Q1: Consider the following lists :
The correct option is
(a) (I) →(P); (II) →(S); (III) →(P); (IV) →(S)
(b) (I) → (P); (II) → (P); (III) → (T); (IV) → (R)
(c) (I) → (Q); (II) →(P); (III) → (T); (IV) → (S)
(d) (I) →(Q); (II) →(S); (III) →(P); (IV) →(R) [JEE Advanced 2022 Paper 1]
Ans: (b)
Solving all question one by one we get,
So,
∴ x has 2 elements → P
So,
∴ x has 2 elements → P
So,
∴ x has 6 elements →T
So,
∴ x has 4 elements →R
Q2: Let α and β be real numbers such that .
If and , then the greatest integer less than or equal to is: [JEE Advanced 2022 Paper 2]
Ans: 1
Given,
and
Let,
= 4/3
Q1: For non-negative integers n, let
Assuming cos−1 x takes values in [0, π], which of the following options is/are correct?
(a) If α = tan(cos−1 f(6)), then α2 + 2α −1 = 0
(b)
(c) sin(7 cos−1 f(5)) = 0
(d) [JEE Advanced 2019 Paper 2]
Ans: (a), (b) & (c)
It is given, that for non-negative integers 'n',
Now,
Now,
Now,
and Now,
Hence, options (a), (b) and (c) are correct.
Q2: Let f(x) = sin(π cos x) and g(x) = cos(2π sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order:
X = {x : f(x) = 0}, Y = {x : f'(x) = 0}
Z = {x : g(x) = 0}, W = {x : g'(x) = 0}
List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets.
here values of sin x, are in an A.P. but corresponding values of x are not in an AP so, (iii) → R.
For W = {x : g'(x) = 0}, x > 0
So, g'(x) = −2 π cos x sin(2π sin x) = 0
⇒ either cos x = 0 or sin(2π sin x) = 0
⇒ (iv) → P, R, S
Hence, option (a) is correct.
Q1: In a ΔPQR = 30∘ and the sides PQ and QR have lengths 10√3 and 10, respectively. Then, which of the following statement(s) is(are) TRUE?
(a) ∠QPR=45∘
(b) The area of the ΔPQR is 25√3 and ∠QRP=120∘
(c) The radius of the incircle of the ΔPQR is 103 − 15
(d) The area of the circumcircle of the ΔPQR is 100 [JEE Advanced 2018 Paper 1]
Ans: (b), (c) & (d)
We have,
In ΔPQR
By cosine rule
Since, PR = QR = 10
Radius of incircle of
and radius of circumcircle
∴ Area of circumcircle of
Hence, option (b), (c) and (d) are correct answer.
Q2: Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the X-axis, Y-axis and Z-axis, respectively, where O(0, 0, 0) is the origin. Let be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p = SP, q = SQ, r = SR and t = ST, then the value of |(p × q) × (r × t)| is _________ [JEE Advanced 2018 Paper 2]
Ans: 0.5
Here, P(1, 0, 0), Q(0, 1, 0), R(0, 0, 1), T = (1, 1, 1) and
Now,
and
Now,
∴
= 1/2
= 0.5
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