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September 3 Shift 2 - Maths
1. If x3dy+xy dx = x2dy+2y dx; y(2) = e and x>1, then y(4) is equal to:
- a) √e/2
- b) (3/2)√e
- c) (1/2)+√e
- d) (3/2)+√e
Solution:
-
(x3-x2)dy = (2-x) ydx
\(\int \frac{dy}{y}=\int \frac{2-x}{x^{2}(x-1)}dx\)\(\int \frac{dy}{y}=-\int \frac{x-2}{x^{2}(x-1)}dx\)\(\int \frac{dy}{y}=-\int (\frac{p}{x}+\frac{q}{x^{2}}+\frac{r}{x-1})dx\)p = 1, q = 2, r = -1
\(\int \frac{dy}{y}=\int \frac{-1}{x}dx-\int\frac{2}{x^{2}}dx-\int\frac{-1}{x-1}dx\)ln|y| = -ln x+(2/x)+ln (x-1)+c
x = 2, y = e
1 = 1-ln 2+c
⇒ c = ln 2
ln|y| = (2/x)-ln |x|+ln |x-1|ln 2
put x = 4
ln|y| = (1/2)-2ln 2+ ln 3+ ln 2
ln y = ln(3/2)+(1/2)
y = (3/2)×e1/2
= (3/2)√e
Answer: b
2. Let A be a 3×3 matrix such that adj A =
- a) (9, 1/81)
- b) (9, 1/9)
- c) (3, 1/81)
- d) (3, 81)
Solution:
-
adj A =
\(\begin{bmatrix} 2 & -1 &1 \\ -1 &0 &2 \\ 1 & -2 & -1 \end{bmatrix}\)⇒ |adj A| = 9
⇒ |A2| = 9
⇒ |A| = 3 = |λ| and |A| = -3 = |λ|
det (adj (adj A)) = ((|A|)(n-1))2
|B| = ((|A|)(3-1))2 = |A|4 = 34 = 81
|(BT)-1| = 1/|(BT)|
= 1/|B|= 1/81=μ
Hence |λ|,μ = (3, 1/81)
Answer: c
3. Let a, b, cεR be such that a2+b2+c2 = 1, if a cos θ = b cos(θ+2π/3) = c cos(θ+4π/3), where θ = π/9, then the angle between the vectors
- a) π/2
- b) 2π/3
- c) π/9
- d) 0
Solution:
-
cos α =
\(\frac{\vec{p}.\vec{q}}{\left | p \right |\left | q \right |}\)= (ab+bc+ca)/a2+b2+c2
= (ab+bc+ca)/1
cos α = ab+bc+ca
cos α = abc[(1/a)+(1/b)+(1/c)] ..(i)
a cos200 = b cos(1400) = c cos(2600) = λ
⇒ (1/a) = cos 200/λ
(1/b) = cos 1400/λ
(1/c) = cos 2600/λ
Put in (i)
⇒ cos α = (abc/λ)(cos 200+ cos 1400+ cos 2600)
⇒ cos α = (abc/λ)(cos 200+ 2cos 2000+ cos 600)
⇒ cos α = (abc/λ)(cos 200- cos 200)
⇒ cos α = 0
⇒ α = π/2
Answer: a
4. Suppose f(x) is a polynomial of degree four, having critical points at (-1,0,1). If T = {xεR f(x) = f(0) }, then the sum of squares of all the elements of T is:
- a) 6
- b) 2
- c) 8
- d) 4
Solution:
-
f’(x) = k(x+1)x(x-1)
f’(x) = k[ x3-x]
Integrating both sides
f(x) = k[(x4/4)- (x2/2)]+c
f(0) = c
f(x) = f(0) ⇒ k[(x4/4)- (x2/2)]+c = c
⇒ k(x2/4)(x2-2) = 0
⇒ x = 0, √2, -√2
Sum of squares of all elements = 02+(√2)2+(-√2)2
= 4
Answer: d
5. If the value of the integral
- a) 2√3+π
- b) 3√2+π
- c) 3√2-π
- d) 2√3-π
Solution:
-
\(\int_{0}^{\frac{1}{2}}\frac{x^{2}}{(1-x^{2})^{\frac{3}{2}}}dx\)
Put x = sinθ
\(\int_{0}^{\frac{\pi }{6}}\frac{\sin ^{2}\theta }{\cos ^{3}\theta }\cos \theta .d\theta\)=
\(\int_{0}^{\frac{\pi }{6}}\tan ^{2} \theta .d\theta = [\tan \theta -\theta]_{0}^{\frac{\pi }{6}}\)⇒ (1/√3)-(π/6) = k/6
(2√3-π)/6 = k/6
k = 2√3-π
Answer: d
6. If the term independent of x in the expansion of ((3/2)x2 -(1/3x))9 is k, then 18 k is equal to:
