Q: If m is the AM of two distinct real numbers l and n(l,n>1) and G1,G2 and G3 are three geometric means between l and n, then G14+2G24+G34 equals

Exp. (b)Given, m is the AM of l and n∴l+n=2mand G1,G2,G3 are geometric means between l and n∴l,G1,G2,G3,n are in GP.Let r be the common ratio of this GP.∴G1=lrG2=lr2G3=lr3n=lr4r=(ln)41 Now,G14+2G24+G34=(lr)4+2(lr2)4+(lr3)4=l4×r4(1+2r4+r8)=l4×r4(r4+1)2=l4×ln(ln+l)2=ln×4m2=4lm2n

Q: limx→0xtan4x(1−cos2x)(3+cosx) is equal to (a) 4 (b) 3 (c) 2 (d) 21

Exp. (c) We have, limx→0xtan4x(1−cos2x)(3+cosx)=limx→0x×4xtan4x×4x2sin2x(3+cosx) =limx→0x22sin2x×limx→04(3+cosx)×limx→04xtan4x1 [∵limθ→0θsinθ=1 and limθ→0θtanθ=1] =2×44×1

Q: If the function g(x)={kx+1mx+2,0≤x≤3,3<x≤5 differentiable, then the value of k+m is (a) 2 (b) 516 (c) 310 (d) 4

Exp. (a) Since, g(x) is differentiable ⇒g(x) must be continuous. g(x)={kx+1mx+2,0≤x≤3,3<x≤5 At x=3,RHL=3m+2 and at x=3,LHL=2k ∴2k=3m+2....(i) Also, g′(x)={2x+1k,m0≤x<3,3<x≤5 ∴L{g′(3)}=4k and R{g′(3)}=m ⇒4k=m i.e. k=4m....(ii) On solving Eqs. (i) and (ii), we get k=58,m=52 ⇒k+m=2

Q: Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2. If limx→0[1+x2f(x)]=3, then f(2) is equal to (a) −8 (b) −4 (c) 0 (d) 4

Exp. (c) Central Idea Any function have extreme values (maximum or minimum) at its critical points, where f′(x)=0 Since, the function have extreme values at x=1 and x=2. ∴f′(x)=0 at x=1 and x=2 ⇒f′(1)=0 and f′(2)=0 Also it is given that limx→0[1+x2f(x)]=3 ⇒1+limx→0x2f(x)=3⇒limx→0x2f(x)=2 ⇒f(x) will be of the form ax4+bx3+2x2, [∵f(x) is of four degree polynomial] Let f(x)=ax4+bx3+2x2 ⇒f′(x)=4ax3+3bx2+4x ⇒f′(1)=4a+3b+4=0........(i) and f′(2)=32a+12b+8=0 ⇒8a+3b+2=0........(ii) On solving Eqs. (i) and (ii), we get a=21,b=−2 ∴f(x)=2x4−2x3+2x2 ⇒f(2)=8−16+8=0

Q: The integral ∫x2(x4+1)43dx equals (a) (x4x4+1)1/4+c (b) (x4+1)1/4+c (c) −(x4+1)1/4+c (d) −(x4x4+1)1/4+c

Exp. (d) ∫x2(x4+1)43dx=∫x5(1+x41)43dx Put 1+x41=t4⇒−x54dx=4t3dt ⇒x5dx=−t3dt=∫t3t3dt=−∫dt=−t+c =−(1+x41)41+c

Q: The integral ∫24logx2+log(36−12x+x2)logx2dx is equal to (a) 2 (b) 4 (c) 1 (d) 6

Exp. (c) Central Idea Apply the property ∫abf(x)dx=∫abf(a+b−x)dx and then add. Let I=∫24logx2+log(36−12x+x2)logx2dx =∫242logx+log(6−x)22logxdx =∫242[logx+log(6−x)]2logxdx ⇒I=∫24[logx+log(6−x)]logxdx......(i) ⇒I=∫24log(6−x)+logxlog(6−x)dx.....(ii) ⇒ On adding Eqs. (i) and (ii), we get ⇒2I=∫24logx+log(6−x)logx+log(6−x)dx ⇒2I=∫24dx=[x]24 ⇒2I=2⇒I=1

