12th Grade > Physics
CALCULUS MCQs
Total Questions : 29
| Page 1 of 3 pages
Answer: Option A. -> e(5x+10)5+c
:
A
I=∫e(5x+10)dx=∫eudu5
= 15eu+c
= 15e(5x+10)+c
:
A
I=∫e(5x+10)dx=∫eudu5
= 15eu+c
= 15e(5x+10)+c
Answer: Option C. -> ex(cosx+sinx)
:
C
y=exsinx.
So dydx=ddx(exsinx)=exddx(sinx)+sinxddx(ex)
=excosx+exsinx=ex(cosx+sinx).
:
C
y=exsinx.
So dydx=ddx(exsinx)=exddx(sinx)+sinxddx(ex)
=excosx+exsinx=ex(cosx+sinx).
Answer: Option A. -> yes
:
A
Yes
:
A
Yes
Answer: Option B. -> ln|2x+3|2+c
:
B
Let u=2x+3
⇒dudx=2
⇒dx=du2
I=∫ydx=∫dx2x+3=∫1u×du2
= 12∫duu
= 12lnu+c
= 12ln|2x+3|+c
:
B
Let u=2x+3
⇒dudx=2
⇒dx=du2
I=∫ydx=∫dx2x+3=∫1u×du2
= 12∫duu
= 12lnu+c
= 12ln|2x+3|+c
Answer: Option A. -> 23(1+y2)32+c
:
A
Let I=∫√1+y2.2ydy
Let u=1+y2, then du=2ydy
I=∫u1/2du=u(1/2)+1(1/2)+1Integrate, using rule no. 3 with n=12
= 23u3/2+C
Simpler form= 23(1+y2)3/2+c (Replace u by 1+y2)
:
A
Let I=∫√1+y2.2ydy
Let u=1+y2, then du=2ydy
I=∫u1/2du=u(1/2)+1(1/2)+1Integrate, using rule no. 3 with n=12
= 23u3/2+C
Simpler form= 23(1+y2)3/2+c (Replace u by 1+y2)
Answer: Option A. -> 5x4
:
A
Given y=x5
Differentiating both sides w.r.t. 'x', Using dxndx=n.xn−1
dydx=ddx[x5]=5x5−1=5x4
:
A
Given y=x5
Differentiating both sides w.r.t. 'x', Using dxndx=n.xn−1
dydx=ddx[x5]=5x5−1=5x4
Answer: Option D. -> none of these
:
D
Differentiating both sides w.r.t. 'x'
dydx=ddx[x2+5x32+2x]
Using the linearity property of the differentiation, we get = ddx[x2]+ddx[5x32]+ddx[2x]
Taking constants out, = ddx[x2]+5d[x32]dx+2ddx[1x]
=2x+5.32x12+2.(−1)x−2=2x+152x12−2x2
:
D
Differentiating both sides w.r.t. 'x'
dydx=ddx[x2+5x32+2x]
Using the linearity property of the differentiation, we get = ddx[x2]+ddx[5x32]+ddx[2x]
Taking constants out, = ddx[x2]+5d[x32]dx+2ddx[1x]
=2x+5.32x12+2.(−1)x−2=2x+152x12−2x2
Answer: Option B. -> 2 units
:
B
I=∫π/2−π/2cosxdx=sinx|π/2−π/2=sinπ2−sin(−π2)=1+1=2∵∫baf(x)dx=I(x)∣∣∣ba=I(b)−I(a)
:
B
I=∫π/2−π/2cosxdx=sinx|π/2−π/2=sinπ2−sin(−π2)=1+1=2∵∫baf(x)dx=I(x)∣∣∣ba=I(b)−I(a)
Answer: Option A. -> ex
:
A
dexdx=ex
dydx for some common functions
ydydxydydxxnnxn−1secxsecxtanxsinxcosxcosecx−cosecxcotxcosx−sinxlnx1xtanxsec2xexexcotx−cosec2x
:
A
dexdx=ex
dydx for some common functions
ydydxydydxxnnxn−1secxsecxtanxsinxcosxcosecx−cosecxcotxcosx−sinxlnx1xtanxsec2xexexcotx−cosec2x
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