Q: The sum of the coefficients of all odd degree terms in the expansion of (x+x3−1)5+(x−x3−1)5,(x>1) is (a) −1 (b) 0 (c) 1 (d) 2

Exp. (d) [ Key idea =(a+b)n+(a−b)n] =2(nC0an+nC2an−2b2+nC4an−4b4…) We have, (x+x3−1)5+(x−x3−1)5,x>1 =2(5C0x5+5C2x3(x3−1)2+5C4x(x3−1)4) =2(x5+10x3(x3−1)+5x(x3−1)2) =2(x5+10x6−10x3+5x7−10x4+5x) Sum of coefficients of all odd degree terms is 2(1−10+5+5)=2

Q: Let a1,a2,a3,…,a49 be in AP such that ∑k=012a4k+1=416 and a9+a43=66. If a12+a22+…+a172=140m, then m is equal to

Exp. (c)We have, a1,a2,a3,…a49 are in AP.∑k=012a4k+1=416 and a9+a43=66Let a1=a and d= common difference∵a1+a5+a9+⋯+a49=416∴a+(a+4d)+(a+8d)+…(a+48d)=416⇒213(2a+48d)=416⇒a+24d=32.....(i) Also, a9+a43=66∴a+8d+a+42d=66⇒2a+50d=66⇒a+25d=33.......(ii)Solving Eqs. (i) and (ii), we geta=8 and d=1Now, a12+a22+a32+⋯+a172=140m82+92+102+…+242=140m⇒(12+22+32+…+242)−(12+22 +32+…+72)=140m⇒624×25×49−67×8×15=140 m⇒63×7×8×5(7×5−1)=140m⇒7×4×5×34=140 m⇒140×34=140 m⇒m=34

Q: If 2y=(cot−1(cosx−3sinx3cosx+sinx))2,x∈(0,2π) then dxdy is equal to (a) 6π−x (b) x−6π (c) 3π−x (d) 2x−3π

Exp. (b) Given expression is 2y=(cot−1(cosx−3sinx3cosx+sinx))2 =(cot−1(cotx−33cotx+1))2 [dividing each term of numerator and denominator by sinx ] =(cot−1(cotx−cot6πcot6πcotx+1))2[∵cot6π=3] =(cot−1(cot(6π−x)))2, [∵cot(A−B)=cotB−cotAcotAcotB+1] ={(6π−x)2,(π+(6π−x))2,0<x<6π6π<x<2π ⎣⎡∵cot−1(cotθ)=⎩⎨⎧π+θ,θ,θ−π,−π<θ<00<θ<ππ<θ<2π⎦⎤ ⇒2y={(6π−x)2,(67π−x)2,0<x<6π6π<x<2π ⇒2dxdy={2(6π−x)(−1),2(67π−x)(−1),0<x<6π6π<x<2π ⇒dxdy={x−6π,x−67π,0<x<6π6π<x<2π

Q: For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→0+x([x1]+[x2]+…+[x15]) (a) is equal to 0 (b) is equal to 15 (c) is equal to 120 (d) does not exist (in R )

Exp. (c) Key idea Use property of greatest integer function [x]=x−{x}. We have, limx→0+x([x1]+[x2]+…+[x15]) We know, [x]=x−{x} ∴[x1]=x1−{x1} Similarly, [xn]=xn−{xn} ∴ Given limit =limx→0+x(x1−{x1}+x2−{x2}+… x15−{x15}) =limx→0+(1+2+3+…+15)−x({x1}+{x2}+…+{x15}) =120−0=120 [∵0≤{xn}<1, therefore 0≤x{xn}<x⇒limx→0+x{xn}=0]

Q: If the curves y2=6x,9x2+by2=16 intersect each other at right angles, then the value of b is (a) 6 (b) 27 (c) 4 (d) 29

Exp. (d) We have, y2=6x ⇒2ydxdy=6 ⇒dxdy=y3 Slope of tangent at (x1,y1) is m1=y13 Also, 9x2+by2=16 ⇒18x+2bydxdy=0 ⇒dxdy=by−9x Slope of tangent at (x1,y1) is m2=by1−9x1 Since, these are intersection at right angle. ∴m1m2=−1⇒by1227x1=1 ⇒6bx127x1=1,[∵y12=6x1] ⇒b=9/2

