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JEE Main 2020 Question Papers
September 3 Shift 1 – Maths
September 3 Shift 1 – Maths
1. The value of (2.1P0-3.2P1+4.3P2-….. up to 51th term) +(1!-2!+3!-…… up to 51th term) is equal to:
a) 1-51(51)!
b) 1+(52)!
c) 1
d) 1+ (51)!
Answer: b
2. Let P be a point on the parabola, y2 = 12x and N be the foot of the perpendicular drawn from P on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 4/3 , then:
a) PN = 4
b) MQ = 1/3
c) PN = 3
d) MQ = 1/4
Q (h, 3t) lie on parabola 9t2 = 12h
h = 3t2/4
Q = (3t2/4, 3t)
Equation of NQ =
y = 3t(x-3t2)/((3t2/4)-3t2)
y = -4(x-3t2)/3t
put x = 0
y = 4t
4t = 4/3
⇒ t = 1/3
PN = 6t
= 6×(1/3)
= 2
M = (1/3, 1), Q = (1/12,1)
MQ = (1/3)-(1/12)
MQ = 1/4
Answer: d
3. If Δ = \(\begin{array}{l}\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ 2x-3&3x-4 & 4x-5\\ 3x-5 & 5x-8 &10x-17 \end{vmatrix}\end{array} \)
= Ax3+Bx2+Cx+D, then B+C is equal to:
a) 1
b) -1
c) -3
d) 9
\(\begin{array}{l}\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ 2x-3&3x-4 & 4x-5\\ 3x-5 & 5x-8 &10x-17 \end{vmatrix}\end{array} \)
= Ax3+Bx2+Cx+D
R2 ? R2 -R1
R3 ? R3 -R2
⇒
\(\begin{array}{l}\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ x-1&x-1 & x-1\\ x-2 & 2(x-2) &6(x-2) \end{vmatrix}\end{array} \)
= Ax3+Bx2+Cx+D
⇒
\(\begin{array}{l}(x-1)(x-2)\begin{vmatrix} x-2 & 2x-3 & 3x-4\\ 1&1 & 1\\ 1 & 2 &6 \end{vmatrix}\end{array} \)
= Ax3+Bx2+Cx+D
⇒ (x-1)(x-2){(x-2)(6- 2)-(2x-3)(6-1)+(3x- 4)(2-1)} = Ax3+Bx2+Cx+D
⇒ (x-1)(x -2){4x-8 -10x+15+3x-4} = Ax3+Bx2+Cx+D
⇒ (x-1)(x-2)(-3x +3) = Ax3+Bx2+Cx+D
-3(x-1)2 (x-2) = Ax3+Bx2+Cx+D
-3x3 +12x2-15x+6 = Ax3+Bx2+Cx+D
Comparing both side
A = -3, B = 12, C = -15
B+C = 12-15 = -3
Answer: c
4. The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points (1,2,3) and (1,1,0) lies on the plane:
a) x-y-2z = 1
b) x-2y+z = 1
c) 2x+y-z = 1
d) x+2y-z = 1
\(\begin{array}{l}\vec{r} = (1,-2,3)+\lambda (0,3,-3)\end{array} \)
\(\begin{array}{l}\overrightarrow{PM} ⊥ \overrightarrow{b}\end{array} \)
\(\begin{array}{l}\overrightarrow{PM} . \overrightarrow{b}= 0\end{array} \)
(-3, 3?-4, -3?).(0,3, -3) = 0
⇒ 0+9?-12+9? = 0
⇒ ?= 12/18 = 2/3
m = (1,0,1) are on 2x+y-z = 1
Answer: c
5. If y2+loge(cos2x) = y, x?(-π/2, π/2), then
a) |y’(0)|+|y’’(0)| = 1
b) |y’’(0)| = 0
c) |y’(0)|+|y’’(0)| = 3
d) |y’’(0)| = 2
yy’ + 2 (-tanx) = y’ ….(1)
diff. w.r.t.x
2y’’+2(y’)2 -2 sec2 x = y” ….(2)
Put x = 0 in given equation we get y = 0, 1
from (1) x = 0, y = 0
⇒ y’(0) = 0
x = 0, y = 1
⇒ y’(0) = 0
from (2) x = 0, y = 0, y’(0) = 0 ⇒ y”(0) = -2
x = 0, y = 1, y’(0) = 0 ⇒ y”(0) = 2
|y”(0)| = 2
Answer: d
6. 2π-(sin-1(4/5)+sin-1(5/13)+sin-1(16/65)) is equal to:
a) 5π/4
b) 3π/2
c) 7π/4
d) π/2
2π-(sin-1(4/5)+sin-1(5/13)+sin-1(16/65)) = 2π-(tan-1(4/3)+tan-1(5/12)+tan-1(16/63))
= 2π-tan-1 [(4/3)+(5/12)/(1-(4/3)(5/12) ]- tan-1(16/63)
= 2π-tan-1(48+15)/(36-20) – tan-1(16/63)
= 2π-tan-1(63/16) – cot-1(63/16)
= 2π-π/2
= 3π/2
Answer: b
7. A hyperbola having the transverse axis of length v2 has the same foci as that of the ellipse 3x2+4y2 = 12, then this hyperbola does not pass through which of the following points ?
