Q: limx→0x2sin(πcos2x) is equal to (a) 2π (b) 1 (c) −π (d) π

Exp. (d) limx→0x2sin(πcos2x) =limx→0x2sinπ(1−sin2x) =limx→0x2sin(π−πsin2x) =limx→0x2sin(πsin2x) =limx→0πsin2xsin(πsin2x)×(π)(x2sin2x),[∵limx→0xsinx=1] =π

Q: The integral ∫(1+x−x1)ex+x1.dx is equal to (a) (x−1)ex+x1+C (b) xex+x1+C (c) (x+1)ex+x1+C (d) −xex+x1+C

Exp. (b) ∫(1+x−x1)ex+x1dx =∫ex+x1dx+∫x(1−x21)ex+x1dx =∫ex+x1dx+xex+x1−∫dxd(x)ex+x1.dx [∵∫(x−x21)ex+x1.dx=ex+x1] =∫ex+x1dx+xex+x1−∫ex+x1.dx =xex+x1+C

Q: The integral ∫0π1+4sin22x−4sin2x.dx is equal to (a) π−4 (b) 32π−4−43 (c) 43−4 (d) 43−4−3π

Exp. (d) ∫0π(1−2sin2x)2dx=∫0π∣∣1−2sin2x∣∣dx =∫03π(1−2sin2x)dx−∫3ππ(1−2sin2x)dx =[x+4cos2x]03π−[x+4cos2x]3πx =43−4−3π

Q: limn→∞∑r=1nn1er/n is equal to (a) e (b) e−1 (c) 1−e (d) e+1

Exp. (b) we have, limn→∞∑r=1nn1er/n Now, limn→∞∑r=1nn1er/n=∫01exdx=[ex]01=e−1

Q: Let the population of rabbits surviving at a time t be governed by the differential equation dtdp(t)=21p(t)−200. If p(0)=100, then p(t) is equal to (a) 400−300et/2 (b) 300−200e−t/2 (c) 600−500et/2 (d) 400−300e−t/2

Exp. (a) dtdp−21p(t)=−200 IF=e∫−(21)dt=e−t/2 Hence, p(t)⋅e−t/2=∫−200⋅e−t/2dt p(t)⋅e−t/2=400e−t/2+K p(0)=100⇒K=−300 ⇒p(t)=400−300et/2

Question 1

If the coefficients of x3 and x4 in the expansion of (1+ax+bx2)(1−2x)18 in powers of x are both zero, then (a,b) is equal to
(a) (16,3251​)
(b) (14,3251​)
(c) (14,3272​)
(d) (16,3272​)

Accepted Answer

Exp. (d)
In the expansion of (1+ax+bx2)(1−2x)18,
Coefficient of x3 in (1+ax+bx2)(1−2x)18
= Coefficient of x3 in (1−2x)18+ Coefficient of x2 in a(1−2x)18+ Coefficient of x in b(1−2x)18
=−18C3​⋅23+a18C2​⋅22−b18C1​⋅2
∵−18C3​⋅23+a18C2​⋅22−b18C1​⋅2=0
⇒3×218×17×16​⋅8+a⋅218×17​22−b⋅18⋅2=0
⇒17a−b=334×16​............................(i)
Similarly, coefficient of x4
18C4​⋅24−a⋅18C3​23+b⋅18C2​⋅22=0
∴32a−32b=240............................(ii)
On solving Eqs. (i) and (ii), we get
a=16 and b=3272​

Question 2

If (10)9+2(11)1(10)8+3(11)2(10)7 +…+10(11)9=K(10)9, then k is equal to

Accepted Answer

Exp. (c)k⋅109=109+2(11)1(10)8+3(11)2(10)7+… +10(11)9k=1+2(1011​)+3(1011​)2+…10(1011​)9… (i) (1011​)k=1(1011​)+2(1011​)2+…+9(1011​)9+10(1011​)10.....(ii)On subtracting Eq. (ii) from Eq. (i), we getk(1−1011​)=1+1011​+(1011​)2+… +(1011​)9−10(1011​)10⇒k(1010−11​)=(1011​−1)1[(1011​)10−1]​−10(1011​)10⇒−k=10[10(1011​)10−10−10(1011​)10]⇒k=100

Question 3

Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is

Accepted Answer

Exp. (d)Let a,ar,ar2 are in GP (r>1).According to the question, a,2ar,ar2 in AP.⇒4ar=a+ar2⇒r2−4r+1=0⇒r=24±16−4​​=2±3​⇒r=2+3​[∵ GP is increasing]

Question 4

limx→0​x2sin(πcos2x)​ is equal to
(a) 2π​
(b) 1
(c) −π
(d) π

Accepted Answer

Exp. (d)
limx→0​x2sin(πcos2x)​
=limx→0​x2sinπ(1−sin2x)​
=limx→0​x2sin(π−πsin2x)​
=limx→0​x2sin(πsin2x)​
=limx→0​πsin2xsin(πsin2x)​×(π)(x2sin2x​),[∵limx→0​xsinx​=1]
=π

