Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

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alok kumar singh

Contributor-Level 10

f' (x)=g (x)
f" (x)=g' (x)
⇒g (x).g' (x)=f' (x).f" (x)
f (x) has five roots
⇒f' (x) has atleast 4 roots.
And f" (x) has atleast of 3 roots
⇒g (x).g' (x)=0 has atleast 7 roots in (a, b)

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alok kumar singh

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General form Σ? ? C? (1/2)? < 1/2
Σ? ? C? < 2?
⇒? C? +? C? +.? C? < 128
⇒256− (? C? +? C? +.? C? ) < 128
⇒? C? +? C? +.? C? > 128
n≥5

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alok kumar singh

Contributor-Level 10

p: there exist M>0
Such that x≥M for all x∈S
Obviously ~p: M>0 such that x. Negation of 'there exists' is 'for all'.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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alok kumar singh

Contributor-Level 10

Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
Equation of plane
x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
Adding (i) and (ii)
α+5−β+12−8=16
α−β+17=24
⇒α−β=7

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alok kumar singh

Contributor-Level 10

Lt? →? x/ (1−sinx)¹/? − (1+sinx)¹/? )
= 2x/ (1−sinx)¹/? − (1+sinx)¹/? ) Multiply by conjugate
= 4x/ (1−sinx)¹/²− (1+sinx)¹/²) Multiply by conjugate
= 8x/ (1−sinx−1−sinx) Multiply by conjugate
= 4x/sinx = −4

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alok kumar singh

Contributor-Level 10

[3¹/²]^ (log? (25? ¹+7)+ [3¹/? ]^ (log? (5? ¹+1) = 180
⇒ 45 (5²? ²+7) / (5? ¹+1) = 180
⇒ (5²? ²+7)/ (5? ¹+1) = 4
Put 5? ¹=t
⇒ (t²+7)/ (t+1) = 4
⇒ t²−4t+3=0
⇒t=1,3
⇒5? ¹=1 or 5? ¹=3
⇒x=1 or x−1=log?3

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alok kumar singh

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S? : |z−2|≤1 is circle with centre (2,0) and radius less than equal to 1.
S? : z (1+i)+z? (1−i)≥4
Put z=x+iy
y≤x−2
Solving with S1
⇒x=2−1/√2, y=-1/√2
Point of intersection P= (2−1/√2, −i/√2)
|z−5/2|² = | (2−1/√2)−i (1/√2)−5/2|² = (5√2+4)/4√2 = (5+2√2)/4

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alok kumar singh

Contributor-Level 10

α=max {2? sin³?2? cos³? }
=max {2? sin³? 2? cos³? }=2¹?
β=min {2? sin³?2? cos³? }=2? ¹?
α¹/? +β¹/? = b/8
⇒4+1/4 = b/8
⇒17/4 = b/8 ⇒ b=-34
Again α¹/? β¹/? =c/8
⇒4*1/4 = c/8
⇒c=8
⇒c−b=8+34=42

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alok kumar singh

Contributor-Level 10

A? =B? (i)
A³B²=A²B³. (ii)
Subtract (i) & (ii)
⇒A³ (A²−B²)=B³ (B²−A²)
⇒ (A²−B²) (A³+B³)=0
A²−B² is invertible matrix
∴A²−B²≠0
⇒A³+B³=0
∴? A³+B³? =0

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