Q: If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is (a) 46 th (b) 59 th (c) 52 nd (d) 58 th

Exp. (d) Clearly, number of words start with A=2!4!=12 Number of words start with L=4!=24 Number of words start with M=2!4!=12 Number of words start with SA=2!3!=3 Number of words start with SL=3!=6 Note that, next word will be "SMALL". Hence, the position of word "SMALL" is 58 th.

Q: If the number of terms in the expansion of (1−x2+x24)n,x=0, is 28 , then the sum of the coefficients of all the terms in this expansion, is (a) 64 (b) 2187 (c) 243 (d) 729

Exp. (d) Clearly, number of terms in the expansion of (1−x2+x24)n is 2(n+2)(n+1) or n+2C2. [assuming x1 and x21 distinct] ∴2(n+2)(n+1)=28 ⇒(n+2)(n+1)=56=(6+1)(6+2)⇒n=6 Hence, sum of coefficients =(1−2+4)6=36 =729

Q: If the sum of the first ten terms of the series (153)2+(252)2+(351)2+42+(454)2+…, is 516m, then m is equal to

Exp. (b)Let S10 be the sum of first ten terms of the series. Then, we haveS10=(153)2+(252)2+(351)2+42+(454)2+… to 10 terms+…+… to 10 terms=(58)2+(512)2+(516)2+42+(524)2 +… to 10 terms)=521(82+122+162+202+242+…=5242(22+32+42+52+… to 10 terms )=5242(22+32+42+52+…+112)=2516((12+22+…+112)−12)=2516(611⋅(11+1)(2⋅11+1)−1)=2516(506−1)=2516×505

Q: Let p=limx→0+(1+tan2x)1/2x, then logp is equal to (a) 2 (b) 1 (c) 21 (d) 41

Exp. (c) Given, p=limx→0+(1+tan2x)2x1(1∞ form ) =elimx→0+2xtan2x=e21limx→0+(xtanx)2=e21 ∴logp=loge21=21

Q: limn→∞(n2n(n+1)(n+2)…3n)1/n is equal to (a) e418 (b) e227 (c) e29 (d) 3 log3−2

Exp. (b) Let I=limn→∞(n2n(n+1)⋅(n+2)…(3n))n1 =limn→∞(n2n(n+1)⋅(n+2)…(n+2n))n1 =limn→∞((nn+1)(nn+2)…(nn+2n))n1 Taking log on both sides, we get logI=limn→∞n1[log{(1+n1)(1+n2)…(1+n2n)}] ⇒logI=limn→∞n1[log(1+n1)+log(1+n2)+…+log(1+n2n)] ⇒logI=limn→∞n1∑r=12nlog(1+nr) ⇒logI=∫02log(1+x)dx ⇒logI=[log(1+x)⋅x−∫1+x1⋅xdx]02 ⇒logI=[log(1+x)⋅x]02−∫021+xx+1−1dx ⇒logI=2⋅log3−∫02(1−1+x1)dx ⇒logI=2⋅log3−[x−log∣1+x∣]02 ⇒logl=2⋅log3−[2−log3] ⇒logI=3⋅log3−2 ⇒logI=log27−2 ⇒I=elog27−2=27⋅e−2=e227

Q: The integral ∫(x5+x3+1)32x12+5x9dx is equal to, (where, C is an arbitrary constant). (a) (x5+x3+1)2−x5+C (b) 2(x5+x3+1)2x10+C (c) 2(x5+x3+1)2x5+C (d) 2(x5+x3+1)2−x10+C

Exp. (b) =∫(x5+x3+1)32x12+5x9dx =∫x15(1+x−2+x−5)32x12+5x9dx =∫(1+x−2+x−5)32x−3+5x−6dx Now, put 1+x−2+x−5=t ⇒(−2x−3−5x−6)dx=dt ⇒(2x−3+5x−6)dx=−dt ∴ I=−∫t3dt=−∫t−3dt =−−3+1t−3+1+C=2t21+C =2(x5+x3+1)2x10+C

Question 1

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is
(a) 46 th
(b) 59 th
(c) 52 nd
(d) 58 th

Accepted Answer

Exp. (d)
Clearly, number of words start with A=2!4!​=12
Number of words start with L=4!=24
Number of words start with M=2!4!​=12
Number of words start with SA=2!3!​=3
Number of words start with SL=3!=6
Note that, next word will be "SMALL".
Hence, the position of word "SMALL" is 58 th.

