Ncert Solutions Maths class 12th

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R
Raj Pandey

Contributor-Level 9

Kindly considere the following Image 

 

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R
Raj Pandey

Contributor-Level 9

f' (x)= (x- (1+x)ln (1+x)/ (x² (1+x). Let h (x)=x- (1+x)ln (1+x).
h' (x)=-ln (1+x). h' (x)>0 for x∈ (-1,0), <0 for x (0, ).
h (0)=0, so h (x)≤0. f' (x)≤0. f is decreasing.

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Raj Pandey

Contributor-Level 9

Normal to plane is n= (-4i+5j+7k).
Plane: -4 (x-3)+5 (y-1)+7 (z-1)=0 ⇒ -4x+5y+7z=0.
Passes through (α, -3,5) ⇒ -4α-15+35=0 ⇒ α=5.

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V
Vishal Baghel

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Let B? be the event where Box-I is selected and B? →where box-II selected
P (B? )=P (B? )=1/2
Let E be the event where selected card is non prime.
For B? : Prime numbers: {2,3,5,7,11,13,17,19,23,29}
For B? : Prime numbers: {31,37,41,43,47}
P (E)=P (B? )*P (E/B? )+P (B? )P (E/B? )
= 1/2*20/30+1/2*15/20
Required probability:
P (B? /E) = (P (E/B? )P (B? )/P (E) = (1/2*20/30)/ (1/2*20/30+1/2*15/20) = (2/3)/ (2/3+3/4) = 8/17

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V
Vishal Baghel

Contributor-Level 10

Two points on the line (say) x/3 = y/2, z=1 are (0,0,1) and (3,2,1)
So dr's of the line is (3,2,0)
Line passing through (1,2,1), parallel to L and coplanar with given plane is r = i+2j+k + t (3i+2j), t∈R (-2,0,1) satisfies the line (for t=-1)
⇒ (-2,0,1) lies on given plane.
Answer of the question is (B)
We can check other options by finding equation of plane
Equation plane: |x-1, y-2, z-1; 1+2, 2-0, 1-1; 2+2, 1-0, 2-1| = 0
⇒ 2 (x-1)-3 (y-2)-5 (z-1)=0
⇒ 2x-3y-5z+9=0

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