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Correct option is: A) Inductive reasoning

Correct option is: A) When diagonals are equal

Correct option is: C) The product of two odd natural numbers is odd.

Correct option is: B) Inductive reasoning

Let the two numbers be a and b. 

Given : 

a + b = 15 

Assume, 

S = a²b³ 

S = a²(15 - a)³ 

\(\frac{dS}{da}\)= 2a(15 - a)³ - 3a²(15 - a)²

Condition maxima and minima is,

⇒\(\frac{dS}{da}\) = 0

⇒ 2a(15 - a)³ - 3a²(15 - a)² = 0 

⇒ a(15 - a)² {2(15 – a) – 3a} = 0 

⇒ a(15 - a)² {30 – 5a} = 0 

⇒ a = 0, 15, 6

\(\frac{d^2S}{da^2}\) = 2(15 - a)³ - 6a(15 - a)² - 6a(15 - a)²+ 3a²(15 - a)

⇒ \(\frac{d^2S}{da^2}\) = (15 - a){2(15 - a)² - 12a(15 - a)+ 3a²}

For S to minimum,

\(\frac{d^2S}{da^2}\) > 0

a = 0 ⇒ \(\frac{d^2S}{da^2}\) = 450 ⇒ minima 

a = 6 ⇒ \(\frac{d^2S}{da^2}\) = - 378 ⇒ maxima 

a = 15 ⇒ \(\frac{d^2S}{da^2}\) = 0 ⇒ point of inflection

Hence, two numbers are 0 and 15

f(x) = |x|

Now to check the maxima and minima at x = 0.

It can be easily seen through the option.

See |x| is x for x > 0 and –x for x < 0

That is no matter if you put a number greater than zero or number less than zero you will get positive answer.

∴ for x = 0 we will get minima.

Correct option is: C) Hypothesis

Correct option is: D) Rama is immortal

Correct option is: D) 11

Correct option is: A) Axiom

Correct option is: A) Hypothesis

Correct option is: D) Not possible

Correct option is: C) odd

f(x) = 1/x+5

domain of function is R-{0}

f'(x) = -1/x²

for all x, f’(x)<0

Hence function is decreasing.

Answer is : B. (0, 1/e)

Given: f(x) = x^x .

⇒ f'(x) = (log x+1) x^x

⇒ keeping f’(x) = 0

We get

⇒ x = 0 or x = 1/e

Now When x > 1/e the function is increasing

x < 0 function is increasing.

But in the interval (0,1/e) the function is decreasing.

It is given that

f(x) = 3^x

By differentiating w.r.t x

f’(x) = 3^x log 3 > 0 for all value of R

Therefore, f(x) is strictly increasing function on R.

It is given that

f(x) = [x – 1/x]

By differentiating w.r.t. x

f’(x) = 1 + 1/x²

On further calculation

f’(x) = (x² + 1)/x²

Here f’(x) ≥ 0 for all x ≠ 0

Therefore, f(x) is increasing function for all x ∈ R, where x ≠ 0.

It is given that

f(x) = x³ – 15x² + 75 x – 50

By differentiating w.r.t x

f’(x) = 3x² – 30x + 75

By taking 3 as common

f’(x) = 3(x² – 10x + 25)

On further simplification

f’(x) = 3(x – 5)²

Here f’(x) ≥ 0 for all x ≥ 0

Therefore, f(x) is increasing function on R.

Equating the corresponding elements, we get

y + 2x = 7 and – x = – 2

⇒ x = 2 and y + 2 × 2 = 7

⇒ y = 7 – 4 = 3

Given,

\(\begin{bmatrix} 3x\,+\,y & -y \\[0.3em] 2y\,-\,x & 3 \\[0.3em] \end{bmatrix} \)= \(\begin{bmatrix} 1 & 2 \\[0.3em] -5 & 3 \\[0.3em] \end{bmatrix} \)