Question 2

If the number of terms in the expansion of (1−x2​+x24​)n,x=0, is 28 , then the sum of the coefficients of all the terms in this expansion, is
(a) 64
(b) 2187
(c) 243
(d) 729

Accepted Answer

Exp. (d)
Clearly, number of terms in the expansion of (1−x2​+x24​)n is 2(n+2)(n+1)​ or n+2C2​. 
[assuming x1​ and x21​ distinct] 
∴2(n+2)(n+1)​=28
 ⇒(n+2)(n+1)=56=(6+1)(6+2)⇒n=6
Hence, sum of coefficients =(1−2+4)6=36 =729

Question 3

If the 2nd,5 th and 9 th terms of a nonconstant AP are in GP, then the common ratio of this GP is

Accepted Answer

Exp. (b)Let a be the first term and d be the common difference. Then, we have a+d,a+4d,a+8d in GP,i.e. (a+4d)2=(a+d)(a+8d)⇒a2+16d2+8ad=a2+8ad+ad+8d2⇒8d2=ad⇒8d=a[∵d=0]Now, common ratio,r=a+da+4d​=8d+d8d+4d​=9d12d​=34​

Question 4

If the sum of the first ten terms of the series (153​)2+(252​)2+(351​)2+42+(454​)2+…, is 516​m, then m is equal to

Accepted Answer

Exp. (b)Let S10​ be the sum of first ten terms of the series. Then, we haveS10​=(153​)2+(252​)2+(351​)2+42+(454​)2+… to 10 terms+…+… to 10 terms=(58​)2+(512​)2+(516​)2+42+(524​)2 +… to 10 terms)=521​(82+122+162+202+242+…=5242​(22+32+42+52+… to 10 terms )=5242​(22+32+42+52+…+112)=2516​((12+22+…+112)−12)=2516​(611⋅(11+1)(2⋅11+1)​−1)=2516​(506−1)=2516​×505

Question 5

Let p=limx→0+​(1+tan2x​)1/2x, then logp is equal to 
(a) 2
(b) 1
(c) 21​
(d) 41​

Accepted Answer

Exp. (c)
 Given, p=limx→0+​(1+tan2x​)2x1​(1∞ form )
=elimx→0+​2xtan2x​​=e21​limx→0+​(x​tanx​​)2=e21​
∴logp=loge21​=21​

Question 6

For x∈R,f(x)=∣log2−sinx∣ and g(x)=f(f(x)), then 
(a) g is not differentiable at x=0
(b) g′(0)=cos(log2)
(c) g′(0)=−cos(log2)
(d) g is differentiable at x=0 and g′(0)=−sin(log2)

Accepted Answer

Exp. (b)
We have, f(x)=∣log2−sinx∣
and g(x)=f(f(x)),x∈R
Note that, for x→0,log2>sinx
∴f(x)=log2−sinx
⇒g(x)=log2−sin(f(x))
=log2−sin(log2−sinx)
Clearly, g(x) is differentiable at x=0 as sinx is differentiable.
Now, g′(x)=−cos(log2−sinx)(−cosx)
=cosx⋅cos(log2−sinx)
⇒g′(0)=1⋅cos(log2)

Question 7

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side =x units and a circle of radius =r units. If the sum of the areas of the square and the circle so formed is minimum, then
(a) 2x=(π+4)r
(b) (4−π)x=πr
(c) x=2r
(d) 2x=r

