Q: Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1−Tn=10, then the value of n is (a) 7 (b) 5 (c) 10 (d) 8

Exp. (b) Given, Tn=nC3 Tn+1=n+1C3 ∴Tn+1−Tn=n+1C3−nC3=10 [given] ⇒nC2+nC3−nC3=10 ⇒nC2=10 ∴n=5

Q: The term independent of x in the expansion of (x2/3−x1/3+1x+1−x−x1/2x−1)10 is (a) 4 (b) 120 (c) 210 (d) 310

Exp. (c) Consider [x2/3−x1/3+1x+1−x−x1/2(x−1)]10 =[x2/3−x1/3+1(x1/3)3+13−x(x−1){(x)2−1}]10 =[x2/3−x1/3+1(x1/3+1)(x2/3+1−x1/3)x(x−1)(x−1)(x+1)]10 =[(x1/3+1)−x(x+1)]10=(x1/3−x−1/2)10 ∴ The general term is Tr+1=10Cr(x1/3)10−r(−x−1/2)r =10Cr(−1)rx310−r−2r For independent of x, put 310−r−2r=0 ⇒20−2r−3r=0 ⇒20=5r⇒r=4 ∴T5=10C4=4×3×2×110×9×8×7=210

Q: limx→0xtan4x(1−cos2x)(3+cosx) is equal to (a) −41 (b) 21 (c) 1 (d) 2

Exp. (d) Let I=limx→0x2(1−cos2x)⋅1(3+cosx)⋅tan4xx =limx→0x22sin2x⋅13+cosx⋅tan4xx =2limx→0(xsinx)2⋅limx→0(3+cosx)⋅limx→04tan4x4x [∵limθ→0θsinθ=1 and limθ→0tanθθ=1] ⇒I=2⋅(1)2⋅(3+cos0∘)⋅41(1) =2⋅1⋅(3+1)⋅41=2⋅4⋅41=2

Q: The intercepts on X-axis made by tangents to the curve, y=∫0x∣t∣dt,x∈R, which are parallel to the line y=2x, are equal to (a) ±1 (b) ±2 (c) ±3 (d) ±4

Exp. (a) Given, y=∫0x∣t∣dt ∴dxdy=∣x∣ Since, tangent to the curve is parallel to the line y=2x. ⇒dxdy=2 ∴x=±2 ∴ Points, y=∫0±2∣t∣dt=±2 ∴ Equation of tangents are y−2=2(x−2) y+2=2(x+2) For x-intercept, put y=0, we get 0−2=2(x−2) ⇒0+2=2(x+2) ⇒x=±1

Q: Statement I. The value of the integral ∫π/6π/31+tanxdx is equal to π/6. Statement II. ∫abf(x)dx=∫abf(a+b−x)dx (a) Statement I is true Statement II is true; Statement II is a correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true

Exp.(d) Let I=∫π/6π/31+tanxdx....(i) ∴I=∫π/6π/31+tan(2π−x)dx=∫π/6π/31+cotxdx ⇒I=∫π/6π/31+tanxtanxdx........(ii) On adding Eqs. (i) and (ii), we get 2l=∫π/6π/3dx⇒2l=[x]π/6π/3 ∴I=21[3π−6π]=12π Hence, Statement I is false. But ∫abf(x)dx=∫abf(a+b−x)dx is a true statement by property of definite integrals.

Q: At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dxdP=100−12x. If the firm employees 25 more workers, then the new level of production of items is (a) 2500 (b) 3000 (c) 3500 (d) 4500

Exp. (c) Given, dxdP=(100−12x) ⇒dP=(100−12x)dx On integrating both sides, we get ∫dP=∫(100−12x)dx ⇒P=100x−8x3/2+C When x=0, then P=2000⇒C=2000 Now, when x=25, then P=100×25−8×(25)3/2+2000 =2500−8×125+2000 =4500−1000=3500

Q: Given A circle, 2x2+2y2=5 and a parabola, y2=45x. Statement I. An equation of a common tangent to these curves is y=x+5.Statement II. If the line, y=mx+m5,(m=0) is the common tangent, then m satisfies m4−3m2+2=0.

Exp. (b)Equation of circle can be rewritten as x2+y2=25Here, centre →(0,0) and radius →25.Let common tangent bey=mx+m5The perpendicular from centre to the tangent is equal to radius.∴1+m2m5=25⇒m1+m2=2⇒m2(1+m2)=2⇒m4+m2−2=0⇒(m2+2)(m2−1)=0⇒m=±1[∵m2+2=0, as m∈R]∴y=±(x+5)Both statements are correct as m=±1 satisfies the given equation of Statement II but statement II is not a correct explanation of statement I.