Q: Let f(x)=x2+x21 and g(x)=x−x1, x∈R−{−1,0,1}. If h(x)=g(x)f(x), then the local minimum value of h(x) is (a) 3 (b) −3 (c) −22 (d) 22

Exp. (d) We have, f(x)=x2+x21 and g(x)=x−x1⇒h(x)=g(x)f(x) ∴h(x)=x−x1x2+x21=x−x1(x−x1)2+2 ⇒h(x)=(x−x1)+x−x12 x−x1>0,(x−x1)+x−x12∈[22,∞) x−x1<0,(x−x1)+x−x12∈(−∞,22] ∴ Local minimum value is 22.

Q: limx→π/2(π−2x)3cotx−cosx equals (a) 241 (b) 161 (c) 81 (d) 41

Exp. (b) limx→π/2(π−2x)3cotx−cosx=limx→π/281⋅sinx(2π−x)3cosx(1−sinx) =limh→081⋅sin(2π−h)(2π−2π+h)3cos(2π−h)[1−sin(2π−h)] =81limh→0cosh⋅h3sinh(1−cosh) =81limh→0cosh⋅h3sinh(2sin22h) =41limh→0h3coshsinh⋅sin2(2h) =41limh→0(hsinh)(2hsin2h)2⋅cosh1⋅41 =41×41=161

Q: The integral ∫(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2sin2xcos2xdx is equal to, (where C is a constant of integration). (a) 3(1+tan3x)1+C (b) 3(1+tan3x)−1+C (c) 1+cot3x1+C (d) 1+cot3x−1+C

Exp. (b) We have, I=∫(sin5x+cos3x⋅sin2x+sin3x⋅cos2x+cos5x)2sin2x⋅cos2xdx =∫{sin3x(sin2x+cos2x)+cos3x(sin2x+cos2x)}2sin2xcos2xdx =∫(sin3x+cos3x)2sin2xcos2xdx=∫cos6x(1+tan3x)2sin2xcos2xdx =∫(1+tan3x)2tan2xsec2xdx Put tan3x=t ⇒3tan2xsec2xdx=dt ∴I=31∫(1+t)2dt ⇒I=3(1+t)−1+C ⇒I=3(1+tan3x)−1+C

Question 1

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf, so that the dictionary is always in the middle.
The number of such arrangements is
(a) at least 1000
(b) less than 500
(c) at least 500 but less than 750
(d) at least 750 but less than 1000

Accepted Answer

Exp. (a)
Given 6 different novels and 3 different dictionaries.
Number of ways of selecting 4 novels from 6 novels is 6C4​=2!4!6!​=15
Number of ways of selecting 1 dictionary is from 3 dictionaries is 3C1​=1!2!3!​=3
∴ Total number of arrangement of 4 novels and 1 dictionary where dictionary is always in the middle, is
15×3×4!=45×24=1080

Question 2

The sum of the coefficients of all odd degree terms in the expansion of (x+x3−1​)5+(x−x3−1​)5,(x>1) is
(a) −1
(b) 0
(c) 1
(d) 2

Accepted Answer

Exp. (d)
[ Key idea =(a+b)n+(a−b)n]
=2(nC0​an+nC2​an−2b2+nC4​an−4b4…)
We have,
(x+x3−1​)5+(x−x3−1​)5,x>1
=2(5C0​x5+5C2​x3(x3−1​)2+5C4​x(x3−1​)4)
=2(x5+10x3(x3−1)+5x(x3−1)2)
=2(x5+10x6−10x3+5x7−10x4+5x)
Sum of coefficients of all odd degree terms is 2(1−10+5+5)=2

Question 3

Let a1​,a2​,a3​,…,a49​ be in AP such that ∑k=012​a4k+1​=416 and a9​+a43​=66. If a12​+a22​+…+a172​=140m, then m is equal to