Equating the corresponding elements, we get

⇒ 3x + y = 1, -y = 2, 2y - x = -5

⇒ y = -2

Putting it in 3x + y = 1 or 2y - x = - 5

we get 3x - 2 = 1 or - 4 - x = - 5

⇒ x = 1 or - x = - 1

⇒ x = 1

Hence, x = 1 and y = - 2

\(\begin{bmatrix}x\,+\,3y & y \\[0.3em]7\,-\,x & 4 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}4 & -1 \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)

Equating the corresponding elements, 

we get,

x + 3y =4, y = -1, 7 - x = 0

⇒ x = 7, y = -1

Given,

\(\begin{bmatrix}9 & -1 & 4 \\[0.3em]-2 & 1 & 3 \\[0.3em]\end{bmatrix}\) = \(A\,+\begin{bmatrix}1 &2 & -1 \\[0.3em]0 & 4 & 9 \\[0.3em]\end{bmatrix}\)

⇒ A = \(\begin{bmatrix}9 & -1 & 4 \\[0.3em]-2 & 1 & 3 \\[0.3em]\end{bmatrix}\)- \(\begin{bmatrix}1 &2 & -1 \\[0.3em]0 & 4 & 9 \\[0.3em]\end{bmatrix}\)

= \(\begin{bmatrix}8 & -3 & 5 \\[0.3em]-2 & -3 & -6 \\[0.3em]\end{bmatrix}\)

\(\begin{bmatrix}x\,+\,3y & y \\[0.3em]7\,-\,x & 4 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}4 & -1 \\[0.3em]0 & 4 \\[0.3em]\end{bmatrix}\)

Equating the corresponding elements, 

we get,

x + 3y =4, y = -1, 7 - x = 0

⇒ x = 7, y = -1

|AA'| = |A|.|A'| = |A|.|A| = |A|² = 2² = 4.

[Note : |AB| = |A|.|B| and |A| = |AT|, where A and B are square matrices.]

Given,

\(A = \begin{bmatrix}cos\,\theta & sin\,\theta \\[0.3em]-sin\,\theta & cos\,\theta\end{bmatrix}\)

∴ \(A^n = \begin{bmatrix}cos\,n\theta & sin\,n\theta \\[0.3em]-sin\,n\theta & cos\,n\theta\end{bmatrix}\)

Least positive integral value of k is 7. 

Since we have,

\( \begin{bmatrix} cos\,\frac{2\pi}{7} & -sin\,\frac{2\pi}{7} \\[0.3em] sin\,\frac{2\pi}{7} & cos\,\frac{2\pi}{7} \end{bmatrix}^7\)

=\( \begin{bmatrix} cos\,2\pi & -sin\,2\pi \\[0.3em] sin\,2\pi & cos\,2\pi \end{bmatrix}\)

= \( \begin{bmatrix} 1 & 0\\[0.3em] 0 & 1 \end{bmatrix}\)

There are many undefined terms for students. They are consistent, because they deal with two different situations. 

(i) When two points A and B ar given, C is in between two. 

(ii) When two points A and B are given take point C which passes through A and B. 

These postulates do not follow from Euclid’s postulates. But they follow Axiom 2.1

Let us consider the LHS

c(a cos B – b cos A)

Now, as LHS contain ca cos B and cb cos A which can be obtained from cosine formulae.

Then, from cosine formula we have

Cos A = (b² + c² – a²)/2bc

bc cos A = (b² + c² – a²)/2 … (i)

Cos B = (a² + c² – b²)/2ac

ac cos B = (a² + c² – b²)/2 … (ii)

Let us subtract equation (ii) from (i) we get,

ac cos B – bc cos A = (a² + c² – b²)/2 – (b² + c² – a²)/2

= a² – b²

∴ c(a cos B – b cos A) = a² – b²

Thus proved.

Let us consider the LHS 

(c - b cos A)/(b - c cos A)

Now, we can observe that we can get terms c – b cos A and b – c cos A from projection formula

From the projection formula we get,

c = a cos B + b cos A

c – b cos A = a cos B …. (i)

And,

b = c cos A + a cos C

b – c cos A = a cos C …. (ii)

On dividing equation (i) by (ii), we get,

(c - b cos A)/(b - c cos A)

= a cos B/a cos C

= cos B/cos C

= RHS

Thus proved.

