Q.Find a point on the curve $y = (x - 3)^2$, where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$.

Statement-I. For the curve $y = 3x^{2}-5$, a tangent at $(t,\,3t^{2}-5)$ has slope $y' = 6t$:

Requiring it to pass through $(1,2)$:

The discriminant is $36 - 84 = -48 < 0$, so no real tangent from $(1,2)$ exists. The proposed line $y = 6x - 4$ meets the parabola where $3x^{2} - 6x - 1 = 0$ (discriminant $48 \neq 0$), i.e. it cuts the curve in two points rather than touching it. Hence Statement-I is not correct.

Statement-II. With slope $m = \dfrac{dy}{dx}$ at a point of ordinate $\beta$: …

Concept & Intuition

Mean deviation (from the mean) measures how spread out the data is, but instead of squaring differences (like variance), it uses absolute values. This makes it less sensitive to extreme values and easier to interpret: it tells you, on average, how far each data point is from the center. The key steps: find the mean, then find the average of the absolute distances from that mean.

Step-by-step solution

1. Compute the arithmetic mean Sum the data:

There are 9 numbers, so the mean is

2. Find the absolute deviations from the mean

   For each data point $x_i$, compute $|x_i - 11|$:

   • $|2 - 11| = 9$
   • $|3 - 11| = 8$
   • $|5 - 11| = 6$
   • $|7 - 11| = 4$
   • $|11 - 11| = 0$
   • $|13 - 11| = 2$
   • $|17 - 11| = 6$
   • $|19 - 11| = 8$
   • $|22 - 11| = 11$

3. Sum the absolute deviations

4. Divide by the number of data points Mean deviation = $\frac{54}{9} = 6$. …

Concept and intuition:

The condition $f(x) > 0$ for every real $x$ means the parabola opens upward (coefficient of $x^2$ is $1 > 0$) and never touches or crosses the $x$-axis. That happens exactly when the discriminant is negative. The discriminant will be an expression in $K$, and setting it $< 0$ yields an open interval $(a, b)$ for $K$.

Once we have that interval, $g(K)$ is a quadratic in $K$ — its own graph is a parabola. On an open interval, a parabola can have a minimum (if the vertex lies inside) but never a maximum, because the endpoints are not included.

1. Write the discriminant condition. For $f(x) = x^2 - 2(4K-1)x + g(K)$, the discriminant is

Factor $4$:

The condition $\Delta < 0$ is equivalent to

2. Substitute $g(K)$. Given $g(K) = 15K^2 - 2K - 7$,

So

The inequality becomes

3. Solve the quadratic inequality. Factor: $K^2 - 6K + 8 = (K-2)(K-4)$. The product is negative when $K$ lies between the roots:

Hence $a = 2$, $b = 4$, and the interval is $(2, 4)$.

4. Analyze $g(K)$ on $(2,4)$. $g(K) = 15K^2 - 2K - 7$ is a quadratic with positive leading coefficient ($15 > 0$), so it opens upward. Its vertex is at $$$$

Rolle's theorem has three conditions: the function must be continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and satisfy $f(a)=f(b)$. If any one fails, the theorem does not apply. Here, the function is defined piecewise, so we must check each condition carefully.

1. Check if $f(x)$ is defined everywhere on $[0,2]$

   The function is given explicitly for $0 \leq x \leq 1$ and $1 \leq x \leq 2$. At $x=1$, both pieces give $f(1)=1$, so it is defined at every point. Option (A) is false.

2. Check continuity

   On $[0,1)$, $f(x)=x$ is continuous. On $(1,2]$, $f(x)=2-x$ is continuous. The only potential trouble is at $x=1$.

   Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.

   Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2-x) = 1$.

   Since both limits equal $f(1)=1$, the function is continuous at $x=1$. So $f$ is continuous on $[0,2]$. Option (B) is false.

3. Check endpoint values

   $f(0)=0$, $f(2)=2-2=0$. So $f(0)=f(2)$. Option (C) is false.

4. Check differentiability on $(0,2)$ …