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Calculation:

Given equation:

$abx^2+bcx+ca=cax^2+abx+bc$

Bring all terms to one side:

⇒ $(ab-ca)x^2+(bc-ab)x+(ca-bc)=0$

Since the roots are equal, the discriminant is zero:

⇒ $B^2=4AC$

⇒ $(bc-ab)^2=4(ab-ca)(ca-bc)$

Divide both sides by $(abc)^2$:

⇒ $\left(\frac{bc-ab}{abc}\right)^2 =4\left(\frac{ab-ca}{abc}\right)\left(\frac{ca-bc}{abc}\right)$

Simplify each fraction:

⇒ $\frac{bc-ab}{abc}=\frac{1}{a}-\frac{1}{c},\quad \frac{ab-ca}{abc}=\frac{1}{c}-\frac{1}{b},\quad \frac{ca-bc}{abc}=\frac{1}{b}-\frac{1}{a}$

So,

⇒ $\left(\frac{1}{a}-\frac{1}{c}\right)^2 =4\left(\frac{1}{c}-\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)$

Rearranging terms, we get:

⇒ $\left(\frac{1}{a}+\frac{1}{c}-\frac{2}{b}\right)^2=0$

Hence,

⇒ $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Hence, the correct answer is Option 4.

Calculation:

 $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., which implies a, b and c are in H.P.

Hence, the correct answer is Option 3.

Calculation:

We are given two arithmetic series:

Series 1: $6 + 10 + 14 + \dots$ up to $m$ terms

Series 2: $1 + 3 + 5 + 7 + \dots$ up to $n$ terms

We are told their sums are equal:

$S_m = S_n$

Sum of Series 1

First term $a_1 = 6$, common difference $d_1 = 4$

Sum formula: $S_m = \frac{m}{2} [2a_1 + (m - 1)d_1]$

$= \frac{m}{2} [12 + 4(m - 1)] = \frac{m}{2} [4m + 8] = 2m(m + 2)$

Sum of Series 2

First term $a_2 = 1$, common difference $d_2 = 2$

Sum formula: $S_n = \frac{n}{2} [2a_2 + (n - 1)d_2]$

$= \frac{n}{2} [2 + 2(n - 1)] = \frac{n}{2} [2n] = n^2$

Equating the sums

$2m(m + 2) = n^2 \Rightarrow n^2 = 2m(m + 2)$

Hence, the correct answer is Option (d):  n² = 2m(m + 2)

Calculation:

From the relation obtained in the first part:

$n^2 = 2m(m+2)$

We need the number of possible integer values of $m$ such that:

$m<25 \quad \text{and} \quad n<25$

Test values of $m$ and check whether $2m(m+2)$ is a perfect square:

$m=1 \Rightarrow n^2=2\cdot1\cdot3=6$ not a perfect square

$m=2 \Rightarrow n^2=2\cdot2\cdot4=16=4^2 \Rightarrow n=4 ;(<25)$

$m=3 \Rightarrow n^2=2\cdot3\cdot5=30$ not a perfect square

$m=4 \Rightarrow n^2=2\cdot4\cdot6=48$ not a perfect square

Similarly for $5 \le m < 25,; 2m(m+2)$ not a perfect square

Hence, only $m=2$ satisfies the condition.

Therefore, the number of possible values of $m$ is one.

Hence, the correct answer is Option 2.

Calculation:

Total number of triangles without restrictions:

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Subtract the degenerate cases:

$\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

Subtract the degenerate triangles from the total:

$\text{Valid triangles} = 56 - 4 = 52$

Hence, the correct answer is 52.

Calculation:

We need to calculate the number of quadrilaterals that can be formed by choosing 4 points from 8 points, out of which 4 are collinear.

The total number of ways to choose 4 points from 8 is given by:

$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$

Now, we subtract the cases where 3 or more points are collinear, as they do not form a quadrilateral. The number of cases where all 4 points are collinear is $\binom{4}{4} = 1$, and the number of cases where exactly 3 points are collinear is $\binom{4}{3} \times \binom{4}{1} = 16$.

Thus, the valid quadrilaterals formed are:

Valid quadrilaterals = 70 - 1 - 16 = 53 

Hence, the correct answer is 53.