- a) 5
- b) 9
- c) 7
- d) 11
Solution:
-
Tr+1 =
\(^{^{9}}C_{r}\left ( \frac{3}{2}x^{2} \right )^{9-r}\left ( \frac{-1}{3x} \right )^{r}\)=
\(^{^{9}}C_{r}\frac{3^{9-r}}{2^{9-r}}(\frac{-1}{3})^{r}x^{18-3r}\)18-3r = 0
⇒ r = 6
=
\(^{^{9}}C_{r}(\frac{3^{-3}}{2^{3}})=k\)⇒ 7/18 = k
⇒ 18k = 7
Answer: c
7. If a triangle ABC has vertices A(-1,7), B(-7,1) and C(5,-5), then its orthocentre has coordinates;
- a) (-3,3)
- b) (-3/5, 3/5)
- c) (3/5, -3/5)
- d) (3,-3)
Solution:
-
mAH. mBC = -1
⇒ ((y-7)/(x+1))(1+5)/(-7-5) = -1
⇒ 2x-y+9 = 0 …..(i)
and mBH. mAC = -1
((y-1)/(x+7))(7--5)/(-1-5) = -1
x-2y+9 = 0 .....(ii)
Solving equation (i) and (ii) we get
(x, y) = (-3,3)
Answer: a
8. Let e1 and e2 be the eccentricities of the ellipse, (x2/25)+(y2/b2) = 1 (b<5) and the hyperbola, (x2/16)-(y2/b2) = 1 respectively satisfying e1e2 = 1. If α and β are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α,β) is equal to:
- a) (8,12)
- b) (24/5,10)
- c) (20/3,12)
- d) (8,10)
Solution:
-
α = 10e1
b2 = 25(1-e12)
β = 8e2
b2 = 16(e22-1)
(e1e2)2 = 1
(1-b2/25)( 1+b2/16) = 1
⇒ 1+(b2/16)-(b2/25)-(b4/400) = 1
⇒ (9b2/16×25)-(b4/400) = 0
⇒ 9b2-b4 = 0
⇒ b2(9-b2) = 0, b ≠ 0
Therefore, 9-b2 = 0
b2 = 9
e1 = 4/5
α = 2ae1
= 10×4/5
= 8
e2 = 5/4
β = 2ae2
= 8×5/4
= 10
(α, β) = (8, 10)
Answer: d
9. If z1, z2 are complex numbers such that Re(z1) = |z1-1|, Re(z2) = |z2-1| and arg (z1-z2) = π/6, then Im(z1+z2) is equal to:
- a) 2√3
- b) 2/√3
- c) 1/√3
- d) √3/2
Solution:
-
z1 = x1+iy1
z2 = x2+iy2
x12 = (x1-1)2+y12
⇒ y12-2x1+1 = 0 ..(i)
x22 = (x2-1)2+y22
⇒ y22-2x2+1 = 0 ..(ii)
From equation (ii)-(i)
(y12-y22)+2(x2-x1) = 0
(y1+y2)(y1-y2) = 2(x1-x2)
(y1+y2) = 2(x1-x2)/ (y1-y2)
arg (z1-z2) = π/6
tan-1(y1-y2)/(x1-x2) = π/6
⇒ (y1-y2)/(x1-x2) = 1/√3
So y1+y2 = 2√3
Answer: a
10. The set of all real values of λ for which the quadratic equations, (λ2+1)x2-4λx+2 = 0 always have exactly one root in the interval (0,1) is:
- a) (-3,-1)
- b) (2,4)
- c) (1,3)
- d) (0,2)
Solution:
-
f(0) f(1) ≤ 0
⇒ (2) [λ2-4λ +3] ≤0
(λ-1)(λ- 3) ≤0
⇒ λ ε [1, 3]
at λ = 1
2x2 -4x+2 = 0
⇒ (x-1)2 = 0
x = 1, 1
⇒λ ε (1, 3]
Answer: c
11. Let the latus rectum of the parabola y2 = 4x be the common chord to the circles C1 and C2 each of them having radius 2√5. Then, the distance between the centres of the circles C1 and C2 is:
- a) 8
- b) 8√5
- c) 4√5
- d) 12
Solution:
-
C1C2 = 2C1A
(C1A)2 + 4 = ( 2√5)2
C1A = 4
C1C2 = 8
Answer: a
12. The plane which bisects the line joining the points (4,-2,3) and (2,4,-1) at right angles also passes through the point:
- a) (0,-1,1)
- b) (4,0,1)
- c) (4,0,-1)
- d) (0,1,-1)
Solution:
-
a = 2, b = -6
c = 4
Equation of plane is
2(x-3)+(-6)(y -1)+ 4(z-1) = 0
⇒ 2x-6y+4z = 4 passes through (4, 0, -1).
