Question

1] Find the Area bounded by the curve y = e⁻ˣ the x - axis and the y - axis.

a) 5

b) 7

c) 3

d) 1

2] The point moves such that it's displacement as a function of time is given by x³ = t³ + 1 . It's acceleration as a function of time t will be

a) $\large\sf{\dfrac{2}{x^5}}$

b) $\large\sf{\dfrac{2t}{x^5}}$

c) $\large\sf{\dfrac{2t}{x^4}}$

d) $\large\sf{\dfrac{2t^2}{x^5}}$

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Answer 1

$\bold\blue{QUESTION \: 1}$

Find the Area bounded by the curve y = e⁻ˣ the x - axis and the y - axis.

GIVEN :

$\sf y \: = \: e^{-x}$

SOLUTION :

When $\sf x \: = \: 0, \: then \: y \: = \: e^{-0}$

x increases and y decreases and only at $\sf x \: = \: \infty, y \: = \: 0$

$\sf Area \: = \: \sf{\int\limits^o_\infty e^{-x}dx = - [ - e^{-x}]^\infty_0 = 1}$

$\bold\blue{QUESTION \: 2}$

The point moves such that it's displacement as a function of time is given by x³ = t³ + 1 . It's acceleration as a function of time t will be

GIVEN :

$\sf x^{3} \: = \: t^{3} \: + \: 1$

SOLUTION :

$\sf x^{3} \: = \: t^{3} \: + \: 1$

=> $\sf x \: = \: (t^{3} \: + \: 1)^{1/3}$

We know that,

$\sf Velocity, \: v \: = \: \dfrac {dx}{dt}$

=> $\sf \dfrac {d}{dt} \: (t^{3} \: + \: 1)^{1/3}$

=> $\sf \dfrac {1}{3} \: (t^{3} \: + \: 1)^{-2/3} \times \: 3t^{2}$

=> $V = t^{2} \times \: (t^{3} \: + \: 1)^{ \frac{ - 2}{3}}$

Now,

The acceleration(a) is given that a = $\sf \dfrac {dv}{dt}$

=> $\sf \dfrac {d}{dt} \: (t^{2} \: (t^{3}) \: + \: 1)^{-2/3}$

=> $\sf 2t \: \times \: (t^{3} \: + \: 1)^{-2/3} \: + \: t^{2} \: \times \: \dfrac {-2}{3} \: (t^{3} \: + \: 1)^{-5/3} \: \times \: 3t^{2}$

=> $\sf 2tx^{-2} \: - \: 2t^{4} \: x^{-5}$

=> $\sf 2t \: (x^{-2} \: - \: t^{3} \: \times \: x \: - \: 5)$

=> $\sf\:2t\:\bigg(\dfrac{1}{x^{2}}\:- {x}^{3} -\dfrac{1}{x^{5}}\bigg)$

=> $\sf a \: = \: \dfrac {2t}{x^3}$

Answer 2

1] Option (d) → 1

2] Option↓

2t² / x³

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