Accepted Answer

Exp. (c)
According to given information, we have
Perimeter of square + Perimeter of circle =2 units
⇒4x+2πr=2
⇒r=π1−2x​...(i)
Now, let A be the sum of the areas of the square and the circle. Then,
A=x2+πr2
=x2+ππ2(1−2x)2​
⇒A(x)=x2+π(1−2x)2​
⇒x=π2−4x​
⇒πx+4x=2
⇒x=π+42​....(ii)
Now, from Eq. (i), we get
r=π1−2⋅π+42​​
=π(π+4)π+4−4​=π+41​.......(iii)
From Eqs. (ii) and (iii), we get
x=2r

Question 8

limn→∞​(n2n(n+1)(n+2)…3n​)1/n is equal to
(a) e418​
(b) e227​
(c) e29​
(d) 3 log3−2

Accepted Answer

Exp. (b)
Let I=limn→∞​(n2n(n+1)⋅(n+2)…(3n)​)n1​
=limn→∞​(n2n(n+1)⋅(n+2)…(n+2n)​)n1​
=limn→∞​((nn+1​)(nn+2​)…(nn+2n​))n1​
Taking log on both sides, we get
logI=limn→∞​n1​[log{(1+n1​)(1+n2​)…(1+n2n​)}]
⇒logI=limn→∞​n1​[log(1+n1​)+log(1+n2​)+…+log(1+n2n​)]
⇒logI=limn→∞​n1​∑r=12n​log(1+nr​)
⇒logI=∫02​log(1+x)dx
⇒logI=[log(1+x)⋅x−∫1+x1​⋅xdx]02​
⇒logI=[log(1+x)⋅x]02​−∫02​1+xx+1−1​dx
⇒logI=2⋅log3−∫02​(1−1+x1​)dx
⇒logI=2⋅log3−[x−log∣1+x∣]02​
⇒logl=2⋅log3−[2−log3]
⇒logI=3⋅log3−2
⇒logI=log27−2
⇒I=elog27−2=27⋅e−2=e227​

Question 9

The integral ∫(x5+x3+1)32x12+5x9​dx is equal to, (where, C is an arbitrary constant).
(a) (x5+x3+1)2−x5​+C
(b) 2(x5+x3+1)2x10​+C
(c) 2(x5+x3+1)2x5​+C
(d) 2(x5+x3+1)2−x10​+C

Accepted Answer

Exp. (b)
=∫(x5+x3+1)32x12+5x9​dx
=∫x15(1+x−2+x−5)32x12+5x9​dx
=∫(1+x−2+x−5)32x−3+5x−6​dx
Now, put 1+x−2+x−5=t
⇒(−2x−3−5x−6)dx=dt
⇒(2x−3+5x−6)dx=−dt
∴ I=−∫t3dt​=−∫t−3dt
=−−3+1t−3+1​+C=2t21​+C
=2(x5+x3+1)2x10​+C

Question 10

If a curve y=f(x) passes through the point (1,−1) and satisfies the differential equation, y(1+xy)dx=xdy, then f(−21​) is equal to
(a) −52​
(b) −54​
(c) 52​
(d) 54​

Accepted Answer

Exp. (d)
Given differential equation is
y(1+xy)dx=xdy⇒ydx+xy2dx=xdy
⇒y2xdy−ydx​=xdx
 ⇒−y2(ydx−xdy)​=xdx
⇒−d(yx​)=xdx
On integrating both sides, we get
−yx​=2x2​+C......(i)
∵ It passes through (1,−1).
∴1=21​+C⇒C=21​
Now, from Eq. (i) −yx​=2x2​+21​
 ⇒x2+1=−y2x​⇒y=−x2+12x​
∴f(−21​)=54​

Question 11

The centres of those circles which touch the circle, x2+y2−8x−8y−4=0, externally and also touch the X-axis, lie on
(a) a circle
(b) an ellipse which is not a circle
(c) a hyperbola
(d) a parabola

Accepted Answer

Exp. (d)
Given equation of circle is
x2+y2−8x−8y−4=0
whose centre is C(4,4) and radius
=42+42+4​=36​=6
Let the centre of required circle be C1​(x,y). 
Now, as it touch the X-axis, therefore its radius =∣y∣. Also, it touch the circle
x2+y2−8x−8y−4=0
 therefore CC1​=6+∣y∣
⇒(x−4)2+(y−4)2​=6+∣y∣
⇒x2+16−8x+y2+16−8y=36+y2+12∣y∣
⇒x2−8x−8y+32=36+12∣y∣
⇒x2−8x−8y−4=12∣y∣

Case I If y>0, then we have
x2−8x−8y−4=12y
⇒x2−8x−20y−4=0
⇒x2−8x−4=20y
⇒(x−4)2−20=20y
⇒(x−4)2=20(y+1)
which is a parabola.