Question 1

Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets of A×B having 3 or more elements is
(a) 256
(b) 220
(c) 219
(d) 211

Accepted Answer

Exp. (c)
Given, n(A)=2 and n(B)=4
∴n(A×B)=8
The number of subsets of A×B having 3 or more elements =8C3​+8C1​+…+8C8​
=28−8C0​−8C1​−8C2​
=256−1−8−28=219, [∵2n=nC0​+nC1​+…+nCn​]

Question 2

Let Tn​ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1​−Tn​=10, then the value of n is
(a) 7
(b) 5
(c) 10
(d) 8

Accepted Answer

Exp. (b)
Given, Tn​=nC3​
Tn+1​=n+1C3​
∴Tn+1​−Tn​=n+1C3​−nC3​=10 [given]
⇒nC2​+nC3​−nC3​=10
⇒nC2​=10
∴n=5

Question 3

The term independent of x in the expansion of (x2/3−x1/3+1x+1​−x−x1/2x−1​)10 is
(a) 4
(b) 120
(c) 210
(d) 310

Accepted Answer

Exp. (c)
Consider [x2/3−x1/3+1x+1​−x−x1/2(x−1)​]10
=[x2/3−x1/3+1(x1/3)3+13​−x​(x​−1){(x​)2−1}​]10
=[x2/3−x1/3+1(x1/3+1)(x2/3+1−x1/3)​x​(x​−1)(x​−1)(x​+1)​]10
=[(x1/3+1)−x​(x​+1)​]10=(x1/3−x−1/2)10
∴ The general term is
Tr+1​=10Cr​(x1/3)10−r(−x−1/2)r
=10Cr​(−1)rx310−r​−2r​
For independent of x, put
310−r​−2r​=0
⇒20−2r−3r=0
⇒20=5r⇒r=4
∴T5​=10C4​=4×3×2×110×9×8×7​=210

Question 4

If x,y and z are in AP and tan−1x,tan−1y and tan−1z are also in AP, then

Accepted Answer

Exp. (a)Since, x,y and z are in AP.∴2y=x+zAlso, tan−1xtan−1y and tan−1z are in AP.∴2tan−1y=tan−1x+tan−1z⇒tan−1(1−y22y​)=tan−1(1−xzx+z​)⇒1−y2x+z​=1−xzx+z​⇒y2=xzSince x,y and z are in AP as well as in GP.∴x=y=z

Question 5

limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to
(a) −41​
(b) 21​
(c) 1
(d) 2

Accepted Answer

Exp. (d)
Let I=limx→0​x2(1−cos2x)​⋅1(3+cosx)​⋅tan4xx​
=limx→0​x22sin2x​⋅13+cosx​⋅tan4xx​
=2limx→0​(xsinx​)2⋅limx→0​(3+cosx)⋅limx→0​4tan4x4x​
[∵limθ→0​θsinθ​=1 and limθ→0​tanθθ​=1]
⇒I=2⋅(1)2⋅(3+cos0∘)⋅41​(1)
=2⋅1⋅(3+1)⋅41​=2⋅4⋅41​=2

Question 6

The intercepts on X-axis made by tangents to the curve, y=∫0x​∣t∣dt,x∈R, which are parallel to the line y=2x, are equal to
(a) ±1
(b) ±2 
(c) ±3
(d) ±4

Accepted Answer

Exp. (a)
Given, y=∫0x​∣t∣dt
∴dxdy​=∣x∣
Since, tangent to the curve is parallel to the line y=2x.
⇒dxdy​=2
∴x=±2
∴ Points, y=∫0±2​∣t∣dt=±2
∴ Equation of tangents are
y−2=2(x−2)
y+2=2(x+2)
For x-intercept, put y=0, we get
0−2=2(x−2)
⇒0+2=2(x+2)
⇒x=±1

Question 7

If ∫f(x)dx=ψ(x), then ∫x5f(x3)dx is equal to
(a) 31​[x3ψ(x3)−∫x2ψ(x3)dx]+C
(b) 31​x3ψ(x3)−3∫x3ψ(x3)dx+C
(c) 31​x3ψ(x3)−∫x2ψ(x3)dx+C
(d) 31​[x3ψ(x3)−∫x3ψ(x3)dx]+C

Accepted Answer

Exp. (c)
Given, ∫f(x)dx=ψ(x)
Let I=∫x5f(x3)dx
Put x3=t
⇒x2dx=3dt​.......(i)
∴I=31​∫tf(t)dt=31​[tψ(t)−∫ψ(t)dt]
[using integration by parts] 
=31​[x3ψ(x3)−3∫x2ψ(x3)dx]+C, [from Eq. (i)]
=31​x3ψ(x3)−∫x2ψ(x3)dx+C

Question 8

Statement I. The value of the integral ∫π/6π/3​1+tanx​dx​ is equal to π/6.
Statement II. ∫ab​f(x)dx=∫ab​f(a+b−x)dx

(a) Statement I is true Statement II is true; Statement II is a correct explanation of Statement I
(b) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true

Accepted Answer

Exp.(d)
Let I=∫π/6π/3​1+tanx​dx​....(i)
∴I=∫π/6π/3​1+tan(2π​−x)​dx​=∫π/6π/3​1+cotx​dx​
⇒I=∫π/6π/3​1+tanx​tanx​dx​........(ii)
On adding Eqs. (i) and (ii), we get
2l=∫π/6π/3​dx⇒2l=[x]π/6π/3​
∴I=21​[3π​−6π​]=12π​
Hence, Statement I is false.
But ∫ab​f(x)dx=∫ab​f(a+b−x)dx is a true statement by property of definite integrals.