Let us consider the LHS

(c² – a² + b²), (a² – b² + c²), (b² – c² + a²)

As we know sine rule in Δ ABC

a/sin A = b/sin B = c/sin C

Now, as LHS contain (c² – a² + b²), (a² – b² + c²) and (b² – c² + a²), which can be obtained from cosine formulae.

From the cosine formula we have:

Cos A = (b² + c² – a²)/2bc

2bc cos A = (b² + c² – a²)

Now, let us multiply both the sides by tan A we get,

2bc cos A tan A = (b² + c² – a²)tan A

2bc cos A (sin A/cos A) = (b² + c² – a²)tan A

2bc sin A = (b² + c² – a²)tan A … (i)

Cos B = (a² + c² – b²)/2ac

2ac cos B = (a² + c² – b²)

Now, let us multiply both the sides by tan B we get,

2ac cos B tan B = (a² + c² – b²)tan B

2ac cos B (sin B/cos B) = (a² + c² – b²)tan B

2ac sin B = (a² + c² – b²)tan B … (ii)

Cos C = (a² + b² – c²)/2ab

2ab cos C = (a² + b² – c²)

Then, let us multiply both the sides by tan C we get,

2ab cos C tan C = (a² + b² – c²)tan C

2ab cos C (sin C/cos C) = (a² + b² – c²)tan C

2ab sin C = (a² + b² – c²)tan C … (iii)

Now, as we are observing that sin terms are being involved therefore let’s use sine formula.

From the sine formula we have,

a/sin A = b/sin B = c/sin C

sin A/a = sin B/b = sin C/c 

Then, let us multiply abc to each of the expression we get,

abc sin A/a = abc sin B/b = abc sin C/c 

bc sin A = ac sin B = ab sin C

2bc sin A = 2ac sin B = 2ab sin C

∴ From the equation (i), (ii) and (iii) we have,

(c² – a² + b²)tan A = (a² – b² + c²)tan B = (b² – c² + a²)tan C

Thus proved.

Data: C lies between A and B such that AC = BC. 

To Prove: AC = \(\frac{1}{2}\)AB 

Proof : In this figure AC = BC. 

Adding AC on both sides, 

AC + AC = BC + AC (equals are added to equals) 

2AC = BC + AC 

2AC = AB (∵ BC + AC = AB) 

∴ AC = AB.

Data: In this figure AC = BD. 

To Prove : AB = CD 

Proof: AC = BD (Data) → (i) 

But, AC = AB + BC 

and BD = BC + CD. 

Substituting value of (i) 

AC = BD 

AB + BC = BC + CD

∴ AB = CD.

Suppose the angles of the triangle be (a – d)^°, a^° and (a + d)^°.

As we know that, the sum of angles of triangle is 180°.

a – d + a + a + d = 180°

3a = 180°

a = 180°/3

= 60^°

Given as

The greatest angle = 5 × least angle

Now, upon cross-multiplication,

The greatest angle/least angle = 5
(a + d)/(a - d) = 5

(60 + d)/(60 - d) = 5

On cross-multiplying we get,
60 + d = 300 – 5d

6d = 240

d = 240/6

= 40

Thus, angles are:

(a – d)° = 60° – 40° = 20°

a° = 60°

(a + d)° = 60° + 40° = 100°

∴ Angles of triangle in radians:

(20 × π/180) rad = π/9
(60 × π/180) rad = π/3

(100 × π/180) rad = 5π/9

(i) False. Number of lines can pass through a single point. 

(ii) False. Because there is a unique line that passes through two points. 

(iii)  True. A terminated line can be produced indefinitely on both the sides. 

(iv)  True. Because circumference of equal circles are equal their distance is same from centre. Hence radii of two circles are equal. 

(v)  True. Because E.g., If AB = 4 cm, PQ = 4 cm., then, AB = PQ = 4 cm. 

PQ = XY = 4 cm. 

∴ AB = PQ = XY 

∴ AB = XY.

Suppose the angles of the triangle be (a – d)°, a° and (a + d)°.

