MATHS IMPORTANT FORMULA Book Jee Mains & Advance 2023

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MATHS Important Formula Book Jee Mains & Advance 2023 he subject of MATHS will help you in preparing for your final exams. Your final exams in MATHS will be held in March or April. These MATHS pdf notes are short, neat and clean, colored, regularly updated, and according to the syllabus set by the CBSE board following NCERT guidelines. The notes are very useful for competitive exams like NEET, AIIMS, and ACT (American College Testing), etc…. Not only our Biology short notes are suitable for CBSE board students, but also for other state board students. If you want to get high marks in MATHS then you must study from these notes. To download MATHS syllabus notes please read this article till the end.

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JEE Main Maths , Relations And Functions Questions With Solutions

Question 1: If A = [(x, y) : x2 + y2 = 25] and B = [(x, y) : x2 + 9y2 = 144], then A ∩ B contains _______ points.

Solution:

A = Set of all values (x, y) : x2 + y2 = 25 = 52

B = [x2 / 144] + [y2 / 16] = 1

i.e., [x2 / (12)2] + [y2 / (4)2] = 1.

Clearly, A ∩ B consists of four points.

Question 2: In a college of 300 students, every student reads 5 newspapers, and every newspaper is read by 60 students. The number of newspapers is ________.

Solution:

Let the number of newspapers be x.

If every student reads one newspaper, the number of students would be x (60) = 60x

Since every student reads 5 newspapers, the number of students = [x * 60] / [5] = 300

x = 25

Question 3: Let R be the relation on the set R of all real numbers defined by a R b if and only if |a − b| ≤ 1. Then R is __________.

Solution:

|a − a| = 0 < 1

Therefore, a R a ∀ a ∈ R

Therefore, R is reflexive.

Again a R b, |a − b| ≤ 1 ⇒ |b − a| ≤ 1 ⇒ b R a

Therefore, R is symmetric.

Again 1 R [½] and [½] R1 but [½] ≠ 1

Therefore, R is not anti-symmetric.

Further, 1 R 2 and 2 R 3, but 1 R 3 is not possible, [Because, |1 − 3| = 2 > 1]

Hence, R is not transitive.

Question 4: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1 o R is ________.

Solution:

First, find R−1.

R−1 = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.

Obtain the elements of R−1 o R.

Pick the element of R and then of R−1.

Since (4, 5) ∈ R and (5, 4) ∈ R−1, we have (4, 4) ∈ R−1 o R

Similarly, (1, 4) ∈ R, (4, 1) ∈ R−1 ⇒ (1, 1) ∈ R−1 o R

(4, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (4, 4) ∈ R−1 o R,

(4, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (4, 7) ∈ R−1 o R

(7, 6) ∈ R, (6, 4) ∈ R−1 ⇒ (7, 4) ∈ R−1 o R,

(7, 6) ∈ R, (6, 7) ∈ R−1 ⇒ (7, 7) ∈ R−1 o R

(3, 7) ∈ R, (7, 3) ∈ R−1 ⇒ (3, 3) ∈ R−1 o R,

Hence, R−1 o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.

Question 5: If

, then f [f { f (x) }] equals ________.
Solution:

f [ f (x) ] = (f (x) − 3)/ ( f (x) + 1)

Now f [f { f (x) }] = f ([3 + x] / [1 − x])

= 4x/4

= x

Therefore, f [f { f (x) }] = x.

Question 6: If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].

Solution:

f (x) = cos (log x)

Let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

y = 0

Question 7: Let f : R → R be defined by f (x) = 2x + |x|, then f (2x) + f (−x) − f (x) = _______.

Solution:

f(x) = 2x + |x|

f(2x) = 2(2x) + |2x| = 4x + 2|x|

f(-x) = -2x + |-x| = -2x + |x|

-f(x) = -2x + (-|x|) = -2x – |x|

Hence, f(2x) + f(-x) – f(x) = 4x + 2|x| – 2x + |x| – 2x – |x|

= 2|x|

= 2x; x ≥ 0 and -2x; x < 0

Question 8: If f (x) = cos [π2] x + cos[−π2] x, where [x] stands for the greatest integer function, then find the function of the right angle.

