IIT JEE 1969 Mathematics Question Paper with Answer and Solution

(A) Let $a^{1/x} = b^{1/y} = c^{1/z} = k$.
Then $a = k^x, b = k^y, c = k^z$.
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Substituting the values of $a, b, c$ in terms of $k$,we get $(k^y)^2 = k^x \cdot k^z$.
This simplifies to $k^{2y} = k^{x+z}$.
Equating the exponents,we get $2y = x + z$.
This condition implies that $x, y, z$ are in $A.P.$

(B) Let $y = \frac{x + 2}{2x^2 + 3x + 6}$.
Then $y(2x^2 + 3x + 6) = x + 2$,which simplifies to $2yx^2 + (3y - 1)x + (6y - 2) = 0$.
If $y = 0$,then $x = -2$,which is a real value.
If $y \neq 0$,for $x$ to be real,the discriminant $D \ge 0$.
$D = (3y - 1)^2 - 4(2y)(6y - 2) \ge 0$.
$9y^2 - 6y + 1 - 48y^2 + 16y \ge 0$.
$-39y^2 + 10y + 1 \ge 0$.
$39y^2 - 10y - 1 \le 0$.
Factoring the quadratic: $(13y + 1)(3y - 1) \le 0$.
This inequality holds for $y \in \left[ -\frac{1}{13}, \frac{1}{3} \right]$.
Since $y = 0$ is included in this interval,the range is $\left[ -\frac{1}{13}, \frac{1}{3} \right]$.

(A) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$.
First,check the slopes of the lines:
Slope of $L_1$ $(m_1)$ = $4/7$.
Slope of $L_3$ $(m_3)$ = $-7/4$.
Since $m_1 \times m_3 = (4/7) \times (-7/4) = -1$,the lines $L_1$ and $L_3$ are perpendicular to each other.
Therefore,the triangle is a right-angled triangle with the right angle at the intersection of $L_1$ and $L_3$.
The orthocentre of a right-angled triangle is the vertex where the right angle is formed.
Solving $4x - 7y = -10$ and $7x + 4y = 15$:
Multiply the first by $4$ and the second by $7$: $16x - 28y = -40$ and $49x + 28y = 105$.
Adding these gives $65x = 65$,so $x = 1$.
Substituting $x = 1$ into $4(1) - 7y = -10$ gives $-7y = -14$,so $y = 2$.
The orthocentre is $(1, 2)$.

(A) The slope of the line $AB$ is $m = \frac{1-0}{3-2} = 1$.
Since $m = \tan \theta = 1$,the angle of inclination is $\theta = 45^\circ$.
When the line is rotated anti-clockwise about point $A$ by $15^\circ$,the new angle of inclination becomes $\theta' = 45^\circ + 15^\circ = 60^\circ$.
The slope of the new line is $m' = \tan 60^\circ = \sqrt{3}$.
The line passes through $A(2,0)$,so the equation is given by $y - 0 = \sqrt{3}(x - 2)$.
$y = \sqrt{3}x - 2\sqrt{3}$.
Rearranging the terms,we get $\sqrt{3}x - y - 2\sqrt{3} = 0$.