Engineering Mathematics Test 5

Explanation:

Given:

f(x) = e^x, g(x) = e^-x

Now,

derivative of f(x) i.e. f^’(x) = e^x

derivative of g(x) i.e. g^’(x) = - e^x

range is given as [2, 3], that is, a = 2 and b = 3

here, f(x) and g(x) are differentiable and continuous and derivative of g(x) is not equal to 0.

They are satisfying the conditions of Cauchy’s mean value theorem.

Now,

Consider it is in interval of [a, b].

\(\frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}}\)

\(\frac{{{e^c}}}{{ - {e^{ - c}}}} = \frac{{{e^b} - {e^a}}}{{{e^{ - b}} - \;{e^{ - a}}}}\)

\(- {e^{2c}} = \frac{{{e^3} - {e^2}}}{{{e^{ - 3}} - \;{e^{ - 2}}}} - \; = \; - {e^{3 + 2}}\)

Concept:

\(\frac{d}{{dx}}\left( {f\left( x \right).g\left( x \right)} \right) = f\left( x \right)g'\left( x \right) + f'\left( x \right)g\left( x \right)\)

Calculation:

f(x) = x cos x   (1)

f'(x) = -xsinx + cosx

f''(x) = -(xcosx + sinx) - sinx 

∴ f''(x) = -xcosx - 2sinx  (2)

g”(x) + g(x) + k sin x = 0

Substituting the value from (1) and (2)

(-xcosx - 2sinx ) +  x cos x +  k sin x = 0

- 2sinx + k sin x = 0

sinx (k - 2) = 0

∴ sin x = 0 OR k - 2 = 0

\(\sin {\rm{x}} = \left( {{\rm{n}} } \right)\left( {\frac{{\rm{\pi }}}{2}} \right){\rm{\;where\;n}} = 0,1,{\rm{\;}}2,3,4 \ldots {\rm{\;}}\)

Since value of k is asked

The following properties are true in calculus:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity
• Continuity does not necessarily imply differentiability.

Hence,

Statement P is wrong.

Statement Q is right

Statement R is Right

f(y) = y3 - 7y2 + 5

The auxiliary function is f₁(y) = y3 - 7y2 = y2(y - 7)

f(0) = f(7)=5.

The function f(x) is everywhere continuous and differentiable because it is a cubic polynomial.

Consequently, it satisfies all the conditions of Rolle’s theorem on the interval [0,7].

Therefore the value q is 7.

Since given function is continuous and differentiable then by Lagrange’s Mean-Value Theorem.

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

\(a = - \frac{{11}}{7},\;b = \frac{{13}}{7}\)

\(f\left( x \right) = {x^3} - 3x - 1\)

\(f'\left( x \right) = 3{x^2} - \;3\)

\({f^{'}}\left( c \right) = 3{c^2} - \;3\)

\(f\left( b \right) = {\left( { - \frac{{11}}{7}} \right)^3} - 3\left( { - \frac{{11}}{7}} \right) - 1 = \; - \frac{{57}}{{343}}\)

\(f\left( a \right) = {\left( {\frac{{13}}{7}} \right)^3} - 3\left( {\frac{{13}}{7}} \right) - 1 = \; - \frac{{57}}{{343}}\)

\(3{c^2} - \;3 = \frac{{ - \;\frac{{57}}{{343}} - \left( { - \;\frac{{57}}{{343}}} \right)}}{{ - \frac{{11}}{7} - \frac{{13}}{7}}}\)

\(3{c^2} - \;3 = 0\)

\({c^2} = 1\)

f(x) = 3x3 – 40.5x2 + 180x + 7

f’(x) = 9x2 – 81x + 180 = 0

put f’(x) = 0 to find the point at which maximum and minimum value exists

9x2 – 81x + 180 = 0

x2 – 9x + 20 = 0

(x – 5)(x – 4 ) = 0

∴ x = 5 or x =4

Interval [4, 5]

f'’(x) = 2x – 9

put x = 4

f’’(x) = -1 < 0 (maximum value might exist)

put x = 5

f’’(x) = 1 > 0 (minimum value might exist)

Also check border values:

The minimum value is 269.5

Rolle's theorem.

To apply Rolle's theorem following 3 conditions should be satisfied:

f(x) should be continuous in interval [a, b],

f(x) should be differentiable in interval (a, b), and

f(a)=f(b)

If these 3 conditions are satisfied simultaneously then, there exists at least one ′x′ such that f′(x)=0

Explanation:

Statement I: There exists \(\theta \in \left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) such that f’(θ) = 0

For the given question, it satisfies all the three conditions, therefore there exists at least one θ that gives f′(θ)=0

Therefore, statement II is true

Statement II: There exists \(\theta \in \left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) such that f’(θ) ≠ 0  

The given function is also not a constant function therefore for some θ, f′(θ) ≠ 0

For Lagrange’s mean value theorem, there exists at least one real number ‘c’ in (a, b) such that

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)     ---(1)

Now,

f(a) = f(0) = 0

\(f\left( b \right) = f\left( {\frac{1}{2}} \right) = \frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) = \frac{3}{8}\)

\(f'\left( x \right) = x\left( {{x^2} - 3x + 2} \right) = {x^3} - 3{x^2} + 2x\)

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

\(3{c^2} - 6c + 2 = \frac{{\frac{3}{8} - 0}}{{\frac{1}{2} - 0}}\)

\(3{c^2} - 6c + 2 = \frac{3}{4}\)

12c² – 24c + 8 = 3

12c² – 24c + 5 = 0

\(c = \frac{{24 \pm \sqrt {{{24}^2} - 12 \times 5 \times 4} }}{{2 \times 12}}\)

c = 1 ± 0.764 = 1.764 or 0.236

But, c = 0.236, since it only lies between 0 and 1/2

By Cauchy's Mean Value theorem we should have

\(\rm \frac {f'(c)}{g'(c)}=\frac {f(b)-f(a)}{g(b)-g(a)}\)

\(\frac{{f'\left( x \right)}}{{g'\left( x \right)}} = \frac{{f\left( 2 \right) - f\left( 1 \right)}}{{g\left( 2 \right) - g\left( 1 \right)}}\)

\(\frac{{8g'\left( x \right)}}{{g'\left( x \right)}} = \frac{{16\; - 4}}{{4\; - g\left( 1 \right)}}\)

\(\frac{{8g'\left( x \right)}}{{g'\left( x \right)}} = \frac{{16\; - 4}}{{4\; - g\left( 1 \right)}}\)

\(8 = \frac{{12}}{{4\; - g\left( 1 \right)}}\)

\(32\; - 8g\left( 1 \right) = 12\)

\(8g\left( 1 \right) = 20\)

\(\therefore g\left( 1 \right) = \frac{5}{2}\)