CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Maxima & Minima.

Q.3. Find the point on the curve y² = 4x which is nearest to the point (2, –8).

Solution :

Let P(t2, 2t) be the point on y² = 4x nearest to the point A(2, –8). Then
AP² = (t² – 2)² + (2t + 8)² = y (say)
Therefore, y = (t² – 2) + (2t + 8)
= (t4 – 4t² + 4) + (4t² + 32t + 64)
= t⁴ + 3²t + 68.
Or, dy/dt = 4t³ + 3² and d²y/dt² = 12t².
For maximum and minimum value of y i.e. of AP, dy/dt = 0
Or, 4(t³ + 8) = 0
Or, (t + 2)(t2 – 2t + 4) = 0
Or, t = –2 = 0
As t2 – 2t + 4 = 0 has imaginary values of t.
For, t = – 2, d²y/dt² = 12 (– 2)² = 48 > 0
Hence y is minimum => AP is minimum.
Thus the point P(t², 2t) = P(4, – 4) is the nearest point to the point (2, – 8).
Therefore, the nearest point is (4, – 4) [Ans.]

Q.4. Find the largest possible area of the right-angled triangle whose hypotenuse is 5 cm.

Solution :

.
Fig.

Let ABC be a right angled triangle whose hypotenuse AC = 5 cm. Then
AB = 5 cos θ; BC = 5 sin θ
Therefore, A = Area of ∆ABC = 1/2×AB×BC = 25/2.cos θsin θ = 25/4 sin2θ
Therefore, dA/dθ = 25/2.cos 2θ and d2A/dθ2 = – 25 sin2θ
For maximum or minimum dA/dθ = 0
=> cos2θ = 0
=> 2θ = π/2
=> θ = π/4.
d²A/dθ2]at θ = π/4 = – 25 < 0.
Therefore, A is maximum, when θ = π/4.
Largest of the triangle = 25/4 sin π/2
= 25/4 sq. units. [Ans.]

Paper By Mr. M.P.Keshari
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