Given:

Equation is (m - 5) x² + 2(m - 10) x + m + 10 = 0

Concept Used:

If ax² + bx + c = 0 is a quadratic equation 

Sum of roots = -b/a

Product of roots = c/a

For real roots: b² - 4ac ≥ 0

For equal roots: b² - 4ac = 0

For imaginary roots: b² - 4ac < 0

Calculation:

We have a quadratic equation below 

(m - 5) x² + 2(m - 10) x + (m + 10) = 0 

For being a quadratic equation, the Coefficient of x² shouldn't be zero

m - 5 ≠ 0

⇒ m ≠ 5       ----(i)

According to the question roots are real

So, [2(m - 10)]² - 4 (m - 5) (m + 10) ≥ 0

⇒ 4 (m² - 20m + 100) - 4 (m² + 5m - 50) ≥ 0

⇒ (m2 - 20m + 100) - (m2 + 5m - 50) ≥ 0

⇒ -25m + 150 ≥ 0

⇒ -25m ≥ -150

⇒ m ≤ 6         ----(ii)

Now, Again according to the question 

Roots are of same sign 

So, We can say the product of roots must be +ve

Product of roots (c/a) > 0

⇒ (m + 10)/(m - 5) > 0

⇒ m < -10 or m > 5      -----(iii)

With the help of number line and equations (i), (ii), and (iii), we get

m < -10 and 5 < m ≤ 6.

∴ The correct answer is option 3.

Concept:

Scaler triple product of the vectors:

Solution:

Therefore,

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

Fractional part function: It is defined as difference between a number and its integral value.

{f} = f - [f]

The binomial expansion of (x + y)n:

(x+ y)n = nC0( xn) + nC1 (xn - 1) y + nC(xn- 2). y2 +....... + nCnyn 

Calculation: 

Given that

R = (5√5 + 11)2n+1      -----(1)

According to question, f is a fractional part of R

⇒ f = R - [R]        ------(2)

Let f₁ =  (5√5 - 11)2n+1    where 0 < f₁ < 1    

R - f₁ =  (5√5 + 11)2n+1 -  (5√5 - 11)2n+1     

[R] + f - f1 =  (5√5 + 11)2n+1 -  (5√5 - 11)2n+1        ------(3)

Expand this Binomially and substract both terms, we will get

(5√5 + 11)2n+1 - (5√5 - 11)2n+1  =

2[^{2n + 1}C₁(5√5)²ⁿ (11)+ 2n + 1C₃(5√5)2n-2(11)³ + ....... +2n +1C_2n+1(11)²ⁿ⁺¹] 

We can see that, √5 has power of multiple 2 hence, 

[R] + f - f1 = even positive integer

We know that, [R] is integer, therefore to make the  [R] + f - f1 integer, 

f - f1 = 0 ⇒ f = f₁

Therefore, 

Rf = Rf₁ =  (5√5 + 11)2n+1.(5√5 - 11)2n+1

⇒ Rf = 4^{2n+ 1}

Hence, option 1 is correct.

Given:

Available flags = 5

Sequence matters

Concept:

Formula:

Explanation:

At least two flags from 5 flags mean ' 2 flags or 3 flags or 4 flags or 5 flags '.

= (4)(5) + (3)(4)(5) + (2)(3)(4)(5) + (1)(2)(3)(4)(5)

= 20 + 60 + 120 + 120

= 320

Formula used :

Calculations:

CONCEPT:

• The angle of intersection between the two curves will be the same as the angle of the intersection or between the tangents on both the curves on a common point.
• If the slope of a tangent at a common point on the first curve is m1 and the slope of a tangent at a common point on the second curve is m2 then the angle of intersection,

CALCULATION:

Given:

First of all, find the point of intersection of these two curves.

⇒ sin x - cos x ⇒ x = π / 4

Now,

Concept:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b) then,

f′(c) = 0 for some c ∈ [a, b].

Calculation:

From Rolle's theorem in (1, 26), f (1) = f (26) = 5,

In a given interval, the function satisfies all the conditions of Rolle's theorem, therefore in [1, 26], at least, there is a point for which f' (x) = 0

Concept:

Calculations:

Two people are required to shake hands.

The total number of handshakes is 66.

Suppose there are n people in the room

⇒ n (n - 1) = 132

⇒ n² -  n - 132 = 0

⇒ (n - 12)(n + 11) = 0

⇒ n = 12 or n = - 11

Here n = - 11 is not possible as the number of persons can't be negative.

⇒ n = 12

∴ The number of people in the room is 12.

Concept:

• The slope of parallel lines is the same. As an example, y = mx + p and y = mx + q are parallel to each other and their slope is m and p and q are their intercept with the y-axis
• The formula for the midpoint of the segment joining two points (a , b) and (c , d) is 

Calculation:

Given, the line parallel to the line 2x - 3y = 1 passes through the midpoint of the segment joining the two points (1 , 3) and (1 , -7).

The equation of the given line can be modified in a way to match the equation y = mx +a to find out the slope of the line as follows,

Here, the slope of the given line is 2/3, therefore the line parallel to this line will also have a slope 2/3 .

Given:

Concept:

To solve such type of differential equations, simply put y = vx.

Calculation:

Putting y = vx

Integrating both sides -

⇒ ln|f(v)| = ln|x| + lnC

⇒ f(v) = Cx

putting v = y/x back,

⇒ f(y/x) = Cx