Which of the following are NOT correct regarding the equation of tangent and normal to the curve $y =\frac{x-11}{(x-2)(x-3)}$ at the point, where it cuts the x-axis?

(A) The point of contact is (11, 0).
(B) The equation of tangent is $x - 72y-11 = 0$.
(C) The equation of normal is $72x+y-11=0$
(D) The slope of the tangent at the given point of contact is $\frac{1}{88}$.

Choose the correct answer from the options given below:

(C) and (D) only

The correct answer is Option (4) → (C) and (D) only

Given curve:

$y=\frac{x-11}{(x-2)(x-3)}$

Where it cuts the x–axis: $y=0$

$\frac{x-11}{(x-2)(x-3)}=0 \Rightarrow x-11=0$

$x=11$

So point of contact = $(11,0)$ → (A) is correct.

Now find slope.

$y=\frac{x-11}{(x-2)(x-3)}$

Differentiate:

Let $u=x-11$, $v=(x-2)(x-3)=x^{2}-5x+6$

$y'=\frac{u'v-uv'}{v^{2}}$

$u'=1$

$v'=2x-5$

$y'=\frac{(x^{2}-5x+6)-(x-11)(2x-5)}{(x^{2}-5x+6)^{2}}$

Evaluate at $x=11$:

Slope of tangent:

$m = \frac{72}{5184}=\frac{1}{72}$

Given option says slope = $\frac{1}{88}$ → (D) is NOT correct.

Tangent at $(11,0)$:

$y = \frac{1}{72}(x-11)$

$72y = x-11$

$x - 72y - 11 = 0$ → matches (B), so (B) is correct.

Normal slope = $-72$

Normal equation:

$y-0 = -72(x-11)$

$y = -72x + 792$

$72x + y - 792 = 0$

Given option says $72x + y - 11 =0$ → wrong constant → (C) is NOT correct.

NOT correct statements: (C) and (D)