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The endometrium is the inner layer that lines the uterus. It is made up of glandular cells that make secretions. The myometrium is the middle and thickest layer of the uterus wall.

The wall of the uterus is made of Three layer .

Auxillary equation of above differential equation is (D²+1)y = 0.

⇒ d²y/dx² + y = 0

⇒ y = A cos x + B sin x

f(x) = (x + 2)(x + 1)(x - 3), \(x \in [2, 3]\)

∵ f(x) is a polynomial of degree 3

∴ f(x) is continuous and differentiable in every domain.

But f(2) = 4 × 3 × -1 = -12

and f(3) = 5 × 4 × 0 = 0

∵ f(2) ≠ f(3)

∴ Rolle's theorem is not applicable in [2, 3].

There is a one mistake

Let x ∈ [-2, 3].

Then f(-2) = 0 × -1 × -4 = 0 and f(3) = 5 × 4 × 0 = 0

∴ f(-2) = f(3).

Hence, Rolle's theorem is applicable for internal [-2, 3].

Then there exist c ∈ (-2, 3) such that f'(c) = 0.

\(\Rightarrow\) 3c² - 7 = 0 

(∵ f(x) = (x + 2) (x + 1)(x - 3)

= (x² + 3x + 2) (x - 3)

= x³ - 7x - 6

∴ f'(x) = 3x² - 7)

\(\Rightarrow\) c² = \(\frac{7}{3}\)

\(\Rightarrow\) \(c = \pm \sqrt{\frac{7}{3}}\) ∈ [-2, 3].

Hence, Rolle's theorem is verified.

Correct answer is a. Functional relationship 

Correct answer is b. Pre-approach .

Correct answer is a. Field Sales Manager 

When you think of plane shapes, you might think of a big jet flying in the sky, but plane shapes didn't get their name because they fly. They're called plane shapes because they're figures that are flat and closed. That's right! Plane shapes don't pop up or out at you. Instead, plane shapes are what you would see in a drawing or a cartoon, but remember they cannot have any gaps or parts that are open. Sometimes, you may even hear them being called two-dimensional or 2-D shapes. This means that you cannot touch them on all sides like a ball or a block. 

Plane shapes can include sides, which are straight lines that make up the shape, and corners, which are where two sides come together. Some examples of plane shapes that you may see every day are stop signs, a sheet of paper, a paper plate, a stamp, or even a tortilla chip. There are many kinds of plane shapes, but we will focus on 5 basic kinds: squares, rectangles, circles, triangles, and octagons.

Given,

AP is 24, 21, 18, 15, …… 

First term of AP is a = 24. 

Common difference of AP is d = a₂ − a₁ 

= 21 − 24 = −3. 

Since,

First term is 24 which is multiple of 3 and common difference is – 3. 

Therefore, 

All terms of AP are multiple of 3. 

∴ First negative term of given AP is – 3. 

Let n^th term of AP is – 3. 

∴ a + (n − 1)d = −3 

(∵ n th term of AP is given by an = a + (n − 1)d) 

⇒ 24 + (n − 1) × −3 = −3 

(∵ a = 24 & d = −3) 

⇒ −3(n − 1) 

= −3 − 24 

= −27

⇒ n −1 = \(\frac{-27}{-3}\) = 9

⇒ n = 1 + 9 = 10.

Hence,

10th term of given AP is the first negative term.

Let first term of AP is a and common difference of AP is d. 

We know that,

n^th term of AP is an = a + (n – 1) d. 

Given that,

8^th term of AP is a₈ = 31 

⇒ a + (8 – 1) d = 31 

⇒ a + 7d = 31 … (1) 

Also, 

Given that,

15th term of AP is 16 more than the 11^th term of AP. 

i.e., a₁₅ = 16 + a₁₁ 

⇒ a + 14d = 16 + a + 10d 

⇒ 14d − 10d = 16 

⇒ d = \(\frac{16}{4}\) = 4. 

From equation (1), 

a + 7 × 4 = 31 

(∵d = 4) 

⇒ a = 31 – 28 = 3. 

Hence, 

The first term of AP is a₁ = a = 3. 

The second term of AP is a₂ = a + d 

= 3 + 4 = 7. 

The third term of AP is a₃ = a + 2d 

= 3 + 8 

= 11. 

Hence, 

The required AP is 3, 7, 11, ……… .

x² − 4x − 3 = 0

⇒ x² − 4x + 4 − 7 = 0

⇒ (x−2)² − 7 = 0

⇒ (x−2)² = 7

∴ x − 2 = −√7​, +√7​

x = 2 − √7​, 2 + √7​

Correct option (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

Explanation:

Let f(x) = (x – a) (x – c) + 2 (x – b) (x – d)

Then f(a) = 2 (a – b) (a – d) > 0

f(b) = (b – a) (b – c) < 0

f(d) = (d – a) (d – b) > 0

Hence a root of f(x) = 0 lies between a & b and another root lies between (b & d).

Hence the roots of the given equation are real and distinct.

Correct option (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

Explanation:

 x² – ax – a = 0

g(1) < 0 => a > 1/2

Correct option (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.

Explanation:

quation can be written as (2^x)² – (a – 4) 2^x – (a – 4) = 0

⇒ 2x = 1 & 2^x = a – 4 Since x ≤ 0 and 2x = a – 4 [∵ x is non positive]

∴ 0 < a – 4 ≤ 1 

⇒ 4 < a ≤ 5 i.e., a∈ (4, 5]

Correct option (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.

Explanation:

x² + x + 1 > 0 ∀x ∈R

a = 1 > 0

b² – 4ac = 1 – 4 = -3 < 0

x² + 2x + 5 > 0 ∀x ∈R

a = 1 > 0 

 b² – 4ac = 4 – 20 = -16 < 0

So x² + x + 1/x² + 2x + 5 > 0 ∀x∈R ‘a’ is correct 

 Correct option (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 

Explanation:

ax² + bx + c = 0 Put x = 1 a + b + c = 0 which is given So clearly ‘1’ is the root of the equation Nothing can be said about the sign of the roots. ‘c’ is correct. 

Correct option (C) Statement – 1 is True, Statement – 2 is False. 

Explanation:

If the coefficients of quadratic equation are not rational then root may be 2 + √3 and 2 + √3.