As we know that, the sum of the angles of a triangle is 180°.

a – d + a + a + d = 180°

3a = 180°

a = 60°

Given as

The number of degrees in the least angle/The number of degrees in the mean angle = 1/120  

(a - d)/a = 1/120
(60 - d)/60 = 1/120

(60 - d)/1 = 1/2

120 - 2d = 1
2d = 119

d = 119/2

= 59.5

∴ The angles are:

(a – d)° = 60° – 59.5° = 0.5°

a° = 60°

(a + d)° = 60° + 59.5° = 119.5°

The angles of triangle in radians

(0.5 × π/180) rad = π/360

(60 × π/180) rad = π/3

(119.5 × π/180) rad = 239π/360

Suppose the number of sides in the first polygon be 2x and

The number of sides in the second polygon be x.

As we know that, angle of an n-sided regular polygon = [(n - 2)/n]π radian 

The angle of the first polygon = [(2x - 2)/2x]π = [(x - 1)/x]π radian

The angle of the second polygon = [(x - 2)/x]π radian  

Hence,

[(x - 1)/x]π/[(x - 2)/x]π = 3/2
(x - 1)/(x - 2) = 3/2

Now, upon cross-multiplication we get,

2x – 2 = 3x – 6

3x - 2x = 6 - 2

x = 4

∴ Number of sides in the first polygon = 2x = 2(4) = 8

Thus, the number of sides in the second polygon = x = 4

Suppose the number of sides in the first polygon be 5x and

The number of sides in the second polygon be 4x.

As we know that, angle of an n-sided regular polygon = [(n - 2)/n]π radian

The angle of the first polygon = [(5x - 2)/5x]180^°

The angle of the second polygon = [(4x - 1)/4x]180^°

Hence,

[(5x - 2)/5x]180^° – [(4x - 1)/4x]180^° = 9

180^°[(4(5x - 2) – 5(4x - 2))/20x] = 9

Now, upon cross-multiplication we get,

(20x – 8 – 20x + 10)/20x = 9/180

2/20x = 1/20

2/x = 1

x = 2

∴Number of sides in the first polygon = 5x = 5(2) = 10

Thus, the number of sides in the second polygon = 4x = 4(2) = 8

B. y = 0

The equation of x-axis is y = 0, since, x-axis is a parallel to itself at a distance 0 from it.

Hence, option (B) is the correct answer.

A. (a, a)

Any point on the line y = x will have x and y coordinate same.

So, any point on the line y = x is of the form (a, a).

Hence, option (A) is the correct answer.

A. 4

Explanation:

We know that,

(2, 0) = (x, y)

Substituting values of x and y in the above equation, we get

2×2 + 3×0 = k

k = 4

Hence, option A is the correct answer.

C. (x, 0)

Any point on the x-axis has its ordinate 0.

So, any point on the x-axis if of the form (x, 0).

Hence, option (C) is the correct answer.

False

Since, every point on the graph of the linear equation represents a solution.

A. (-9/2 , m) 

Explanation:

Solving the above equation we get,

2x = -9

x = -9/2

As the coefficient of y is 0, therefore, y can take any value and will not affect our answer.

A. x = -9/2

y = any value

B. x = n

C. x = 0

D. x = -9

Hence, option A is the correct answer.

B. 1. x + 0. y = 7

A. Simplifying the equation we get x + y = 7

B. Simplifying the equation we get x + 0y = 7 which is equal to x = 7

C. Simplifying the equation we get y = 7

D. simplifying the equation we get 0x + 0y = 7 which is not possible.

Hence, option (B) is the correct answer.

False

Since, the graph of a linear equation in two variables always represent a line.

D. (0, 2)

Let 2x + 3y = 6 cut the y-axis at P. therefore at P x-coordinate = 0.

Substituting x = 0, we get

2(0) + 3y = 6

3y = 6

y = 2

Hence the coordinates are (0, 2).

A. (2, 0) is wrong because it has x = 2

B. (0, 3) is wrong because it has y = 3

C. (3, 0) is wrong because it has x = 3

D. (0, 2) is right because it has x = 0 and y = 2 which is equal to the coordinates (0,2)

Hence, option D is the correct answer.