Question

please solve this question

Answer 1

Question 1:

Let 'm' be the any integer,
Now 'm' can be even or odd
So option (a) can't always be even.

Now if 'm' is even then m+1 will be odd and if 'm' is odd then m+1 will be even, so option (b) is also wrong.

Now in option (c), 2m will always be an even integer whether 'm' is even or odd. So option (c) is correct.

And in option (d) 2m + 1 will always be odd integer whether 'm' is even or odd. So option (d) is wrong.

>>> Correct option is option (c): [2m]

Question 2:

One of the zeros of the given polynomial is -3, which means that substituting the value of x as -3, we will get 0 (zero)

Now let's substituting value of x as -3 and putting the polynomial expression as zero, so we will get,

9(k - 1) - 3 k + 1 = 0
9k - 9 - 3k + 1 = 0
6k = 9 - 1
6k = 8
3k = 4

$k \: = \: \frac{4}{3}$

Answer is 4/3.

"Thanks"