Q: The area (in sq. units) of the region described by {(x,y):y2≤2x and y≥4x−1} is (a) 327 (b) 645 (c) 6415 (d) 329

Exp. (d) Given region is {(x,y):y2≤2x and y≥4x−1} y2≤2x represents a region inside the parabola y2=2x..........(i) and y≥4x−1 represents a region to the left of the line y=4x−1..................(ii) The point of intersection of the curve (i) and (ii) is (4x−1)2=2x ⇒16x2+1−8x=2x ⇒16x2−10x+1=0 ⇒x=21,81 ∴ The points where these curves intersect, are (21,1) and (81,−21)

Q: Let y(x) be the solution of the differential equation (xlogx)dxdy+y=2xlogx,(x≥1). Then, y(e) is equal to (a) e (b) 0 (c) 2 (d) 2e

Exp.(c) Given differential equation is (xlogx)dxdy+y=2xlogx,(x≥1) ⇒dxdy+xlogxy=2 This is a linear differential equation. ∴IF=e∫xlogx1dx=elog(logx)=logx Now, the solution of given differential equation is given by y⋅logx=∫logx⋅2dx ⇒y⋅logx=2∫logxdx ⇒y⋅logx=2[xlogx−x]+c At x=1,c=2 ⇒y⋅logx=2[xlogx−x]+2 At x=e,y=2(e−e)+2⇒y=2

Question 1

Let A and B be two sets containing four and two elements respectively. Then, the number of subsets of the set A×B, each having at least three elements are
(a) 219
(b) 256
(c) 275
(d) 510

Accepted Answer

Exp. (a)
Given,
n(A)=4,n(B)=2
⇒n(A×B)=8
Total number of subsets of set (A×B)=28
Number of subsets of set A×B having no element  (i.e. ϕ)=1
Number of subsets of set A×B having one element =8C1​
Number of subsets of set A×B having two elements =8C2​
∴ Number of subsets having atleast three elements
=28−(1+8C1​+8C2​)
=28−1−8−28
=28−37
=256−37=219

Question 2

If m is the AM of two distinct real numbers l and n(l,n>1) and G1​,G2​ and G3​ are three geometric means between l and n, then G14​+2G24​+G34​ equals

Accepted Answer

Exp. (b)Given, m is the AM of l and n∴l+n=2mand G1​,G2​,G3​ are geometric means between l and n∴l,G1​,G2​,G3​,n are in GP.Let r be the common ratio of this GP.∴G1​=lrG2​=lr2G3​=lr3n=lr4r=(ln​)41​ Now,G14​+2G24​+G34​=(lr)4+2(lr2)4+(lr3)4=l4×r4(1+2r4+r8)=l4×r4(r4+1)2=l4×ln​(ln+l​)2=ln×4m2=4lm2n

Question 3

The sum of first 9 terms of the series 113​+1+313+23​+1+3+513+23+33​+… is

Accepted Answer

Exp. (b)Central Idea Write the nth term of the given series and simplify it to get its lowest form. Then, apply, Sn​=ΣTn​.Given series is1t3​+1+3t3+23​+1+3+5t3+23+33​+…∞Let Tn​ be the nth term of the given series.∴Tn​ =1+3+5+…+ to n terms 13+23+33+…+n3​=n2{2n(n+1)​}2​=4(n+1)2​S9​=∑n=19​4(n+1)2​=41​[(22+32+…+102)+12−12]=41​[610(10+1)(20+1)​−1]=4384​=96

Question 4

limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to
(a) 4
(b) 3
(c) 2
(d) 21​

Accepted Answer

Exp. (c)
We have,
limx→0​xtan4x(1−cos2x)(3+cosx)​=limx→0​x×4xtan4x​×4x2sin2x(3+cosx)​
=limx→0​x22sin2x​×limx→0​4(3+cosx)​×limx→0​4xtan4x​1​
[∵limθ→0​θsinθ​=1 and limθ→0​θtanθ​=1]
=2×44​×1

