TN State Board 10th Maths Question Paper 2024

Since $n(A\times B)=n(A)\cdot n(B)$. Given $A=\{1,3\}$, so $n(A)=2$. Also $n(A\times B)=6$. Hence $2\cdot n(B)=6$, so $n(B)=3$. Therefore, the correct choice is (b).

A bijective function is both one-one and onto, so the two sets must have the same number of elements. Therefore, if $n(B)=7$, then $n(A)=7$. Hence the correct choice is (a).

The least number divisible by all the numbers from 1 to 10 is the LCM of $1,2,3,4,5,6,7,8,9,10$. Prime factors with highest powers: $2^3, 3^2, 5, 7$ So, $$\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520.$$ Hence the correct choice is (d).

In an A.P. with 31 terms, the middle term is the $16^{\text{th}}$ term. The sum of an odd number of terms in an A.P. equals the number of terms multiplied by the middle term. So, $$S_{31} = 31\times m = 31m.$$ Therefore, the correct choice is (c), not (b).

To make $\dfrac{x^4}{16}$ a perfect square, we need to add a term so that the expression becomes a square of a binomial. Now, $$\frac{x^4}{16} = \left(\frac{x^2}{4}\right)^2.$$ The nearest perfect-square form is $$\left(\frac{x^2}{4}+2x\right)^2 = \frac{x^4}{16} + x^3 + 4x^2,$$ which is not among the options, so we check the intended standard form: To complete a square for $\dfrac{x^4}{16}$ with a term in $x^2$, we use $$\left(\frac{x^2}{4}+2x\right)^2 \text{ or } \left(\frac{x^2}{4}-2x\right)^2,$$ which requires adding $16x^2$. Hence the correct choice is (b).