a) (v(3/2), 1/v2)
b) (1, -1/v2)
c) (1/v2, 0)
d) (-v(3/2),1)
(x2/4)+(y2/3) = 1
b12 = a12(1-e12)
3 = 4(1-e12)
e1 = 1/2
focus = (± a1e1, 0)
= (±1, 0)
Length of the transverse axis 2a2 = v2
a2 = 1/v2
a2e2 = 1
So e2 = v2
b22 = a22(e22-1)
b22 = (1/2)(2-1)
b22 = 1/2
Equation of hyperbola is x2-y2 = 1/2.
(v(3/2), 1/v2) does not lie on it.
Answer: a
8. For the frequency distribution
Variate (X): x1 x2 x3…x15
Frequency (f): f1 f2 f3…f15
where 0<x1<x2< x3<… x15 = 10 and
\(\begin{array}{l}\sum_{i=1}^{15}f_{i}> 0\end{array} \)
, the standard deviation cannot be:
a) 1
b) 4
c) 6
d) 2
s2 = (1/4)(M-m)2
M = upper bound of value of any random variable
m = Lower bound of value of any random variable
s2 = (1/4)(10-0)2
s2 = 25
-5< s < 5
s ≠ 6
Answer: c
9. A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is:
a) 1/3
b) 1/4
c) 1/8
d) 1/9
Total Possibilities = (1, 3), (3, 1), (2, 2), (2, 6), (6, 2) (4, 4), (3, 5), (5, 3) (6, 6)
(A n B) = (4,4)
n(A n B) = 1
Required probability = P(B/A)
= P(A n B)/P(A)
= 1/9
Answer: d
10. If the number of integral terms in the expansion of (31/2 +51/8)n is exactly 33, then the least value of n is:
a) 128
b) 248
c) 256
d) 264
Tr+1 = nCr(31/2)n-r(51/8)r
(n-r)/2 -> n-r = 0, 2, 4, 6, 8,…
r/8 -> r = 0, 8, 16, 24….
common ratio = 0, 8, 16, 24…….
no. of integral term = 33.
L = 0+(33-1)×8
L = 32×8
= 256
Answer: c
11. \(\begin{array}{l}\int_{-\pi }^{\pi }\left | \pi -\left | x \right | \right |dx\end{array} \)
is equal to:
a) π2
b) π2/2
c) v2π2
d) 2π2
\(\begin{array}{l}\int_{-\pi }^{\pi }\left | \pi -\left | x \right | \right |dx\end{array} \)
even function
\(\begin{array}{l}2\int_{0 }^{\pi }\left | \pi -x \right |dx\end{array} \)
= 2[πx-x2/2]0π
= 2[π2/2]
= π2
Answer: a
12. Consider the two sets:
A = {m ∈ R : both the roots of x2-(m+1)x+m+4 = 0 are real} and B = [-3,5).
Which of the following is not true ?
a) A-B = (-∞,-3) ? (5,∞)
b) A n B = {-3}
c) B-A = (-3,5)
d) A U B = R
D = 0
(m+1)2 -4(m+4) = 0
m2-2m-15 = 0
(m-5)(m+3) = 0
m ∈(-∞, -3] U [5, ∞)
A = (-∞, -3] U [5, ∞)
B = [-3, 5]
A-B = (-∞, -3] U [5, ∞)
A U B = R
Only first option is not true.
