For option (A),

\(f\) is a polynomial function, it is continuous and differentiable every where.

\(\therefore\) LMVT is applicable in any interval and hence in [–2, 5].

For option (B),

\(f\) is continuous and differentiable for all values of \(x\) having LMVT applicable in [5, 7].

For option (C),

\(f\) is non differentiable and discontinuous at \(x=\dfrac {\pi}{2}\)

\(\therefore\) discontinuous in interval \([0, \pi]\).

\(\therefore\) LMVT is not applicable.

For option (D),

\(f\) is discontinuous at  \(x=0\), but is continuous and differentiable in [5, 9]

\(\therefore\) LMVT is applicable.

Step 1 : Find \(f'(x)\) i.e. derivative of \(f(x)\).

Step 2 : Put\( f ' (x) = \dfrac {f(b) - f(a)}{b-a}\)

Step 3 : Find roots of the above equation, these will be the values of 'c'

\(f(x) = x^2 -2x + 4\) in [1, 5]

To find c, solve

\( f ' (x) = \dfrac {f(5) - f(1)}{5-1}\)

\(\Rightarrow 2x-2=\dfrac {(25-10+4)-(1-2+4)}{4}\)

\(\Rightarrow 2x-2=\dfrac {16}{4}\)

\(\Rightarrow 2x-2=4\)

\(\Rightarrow x=3\)

Since, \(3\in (1, 5)\)

\(\Rightarrow c = 3\)

Apply Newton's Method,

\(\Rightarrow\;x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)} \)

\(f(x)=3x^3+2x-5 \)

\(\Rightarrow f'(x)=9x^2+2\)

\(\therefore\;x_2=x_1-\dfrac {f(x_1)}{f'(x_1)} \) \(\Rightarrow\;x_2=2-\dfrac {f(2)}{f'(2)}\)

\(\Rightarrow\;x_2=2-\dfrac {23}{38}=2-.6052=1.3948\)

\(\therefore\;x_3=x_2-\dfrac {f(x_2)}{f'(x_2)} \)

\(\therefore\;x_3=1.3948-\dfrac {f(1.3948)}{f'(1.3948)} \)

\(\Rightarrow\;x_3=1.3948-\dfrac {5.9302}{19.5092}=1.3948-.3039=1.0909\)

Apply Newton's Method,

\(\Rightarrow\;x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)} \)

To find the value of \(\sqrt [5]{10}\)

Let \(x=10^{1/5}\) , solve the equation 

\(x^5-10=0\)

\(f(x)=x^5-10\\\Rightarrow f(x)=5x^4\)

\(\therefore\;x_2=x_1-\dfrac {f(x_1)}{f'(x_1)} \) (Choose \(x_1=1\))

\(\Rightarrow\;x_2=1-\dfrac {f(1)}{f'(1)}\\ =1-\dfrac {(-9)}{5}\\=\dfrac {14}{5}\\=2.8\)

\(\therefore\;x_3=x_2-\dfrac {f(x_2)}{f'(x_2)} \)

\(=2.8-\dfrac {f(2.8)}{f'(2.8)} \)

\(=2.8-\dfrac {(162.1037)}{307.328}\)

\(=2.8-0.5275\\=2.2725\)

\(\therefore\;x_4=x_3-\dfrac {f(x_3)}{f'(x_3)} \)

\(=2.2725-\dfrac {50.6065}{133.3477}\)

\(=2.2725-0.3795\\=1.893\)

\(f(x) = (x-1)(x-2)(x-3)\) in [0, 4]

To find c, solve

\( f ' (x) = \dfrac {f(4) - f(0)}{4-0}\)

\(\Rightarrow f'(x)=\dfrac {(3×2×1)-(-1×-2×-3)}{4-0}\)

\(\Rightarrow f'(x)=\dfrac {12}{4}\)

\(\Rightarrow f'(x)=3\)

Now, \(f '(x) = (x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3)\)

(Apply product rule for three functions)

\(\Rightarrow\;f '(x) = 3x^2-12x+11\)

\(\therefore\) \( 3x^2-12x+11=3\)

\(\Rightarrow 3x^2-12x+8=0\)

\(\Rightarrow x=\dfrac {12\pm\sqrt {144-4×3×8}}{2×3}\)

\(\Rightarrow x= {2\pm }\dfrac{2}{\sqrt 3}\)

\( {2+}\dfrac{2}{\sqrt 3}\in(0,4)\) and \( {2-}\dfrac{ 2}{\sqrt 3}\in(0,4)\)

Hence, two values of \(c= {2\pm}\dfrac{ 2}{\sqrt 3}\)

By LMVT

\(f'(c)= \dfrac {f(b)-f(a)}{b-a}\),               for some \(c\in(a,b)\)

 \(\Rightarrow \)\(f'(c)= \dfrac {f(8)-f(2)}{8-2}\), for some \(c\in(2,8)\)

\(\Rightarrow\) \(f'(c)= \dfrac {f(8)-7}{6}\), for some \(c\in(2,8)\)

\(\Rightarrow\)\(f(8)=6f'(c)+7\) 

 for some \(c\in(2,8)\)

Now,

\(\Rightarrow f ' (x) \geq 5\\ \Rightarrow\; f ' (c) \geq 5 \\\Rightarrow\; 6 f' (c) \geq 30\)

\(\therefore \; f(8)\geq30+7\)

\(\Rightarrow f(8) \geq 37\)

\(\therefore\) least possible value of \(f(8)\) is 37

If \(x _1= 2\), then \(x _ 2=x_1-\dfrac {f(x_1)}{f'(x_1)}\)

Now,

 \(f'(x_1)=0\)

So, \(x_2\) is undefined.

\(\therefore\)  We cannot prove further in general if any initial approximation is the point of horizontal tangency, we cannot proceed by Newton's Method.

If \('f'\) is a function defined in \([a,b]\) then LMVT is applicable if 

1. It is continuous in  \([a,b]\).

2. It is differentiation in (a,b).

In this case,

 \(f(x) = mx+5\)   if  \(x \leq1\) 

 \(= 4x^2+x+n\)    if \(x\geq 1\) 

This  \(f\) is continuous and differential for all values of \(x \), except perhaps at \(x = 1\) .

(It is a polynomial everywhere)

For continuity at  \(x = 1 \to\)  \(R.H.L. = L.H.L. = f(1)\)

\(R.H.L. = \lim\limits_{x\to1^+} f(x) = \lim\limits_{x\to 1^+} 4x^2+x+n = 5+n\)

\(L.H.L. = \lim\limits_{x\to1^-} f(x) = \lim\limits_{x\to 1^-} (mx+5) = m+5\)

\(\Rightarrow f(1) = m+5\)

\(\therefore 5+n = m+5\\ \Rightarrow n=m\)

For differentiable  at \(x =1, \,\,R.H.D. = L.H.D.\)

\(R.H.D. = 8+1=9\)

\(L.H.D. = m\)

\(\therefore 9=m\)

\(\therefore n=m=9\)