Answer: c
13.
- a) (2/9)4/3
- b) (2/3)4/3
- c) (2/3)(2/9)1/3
- d) (2/9)(2/3)1/3
Solution:
-
Apply L hospital rule
⇒ [(2/3)(3a)-2/3-1/32/3(a-2/3)]/[(1/3)(4a)-2/3-(1/3)41/3a-2/3]
= (2/3)×(2/9)1/3
Answer: c
14. Let xi (1≤ i ≤10)be ten observations of a random variable X. If
where 0 ≠ p εR , then the standard deviation of these observations is :
- a) 7/10
- b) 9/10
- c) √(3/5)
- d) 4/5
Solution:
-
Standard deviation is free from shifting of origin
S.D = √variance
= √[(9/10)-(3/10)2]
= √[(9/10)-(9/100)]
= √(81/100)
= 9/10
Answer: b
15. The probability that a randomly chosen 5 digit number is made from exactly two digits
is :
- a) 134/104
- b) 121/104
- c) 135/104
- d) 50/104
Solution:
-
Total case = 9(104)
First case : Choose two non-zero digits = 9C2
Now, number of 5-digit numbers containing both digits = 25-2
Second case : Choose one non-zero and one zero as digit = 9C1
Number of 5-digit numbers containing one non-zero and one zero both = 24-1.
fav. case = 9C2(25-2)+ 9C1(24-1)
= 1080+135
= 1215
Probability = 1215/9×104
= 135/104
Answer: c
16. If
- a) (x+1, -√x)
- b) (x-1, -√x)
- c) (x+1, √x)
- d) (x-1, √x)
Solution:
-
Given
\(\int \sin ^{-1}\left ( \sqrt{\frac{x}{1+x}} \right )dx= A(x)\tan ^{-1}(\sqrt{x})+B(x)+C\)∫tan-1√x. 1 dx
Here tan-1√x is the first function and 1 dx is the second function.
Integrating we get
(tan-1√x)x-∫(x/1+x)×(1/2√x)dx
Put x = t2
dx = 2t dt
(tan-1√x)x-∫(x/1+x)×(1/2√x)dx = x(tan-1√x)- ∫(t2)(2tdt)/(1+t2)2t
= x(tan-1√x)-t+tan-1t+c
= x(tan-1√x)-√x+tan-1√x+c
= tan-1√x(x+1)-√x+c
A(x) = x+1
B(x) = -√x
Answer: a
17. If the sum of the series
- a) n = 60
- b) n = 41
- c) nth term is -4
- d) nth term is \(-4\frac{2}{5}\)
Solution:
-
\(20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+...\)
Sn = 488
⇒ (n/2)(2×20+(n-1)(-2/5)) = 488
⇒ 20n-(n2/5)+(n/5) = 488
⇒ 100n-n2+n = 2440
⇒ n2-101n+2440 = 0
⇒ n = 61 or 40
For n = 40, Tn = 20+39(-2/5) = +ve
n = 61, Tn = 20+60(-2/5) =20-24 = -4
Answer: c
18. Let p, q, r be three statements such that the truth value of (p˄q) -> (∼q˅r) is F. Then the truth values of p, q, r are respectively:
- a) F, T, F
- b) T, F, T
- c) T, T, F
- d) T, T, T
Solution:
-
(p˄q) -> (∼q˅r) possible when p˄q -> T
∼q˅r -> F
p -> T ; p˄q⇒ T
q -> T ; ∼q˅r -> F˅F ⇒ F
r -> F ; T F ⇒ F
Answer: c
19. If the surface area of a cube is increasing at a rate of 3.6 cm2/sec, retaining its shape; then the rate of change of its volume (in cm3/sec), when the length of a side of the cube is 10cm, is :
- a) 9
- b) 10
- c) 18
- d) 20
Solution:
-
A = 6a2
A is the side of cube.
