JEE Advance - Mathematics (2003)
- 2(i) Find the equation of the plane passing through the points $$(2, 1, 0), (5, 0, 1)$$ and $$(4, 1, 1).$$
(ii) If $$P$$ is the point $$(2, 1, 6)$$ then find the point $$Q$$ such that $$PQ$$ is perpendicular to the plane in (i) and the mid point of $$PQ$$ lies on it.Відповідь(A)Plane: $$x + y - 2z = 3$$, Point Q: $$(6, 5, -2)$$ - 3If $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w ,$$ are three non-coplanar unit vectors and $$\alpha ,\beta ,\gamma $$ are the angles between $$\overrightarrow u $$ and $$\overrightarrow v $$ and $$\overrightarrow w ,$$ $$\overrightarrow w $$ and $$\overrightarrow u $$ respectively and $$\overrightarrow x ,\overrightarrow y ,\overrightarrow z ,$$ are unit vectors along the bisectors of the angles $$\alpha ,\,\,\beta ,\,\,\gamma $$ respectively. Prove that $$\,\left[ {\overrightarrow x \times \overrightarrow y \,\,\overrightarrow y \times \overrightarrow z \,\,\overrightarrow z \times \overrightarrow x } \right] = {1 \over {16}}{\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w } \right]^2}\,{\sec ^2}{\alpha \over 2}{\sec ^2}{\beta \over 2}{\sec ^2}{\gamma \over 2}.$$Відповідь(A)The statement is always true.
- 4For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is $$p.$$ If he fails in one of the exams then the probability of his passing in the next exam is $${p \over 2}$$ otherwise it remains the same. Find the probability that he will qualify.Відповідь(A)p^2(3 - 2p)
- 7If $${z_1}$$ and $${z_2}$$ are two complex numbers such that $$\,\left| {{z_1}} \right| < 1 < \left| {{z_2}} \right|\,$$ then prove that $$\,\left| {{{1 - {z_1}\overline {{z_2}} } \over {{z_1} - {z_2}}}} \right| < 1$$.Відповідь(E)The inequality holds because |z1| < 1 and |z2| > 1 imply |1 - z1 * conjugate(z2)| > |z1 - z2|.
- 9If the function $$f:\left[ {0,4} \right] \to R$$ is differentiable then show that
(i)$$\,\,\,\,\,$$ For $$a, b$$$$\,\,$$$$ \in \left( {0,4} \right),{\left( {f\left( 4 \right)} \right)^2} - {\left( {f\left( 0 \right)} \right)^2} = gf'\left( a \right)f\left( b \right)$$
(ii)$$\,\,\,\,\,$$ $$\int\limits_0^4 {f\left( t \right)dt = 2\left[ {\alpha f\left( {{\alpha ^2}} \right) + \beta \left( {{\beta ^2}} \right)} \right]\forall 0 < \alpha ,\beta < 2} $$ВідповідьABD - 10Tangent is drawn to ellipse
$${{{x^2}} \over {27}} + {y^2} = 1\,\,\,at\,\left( {3\sqrt 3 \cos \theta ,\sin \theta } \right)\left( {where\,\,\theta \in \left( {0,\pi /2} \right)} \right)$$.Then the value of $$\theta $$ such that sum of intercepts on axes made by this tangent is minimum, is
Відповідь(C)$$\pi /8$$ - 12If $${I_n}$$ is the area of $$n$$ sided regular polygon inscribed in a circle of unit radius and $${O_n}$$ be the area of the polygon circumscribing the given circle, prove that $$${I_n} = {{{O_n}} \over 2}\left( {1 + \sqrt {1 - {{\left( {{{2{I_n}} \over n}} \right)}^2}} } \right)$$$Відповідь(B)The formula can be derived using trigonometric identities and geometric relationships between inscribed and circumscribed polygons.
- 16Prove that
$${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$$
$$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$$Відповідь(A)$$\left( {\matrix{ n \\ k \\ } } \right)$$