First slide

Theory of expressions

Question

Let $f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ such that minimum value of f(x) for $x \in [0, 2]$ is equal to 3.

Moderate

Question

Number of values of a for which global minimum value, that is equal to 3 for $x \in [0, 2]$ , occurs at the end point of interval [0, 2] is

A

1
B

2
C

3
D

0

Solution

$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ represents a parabola whose vertex is at $(\frac{a}{2}, 2 - 2 a)$ .
Case I: $0 < \frac{a}{2} < 2$
In this case, f(x) will attain the minimum value at $x = \frac{a}{2}$ .
Thus, $f (\frac{a}{2}) = 3$ .
$\begin{array}{l} \Rightarrow 3 = - 2 a + 2 \\ \Rightarrow a = - \frac{1}{2} (Rejected) \end{array}$
Case II: $\frac{a}{2} \geq 2$
In this case, f(x) attains the global minimum value at x = 2.
Thus, f(2) = 3.
$\begin{array}{l} \Rightarrow 3 = 16 - 8 a + a^{2} - 2 a + 2 \\ \Rightarrow a = 5 \pm \sqrt{10} \\ So, a = 5 + \sqrt{10} \end{array}$
Case III: $\frac{a}{2} \leq 0$
In this case, f(x) attains the global minimum value at x = 0.
Thus, f(0) = 3.
$\begin{array}{l} \Rightarrow 3 = a^{2} - 2 a + 2 \\ \Rightarrow a = 1 \pm \sqrt{2} \\ So, a = 1 - \sqrt{2} \end{array}$
Hence, permissible values of a are $1 - \sqrt{2} and 5 + \sqrt{10}$ .
$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ is monotonic in [0, 2].
Hence, point of minima of function should not lie in [0, 2].
Now, f'(x) = 0
$\begin{array}{l} \Rightarrow 8 x - 4 a = 0 \\ \Rightarrow x = a / 2 \end{array}$
If $\frac{a}{2} \in [0, 2], then a \in [0, 4]$ .
For f(x) to be monotonic in $[0, 2], a \notin [0, 4]$ . So, $a \leq 0 or a \geq 4$ .

Question

Number of values of a for which global minimum value, that is equal to 3 for $x \in [0, 2]$ , occurs for the value of x lying in (0, 2) is

A

1
B

2
C

3
D

0

Solution

$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ represents a parabola whose vertex is at $(\frac{a}{2}, 2 - 2 a)$ .
Case I: $0 < \frac{a}{2} < 2$
In this case, f(x) will attain the minimum value at $x = \frac{a}{2}$ .
Thus, $f (\frac{a}{2}) = 3$ .
$\begin{array}{l} \Rightarrow 3 = - 2 a + 2 \\ \Rightarrow a = - \frac{1}{2} (Rejected) \end{array}$
Case II: $\frac{a}{2} \geq 2$
In this case, f(x) attains the global minimum value at x = 2.
Thus, f(2) = 3.
$\begin{array}{l} \Rightarrow 3 = 16 - 8 a + a^{2} - 2 a + 2 \\ \Rightarrow a = 5 \pm \sqrt{10} \\ So, a = 5 + \sqrt{10} \end{array}$
Case III: $\frac{a}{2} \leq 0$
In this case, f(x) attains the global minimum value at x = 0.
Thus, f(0) = 3.
$\begin{array}{l} \Rightarrow 3 = a^{2} - 2 a + 2 \\ \Rightarrow a = 1 \pm \sqrt{2} \\ So, a = 1 - \sqrt{2} \end{array}$
Hence, permissible values of a are $1 - \sqrt{2} and 5 + \sqrt{10}$ .
$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ is monotonic in [0, 2].
Hence, point of minima of function should not lie in [0, 2].
Now, f'(x) = 0
$\begin{array}{l} \Rightarrow 8 x - 4 a = 0 \\ \Rightarrow x = a / 2 \end{array}$
If $\frac{a}{2} \in [0, 2], then a \in [0, 4]$ .
For f(x) to be monotonic in $[0, 2], a \notin [0, 4]$ . So, $a \leq 0 or a \geq 4$ .

Question

Values of a for which f(x) is monotonic for $x \in [0, 2]$ are given by

A

$a \leq 0 or a \geq 4$
B

$0 \leq a \leq 4$
C

a > 0
D

none of these

Solution

$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ represents a parabola whose vertex is at $(\frac{a}{2}, 2 - 2 a)$ .
Case I: $0 < \frac{a}{2} < 2$
In this case, f(x) will attain the minimum value at $x = \frac{a}{2}$ .
Thus, $f (\frac{a}{2}) = 3$ .
$\begin{array}{l} \Rightarrow 3 = - 2 a + 2 \\ \Rightarrow a = - \frac{1}{2} (Rejected) \end{array}$
Case II: $\frac{a}{2} \geq 2$
In this case, f(x) attains the global minimum value at x = 2.
Thus, f(2) = 3.
$\begin{array}{l} \Rightarrow 3 = 16 - 8 a + a^{2} - 2 a + 2 \\ \Rightarrow a = 5 \pm \sqrt{10} \\ So, a = 5 + \sqrt{10} \end{array}$
Case III: $\frac{a}{2} \leq 0$
In this case, f(x) attains the global minimum value at x = 0.
Thus, f(0) = 3.
$\begin{array}{l} \Rightarrow 3 = a^{2} - 2 a + 2 \\ \Rightarrow a = 1 \pm \sqrt{2} \\ So, a = 1 - \sqrt{2} \end{array}$
Hence, permissible values of a are $1 - \sqrt{2} and 5 + \sqrt{10}$ .
$f (x) = 4 x^{2} - 4 ax + a^{2} - 2 a + 2$ is monotonic in [0, 2].
Hence, point of minima of function should not lie in [0, 2].
Now, f'(x) = 0
$\begin{array}{l} \Rightarrow 8 x - 4 a = 0 \\ \Rightarrow x = a / 2 \end{array}$
If $\frac{a}{2} \in [0, 2], then a \in [0, 4]$ .
For f(x) to be monotonic in $[0, 2], a \notin [0, 4]$ . So, $a \leq 0 or a \geq 4$ .

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