Question

Two lines drawn through the point P(4, 0) divide the area bounded by the curves $y = \sqrt { 2 } \sin \frac { \pi x } { 4 }$ and x-axis , between the lines x = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to

• A. $- \frac { 2 \sqrt { 2 } } { \pi }$
• B. $- \frac { \sqrt { 2 } } { \pi }$
• C. $- \frac { 2 } { \pi }$
• D. $- \frac { 4 \sqrt { 2 } } { \pi }$

Answer 1

Answer: (A)

$- \frac { 2 \sqrt { 2 } } { \pi }$

Answer 2

Area bounded by $y = \sqrt { 2 } \cdot \sin \frac { \pi x } { 4 }$ and x-axis between the lines x = 2 and x = 4,

$\Delta = \sqrt { 2 } \int _ { 2 } ^ { 4 } \sin \frac { \pi x } { 4 } d x = - \left. \frac { 4 \sqrt { 2 } } { \pi } \cdot \cos \frac { \pi x } { 4 } \right| _ { 2 } ^ { 4 }$

$= \frac { 4 \sqrt { 2 } } { \pi }$ sq. units.

Let the drawn lines are L₁ : y − m₁(x − 4) = 0 and L₂ : y − m₂(x − 4) = 0, meeting the line x = 2 at the points A and B respectively.

Clearly A ≡ (2, -2m,), B ≡ (2, -2m₂) Now ∆_ACD = $\frac { \Delta } { 3 }$

⇒ $\frac { 4 \sqrt { 2 } } { 3 \pi } = \frac { 1 } { 2 }$ .2. − 2m₁

⇒ m₁ = $- \frac { 2 \sqrt { 2 } } { 3 \pi }$ Also A_BCD = $\frac { 2 \Delta } { 3 }$

⇒ $\frac { 8 \sqrt { 2 } } { 3 \pi } = \frac { 1 } { 2 }$.2.−2m₂

⇒ m₂ = −$\frac { 4 \sqrt { 2 } } { 3 \pi }$. Required sum = $- \frac { 2 \sqrt { 2 } } { \pi }$