dA/dt = 6×2a da/dt
⇒ 3.6 = 12×10 (da/dt)
⇒ (da/dt) = 3/100
V = a3
dV/dt = 3a2da/dt
= 3×100×3/100
= 9 cm3/sec
Answer: a
20. Let R1 and R2 be two relations defined as follows:
R1 = {(a,b) ∈ R2: a2+b2 ∈ Q} and
R2 = {(a,b) ∈ R2: a2+b2 ∉Q}, where Q is the set of all rational numbers. Then :
- a) R1 is transitive but R2 is not transitive
- b) R1 and R2 are both transitive
- c) R2 is transitive but R1 is not transitive
- d) Neither R1 nor R2 is transitive
Solution:
-
For R1
Let a = 1+√2, b = 1-√2, c = 81/4
aR1b ⇒ a2+b2 = (1+√2)2+(1-√2)2 = 6 ∈ Q
bR1c ⇒ b2+c2 = (1-√2)2+(81/4)2 = 3 ∈ Q
aR1c ⇒ a2+c2 = (1+√2)2+(81/4)2 = 3+4√2 ∉Q
R1 is not transitive
For R2
Let a = 1+√2, b = √2, c = 1-√2
aR2b ⇒ a2+b2 = 5+2√2 ∉Q
bR2c ⇒ b2+c2 = 5-2√2 ∉Q
aR2c ⇒ a2+c2 = 6 ∈ Q
R2 is not transitive
Answer: d
21. If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to
Solution:
-
3, …., 243 (m A.M)
d = (b-a)/(n+1)
= (243-3)/(m+1)
= 240/(m+1)
4th AM = 3+4d
= 3+4(240/m+1)
3, …, 243 (3G.M)
243 = 3r4
r = 3
2nd G.M = ar2 = 27
Given that 4th A.M. is equal to 2nd G.M.
3+(960/m+1) = 27
= 960/m+1 = 24
⇒ m = 39
Answer: 39
22. Let a plane P contain two lines
Solution:
-
\(\vec{r} = \hat{i}+\lambda (\hat{i}+\hat{j})\)\(\vec{r} = -\hat{j}+\mu (\hat{j}-\hat{k})\)\(\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1 & 1 & 0\\ 0& 1 & -1 \end{vmatrix}\)
= (-1, 1 ,1)
Equation of plane
-1(x -1)+1(y-0) +1(z-0) = 0
⇒x-y-z-1 = 0
Foot of the perpendicular from m (1,0,1)
(x-1)/1 = (y-0)/-1 = (z-1)/-1 = -(1-0-1-1)/3
(x-1) = 1/3
y/-1 = 1/3
(z-1)/-1 = 1/3
x = 4/3
y = -1/3
z = 2/3
⇒α = 4/3
β = -1/3
γ= 2/3
α+β+ γ= (4/3)-(1/3)+(2/3) = 5/3
3(α+β+γ) = 5
Answer: 5
23. Let S be the set of all integer solutions, (x, y, z), of the system of equations
x-2y+5z = 0
-2x+4y+z = 0
-7x+14y+9z = 0
such that 15≤x2+y2+z2≤150. Then, the number of elements in the set S is equal to:
Solution:
-
x-2y+5z = 0.....(i)
-2x+4y+z = 0....(ii)
-7x+14y+9z = 0.....(iii)
2.(i) + (ii) we get z = 0, x = 2y
15 ≤4y2+y2 ≤150
⇒ 3≤y2≤30
y ∈ [-√30, -√3] U [√3, √30]
y = 2, 3, 4, 5, -2, -3, -4, -5
Number of elements in S is 8.
Answer: 8
24. The total number of 3 digit numbers, whose sum of digits is 10, is:
Solution:
-
Let xyz be 3 digit number
x+y+z = 10 where x ≥1, y ≥0, z ≥0
x-1≥0
Put x-1 = t
t≥0
⇒ t+ y+z = 9
9+3-1C3-1 = 11C2 = 55
But for t = 9, x = 10 not possible
Total numbers = 55 - 1 = 54
Answer: 54
25. If the tangent to the curve, y = ex at a point (c, ec) and the normal to the parabola, y2 = 4x at the point (1,2) intersect at the same point on the x-axis, then the value of c is:
Solution:
-
Tangent at (c, ec)
y-ec = ec(x-c) ....(i)
normal to parabola y-2 = -1(x-1)
x+y = 3 ...(ii)
At x-axis y = 0
in (i), x = c-1
in (ii), x = 3
c-1 = 3
⇒ c = 4
Answer: 4
Video Lessons - September 3 Shift 2 Maths
JEE Main 2020 Maths Paper With Solutions Shift 2 September 3