Accepted Answer

Exp. (c)We have, a1​,a2​,a3​,…a49​ are in AP.∑k=012​a4k+1​=416 and a9​+a43​=66Let a1​=a and d= common difference∵a1​+a5​+a9​+⋯+a49​=416∴a+(a+4d)+(a+8d)+…(a+48d)=416⇒213​(2a+48d)=416⇒a+24d=32.....(i) Also, a9​+a43​=66​∴a+8d+a+42d=66⇒2a+50d=66​⇒a+25d=33.......(ii)Solving Eqs. (i) and (ii), we geta=8 and d=1Now, a12​+a22​+a32​+⋯+a172​=140m82+92+102+…+242=140m⇒(12+22+32+…+242)−(12+22 +32+…+72)=140m⇒624×25×49​−67×8×15​=140 m⇒63×7×8×5​(7×5−1)=140m⇒7×4×5×34=140 m⇒140×34=140 m⇒m=34

Question 4

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12+2⋅22+32+2⋅42+52+2⋅62+…. If B−2A=100λ, then λ is equal to

Accepted Answer

Exp. (b)We have,12+2⋅22+32+2⋅42+52+2⋅62+…A= sum of first 20 termsB= sum of first 40 terms∴A=12+2⋅22+32+2⋅42+52+2⋅62+…+2⋅202A=(12+22+32+…+202)+(22+42 +62+…+202)A=(12+22+32+…+202) +4(12+22+32+…+102)A=620×21×41​+64×10×11×21​A=620×21​(41+22)=620×41×63​ B=(12+22+32+…+402) +4(12+22+…+202)B=640×41×81​+64×20×21×41​B=640×41​(81+42)=640×41×123​Now, B−2A=100λ∴640×41×123​−62×20×21×63​=100λ⇒640​(5043−1323)=100λ⇒640​×3720=100λ⇒40×620=100λ⇒λ=10040×620​=248

Question 5

If 2y=(cot−1(cosx−3​sinx3​cosx+sinx​))2,x∈(0,2π​) then dxdy​ is equal to
(a) 6π​−x
(b) x−6π​
(c) 3π​−x
(d) 2x−3π​

Accepted Answer

Exp. (b)
Given expression is
2y=(cot−1(cosx−3​sinx3​cosx+sinx​))2
=(cot−1(cotx−3​3​cotx+1​))2
[dividing each term of numerator and denominator by sinx ]
=(cot−1(cotx−cot6π​cot6π​cotx+1​))2[∵cot6π​=3​]
=(cot−1(cot(6π​−x)))2, [∵cot(A−B)=cotB−cotAcotAcotB+1​]
={(6π​−x)2,(π+(6π​−x))2,​0<x<6π​6π​<x<2π​​
⎣⎡​∵cot−1(cotθ)=⎩⎨⎧​π+θ,θ,θ−π,​−π<θ<00<θ<ππ<θ<2π​⎦⎤​
⇒2y={(6π​−x)2,(67π​−x)2,​0<x<6π​6π​<x<2π​​
⇒2dxdy​={2(6π​−x)(−1),2(67π​−x)(−1),​0<x<6π​6π​<x<2π​​
⇒dxdy​={x−6π​,x−67π​,​0<x<6π​6π​<x<2π​​

Question 6

For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→0+​x([x1​]+[x2​]+…+[x15​]) 
(a) is equal to 0
(b) is equal to 15
(c) is equal to 120
(d) does not exist (in R )

Accepted Answer

Exp. (c)
Key idea Use property of greatest integer
function [x]=x−{x}.
We have,
limx→0+​x([x1​]+[x2​]+…+[x15​])
We know, [x]=x−{x}
∴[x1​]=x1​−{x1​}
Similarly, [xn​]=xn​−{xn​}
∴ Given limit =limx→0+​x(x1​−{x1​}+x2​−{x2​}+… x15​−{x15​})
=limx→0+​(1+2+3+…+15)−x({x1​}+{x2​}+…+{x15​})
=120−0=120
[∵0≤{xn​}<1, therefore 0≤x{xn​}<x⇒limx→0+​x{xn​}=0]

Question 7

If the curves y2=6x,9x2+by2=16 intersect each other at right angles, then the value of b is
(a) 6
(b) 27​
(c) 4
(d) 29​

