JEE Main Question Paper solutions for August 26 Shift 1 Maths are available here. Candidates can find the JEE Main 2021 August 26 Shift 1 Maths Question Paper with Solutions to evaluate their performance. Students can directly download the PDF version of the question paper to check solutions offline. This set of solved questions given in JEE Main 2021 session 4 Maths question paper for August 26 will help you to formulate the correct answers. The JEE Main 2021 Shift 1 Maths question paper for August 26 given here will aid students to eventually be well prepared to face the upcoming exam.
Question 1. f(x) =
Solution:
Answer: (b)
f(x) =
f(x) =
= cos((2 tan-1(√x)
= cos((cos-1 (((1 - x)/(1 + x)) ) = (1 - x)/(1 + x)
f'(x) = -2/(1+x)2
2(f(x))2 = 2((1-x)/(1+x))2
(1 - x) 2 f'(x) + 2(f(x)) 2= 0
Question 2. How many 3 digit numbers by using the digits 0,1,3,4,6,7 can be possible if repetition are allowed?
Solution:
Answer: 180
digits 0,1,3,4,6
So, 5 x 6x 6 = 180
Question 3. The sum of the square of the distance of a point from (0, 0), (0, 1), (1, 1), (1, 0) is 18 and its locus is a circle. If d is the diameter of the circle, then find d2.
Solution:
Answer: 16
(OP) 2 + (AP) 2 + (PC) 2 + (PB) 2=18
=> x2 + y2 + x2 + (y - 1) 2 + (x - 1) 2 + y2 + (x - 1) 2 + (y - 1) 2 =18
=> 4x2 + 4y2 - 4x - 4y=14
=> x2 + y2 - x - y=7/2
=> r=2
=> d = 4 => d2 = 16
Question 4: If loge (x+y) = 4xy. Find (d2y)/(dx2) at x = 0.
Solution:
Answer: 40
loge (x+y) = 4xy
or 1/((x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]
or 1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)
If x = 0, then y = 1
From (i)
1 + dy/dx = 4(0 + 1)[0 + 1] = 4
Or dy/dx = 3
From (i), again differentiate w.r.t. x
0 + (d2 y)/(dx2 ) = 4(x + y)[x (d2y)/(dx2 ) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))
At x = 0, y = 1, dy/dx = 3
(d2 y)/(dx2) = 4(0 + 1)[0 + 2x3]+4[0 + 1](1 + 3) = 40
Therefore, (d2 y)/(dx2 ) = 40
Question 5: Find value of (5 - e2 ) for ellipse x2/8 + y2/4 = 1
Solution:
Answer: 9/2
x2/8 + y2/4 = 1
a2 = 8, b2 = 4
e = c/a = √((a2 - b2)/a2 ) = √(1-4/8) = √(1/2)
(5 - e2 ) = 5 - 1/2 = 9/2
Question 6: In a hospital, 89% has disease A and 98% has disease B, K% is the number of patients who has both, then K can’t be the subset of ____.
Solution:
Answer: Least k =87, Max k = 89
Let total =100
n(A)=89, n(B)=98
max (n(A),n(B)) ≤ n(AUB) ≤ 100
98 ≤ n(A)+n(B)-n(A n B) ≤ 100
=> 98 ≤ 98 + 89- k ≤ 100
Least k =87
Max k = 89
Question 7: P(A) = p, P(B) = 2p, P(Exactly one out of A and B occurs) = 5/9. What will be the maximum value of p?
Solution:
Answer: 5/9
Given: P(A) = p and P(B) = 2p
P(A)-P(A∩B)+P(B)-P(A∩B) = 5/9
3p-2P(A∩B) = 5/9
0 ≤ P(A∩B) ≤ p
For maximum value: 3p - 2p = 5/9
So, p = 5/9
Question 8: Evaluate
Solution:
Answer: 4 ln ½
Question 9: If cosx/(1+sinx) = |tanx|, then find the number of solutions in [0, 2π].
Solution:
Answer: x = π/6
Case 1: tan x ≥ 0
cosx/(1+sinx ) = sinx/cosx
⇒cos2x = sin2x+sinx
⇒1-sin2x = sin2x + sinx
⇒2 sin2x + sin x – 1 = 0
⇒sin x = -1 ,1/2
sin x = 1/2 ⇒ x = π/6
sin x = -1 (not possible)
Case 2: tan x < 0
(Cos x)/(1+sinx) = -sin x/cos x
⇒cos2x = -sin2x - sinx
⇒1-sin2x = -sin2x - sinx
Sin x = -1 (Not possible)
Only one solution exists x = π/6
Question 10: Sum of infinite terms A, AR, AR2, AR3,.... is 15. If the sum of the square of these terms is 150, then find the sum of AR2, AR4, AR6, ⋯.
Solution:
Answer: ½
A + AR + AR2 + ⋯ = 15
⇒A/(1-R) = 15 ⋯(i)
Now, A2 + A2 R2 + A2 R4 + ⋯ = 150
⇒A2/(1-R2 ) = 150
⇒(A⋅A)/((1-R)(1 + R)) = 150
⇒15A/(1 + R) = 150
⇒A = 10(1 + R)⋯(ii)
Solving (i) & (ii):
10(1 + R)/(1-R) = 15
⇒R = 1/5
So, A = 12
AR2 + AR4 + AR6 + ⋯ = (AR^2)/(1-R2 ) = (12⋅1/25)/(1-1/25) = 1/2
Question 11: If
Solution:
Answer: √2
Question 12: A wire of length 36 units is cut into two parts which are bent respectively to form a square of side x units and a circle of radius of r units. If the sum of areas of the square and the circle so formed is minimum, then find the circumference of the circle.
Solution:
Answer: 2π x 18/(4 + π) = 36π/(4 + π)
Sum of perimeter of square and circle = 36
4x + 2πr = 36⇒2x + πr = 18
x = (18 - πr)/2 = 9 - πr/2 .....(i)
Sum of areas: S = x2 + πr2 ....(ii)
S = πr2 + (9 - πr/2) 2 , diff. w.r.to r
dS/dr = 2πr + 2(9 - πr/2)( - π/2)
dS/dr = 0
⇒r - 9/2 + πr/4
⇒r(π + 4)/4 = 9/2
⇒r = 18/(4 + π)
Circumference of circle = 2π x 18/(4 + π) = 36π/(4 + π)
Question 13: Solve
Solution:
Answer: [(1/2)] tan-14 - 0
= [(1/2)] tan-14 - 0
Question 14: If
Solution:
Answer:
Given
This implies,
or
Question 15: Evaluate
Solution:
Answer: 20 ⋅ 21 ⋅ 218
Question 16: Solve (1 + y) tan2x + tan x (dy/dx) + y = 0
Solution:
Answer: y tan x = -tan x + x + c
(1 + y) tan2x + tan x (dy/dx) + y = 0
tan x(dy/dx) + (tan2x + 1)y = -tan2x
dy/dx + (tan x + cot x )y = -tan x
I. F. =
= e ln sec x + ln sin x
= e ln sec x . sin x
= tan x
Question 17: If
Solution:
Answer: