A student Arun is running on a playground along the curve given by y=x+7. Another student Manita standing
at point (3, 7) on playground wants to hit Arun by paper ball when Arun is nearest to Manita.
3.7
Based on above information, answer the following questions
(1) Arun's position at any value of x will be
(a) (xy-7)
(b) (y + 7) (c)(x +7 (d) (x-7)
(ii) Distance (say D) between Arun and Manita will be
(a) (x - 1)(2x + 2x + 3)
(b) (x - 3)2 + x
(c) √(x-3)² + x
(d) VG-1)(2x +2x+3)
(ii) For which real value(s) of x, first derivative of D' wil Y will Vanish?
(a) 1
(b) 2
(c) 3
(d) 4
(iv) Find the position of Arun when Manita will hit the paper hall.
(a) (5, 32)
(b) (1,8)
(c) (3,7)
(d) (3, 16)
(v) The minimum value of D is
(a) 3
(c) s
(d) V5
(b) V3
Answers
Solution :-
given that, arun is running on a playground along the curve given by y = x² + 7.
So,
→ Arun's position at any value of x will be = (x, x² + 7) (c)
now,
→ Arun's position = (x, x² + 7)
→ Manita's postion = (3, 7)
then,
→ Distance (say D) between Arun and Manita will be = √{(x1 - x2)² + (y1 - y2)²}
→ D = √{(x - 3)² + (x² + 7 - 7)²}
→ D = √{(x - 3)² + x⁴} (c)
now,
→ D² = (x - 3)² + x⁴
→ D² = x⁴ + x² - 6x + 9
So, differentiate the following w.r.t. x,
→ d/dx(D²) = 4x³ + 2x - 6
putting first derivative equal to zero we get,
→ 4x³ + 2x - 6 = 0
at x = 1,
→ 4*1 + 2*1 - 6 = 0
→ 6 - 6 = 0
→ 0 = 0
therefore, we can conclude that, at x = 1, first derivative of D² w.r.t x will vanish .
now,
→ The position of Arun when Manita will hit the paper hall = (x, x² + 7) = (1, 1² + 7) = (1 , 8) (b)
and,
→ Minimum value of D (at x = 1) = √(x⁴ + x² - 6x + 9) = √(1 + 1 - 6 + 9) = √5 (d)
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