JEE Advance - Mathematics (1980)
- 4$$ABC$$ is a triangle with $$AB=AC$$. $$D$$ is any point on the side $$BC$$. $$E$$ and $$F$$ are points on the side $$AB$$ and $$AC$$, respectively, such that $$DE$$ is parallel to $$AC$$, and $$DF$$ is parallel to $$AB$$. Prove that $$$DF + FA + AE + ED = AB + AC$$$回答(A)The statement is always true.
- 9Given $$A = \left\{ {x:{\pi \over 6} \le x \le {\pi \over 3}} \right\}$$ and
$$f\left( x \right) = \cos x - x\left( {1 + x} \right);$$ find $$f\left( A \right).$$回答(C)$$\left[ {{1 \over 2} - {\pi \over 3}\left( {1 + {\pi \over 3}} \right),,{{\sqrt 3 } \over 2} - {\pi \over 6}\left( {1 + {\pi \over 6}} \right)} \right]$$ - 10For all $$\theta $$ in $$\left[ {0,\,\pi /2} \right]$$ show that, $$\cos \left( {\sin \theta } \right) \ge \,\sin \,\left( {\cos \theta } \right).$$回答(D)The inequality holds because \(\cos(x) \ge \sin(x)\) for all \(x \in [0, \pi/4]\) and because of the properties of sine and cosine functions in the interval \([0, \pi/2]\).
- 19The point $$\,\left( {4,\,1} \right)$$ undergoes the following three transformations successively.
Reflection about the line $$y=x$$.
Translation through a distance 2 units along the positive direction of x-axis.
Rotation through an angle $$p/4$$ about the origin in the counter clockwise direction.
Then the final position of the point is given by the coordinates.回答(C)$$\left( { - {1 \over {\sqrt 2 }},{7 \over {\sqrt 2 }}} \right)$$ - 27$$ABC$$ is a triangle. $$D$$ is the middle point of $$BC$$. If $$AD$$ is perpendicular to $$AC$$, then prove that $$$\cos A\,\cos C = {{2\left( {{c^2} - {a^2}} \right)} \over {3ac}}$$$回答(C)Applying the Law of Cosines and the properties of medians to express \(\cos A\) and \(\cos C\), and using the given perpendicularity to simplify to the desired form. Considering Stewart's theorem.
- 28(ii) $$AB$$ is vertical pole. The end $$A$$ is on the level ground. $$C$$ is the middle point of $$AB$$. $$P$$ is a point on the level ground. The portion $$CB$$ subtends an angle $$\beta $$ at $$P$$. If $$AP = n\,AB,$$ then show that tan$$\beta $$ $$ = {n \over {2{n^2} + 1}}$$回答(A)The height of the tower is abc tan(θ) / 4Δ and tan(β) = n / (2n^2 + 1)