Answer: a
13. The proposition p -> ∼(p^q) is equivalent to:
a) (∼p) ?(∼ q)
b) (∼ p) ^q
c) q
d) (∼ p) ?q
∼(p^ ∼q) ? ∼ p?q
p ? (p~?q)
⇒ ~ p?(~p?q)
⇒ ~ p?q
Answer: d
14. The function, f(x) = (3x-7)x2/3, x ∈ R is increasing for all x lying in:
a) (-∞, -14/15) U (0,∞)
b) (-∞, 14/15)
c) (-∞,0)U(14/15, ∞)
d) (-∞,0)U(3/7, ∞)
f’(x) = (3x-7)(2/3x1/3)+(x2/3×3)
= (6x-14+9x)/3x1/3
= (15x-14)/3x1/3
f(x) >0 ⇒ x ∈ (-∞,0)U(14/15, ∞)
Answer: c
15. If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is:
a) 1/6
b) 1/5
c) 1/4
d) 1/7
a = 3
(25/2)(2a+24d) = (15/2)(2(a+25d)+14d)
50a+600d = 15[2a+50d+14d]
20a+600d = 960d
60 = 360d
d = 1/6
Answer: a
16. The solution curve of the differential equation, (1+e-x)(1+y2)dy/dx = y2, which passes through the point (0,1), is:
a) y2 = 1+y loge(1+e-x)/2
b) y2 +1 = y ((loge(1+e-x)/2 )+2)
c) y2 +1 = y ((loge(1+ex)/2 )+2)
d) y2 = 1+y loge(1+ex)/2
\(\begin{array}{l}\int \left ( \frac{1+y^{2}}{y^{2}} \right )dy=\int \left ( \frac{1}{1+e^{-x}} \right )dx\end{array} \)
\(\begin{array}{l}\int \frac{1}{y^{2}}dy+\int dy=\int \frac{e^{x}}{e^{x}+1}dx\end{array} \)
⇒ (-1/y)+y = lnex+1 + C
x = 0, y = 1
⇒ -1+1 = ln 2+C
⇒ C = -ln 2
⇒ (-1/y)+y = lnex+1 – ln 2
⇒ y2 = 1+y[ln(ex+1)/2]
Answer: d
17. The area (in sq. units) of the region {(x,y):0 =y =x2+1, 0 =y =x+1, 1/2 =x=2} is
a) 23/16
b) 79/16
c) 23/6
d) 79/24
A=
\(\begin{array}{l}\int_{\frac{1}{2}}^{1}(x^{2}+1)dx+\int_{1}^{2}(x+1)dx\end{array} \)
=
\(\begin{array}{l}\left ( \frac{x^{3}}{3}+x\right )_{\frac{1}{2}}^{1}+\left ( \frac{x^{2}}{2}+x \right )_{1}^{2}\end{array} \)
= (1/3)+1-[(1/24)+(1/2)]+[2+2-(3/2)]
= (4/3)-(13/24)+(5/2)
= (19+60)/24
= 79/24
Answer: d
18. If α and β are the roots of the equation x2+px+2 = 0 and (1/α) and (1/β) are the roots of the equation 2x2+2qx+1 = 0, then (α-1/α)(β-1/β)(α+1/α)(β+1/β) is equal to :
a) (9/4)(9+p2)
b) (9/4)(9+q2)
c) (9/4)(9-p2)
d) (9/4)(9-q2)
α+β = -p
αβ = 2
(1/α)+(1/β) = -q
1/αβ = 1/2
(α+1/β)(β+1/α) = αβ+(1/αβ)+2
= 2+(1/2)+2
= 9/2
(α-1/α)(β-1/β) = αβ+(1/αβ)-(α/β)-(β/α)
= 2+(1/2)-(α2+β2)/αβ
= (5/2)-[((α+β)2-2αβ)/αβ)]
= (5/2)-[(p2-4)/2]
= (9-p2)/2
(α-1/α)(β-1/β)(α+1/α)(β+1/β) = ((9-p2)/2)(9/2)
= (9/4)(9-p2)
Answer: c
19. The lines \(\begin{array}{l}\vec{r}= (\hat{i}-\hat{j})+l(2\hat{i}+\hat{k})\end{array} \)
and \(\begin{array}{l}\vec{r}= (2\hat{i}-\hat{j})+m(\hat{i}+\hat{j}-\hat{k})\end{array} \)
a) do not intersect for any values of l and m
b) intersect when l = 1 and m = 2
c) intersect when l = 2 and m = 1/2
d) intersect for all values of l and m
\(\begin{array}{l}\vec{r}= (\hat{i}-\hat{j})+l(2\hat{i}+\hat{k})\end{array} \)
..(i)
\(\begin{array}{l}\vec{r}= (2\hat{i}-\hat{j})+m(\hat{i}+\hat{j}-\hat{k})\end{array} \)
…(ii)
Let two lines
\(\begin{array}{l}\vec{r}= \vec{a}+t\vec{b}\end{array} \)
and
\(\begin{array}{l}\vec{r}= \vec{c}+t\vec{d}\end{array} \)
= 1/v(-1)2+(-1)2+22
= 1/v6
Since shortest distance between these lines exist
Hence lines do not intersect.