Question 5

If the function g(x)={kx+1​mx+2​,0≤x≤3,3<x≤5​ differentiable, then the value of k+m is
(a) 2
(b) 516​
(c) 310​
(d) 4

Accepted Answer

Exp. (a)
Since, g(x) is differentiable ⇒g(x) must be continuous.
g(x)={kx+1​mx+2​,0≤x≤3,3<x≤5​
At x=3,RHL=3m+2
and at x=3,LHL=2k
∴2k=3m+2....(i)
Also, g′(x)={2x+1​k​,m​0≤x<3,3<x≤5​
∴L{g′(3)}=4k​ and R{g′(3)}=m
⇒4k​=m i.e. k=4m....(ii)
On solving Eqs. (i) and (ii), we get
k=58​,m=52​
⇒k+m=2

Question 6

Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2. If limx→0​[1+x2f(x)​]=3, then f(2) is equal to
(a) −8
(b) −4
(c) 0
(d) 4

Accepted Answer

Exp. (c)
Central Idea Any function have extreme values (maximum or minimum) at its critical points, where f′(x)=0

Since, the function have extreme values at x=1 and x=2.
∴f′(x)=0 at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Also it is given that
limx→0​[1+x2f(x)​]=3
⇒1+limx→0​x2f(x)​=3⇒limx→0​x2f(x)​=2
⇒f(x) will be of the form ax4+bx3+2x2, [∵f(x) is of four degree polynomial]
Let f(x)=ax4+bx3+2x2
⇒f′(x)=4ax3+3bx2+4x
⇒f′(1)=4a+3b+4=0........(i)
and f′(2)=32a+12b+8=0
⇒8a+3b+2=0........(ii)
On solving Eqs. (i) and (ii), we get
a=21​,b=−2
∴f(x)=2x4​−2x3+2x2
⇒f(2)=8−16+8=0

Question 7

The integral ∫x2(x4+1)43​dx​ equals
(a) (x4x4+1​)1/4+c
(b) (x4+1)1/4+c
(c) −(x4+1)1/4+c
(d) −(x4x4+1​)1/4+c

Accepted Answer

Exp. (d)
∫x2(x4+1)43​dx​=∫x5(1+x41​)43​dx​
Put 1+x41​=t4⇒−x54​dx=4t3dt
⇒x5dx​=−t3dt=∫t3t3dt​=−∫dt=−t+c
=−(1+x41​)41​+c

Question 8

The integral ∫24​logx2+log(36−12x+x2)logx2​dx is equal to
(a) 2
(b) 4
(c) 1
(d) 6

Accepted Answer

Exp. (c)
Central Idea Apply the property
∫ab​f(x)dx=∫ab​f(a+b−x)dx and then add.

Let I=∫24​logx2+log(36−12x+x2)logx2​dx
=∫24​2logx+log(6−x)22logx​dx
=∫24​2[logx+log(6−x)]2logxdx​
⇒I=∫24​[logx+log(6−x)]logxdx​......(i)
⇒I=∫24​log(6−x)+logxlog(6−x)​dx.....(ii)
⇒ On adding Eqs. (i) and (ii), we get 
⇒2I=∫24​logx+log(6−x)logx+log(6−x)​dx
⇒2I=∫24​dx=[x]24​
⇒2I=2⇒I=1

Question 9

The area (in sq. units) of the region described by {(x,y):y2≤2x and y≥4x−1} is
(a) 327​
(b) 645​
(c) 6415​
(d) 329​

Accepted Answer

Exp. (d)
Given region is {(x,y):y2≤2x and y≥4x−1} y2≤2x 
represents a region inside the parabola 
y2=2x..........(i)
and y≥4x−1 represents a region to the left of the line
y=4x−1..................(ii)
The point of intersection of the curve (i) and (ii) is
(4x−1)2=2x
⇒16x2+1−8x=2x
⇒16x2−10x+1=0
⇒x=21​,81​
∴ The points where these curves intersect, are (21​,1) and (81​,−21​)

