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JEE Mathematics Solved Question Paper 2011

Short Answer Type

81.

The harmonic mean of two numbers is 4. Their arithmetic mean A and geometric mean G satisfy the relation 2A +G² = 27. Find the numbers.

82.

Find the image of the point (- 8, 12) with respect to the line 4x + 7y + 13 = 0.

83.

How many triangles can be formed by joining 6 points lying on a circle?

84.

If r² = x² + y² + z², then prove that

85.

Determine the sum of imaginary roots of the equation (2x + x - 1) ( 4x² + 2x - 3) = 6

86.

If cos(A) + cos(B) + cos(C) = 0, prove that
cos(3A) + cos(3B) + cos(3C) = 12 cos(A) cos(B)cos(C)

87.
If the area of a rectangle is 64 sq unit, find the minimum value possible for its perimeter

Let the dimensions be a and b.

$∴ Area = ab \Rightarrow 64 = ab$

Perimeter = 2(a + b)

$∴ P = 2 (a + \frac{64}{a}) \Rightarrow \frac{d P}{d a} = 2 (1 - \frac{64}{a^{2}})$

For maxima and minima, put $\frac{d P}{d a} = 0$

$∴ 2 (1 - \frac{64}{a^{2}}) = 0 \Rightarrow a^{2} = 64 \Rightarrow a = \pm 8 Now, \frac{d^{2} P}{d a^{2}} = 2 (+ \frac{128}{a^{3}}) At a = 8, \frac{d^{2} P}{d a^{2}} = 2 (\frac{128}{8^{3}}) = \frac{1}{2} > 0, minima ∴ Minimum value is P (8) = 2 (8 + \frac{64}{8}) = 32$