Question 5

If f and g are differentiable functions in (0,1) satisfying f(0)=2=g(1),g(0)=0 and f(1)=6, then for some c∈]0,1[
(a) 2f′(c)=g′(c)
(b) 2f′(c)=3g′(c)
(c) f′(c)=g′(c)
(d) f′(c)=2g′(c)

Accepted Answer

Exp. (d)
Here, f(0)=2=g(1),g(0)=0 and f(1)=6
∵f and g are differentiable in (1,0).
Let h(x)=f(x)−2g(x)
h(0)=f(0)−2g(0)
h(0)=2−0
h(0)=2
Now, h(1)=f(1)−2g(1)=6−2(2)
h(1)=2,h(0)=h(1)=2
Hence, using Rolle's theorem,
There exists c∈]0,1[, such that
h′(c)=0
⇒f′(c)−2g′(c)=0, for some c∈]0,1[
⇒f′(c)=2g′(c)

Question 6

If x=−1 and x=2 are extreme points of f(x)=αlog∣x∣+βx2+x, then 
(a) α=−6,β=21​
(b) α=−6,β=−21​
(c) α=2,β=−21​
(d) α=2,β=21​

Accepted Answer

Exp. (c)
Here, x=−1 and x=2 are extreme points of f(x)=αlog∣x∣+βx2+x, then
f′(x)=xα​+2βx+1
∴f′(−1)=−α−2β+1=0....(i)
 and f′(−2)=2α​+4β+1=0.....(ii)
On solving Eqs. (i) and (ii), we get
α=2,β=−21​

Question 7

The integral ∫(1+x−x1​)ex+x1​.dx is equal to
(a) (x−1)ex+x1​+C
(b) xex+x1​+C
(c) (x+1)ex+x1​+C
(d) −xex+x1​+C

Accepted Answer

Exp. (b)
∫(1+x−x1​)ex+x1​dx
=∫ex+x1​dx+∫x(1−x21​)ex+x1​dx
=∫ex+x1​dx+xex+x1​−∫dxd​(x)ex+x1​.dx
[∵∫(x−x21​)ex+x1​.dx=ex+x1​]
=∫ex+x1​dx+xex+x1​−∫ex+x1​.dx
=xex+x1​+C

Question 8

The integral ∫0π​1+4sin22x​−4sin2x​​.dx is equal to
(a) π−4
(b) 32π​−4−43​
(c) 43​−4
(d) 43​−4−3π​

Accepted Answer

Exp. (d)
∫0π​(1−2sin2x​)2​dx=∫0π​∣∣​1−2sin2x​∣∣​dx
=∫03π​​(1−2sin2x​)dx−∫3π​π​(1−2sin2x​)dx
=[x+4cos2x​]03π​​−[x+4cos2x​]3π​x​
=43​−4−3π​

Question 9

limn→∞​∑r=1n​n1​er/n is equal to

(a) e
(b) e−1
(c) 1−e
(d) e+1

Accepted Answer

Exp. (b)
we have,
limn→∞​∑r=1n​n1​er/n
Now, 
limn→∞​∑r=1n​n1​er/n=∫01​exdx=[ex]01​=e−1

Question 10

Let the population of rabbits surviving at a time t be governed by the differential equation dtdp(t)​=21​p(t)−200. If p(0)=100, then p(t) is equal to
(a) 400−300et/2
(b) 300−200e−t/2
(c) 600−500et/2
(d) 400−300e−t/2

Accepted Answer

Exp. (a)
dtdp​−21​p(t)=−200
IF=e∫−(21​)dt=e−t/2
Hence, p(t)⋅e−t/2=∫−200⋅e−t/2dt
p(t)⋅e−t/2=400e−t/2+K
p(0)=100⇒K=−300
⇒p(t)=400−300et/2

Question 11

Let a,b,c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

Accepted Answer

Exp. (c)Let coordinate of the point be (α,−α).Since, (α,−α) lie on 4ax+2ay+c=0and 5bx+2by+d=0.∴4aα−2aα+c=0⇒α=2a−c​.......(i)Also, 5bα−2bα+d=0⇒α=3b−d​........(ii)From Eqs. (i) and (ii), 2a−C​=3b−d​∴3bc=2ad

Question 12

The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it, is
(a) (x2−y2)2=6x2+2y2
(b) (x2−y2)2=6x2−2y2
(c) (x2+y2)2=6x2+2y2
(d) (x2+y2)2=6x2−2y2

Accepted Answer

Exp. (c)
Equation of ellipse is x2+3y2=6
Equation of the tangent is axcosθ​+bysinθ​=1.
Let (h,k) be any point on the locus, then
ah​cosθ+bk​sinθ=1.....(i)
Slope of the tangent line is a−b​cotθ.
Slope of perpendicular drawn from centre (0,0) to (h,k) is hk​.
Since, both the lines are perpendicular.
∴(hk​)×(−ab​cotθ)=−1
⇒hacosθ​=kbsinθ​=α
⇒cosθ=αha,sinθ=αkb
From Eq. (i),
ah​(αha)+bk​(αkb)=1
⇒⇒​​h2α+k2αα=h2+k21​​=1
Also, sin2θ+cos2θ=1
⇒(αkb)2+(αha)2=1
⇒α2k2b2+α2h2a2=1
⇒(h2+k2)2k2b2​+(h2+k2)2h2a2​=1
⇒(h2+k2)22k2​+(h2+k2)26h2​=1
[∵a2=6 and b2=2]
⇒6x2+2y2=(x2+y2)2

