Given \(f(x)=|\frac{1}{x}|, x\in [-1,1]\)
Rolle's theorem is not applicable since \(f(x)=(\frac{1}{x})\) is not continuous at x = 0 in [-1, 1] and not differentiable in (-1, 1)

Given x = a cos³ t; y = b sin³ t
\(\frac { dx }{ dt } =-3a \ { cos }^{ 2 } \ t \ sin \ t\)
\(\frac { dy }{ dt } =3b \ { sin }^{ 2 } \ t \ cot \ t\)
\(\therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\(\frac { 3b{ sin }^{ 2 }tcost }{ 3a{ cos }^{ 2 }tsint } =\frac { -b }{ a } \) tan t
Slope of the tangent at t = \(\frac { \pi }{ 2 } \) is
\({ m=\left( \frac { dy }{ dx } \right) }_{ t=\frac { \pi }{ 2 } }\)
\(\frac { -b }{ a } tan\frac { \pi }{ 2 } =\frac { -b }{ a } \times \infty =\infty \)
\(\therefore m=\infty \)

By the mean value theorem we have, there exists 'c'∈(2, 7) such that,
\(\frac { f(7)-f(2) }{ 7-2 } \) = f'(c) ≤ 29
Hence, f(7) ≤ 5× 29 +17 = 162
Therefore, the maximum value of f (7) is 162.

Let f (t) be the distance travelled by the trucker in 't' hours. This is a continuous function in [0, 2] and differentiable in (0, 2). Now, f (0) = 0 and f (2) =164. By an application of the Mean Value Theorem, there exists a time c such that, \(f'(c)=\frac{164-0}{2-0}=82>80\)
Therefore at some point of time, during the travel in 2 hours the trucker must have travelled with a speed more than 80 km which justifies the issuance of a speed violation ticket.

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).