Solution:

f (x) = cos [π2] x + cos[−π2] x

f (x) = cos (9x) + cos (−10x) {since π = 3.14}

= cos (9x) + cos (10x)

= 2 cos (19x / 2) cos (x / 2)

Now, right angle = π/2

So, f (π / 2) = 2 cos (19π / 4) cos (π / 4)

f (π / 2) = 2 *(−1 / √2) * (1/ √2)

= −1

Question 9: If

, for every real number, then what is the minimum value of f?
Solution:

Let

= 1 − (2 / [x2 + 1]) [Because [x2 + 1] > 1 also (2 / [x2 + 1]) ≤ 2]

So 1 − [2 / [x2 + 1]] ≥ 1 − 2;

−1 ≤ f (x) < 1

Thus, f (x) has a minimum value equal to -1.

Question 10: The function f : R → R defined by f (x) = ex is ________.

Solution:

Function f: R → R is defined by f (x) = ex.

Let x1, x2 ∈ R and f (x1) = f (x2) or ex1 = ex2 or x1 = x2.

Therefore, f is one-one.

Let f (x) = ex = y.

Taking log on both sides, we get x = log y.

As we know, negative real numbers have no pre-image, or the function is not onto, and zero is not the image of any real number.

Therefore, function f is one-one and into.

Question 11: If f: R → S defined by f (x) = sin x − √3 cos x + 1 is onto, then what is the interval of S?

Solution:

Given,

f (x) = sin x − √3 cos x + 1

As we know, the range of the function f(x) = a cos x + b sin x + c is given by:

c – √(a2 + b2) ≤ f(x) ≤ c + √(a2 + b2)

− √[1 + (√−3)2] ≤ (sin x − √3 cos x) ≤ √[1 + (√−3)2]

−2 ≤ (sin x − √3 cos x) ≤ 2

−2 + 1 ≤ (sin x − √3 cos x + 1) ≤ 2 + 1

−1 ≤ (sin x − √3 cos x + 1) ≤ 3

i.e., range = [−1, 3]

For f to be onto, the interval of S = [−1, 3].

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JEE Main Maths , Relations And Functions Questions With Solutions

Question 12: What is the domain of the function

?
Solution:

Given,

Let g (x) = sin−1 (3 − x)

−1 ≤ 3 −x ≤ 1

Domain of g(x) is [2, 4] and let h (x) = log [|x| − 2]

|x|− 2 > 0

|x| > 2

x < −2 or x > 2

(−∞, −2) ∪ (2, ∞)

Also, log(|x| – 2) ≠ 0

|x| – 2 ≠ 1

|x| ≠ 3

We know that (f / g) (x) = f(x) / g(x) ∀ x ∈ D1 ∩ D2 − {x ∈ R : g (x) = 0}

Domain of f (x) = (2, 4] − {3} = (2, 3) ∪ (3, 4].

Question 13: If f (x) = a cos (bx + c) + d, then what is the range of f (x)?

Solution:

f (x) = a cos (bx + c) + d ..(i)

As we know, -1 ≤ cos θ ≤ 1

For minimum, cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum, cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Alternatively,

-1 ≤ cos(bx + c) ≤1

-a ≤ a cos(bx + c) ≤ a

-a + d ≤ a cos(bx + c) + d ≤ a + d

Range of f(x) = [d – a, a + d]

Question 14: The function f: R → R is defined by f (x) = cos2 x + sin4x for x ∈ R, then what is f (R)?

Solution:

f (x) = cos2 x + sin4x

y = f (x) = cos2 x + sin2x (1 − cos2x)

y = cos2 x + sin2x − sin2x cos2x

y = 1 − sin2x cos2x

y = 1 − [1 / 4] * [sin22x]

3 / 4 ≤ f (x) ≤ 1, (Because 0 ≤ sin22x ≤ 1)

f (R) ∈ [3/4, 1]

Question 15: If f (x) = 3x − 5, then f−1(x) is _____________.

Solution:

Let f (x) = y ⇒ x = f−1 (y).

Hence, f (x) = y = 3x − 5

⇒f−1 (y) = x

f is one-one and onto, so f−1 exists and is given by f−1 (x) = [x + 5] / [3]

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MATHS IMPORTANT FORMULA Book Jee Mains & Advance 2023