Accepted Answer

Exp. (d)
We have, y2=6x
⇒2ydxdy​=6
⇒dxdy​=y3​
Slope of tangent at (x1​,y1​) is m1​=y1​3​
Also, 9x2+by2=16
⇒18x+2bydxdy​=0
⇒dxdy​=by−9x​
Slope of tangent at (x1​,y1​) is m2​=by1​−9x1​​
Since, these are intersection at right angle.
∴m1​m2​=−1⇒by12​27x1​​=1
⇒6bx1​27x1​​=1,[∵y12​=6x1​]
⇒b=9/2

Question 8

Let f(x)=x2+x21​ and g(x)=x−x1​, x∈R−{−1,0,1}. If h(x)=g(x)f(x)​, then the local minimum value of h(x) is
(a) 3
(b) −3
(c) −22​
(d) 22​

Accepted Answer

Exp. (d)
We have,
f(x)=x2+x21​ and g(x)=x−x1​⇒h(x)=g(x)f(x)​
∴h(x)=x−x1​x2+x21​​=x−x1​(x−x1​)2+2​
⇒h(x)=(x−x1​)+x−x1​2​
x−x1​>0,(x−x1​)+x−x1​2​∈[22​,∞)
x−x1​<0,(x−x1​)+x−x1​2​∈(−∞,22​]
∴ Local minimum value is 22​.

Question 9

limx→π/2​(π−2x)3cotx−cosx​ equals
(a) 241​
(b) 161​
(c) 81​
(d) 41​

Accepted Answer

Exp. (b)
limx→π/2​(π−2x)3cotx−cosx​=limx→π/2​81​⋅sinx(2π​−x)3cosx(1−sinx)​
=limh→0​81​⋅sin(2π​−h)(2π​−2π​+h)3cos(2π​−h)[1−sin(2π​−h)]​
=81​limh→0​cosh⋅h3sinh(1−cosh)​
=81​limh→0​cosh⋅h3sinh(2sin22h​)​
=41​limh→0​h3coshsinh⋅sin2(2h​)​
=41​limh→0​(hsinh​)(2h​sin2h​​)2⋅cosh1​⋅41​
=41​×41​=161​

Question 10

The integral ∫(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2sin2xcos2x​dx is equal to, (where C is a constant of integration).
(a) 3(1+tan3x)1​+C
(b) 3(1+tan3x)−1​+C
(c) 1+cot3x1​+C
(d) 1+cot3x−1​+C

Accepted Answer

Exp. (b)
We have,
I=∫(sin5x+cos3x⋅sin2x+sin3x⋅cos2x+cos5x)2sin2x⋅cos2x​dx
=∫{sin3x(sin2x+cos2x)+cos3x(sin2x+cos2x)}2sin2xcos2x​dx
=∫(sin3x+cos3x)2sin2xcos2x​dx=∫cos6x(1+tan3x)2sin2xcos2x​dx
=∫(1+tan3x)2tan2xsec2x​dx
Put tan3x=t
⇒3tan2xsec2xdx=dt
∴I=31​∫(1+t)2dt​
⇒I=3(1+t)−1​+C
⇒I=3(1+tan3x)−1​+C

Question 11

The value of ∫−2π​2π​​1+2xsin2x​dx is
(a) 8π​
(b) 2π​
(c) 4π
(d) 4π​

Accepted Answer

Exp. (d)
Key idea
Use property =∫ab​f(x)dx=∫ab​f(a+b−x)dx

Let I=∫−π/2π/2​1+2xsin2x​dx
⇒I=∫−π/2π/2​1+2−2π​+2π​−xsin2(−2π​+2π​−x)​dx
[∵∫ab​f(x)dx=∫ab​f(a+b−x)dx]
⇒I=∫−π/2π/2​1+2−xsin2x​dx
⇒I=∫−π/2π/2​2x+12xsin2x​dx
⇒2I=∫−π/2π/2​sin2x(2x+12x+1​)dx
⇒2I=∫−π/2π/2​sin2xdx⇒2I=2∫0π/2​sin2xdx
⇒I=∫0π/2​sin2xdx
⇒I=∫0π/2​cos2xdx[∵∫0a​f(x)dx=∫0a​f(a−x)dx]
⇒2I=∫0π/2​dx⇒2I=[x]0π/2​⇒I=4π​

