IIT JEE 1960 Mathematics Question Paper with Answer and Solution

(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.
Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.
The denominator is of the form $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$.
Here,$x = 3$ and $y = 2$,so the denominator is $(3 + 2)^6 = 5^6$.
Since $5^6 = (5^2)^3 = 25^3$,the denominator is $25^3$.
Therefore,the value of the expression is $\frac{25^3}{25^3} = 1$.

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
For $\Delta ABC$ with $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$:
$\text{Area}(\Delta ABC) = \frac{1}{2} |6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5)| = \frac{1}{2} |6(7) + (-3)(-5) + 4(-2)| = \frac{1}{2} |42 + 15 - 8| = \frac{49}{2}$.
For $\Delta DBC$ with $D(x, 3x)$,$B(-3, 5)$,$C(4, -2)$:
$\text{Area}(\Delta DBC) = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - 3x) + 4(3x - 5)| = \frac{1}{2} |7x + 6 + 9x + 12x - 20| = \frac{1}{2} |28x - 14| = |14x - 7|$.
Given $\frac{\text{Area}(\Delta DBC)}{\text{Area}(\Delta ABC)} = \frac{1}{2}$,we have $2 \times \text{Area}(\Delta DBC) = \text{Area}(\Delta ABC)$.
$2 |14x - 7| = \frac{49}{2} \Rightarrow |14x - 7| = \frac{49}{4}$.
Case $1$: $14x - 7 = \frac{49}{4}$ $\Rightarrow 14x = \frac{49}{4} + 7 = \frac{77}{4}$ $\Rightarrow x = \frac{77}{56} = \frac{11}{8}$.
Case $2$: $14x - 7 = -\frac{49}{4}$ $\Rightarrow 14x = 7 - \frac{49}{4} = -\frac{21}{4}$ $\Rightarrow x = -\frac{21}{56} = -\frac{3}{8}$.
Since $\frac{11}{8}$ is in the options,the correct value is $\frac{11}{8}$.