Manipal MET2015MathematicsApplication of Derivatives

The approximate value of (1.0002)³⁰⁰⁰ is

Options

A1.2
B1.4
C1.6
D1.8

Correct answer

C. 1.6

Step-by-step solution

Let y=x³⁰⁰⁰, x=1 and x+ x=1.0002 Then, x=1.0002-1=0.0002 Also, y=1 , when x=1 Now, y=x³⁰⁰⁰ d y d x =3000 x²⁹⁹⁹ [ d y d x ]_ x=1 =3000 y= d y d x x y=3000 0.0002= 6 10 =0.6 Hence, (1.0002)³⁰⁰⁰=y+ y=1+0.6=1.6

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025 The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025 If f(x)=x e^ x(1-x) , x R , then f(x) is 2025 The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025 If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025 If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025 If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025 If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All Manipal MET PYQs