`|{:(,"List-I(Orbitals)",,"List-II"),(,"(Approach axis is z-axis)",,"(Type of bond)"),((P),p_(y)+p_(y),(1),sigma-bond),((Q),p_(y)+p_(x),(2),pi-bond),((R ),p_(z)+p_(z),(3),delta-("delta")bond),((S),d_(x^(2)-y^(2))+d_(x^(2)-y^(2)),(4),"No bond will form"):}|`
Code :

Options

A. Number of roots of equation in `[-2pi,2pi]is 2 for m=1`
B. Number of roots of equation in `[-2pi,2pi]is 1 for m=9`
C. Roots are real for `m le 1`
D. If sin x =t then `t^(2)-(m-3)t+m=0` can not have both roots in [-1,1]`

Answer

To solve the problem, we need to find the integer pairs \((a, b)\) that satisfy the equation \(2a^2 + 3b^2 = 35\).

### Step-by-Step Solution:

1. **Start with the equation:**
\[ 2a^2 + 3b^2 = 35 \]

2. **Rearranging the equation:**
We can express \(3b^2\) in terms of \(a\):
\[ 3b^2 = 35 - 2a^2 \]

3. **Finding integer values for \(a\):**
Since \(b^2\) must be a non-negative integer, \(35 - 2a^2\) must be non-negative:
\[ 35 - 2a^2 \geq 0 \implies 2a^2 \leq 35 \implies a^2 \leq 17.5 \]
Thus, \(a\) can take values from \(-4\) to \(4\) (since \(a\) is an integer).

4. **Testing integer values for \(a\):**
- For \(a = -4\):
\[ 2(-4)^2 + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1 \]
Pairs: \((-4, 1), (-4, -1)\)

- For \(a = -3\):
\[ 2(-3)^2 + 3b^2 = 35 \implies 18 + 3b^2 = 35 \implies 3b^2 = 17 \implies b^2 = \frac{17}{3} \text{ (not an integer)} \]

- For \(a = -2\):
\[ 2(-2)^2 + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3 \]
Pairs: \((-2, 3), (-2, -3)\)

- For \(a = -1\):
\[ 2(-1)^2 + 3b^2 = 35 \implies 2 + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11 \text{ (not an integer)} \]

- For \(a = 0\):
\[ 2(0)^2 + 3b^2 = 35 \implies 3b^2 = 35 \implies b^2 = \frac{35}{3} \text{ (not an integer)} \]

- For \(a = 1\):
\[ 2(1)^2 + 3b^2 = 35 \implies 2 + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11 \text{ (not an integer)} \]

- For \(a = 2\):
\[ 2(2)^2 + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3 \]
Pairs: \((2, 3), (2, -3)\)

- For \(a = 3\):
\[ 2(3)^2 + 3b^2 = 35 \implies 18 + 3b^2 = 35 \implies 3b^2 = 17 \implies b^2 = \frac{17}{3} \text{ (not an integer)} \]

- For \(a = 4\):
\[ 2(4)^2 + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1 \]
Pairs: \((4, 1), (4, -1)\)

5. **Collecting all pairs:**
The valid pairs \((a, b)\) are:
- \((-4, 1)\)
- \((-4, -1)\)
- \((-2, 3)\)
- \((-2, -3)\)
- \((2, 3)\)
- \((2, -3)\)
- \((4, 1)\)
- \((4, -1)\)

6. **Counting the pairs:**
We have a total of 8 valid pairs.

### Final Answer:
The number of elements satisfying the equation is **8**.