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1
IIT-JEE 1990
MCQ (Single Correct Answer)
+2
-0.5
The equation $$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0\,$$ In the variable x, has real roots. Then p can take any value in the interval
A
$$\left( {0,2\pi } \right)\,$$
B
$$\left( { - \pi ,0} \right)\,\,\,$$
C
$$\left[ { - {\pi \over 2},{\pi \over 2}} \right]\,$$
D
$$\left( {0,\pi } \right)$$
2
IIT-JEE 1990
Fill in the Blanks
+2
-0
If $$\int {{{4{e^x} + 6{e^{ - x}}} \over {9{e^x} - 4{e^{ - x}}}}\,dx = Ax + B\,\,\log \left( {9{e^{2x}} - 4} \right) + C,} $$ then
$$A = .....,B = .....$$ and $$C = .....$$
3
IIT-JEE 1990
Subjective
+4
-0
Let $${z_1}$$ = 10 + 6i and $${z_2}$$ = 4 + 6i. If Z is any complex number such that the argument of $${{(z - {z_1})} \over {(z - {z_2})}}\,is{\pi \over 4}$$ , then prove that $$\left| {z - 7 - 9i} \right| = 3\sqrt 2 $$.
4
IIT-JEE 1990
Fill in the Blanks
+2
-0
If $$\,x < 0,\,\,y < 0,\,\,x + y + {x \over y} = {1 \over 2}$$ and $$(x + y)\,{x \over y} = - {1 \over 2}$$, then x =..........and y =.........

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