Case II If y<0, then we have
x2−8x−8y−4=−12y⇒x2−8x−8y−4+12y=0⇒x2−8x+4y−4=0⇒x2−8x−4=−4y⇒(x−4)2=20−4y⇒(x−4)2=−4(y−5) which is again a parabola. ​

Question 12

Let P be the point on the parabola, y2=8x, which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1. Then, the equation of the circle, passing through C and having its centre at P is

Accepted Answer

Exp. (a)Centre of circle x2+(y+6)2=1 is C(0,−6).Let the coordinates of point P be (2t2,4t).Now, let D=CP=(2t2)2+(4t+6)2​ ⇒D=4t4+16t2+36+48t​Squaring on both side⇒D2(t)=4t4+16t2+48t+36 Let F(t)=4t4+16t2+48t+36For minimum, F′(t)=0⇒16t3+32t+48=0⇒t3+2t+3=0⇒(t+1)(t2−t+3)=0⇒t=−1Thus, coordinate of point P are (2,−4).Now, CP=22+(−4+6)2​=4+4​=22​Hence, required equation of circle is(x−2)2+(y+4)2=(22​)2⇒x2+4−4x+y2+16+8y=8⇒x2+y2−4x+8y+12=0

Question 13

The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is

Accepted Answer

Exp. (c)We have,a2b2​=8 and 2b=ae⇒b2=4a and 2b=aeConsider, 2b=ae⇒4b2=a2e2⇒4a2(e2−1)=a2e2⇒4e2−4=e2[∵a=0]⇒3e2=4⇒e=3​2​[∵e>0]

Question 14

The distance of the point (1,−5,9) from the plane x−y+z=5 measured along the line x=y=z is
(a) 310​
(b) 103​
(c) 3​10​
(d) 320​

Accepted Answer

Exp. (b)
Equation of line passing through (1,−5,9) and parallel to x=y=z is
1x−1​=1y+5​=1z−9​=λ
Thus, any point on this line is of the form (λ+1,λ−5,λ+9).
Now, if P(λ+1,λ−5,λ+9) is the point of intersection of line and plane, then
(λ+1)−(λ−5)+λ+9=5
⇒λ+15=5
⇒λ=−10
∴ Coordinates of point P are (−9,−15,−1).
Hence, the required distance
=(1+9)2+(−5+15)2+(9+1)2​
=102+102+102​=103​

Question 15

If the line, 2x−3​=−1y+2​=3z+4​ lies in the plane, lx+my−z=9, then l2+m2 is equal to
(a) 26
(b) 18
(c) 5
(d) 2

Accepted Answer

Exp. (d)
Since, the line 2x−3​=−1y+2​=3z+4​ lies in the plane lx+my−z=9, 
therefore we have
2l−m−3=0, [ ∵ normal will be perpendicular to the line]
⇒2l−m=3...(i) and 3l−2m+4​=9​,
[∵ point (3,−2,−4) lies on the plane] 
⇒3l−2m=5......(ii)
On solving Eqs. (i) and (ii), we get
l=1 and m=−1
∴l2+m2=2

Question 16

Let a,b and c be three unit vectors such that a×(b×c)=23​​(b+c). If b is not parallel to c, then the angle between a and b is

Accepted Answer

Given, ∣a^∣=∣b^∣=∣c^∣=1and a×(b^×c^)=23​​(b^+c^)Now, consider a^×(b^×c^)=23​​(b^+c^)⇒(a^⋅c^)b^−(a^⋅b^)c^=23​​b^+23​​c^On comparing, we geta^⋅b^=−23​​⇒∣a^∣∣b^∣cosθ=−23​​⇒cosθ=−23​​[∵∣a^∣=∣b^∣=1]⇒cosθ=cos(π−6π​)⇒θ=65π​