Question 12

Let g(x)=cosx2,f(x)=x​ and α,β(α<β) be the roots of the quadratic equation 18x2−9πx+π2=0. Then, the area (in sq units) bounded by the curve y=(gof)(x) and the lines x=α,x=β and y=0, is
(a) 21​(3​−1)
(b) 21​(3​+1)
(c) 21​(3​−2​)
(d) 21​(2​−1)

Accepted Answer

Exp. (a)
We have,
⇒18x2−9πx+π2=0
⇒18x2−6πx−3πx+π2=0
⇒(6x−π)(3x−π)=0
x=6π​,3π​
Now, α<β
∴α=6π​,β=3π​
Given, g(x)=cosx2 and f(x)=x​
y=gof(x)
y=g(f(x))=cosx
Area of region bounded by x=α,x=β,y=0 and curve y=g(f(x)) is
A=∫π/6π/3​cosxdx
A=[sinx]π/6π/3​
A=sin3π​−sin6π​=23​​−21​
A=(23​−1​)

Question 13

Let y=y(x) be the solution of the differential equation sinxdxdy​+ycosx=4x,x∈(0,π). If y(2π​)=0, then y(6π​) is equal to
(a) 93​4​π2
(b) 93​−8​π2
(c) −98​π2
(d) −94​π2

Accepted Answer

Exp. (c)
We have,
sinxdxdy​+ycosx=4x⇒dxdy​+ycotx=4xcosecx
This is a linear differential equation of form
dxdy​+Py=Q
where P=cotx,Q=4xcosecx
Now, IF =e∫Pdx=e∫cotxdx=elogsinx=sinx
Solution of the differential equation is
y⋅sinx=∫4xcosecxsinxdx+C
⇒ysinx=∫4xdx+C=2x2+C
Put x=2π​,y=0, we get
C=−2π2​⇒ysinx=2x2−2π2​
Put x=6π​
∴y(21​)=2(36π2​)−2π2​
⇒y=9π2​−π2⇒y=−98π2​

Alternate Method
We have, sinxdxdy​+ycosx=4x, which can be written as dxd​(sinx⋅y)=4x
On integrating both sides, we get
∫dxd​(sinx⋅y)⋅dx=∫4x⋅dx
⇒y⋅sinx=24x2​+C⇒y⋅sinx=2x2+C
Now, as y=0 when x=2π​
∴C=−2π2​⇒y⋅sinx=2x2−2π2​
Now, putting x=6π​, we get
y(21​)=2(36π2​)−2π2​⇒y=9π2​−π2=−98π2​

Question 14

If the tangent at (1,7) to the curve x2=y−6 touches the circle x2+y2+16x+12y+c=0, then the value of c is

Accepted Answer

Exp. (d)Key Idea Equation of tangent to the curve x2=4ay at (x1​,y1​) is xx1​=4a(2y+y1​​)Tangent to the curve x2=y−6 at (1,7) isx=2y+7​−6⇒2x−y+5=0.......(i)Equation of circle is x2+y2+16x+12y+c=0Centre (−8,−6)r=82+62−c​=100−c​Since, line 2x−y+5=0 also touches the circle.∴100−c​=∣∣​22+12​2(−8)−(−6)+5​∣∣​⇒100−c​=∣∣​5​−16+6+5​∣∣​⇒100−c​=∣−5​∣⇒100−c=5⇒c=95

Question 15

If L1​ is the line of intersection of the planes 2x−2y+3z−2=0,x−y+z+1=0 and L2​ is the line of intersection of the planes x+2y−z−3=0,3x−y+2z−1=0, then the distance of the origin from the plane, containing the lines L1​ and L2​ is
(a) 42​1​
(b) 32​1​
(c) 22​1​
(d) 2​1​

