64,0,4α,9β
Variance σ²=Σf? d? ²/Σf? =6.8
⇒ (64+4α+9β)/ (8+α+β)=6.8
⇒2.8α+ (−2.2β)=9.6
⇒28α−22β=96
14α−11β=48 . (ii)
Solving (i) and (ii), 
⇒β=2, α=5
New mean=Σx? f? /Σf? =85/15=17/3
New answer posted
a month agoContributor-Level 10
2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)
Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0
New answer posted
a month agoContributor-Level 10
f (x)= {sinx,  0≤x2; 1, /2x 2+cosx, x>π}
f' (x)= {cosx,  0
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 0
⇒ f (x) is differentiable in (0, ∞)
New answer posted
a month agoContributor-Level 10
. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12
New answer posted
a month agoContributor-Level 10
for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
 (x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
 (y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric
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