Question 10

Let y(x) be the solution of the differential equation (xlogx)dxdy​+y=2xlogx,(x≥1). Then, y(e) is equal to
(a) e
(b) 0
(c) 2
(d) 2e

Accepted Answer

Exp.(c)
Given differential equation is
(xlogx)dxdy​+y=2xlogx,(x≥1)
⇒dxdy​+xlogxy​=2
This is a linear differential equation.
∴IF=e∫xlogx1​dx=elog(logx)=logx
Now, the solution of given differential equation is given by
y⋅logx=∫logx⋅2dx
⇒y⋅logx=2∫logxdx
⇒y⋅logx=2[xlogx−x]+c
 At x=1,c=2
⇒y⋅logx=2[xlogx−x]+2
 At x=e,y=2(e−e)+2⇒y=2

Question 11

The number of common tangents to the circles x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0 is

Accepted Answer

Exp. (c)Central Idea Number of common tangents depend on the position of the circle with respect to each other.(i) If circles touch externally ⇒C1​C2​=r1​+r2​,3 common tangents(ii) If circles touch internally ⇒C1​C2​=r2​−r1​,1 common tangent(iii) If circles do not touch each other, 4 common tangentsGiven equations of circles arex2+y2−4x−6y−12=0.....(i) and x2+y2+6x+18y+26=0.....(ii) Centre of circle (i) is C1​(2,3) and radius =4+9+12​=5(r1​) Centre of circle (ii) is C2​(−3,−9) and radius=9+81−26​=8(r2​), (say)Now, C1​C2​=(2+3)2+(3+9)2​⇒⇒​C1​C2​=52+122​C1​C2​=25+144​=13​∴ Also, ​r1​+r2​=5+8=13C1​C2​=r1​+r2​​Thus, both circles touch each other externally.Hence, there are three common tangents.

Question 12

The distance of the point (1,0,2) from the point of intersection of the line 3x−2​=4y+1​=12z−2​ and the plane x−y+z=16 is
(a) 214​
(b) 8
(c) 321​
(d) 13

Accepted Answer

Exp. (d)
Given equation of line is
3x−2​=4y+1​=12z−2​=λ.....(i)
and equation of plane is
x−y+z=16.....(ii)
Any point on the line (i) is
(3λ+2,4λ−1,12λ+2)
Let this point be point of intersection of the line and plane.
(3λ+2)−(4λ−1)+(12λ+2)=16
⇒11λ+5=16
⇒11λ=11
⇒λ=1
∴ Point of intersection is (5,3,14).
=(5−1)2+(3−0)2+(14−2)2​
=16+9+144​
=169​=13

Question 13

The equation of the plane containing the line 2x−5y+z=3,x+y+4z=5 and parallel to the plane x+3y+6z=1 is 
(a) 2x+6y+12z=13
(b) x+3y+6z=−7
(c) x+3y+6z=7
(d) 2x+6y+12z=−13

Accepted Answer

Exp. (c)
Let equation of plane containing the lines
2x−5y+z=3 and x+y+4z=5 be 
(2x−5y+z−3)+λ(x+y+4z−5)=0
⇒(2+λ)x+(λ−5)y+(4λ+1)z−3−5λ=0......(i)
This plane is parallel to the plane x+3y+6z=1. 
∴12+λ​=3λ−5​=64λ+1​
On taking first two equalities, we get
6+3λ=λ−5
⇒2λ=−11
⇒λ=−211​
On taking last two equalities, we get
6λ−30=3+12λ
⇒−6λ=33⇒λ=−211​
So, the equation of required plane is
(2−211​)x+(2−11​−5)y+(−244​+1)z−3+5×211​=0
⇒−27​x−221​y−242​z+249​=0
⇒x+3y+6z−7=0

Question 14

The angle between the lines whose direction cosines satisfy the equation l+m+n=0 and l2=m2+n2 is
(a) 3π​
(b) 4π​
(c) 6π​
(d) 2π​