Calculation:

Given the quadratic polynomial function:

$f(x) = ax^2 + bx + c$

We are provided with the following conditions:

$f(1) = 2$

$f(4) = 2$

2 is a root of $f(x) = 0$

Since 2 is a root of the polynomial, we know that f(2) = 0  This gives us a condition:

$f(1) = 2 \Rightarrow a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2$

$f(4) = 2 \Rightarrow a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2$

$f(2) = 0 \Rightarrow a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0$

Now, we have the system of equations:

$a + b + c = 2$

$16a + 4b + c = 2$

$4a + 2b + c = 0$

Subtract equation (1) from equation (2):

$(16a + 4b + c) - (a + b + c) = 2 - 2 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0$

This gives us equation (4): $5a + b = 0$

Now, subtract equation (1) from equation (3):

$(4a + 2b + c) - (a + b + c) = 0 - 2 \Rightarrow 3a + b = -2$

This gives us equation (5): $3a + b = -2$

From equation (4), $b = -5a$. Substituting this into equation (5):

$3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1$

Substitute $a = 1$ into $b = -5a$:

$b = -5(1) = -5$

Substitute $a = 1$ and $b = -5$ into $a + b + c = 2$:

$1 - 5 + c = 2 \Rightarrow c = 6$

The quadratic equation is now:

$f(x) = x^2 - 5x + 6$

Since 2 is a root, we factor the quadratic:

$f(x) = (x - 2)(x - 3)$

The other root is 3.

Hence, the correct answer is 3.

Calculation:

Given the quadratic equation f(x) = $ax^2 + bx + c$, we are provided with the following conditions:

Condition 1: f(1) = 2 , which gives:

⇒ $a(1)^2 + b(1) + c = 2 \Rightarrow a + b + c = 2$....(1)

Condition 2: f(4) = 2 , which gives:

⇒ $a(4)^2 + b(4) + c = 2 \Rightarrow 16a + 4b + c = 2$......(2)

Condition 3: Since 2 is a root of f(x) = 0, f(2) = 0 , which gives:

⇒ $a(2)^2 + b(2) + c = 0 \Rightarrow 4a + 2b + c = 0$

Now, we have the system of equations:

⇒ $a + b + c = 2$

⇒ $16a + 4b + c = 2$

⇒ $4a + 2b + c = 0$.......(3)

Subtract Equation 1 from Equation 2:

⇒ $(16a + 4b + c) - (a + b + c) = 0 \Rightarrow 15a + 3b = 0 \Rightarrow 5a + b = 0$......(4)

Subtract Equation 1 from Equation 3 to eliminate c

⇒ (4a + 2b + c) − (a + b + c) = 0 − 2

⇒ 3a + b = −2 (........ 5)

From Equation 4, we get:

⇒ $b = -5a$

Substitute b = -5a into Equation 5:

⇒ $3a + (-5a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1$

Now substitute a = 1 into b = -5a :

⇒ $b = -5$

Substitute  a = 1 and b = -5 into Equation 1:

⇒ $1 - 5 + c = 2 \Rightarrow c = 6$

Thus, the value of  a + b + c  is:

⇒ $a + b + c = 1 - 5 + 6 = 2$

Hence the correct answer is 2.

Calculation:

Given Matrix A:

$A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix}$

⇒ $\text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)$

⇒ $\text{det}(A) = \cos^2 \theta + \sin^2 \theta$

⇒ $\text{det}(A) = 1$ (Using the identity $\cos^2 \theta + \sin^2 \theta = 1$ 

⇒ $\text{det}(A^4) = (\text{det}(A))^4 = 1^4 = 1$

Hence, the correct answer is Option 2.

Calculation:

Given Matrix A:

$A = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix}$

This is a rotation matrix, and rotation matrices have special properties.

$\text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)$

$\text{det}(A) = \cos^2 \theta + \sin^2 \theta$

$\text{det}(A) = 1$ Using the identity $\cos^2 \theta + \sin^2 \theta = 1$

Since det(A) = 1, the inverse of a matrix is related to its adjugate:

$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \text{adj}(A)$

Therefore:

$[\text{adj } A]^{-1} = A$

Hence, the correct answer is option 3.