Question 13

The slope of the line touching both the parabolas y2=4x and x2=−32 is,

Accepted Answer

Exp. (a)For parabola, y2=4xLet y=mx+m1​ is tangent line and it touches the parabola x2=−32y.∴x2=−32(mx+m1​)⇒x2+32mx+m32​=0∵D=0∴(32m)2−4⋅(m32​)=0⇒m3=81​⇒m=21​

Question 14

The image of the line 3x−1​=1y−3​=−5z−4​ in the plane 2x−y+z+3=0 is the line
(a) 3x+3​=1y−5​=−5z−2​ 
(b) −3x+3​=−1y−5​=5z+2​
(c) 3x−3​+1y+5​=−5z−2​ 
(d) −3x−3​=−1y+5​=5z−2​

Accepted Answer

Exp. (a)
Plane and line are parallel to each other. Equation of normal to the plane through the point (1,3,4) is
2x−1​=−1y−3​=1z−4​=k
Any point in this normal is (2k+1,−k+3,4+k)
⇒(22k+1+1​,23−k+3​,24+k+4​) lies on plane 
⇒2(k+1)−(26−k​)+(28+k​)+3=0
⇒k=−2
Hence, point through which this image pass is
 i.e.,(2k+1,3−k,4+k)
Hence, equation of image line is
3x+3​=1y−5​=−5z−2​

Question 15

Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is
(a) 23​
(b) 25​
(c) 27​
(d) 29​

Accepted Answer

Given planes are
2x+y+2z−8=0
and 2x+y+2z+25​=0
Distance between two planes
=a2+b2+c2​∣c1​−c2​∣​=∣∣​22+12+22​−8−25​​∣∣​=3221​​=27​

Question 16

Let A and B be two events such that P(A∪B)=61​,P(A∩B)=41​ and P(Aˉ)=41​, where Aˉ stands for the complement of the event A. Then, the events A and B are
(a) independent but not equally likely
(b) independent and equally likely
(c) mutually exclusive and independent
(d) equally likely but not independent

Accepted Answer

Exp. (a)
P(A∪B)=61​,P(A∩B)=41​,P(Aˉ)=41​
∴P(A∪B)=1−P(A∪B)=1−61​=65​
and P(A)=1−P(Aˉ)=1−41​=43​
P(A∪B)=P(A)+P(B)−P(A∩B)
⇒65​=43​+P(B)−41​⇒P(B)=31​
⇒A and B are not equally likely.
Also, P(A∩B)=P(A)⋅P(B)=41​
So, events are independent.

Question 17

The variance of first 50 even natural numbers is
(a) 4833​
(b) 833
(c) 437
(d) 4437​

Accepted Answer

Exp. (b)
Variance, σ2=n1​Σxi2​−(xˉ)2
Xˉ=nΣxi​​ =502+4+6+8+…+100​
=5050×51​=51
σ2=501​(22+42+…+1002)−(51)2
=833

Question 18

If fk​(x)=k1​(sinkx+coskx), where x∈R, k≥1, then f4​(x)−f6​(x) is equal to
(a) 61​
(b) 31​
(c) 41​
(d) 121​

Accepted Answer

Exp. (d)
fk​(x)=k1​(sinkx+coskx), where x∈R and k≥1
Now, f4​(x)−f6​(x)
=41​(sin4x+cos4x)−61​(sin6x+cos6x)
=41​(1−2sin2x⋅cos2x)−61​(1−3sin2x⋅cos2x)
=41​−61​=121​

Question 19

Let α and β be the roots of equation px2+qx+r=0,p=0. If p,q and r are in AP and α1​+β1​=4, then the value of ∣α−β∣ is
(a) 961​​
(b) 9217​​
(c) 934​​
(d) 9213​​

Accepted Answer

Exp. (d)
Since, α and β are roots of px2+qx+r=0,p=0. 
∴α+β=p−q​,αβ=pr​
Since, p,q and r are in AP.
∴2q=p+r
 Also, α1​+β1​=4⇒αβα+β​=4
⇒α+β=4αβ⇒p−q​=p4r​
2q=p+r
⇒2(−4r)=p+r⇒p=−9r
α+β=p−q​=p4r​=−9r4r​=−94​
 and αβ=pr​=−9rr​=−91​
∴(α−β)2=(α+β)2−4αβ
=8116​+94​=8116+36​
⇒(α−β)2=8152​
⇒∣α−β∣=3213​​

Filters

Filters

exam

year

subject

difficulty

Solutions For Exams JEE Mains, 2014

Found the solution, but did not understand the concept?

Ask my first question