Accepted Answer

Exp. (a)
Given, l+m+n=0
⇒l=−(m+n)
⇒(m+n)2=l2
⇒m2+n2+2mn=m2+n2
[∵l2=m2+n2, given ]
⇒2mn=0
Case I When m=0, then
l=−n
Hence, (l,m,n) is (1,0,−1).
Case II When n=0, then
l=−m
Hence, (l,m,n) is (1,−1,0).
∴cosθ=2​×2​1+0+0​=21​
⇒θ=3π​

Question 15

Let a, b and c be three non-zero vectors such that no two of them are collinear and (a×b)×c=31​∣b∥c∣a. If θ is the angle between vectors band c, then a value of sinθ is

Accepted Answer

Given, (a×b)×c=31​∣b∣∣c∣a⇒−c×(a×b)=31​∣b∣∣c∣a⇒−(c⋅b)⋅a+(c⋅a)b=31​∣b∣∣c∣a[31​∣b∣∣c∣+(c⋅b)]a=(c⋅a)bSince, a and b are not collinear.c⋅b+31​∣b∣∣c∣=0 and c⋅a=0⇒∣c∣∣b∣cosθ+31​∣b∣∣c∣=0⇒∣b∣c∣(cosθ+31​)=0⇒cosθ+31​=0⇒cosθ=−31​⇒sinθ=38​​=322​​

Question 16

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls, is
(a) 355​(32​)11
(b) 55(32​)10
(c) 220(31​)12
(d) 22(31​)11

Accepted Answer

Exp. (a)
There seems to be ambiguity in this question. It should be mentioned that boxes are different and one particular box has 3 balls.
Then, number of ways =31212C3​×29​=355​(32​)11
According to the question,
3123C1​×12C3​29−3C2​12C3​9C3​+3!3!6!3!12!×3!​​

Question 17

The mean of the data set comprising of 16 observations is 16 . If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data is
(a) 16.8
(b) 16.0
(c) 15.8
(d) 14.0

Accepted Answer

Exp. (d)
Given, 16x1​+x2​+x3​+…+x16​​=16
⇒∑i=116​xi​=16×16
Sum of new observations
=∑i=118​yi​=(16×16−16)+(3+4+5)=252
Number of observations =18
∴ New mean =18∑i=118​yi​​=18252​=14

Question 18

Let tan−1y=tan−1x+tan−1(1−x22x​), where ∣x∣<3​1​. Then, a value of y is
(a) 1−3x23x−x3​
(b) 1−3x23x+x3​
(c) 1+3x23x−x3​
(d) 1+3x23x+x3​

Accepted Answer

Exp. (a) Given, tan−1y=tan−1x+tan−1(1−x22x), where ∣x∣<31 ⇒tan−1y=tan−1{1−x(1−x22x)x+1−x22x} [∵tan−1x+tan−1y=tan−1(1−xyx+y),x>0,y>0,xy<1] =tan−1(1−x2−2x2x−x3+2x) ⇒tan−1y=tan−1(1−3x23x−x3) y=1−3x23x−x3 Aliter ∣x∣<31⇒−31

Question 19

The negation of ∼s∨(∼r∧s) is equivalent to
(a) s∧∼r
(b) s∧(r∧∼s)
(c) s∨(r∨∼s)
(d) s∧r

Accepted Answer

Exp. (d)
∼(∼s∨(∼r∧s))
≡s∧(∼(∼r∧s))≡s∧(r∨∼s)≡(s∧r)∨(s∧∼s)
≡(s∧r)∨F, (∵s∧∼s is false)
≡s∧r

Question 20

The number of integers greater than 6000 that can be formed, using the digits 3,5,6,7 and 8 without repetition, is
(a) 216
(b) 192
(c) 120
(d) 72

Accepted Answer

Exp. (b)
The integer greater than 6000 may be of 4 digit or 5 digit. So, here two cases arise.
Case I When number is of 4 digit. Four digit number can starts from 6,7 or 8
Thus, total number of 4 digit number, which are greater than 6000=3×4×3×2=72
Case II When number is of 5 digit. Total number of five digit number which are greater than 6000=5 ! =120
∴ Total number of integers =72+120=192

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Solutions For Exams JEE Mains, 2015

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