Accepted Answer

Exp. (b)
L1​ is the line of intersection of the plane 2x−2y+3z−2=0 and x−y+z+1=0 and L2​ is the line of intersection of the plane x+2y−z−3=0 and 3x−y+2z−1=0
Since L1​ is parallel to ∣∣​i^21​j^​−2−1​k^31​∣∣​=i^+j^​
L2​ is parallel to ∣∣​i^13​j^​2−1​k^−12​∣∣​=3i^−5j^​−7k^
Also, L2​ passes through (75​,78​,0), [put z=0 in last two planes]
So, equation of plane is  ∣∣​x−75​13​y−78​1−5​z0−7​∣∣​=0
 ⇒7x−7y+8z+3=0
Now, perpendicular distance from origin is
∣∣​72+72+82​3​∣∣​=162​3​=32​1​

Question 16

Let u be a vector coplanar with the vectors a=2i^+3j^​−k^ and b=j^​+k^. If u is perpendicular to a and u⋅b=24, then ∣u∣2 is equal to

Accepted Answer

Here, u is coplanar with a and b.∴u=λa+μbDot product with a, we getu⋅a=λ(a⋅a)+μ(b⋅a)⇒0=14λ+2μ… (i) [∵a=2i^+3j^​−k^,b=j^​+k^,u⋅a=0]Dot product with b, we getu⋅b =λ(a⋅b)+μ(b⋅b)24=2λ+2μ… (ii) [∵u⋅b=24]Solving Eqs. (i) and (ii), we getλ=−2,μ=14Dot product with u, we get∣u∣2=λ(u⋅a)+μ(u⋅b)∣u∣2=−2(0)+14(24)⇒∣u∣2=336

Question 17

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is 
(a) 103​
(b) 52​
(c) 51​
(d) 43​

Accepted Answer

Exp. (b)
 Key idea Use the theorem of total probability 
 Let E1​= Event that first ball drawn is red 
E2​= Event that first ball drawn is black 
A= Event that second ball drawn is red 
P(E1​)=104​,P(E1​A​)=126​
⇒P(E2​)=106​,P(E2​A​)=124​
 By law of total probability 
P(A)=P(E1​)×P(E1​A​)+P(E2​)×P(E2​A​)
=104​×126​+106​×124​=12024+24​=12048​=52​

Question 18

If ∑i=19​(xi​−5)=9 and ∑i=19​(xi​−5)2=45, then the standard deviation of the 9 items x1​,x2​,…,x9​ is
(a) 9
(b) 4
(c) 2
(d) 3

Accepted Answer

Exp.(c)
Key idea Standard deviation is remain unchanged, if observations are added or subtracted by a fixed number.
We have,
∑i=19​(xi​−5)=9 and ∑i=19​(xi​−5)2=45
SD=9∑i=19​(xi​−5)2​−(9∑i=19​(xi​−5)​)2​
⇒SD=945​−(99​)2​
⇒SD=5−1​=4​=2

Question 19

If sum of all the solutions of the equation 8cosx⋅(cos(6π​+x)⋅cos(6π​−x)−21​) =1 in [0,π] is kπ, then k is equal to
(a) 32​
(b) 913​
(c) 98​
(d) 920​

Accepted Answer

Exp. (b)
We have,
8cosx(cos(6π​+x)cos(6π​−x)−21​)=1
⇒8cosx(cos26π​−sin2x−21​)=1
⇒8cosx(43​−sin2x−21​)=1
⇒8cosx(43​−21​−1+cos2x)=1
⇒8cosx(4−3+4cos2x​)=1
⇒2(4cos3x−3cosx)=1
⇒2cos3x=1⇒cos3x=21​
⇒3x=3π​,35π​,37π​[0≤3x≤3π]
⇒x=9π​,95π​,97π​
Sum =9π​+95π​+97π​=913π​⇒kπ=913π​
Hence, k=913​

Question 20

The boolean expression ∼(p∨q)∨(∼p∧q) is equivalent to
(a) ∼p
(b) p
(c) q
(d) ∼q

Accepted Answer

Exp. (a)
Key Idea Use DeMorgan's and distributive law.
We have, ∼(p∨q)∨(∼p∧q)
≡(∼p∧∼q)∨(∼p∧q)
[∵ By DeMorgan's law ∼(p∨q)=(∼p∧∼q)]
≡∼p∧(∼q∨q)[ By distributive law] 
≡∼p∧t[∼q∨q=t]
≡∼p

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Solutions For Exams JEE Mains, 2018

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