Maths NCERT Exemplar Solutions Class 11th Chapter Four

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A
alok kumar singh

Contributor-Level 10

|z|-Re (z)≤1 ⇒ √ (x²+y²)-x≤1 ⇒ √ (x²+y²)≤1+x ⇒ y²≤1+2x=2 (x+1/2).

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alok kumar singh

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Locus is auxiliary circle x²+y²=a²=4. (-1, √3) satisfies.

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alok kumar singh

Contributor-Level 10

m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies

 

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alok kumar singh

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Required probability = (? C? +? C? )/¹¹C? = 25/165=5/33.

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alok kumar singh

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f (x+y)=f (x)f (y). f (n)=f (1)?
Σf (x)=f (1)/ (1-f (1)=2 ⇒ f (1)=2/3.
f (4)/f (2) = (f (1)? )/ (f (1)²) = f (1)² = 4/9.

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alok kumar singh

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Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

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alok kumar singh

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√ (1+x²) (1+y²) + xy (dy/dx)=0.
√ (1+x²)/x dx + √ (1+y²)/y dy = 0.
√ (1+x²)+½ln| (√ (1+x²)-1)/ (√ (1+x²)+1)|+√ (1+y²)=C.

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alok kumar singh

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Required area = 4 [∫? ¹/² 2y²dy + ½*½*1] = 5/6.

 

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alok kumar singh

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α+β=64; αβ=256=2?
(α³/β? )¹/? + (β³/α? )¹/? = (α+β)/ (αβ)? /? = 64/32=2.

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