Question 17

If the standard deviation of the numbers 2 , 3,a and 11 is 3.5, then which of the following is true?
(a) 3a2−26a+55=0
(b) 3a2−32a+84=0
(c) 3a2−34a+91=0
(d) 3a2−23a+44=0

Accepted Answer

Exp. (b)
We know that, if x1​,x2​,…,xn​ are n observations, then their standard deviation is given by
n1​Σxi2​−(nΣxi​​)2​
We have, (3.5)2=4(22+32+a2+112)​−(42+3+a+11​)2
⇒449​=44+9+a2+121​−(416+a​)2
⇒449​=4134+a2​−16256+a2+32a​
⇒449​=164a2+536−256−a2−32a​
⇒49×4=3a2−32a+280
⇒3a2−32a+84=0

Question 18

Let two fair six-faced dice A and B be thrown simultaneously. If E1​ is the event that die A shows up four, E2​ is the event that die B shows up two and E3​ is the event that the sum of numbers on both dice is odd, then which of the following statements is not true?
(a) E1​ and E2​ are independent
(b) E2​ and E3​ are independent
(c) E1​ and E3​ are independent
(d) E1​,E2​ and E3​ are not independent

Accepted Answer

Exp⋅(d)
Clearly, E1​={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}
E2​={(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)}
and E3​={(1,2),(1,4),(1,6),(2,1),(2,3),
(2,5),(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,2),(5,1),(5,6),(6,1),(6,3),(6,5)}

⇒P(E1​)=366​=61​,P(E2​)=366​=61​
and P(E3​)=3618​=21​

Now, P(E1​∩E2​)=P(Probability of getting 4 on die A and 2 on die B )
=361​=P(E1​)⋅P(E2​)

P(E2​∩E3​)=P (Probability of getting 2 on die B and sum of numbers on both dice is odd)
=363​=P(E2​)⋅P(E3​)

P(E1​∩E3​)=P (getting 4 on die A and sum of numbers on both dice is odd)
=363​=P(E1​)⋅P(E3​)

and P(E1​∩E2​∩E3​)=P (getting 4 on die A,2 on die B and sum of numbers is odd)
=P (impossible event )=0
Hence, E1​,E2​ and E3​ are not independent.

Question 19

If 0≤x<2π, then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is
(a) 3
(b) 5
(c) 7
(d) 9

Accepted Answer

Exp. (c)
Given equation is
cosx+cos2x+cos3x+cos4x=0
⇒(cosx+cos3x)+(cos2x+cos4x)=0
⇒2cos2xcosx+2cos3xcosx=0
⇒2cosx(cos2x+cos3x)=0
⇒2cosx(2cos25x​cos2x​)=0
⇒cosx⋅cos25x​⋅cos2x​=0
⇒cosx=0 or cos25x​=0 or cos2x​=0
Now, cosx=0⇒x=2π​,23π​[∵0≤x<2π]
cos25x​=0⇒25x​=2π​,23π​,25π​,27π​29π​,211π​
⇒x=5π​,53π​,π,57π​,59π​[∵0≤x<2π]
and cos2x​=0⇒2x​=2π​,23π​,25π​,…
⇒x=π[∵0≤x<2π]
Hence, x=2π​,23π​,π,5π​,53π​,57π​,59π​

Question 20

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

Accepted Answer

Exp. (c)
Consider, (p∧∼q)∨q∨(∼p∧q)
≡[(p∧∼q)∨q]∨(∼p∧q)
≡[(p∨q)∧(∼q∨q)]∨(∼p∧q)
≡[(p∨q)∧t]∨(∼p∧q)
≡(p∨q)∨(∼p∧q)
≡(p∨q∨∼p)∧(p∨q∨q)
≡(q∨t)∧(p∨q)≡t∧(p∨q)≡p∨q

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Solutions For Exams JEE Mains, 2016

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