Ncert Solutions Maths class 11th
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New answer posted
3 weeks agoContributor-Level 10
e? +e³? −4e²? −e? +1=0
Divide by e²?
⇒ (e²? +e? ²? )− (e? +e? )−4=0
Put e? +e? =t>0
e²? +e? ²? +2=t²
t²−2−t−4=0
⇒t²−t−6=0
⇒t=3, t=−2 but t≠−2
⇒t=3
⇒e? +e? =3
Number of solution=2
New answer posted
3 weeks agoContributor-Level 10
p: there exist M>0
Such that x≥M for all x∈S
Obviously ~p: M>0 such that x
New answer posted
3 weeks agoContributor-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
New answer posted
3 weeks agoContributor-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
New answer posted
3 weeks agoContributor-Level 10
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
New answer posted
3 weeks agoContributor-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
New answer posted
3 weeks 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
3 weeks 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, ∞)
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