Question 9

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dxdP​=100−12x​. If the firm employees 25 more workers, then the new level of production of items is
(a) 2500
(b) 3000
(c) 3500
(d) 4500

Accepted Answer

Exp. (c)
Given, dxdP​=(100−12x​)
 ⇒dP=(100−12x​)dx
On integrating both sides, we get
∫dP=∫(100−12x​)dx
⇒P=100x−8x3/2+C
When x=0, then P=2000⇒C=2000
Now, when x=25, then
P=100×25−8×(25)3/2+2000
=2500−8×125+2000
=4500−1000=3500

Question 10

Given A circle, 2x2+2y2=5 and a parabola, y2=45​x. Statement I. An equation of a common tangent to these curves is y=x+5​.Statement II. If the line, y=mx+m5​​,(m=0) is the common tangent, then m satisfies m4−3m2+2=0.

Accepted Answer

Exp. (b)Equation of circle can be rewritten as x2+y2=25​Here, centre →(0,0) and radius →25​​.Let common tangent bey=mx+m5​​The perpendicular from centre to the tangent is equal to radius.∴1+m2​m5​​​=25​​⇒m1+m2​=2​⇒m2(1+m2)=2⇒m4+m2−2=0⇒(m2+2)(m2−1)=0⇒m=±1[∵m2+2=0, as m∈R]∴y=±(x+5​)Both statements are correct as m=±1 satisfies the given equation of Statement II but statement II is not a correct explanation of statement I.

Question 11

If the lines 1x−2​=1y−3​=−kz−4​ and kx−1​=2y−4​=1z−5​ are coplanar, then k can have
(a) any value
(b) exactly one value
(c) exactly two values
(d) exactly three values

Accepted Answer

Exp.(c)
The given lines are
 1x−2​=1y−3​=−kz−4​.....(i)
 and kx−1​=2y−4​=1z−5​....(ii)
Condition for two lines are coplanar.
∣∣​x1​−x2​l1​l2​​y1​−y2​m1​m2​​z1​−z2​n1​n2​​∣∣​=0
where, (x1​,y1​,z1​) and (x2​,y2​,z2​) are any points on the lines (i) and (ii), respectively and (l1​,m1​,n1​) and (l2​,m2​,n2​) are direction cosines of lines (i) and (ii), respectively.
∣∣​2−11k​3−412​4−5−k1​∣∣​=0
⇒∣∣​11k​−112​−1−k1​∣∣​=0
⇒1(1+2k)+(1+k2)−(2−k)=0
⇒k2+2k+k=0
⇒k2+3k=0
⇒k=0,−3

Question 12

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
(a) 3517​
(b) 3513​
(c) 3511​
(d) 3510​

Accepted Answer

Exp. (c)
Probability of guessing a correct answer, p=31​ 
and probability of guessing a wrong answer, q=32​ 
So, the probability of guessing 4 or more correct answers
=5C4​(31​)4⋅32​+5C5​(31​)5=5⋅352​+351​=3511​

Question 13

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measure will not change even after the grace marks were given?
(a) Mean
(b) Median
(c) Mode
(d) Variance

Accepted Answer

Exp.(d)
If initially all marks were xi​, then
σi2​=NΣ(xi​−xˉ)2​
Now, each is increased by 10.
∴ New σi2​=NΣ[xi​+10)−(xˉ+10)]2​=σi2​
So, variance will not change whereas mean, median and mode will increase by 10 .

Question 14

The expression 1−cotAtanA​+1−tanAcotA​ can be written as-
(a) sinAcosA+1
(b) secAcosecA+1
(c) tanA+cotA
(d) secA+cosecA

Accepted Answer

Exp. (b)
Given expression is
1−cotAtanA​+1−tanAcotA​=cosAsinA​×sinA−cosAsinA​+sinAcosA​×cosA−sinAcosA​
=sinA−cosA1​{cosAsinA​sin3A−cos3A​}
=sinAcosAsin2A+sinAcosA+cos2A​
=sinAcosA1+sinAcosA​
=1+secAcosecA

Question 15

If y=sec(tan−1x), then dxdy​ at x=1 is equal to
(a) 2​1​
(b) 21​
(c) 1
(d) 2​

Accepted Answer

Exp. (a)
Given, y=sec(tan−1x)

Let tan−1x=θ⇒x=tanθ
∴y=secθ=1+x2​
On differentiating w.r.t. x, we get
dxdy​=21+x2​1​⋅2x
At x=1,dxdy​=2​1​

Question 16

The area (in sq. units) bounded by the curves y=x​,2y−x+3=0,X-axis and lying in the first quadrant is
(a) 9
(b) 36
(c) 18
(d) 427​

Accepted Answer

Exp. (a)
Given curves are
y=x​.......(i)
and 2y−x+3=0......(ii)
On solving Eqs. (i) and (ii), we get
2x​−(x​)2+3=0
(x​)2−2x​−3=0

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Solutions For Exams JEE Mains, 2013

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