Answer: a
20. Let [t] denote the greatest integer =t. If for some ? ∈ R -{0, 1}, \(\begin{array}{l}\lim_{x\to0}\left | \frac{1-x+\left | x \right |}{\lambda -x+[x]} \right |= L\end{array} \)
Then L is equal to:
a) 0
b) 2
c) 1/2
d) 1
\(\begin{array}{l}\lim_{x\to0}\left | \frac{1-x+\left | x \right |}{\lambda -x+[x]} \right |= L\end{array} \)
RHL:
\(\begin{array}{l}\lim_{x\to0^{+}}\left | \frac{1-x+x}{\lambda -x+[x]} \right |\end{array} \)
= 1/(-1)
Since |?| = |?-1|, [-h] = -1
?2 = ?2-2?+1
⇒ ?= 1/2
L = 2
Answer: b
21. If \(\begin{array}{l}\lim_{x\to0}\left \{ \frac{1}{x^{8}}\left ( 1-\cos \frac{x^{2}}{2} -\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2}\cos \frac{x^{2}}{4}\right ) \right \}= 2^{-k}\end{array} \)
then the value of k is:
\(\begin{array}{l}\lim_{x\to0}\frac{(1-\cos \frac{x^{2}}{2})(1-\cos \frac{x^{2}}{4})(\frac{x^{2}}{2})^{2}(\frac{x^{2}}{4})^{2}}{(\frac{x^{2}}{2})^{2}(\frac{x^{2}}{4})^{2}x^{8}}\end{array} \)
⇒
\(\begin{array}{l}\lim_{x\to0}\frac{1}{4}\times \frac{1}{4}\times \frac{1}{16}\end{array} \)
⇒ 1/256 = 2-k
⇒ 2-8 = 2-k
⇒ k = 8
Answer: 8
22. The diameter of the circle, whose centre lies on the line x+y = 2 in the first quadrant and which touches both the lines x = 3 and y = 2, is ……
p = r
for y = 2
r = |(2-α-2)/1| = |α|
for x = 3
r = |(α-3)/1| = |α-3|
| α | = |α – 3 |
⇒ α2 = α2-6α+9
⇒ α = 3/2
2α = 3 = 2r
Answer: 3
23. The value of \(\begin{array}{l}0.16^{log_{2.5}\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+…\infty }\end{array} \)
is equal to:
(1/3)+(1/32)+(1/33)+….∞ = 1/3(1-1/3) = 1/2
log2.5(1/2) ⇒ log5/2 (1/2)
0.16 = 16/100 = 4/25 = (2/5)2
⇒
\(\begin{array}{l}\left (\frac{2}{5} \right )^{2log_{\frac{5}{2}}\frac{1}{2}} = \left (\frac{5}{2} \right )^{-2log_{\frac{5}{2}}\frac{1}{2}}\end{array} \)
⇒
\(\begin{array}{l}\left (\frac{5}{2} \right )^{log_{\frac{5}{2}}(\frac{1}{2})^{-2}}\end{array} \)
= 4
Answer: 4
24. Let A = \(\begin{array}{l}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\end{array} \)
, x?R and A4 = [aij]. If a11 = 109, then a22 is equal to:
A =
\(\begin{array}{l}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\end{array} \)
A2 =
\(\begin{array}{l}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\end{array} \)
= \(\begin{array}{l}\begin{bmatrix} x^{2} +1&x\\ x & 1 \end{bmatrix}\end{array} \)
A3 =
\(\begin{array}{l}\begin{bmatrix} x^{2} +1&x\\ x & 1 \end{bmatrix}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\end{array} \)
=
\(\begin{array}{l}\begin{bmatrix} x^{3} +x+x&x^{2}+1\\ x ^{2}+1& x \end{bmatrix}\end{array} \)
A4 =
\(\begin{array}{l}\begin{bmatrix} x^{3} +2x&x^{2}+1\\ x ^{2}+1& x \end{bmatrix}\begin{bmatrix} x &1 \\ 1 & 0 \end{bmatrix}\end{array} \)
=
\(\begin{array}{l}\begin{bmatrix} x^{4} +2x^{2}+x^{2}+1&x^{3}+2x\\ x ^{3}+x+x& x^{2} +1\end{bmatrix}\end{array} \)
a11 ⇒ x4+3x2+1 = 109
x4 +3x2-108 = 0
⇒ (x2+12)(x2-9) = 0
x = ±3
a22 = x2+1 = 10
Answer: 10
25. If ((1+i)/(1-i))m/2 = ((1+i)/(i-1))n/3 = 1, (m, n∈N) then the greatest common divisor of the least values of m and n is :
[(1+i)(1+i)/(1+i)(1-i)]m/2 = [(1+i)(-1-i)/(-1+i)(-1-i)]n/3 = 1 (i = v-1)
Solving LHS of above equation, we get
(2i/2)m/2 = 1
⇒ m = 8
Solving RHS,
[(-1-i-i+1)/(1+1)]n/3 = 1
(-2i/2)n/3 = 1
(-i)n/3 = 1
⇒ n = 12
Greatest common divisor of m and n is 4.
Answer: 4
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