Class 12th
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New answer posted
2 months agoContributor-Level 10
f (x)= {sinx, 0? x /2; 1? /2? x? 2+cosx, x>? }
f' (x)= {cosx, 0
New answer posted
2 months 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
2 months agoContributor-Level 10
⇒3αβ−2αβ=−1
⇒2αβ=4⇒αβ=2 . (i)
b.c=10
⇒−3α−2β−α=10
⇒4α+2β=−10
⇒2α+β=−5 . (ii)
From (i) and (ii)
α=−1/2, α=−2
β=−4, β=−1
a=i−2j−k
b=3i−2j+2k
c=2i−2j+k
a. (b*c)=9
New answer posted
2 months agoContributor-Level 10
e? dy = e? /α dx
⇒−e? =e? /α+c
y (ln2)=ln2 and y (0)=−ln2
⇒−2=−1/α+c
⇒c=−2−1/α
⇒e? = 1/α e? −2−1/α
⇒−e? ² = 1/α e? ²−2−1/α
⇒2? ¹=3/α
⇒α=2
New answer posted
2 months agoContributor-Level 10
A² = [1 2 3; 0 1 2; 0 1]
A³=A².A= [1 3 6; 0 1 3; 0 1]
A²? = [1 20 1+2+3.20; 0 1 20; 0 1] = [1 20 210; 0 1 20; 0 1]
M= [20 210 520; 0 20 210; 0 20]
M (a? )=T? =n (n+1)/2
S? = 1/2 [ n (n+1) (2n+1)/6 + n (n+1)/2 ]
⇒S? =1540
⇒M=2020
New answer posted
2 months agoContributor-Level 10
∫? ^ (π/2) sin³x e? sin²? dx
=∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
=2∫? ^ (π/2) (1−cos²x)sinx e? (¹? cos²? )dx
Let cos²x=t⇒sin2xdx=−dt
=−2∫? (1−t)e? (¹? ) dt/ (−2cosx)
=1/e ∫? e? dt −∫? te? dt
=1/e [e? ]? − [te? −e? ]?
=2e−3∫? ¹ √t e? dt
⇒α=2, β=3
α+β=5
New answer posted
2 months agoContributor-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)
New answer posted
2 months agoContributor-Level 10
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
New answer posted
2 months agoContributor-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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