MATHS M.C.Q (1 Marks)

1. Increasing.

3. $\text{a}\leq-\frac{1}{2}$

2. Odd and increasing.

3. $\lambda>\frac{1}{2}$

$f(x) = 3x^2 - 27x$
$\Rightarrow f'(x) = x^3 - 27x + 5$
$\Rightarrow f'(x) = 3(x^2 - 9)$
Function is increasing,
$3\big(\text{x}^2-9\big)\geq0$
$\Rightarrow\text{x}^2\geq9$
$\Rightarrow|\text{x}|\geq3$

$f(x) = kx^3 - 9x^2 + 9x + 3$
$f'(x) = kx^2 - 27$
$= 3(x^2 - 9)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$\Rightarrow 3(x^2 - 9) > 0$
$\Rightarrow (x^2 - 9) > 0 [$Since, $3 > 0, 3(x^2 - 9) > 0\Rightarrow (x^2 - 9) > 0]$
$\Rightarrow (x + 3)(x - 3) > 0$
$\Rightarrow x < -3$ or $x > 3$
$\Rightarrow |x| > 3$

2. a > 1

2. (1, 3)

4. $\lambda>2$

3. $\text{a}=\frac{1}{2}$

$f(x) = 2x^3 - 9x^2 + 12x + 29$
$\Rightarrow f'(x) = 6x^2 - 18x + 12$
$\Rightarrow f'(x) = 6(x^2 - 3x + 2)$
$\Rightarrow f'(x) = 6(x - 1)(x - 2)$
For $f(x)$ to be decreasing, we must have
$ f'(x) < 0$
$\Rightarrow 6(x - 1)(x - 2) < 0$
$\Rightarrow (x - 1)(x - 2) < 0$
$[$Since, $6 > 0, 6(x - 1)(x - 2) < 0 \Rightarrow (x - 1)(x - 2) < 0]$
$\Rightarrow 1 < x < 2$
So, $f(x)$ is decreasing for $1 < x < 2.$

3. $(-\infty,\infty)$

3. $\text{x}\in\text{R}$

Given, $\text{f}(\text{x})=\text{x}^{\text{x}}$
Applying log with base e on both sides, we get
$\log(\text{f}(\text{x}))=\text{x}\log_\text{e}\text{x}$
$\frac{\text{f}'(\text{x})}{\text{f}(\text{x})}=1+\log_{\text{e}}\text{x}$
$\text{f}'(\text{x})=\text{f}(\text{x})(1+\log_\text{e}\text{x})$
$=\text{x}^\text{x}(1+\log_\text{e}\text{x})$
For $f(x)$ to be decreasing, we must have
$\text{f}'(\text{x})<0$
$\Rightarrow\text{x}^\text{x}(1+\log_\text{e}\text{x})<0$
Here, logarithmic function is defined for positive values of $x.$
$\Rightarrow\text{x}^\text{x}>0$
$\Rightarrow1+\log_\text{e}\text{x}<0$
$[\text{Since }\text{x}^\text{x}(1+\log_\text{e}\text{x})<0\Rightarrow1+\log_\text{e}\text{x}<0]$
$\Rightarrow\log_\text{e}\text{x}<-1$
$\Rightarrow\text{x}<\text{e}^{-1}$
$\big[\because\ \log_\text{a}\text{x}<\text{N}\Rightarrow\text{a}^\text{N}\text{ for }\text{a}>1\big]$
Here,
$\text{e}>1$
$\Rightarrow\log_\text{e}\text{x}<-1$
$\Rightarrow\text{x}<\text{e}^{-1}$
$\Rightarrow\text{x}\in\big(0,\text{e}^{-1}\big)$
So, $f(x)$ is decreasing on $\Big(0,\frac{1}{\text{e}}\Big).$

$\text{f}(\text{x})=\text{x}^9+3\text{x}^7+64$
$\text{f}'(\text{x})=9\text{x}^8+21\text{x}^6>0,\forall\ \text{x}\in\text{R}$
So, $f(x)$ is increasing on $R.$

2. Monotonically decreasing.

1. Stritcly increasing.

1. Increasing on $\Big(0,\frac{\pi}{2}\Big)$

$\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$
$\text{f}'(\text{x})=3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})$
Given, $f(x)$ is increasing on $R.$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})>0,\forall\ \text{x}\in\text{R}$
Since, the quadratic function is $ > 0,$ its discriminant is $ < 0.$
$\Rightarrow(2\text{a})^2-4(3)(\text{b}+5\sin2\text{x})<0$
$\Rightarrow4\text{a}^2-12\text{b}-60\sin2\text{x}<0$
$\Rightarrow\text{a}^2-3\text{b}-15\sin2\text{x}<0$
We know that the minimum value of $\sin2\text{x}$ is $ -1.$
$\therefore\ \text{a}^2-3\text{b}-15<0$

$\text{f}(\text{x})=\text{a}^\text{x}$
$\text{f}'(\text{x})=\text{a}^\text{x}\log\text{a}$
Given: $f(x)$ is increasing on $R.$
$\Rightarrow\text{f}'(\text{x})>0$
$\Rightarrow\text{a}^\text{x}\log\text{a}>0$
$\Rightarrow\text{a}^\text{x}>0$
$($Logarithmic function is defined for positive value of $a)$
We know,
$\Rightarrow\text{a}^\text{x}\log\text{a}>0$
It can be possible when $\text{a}^\text{x}>0$ and $\log\text{a}>0$ or $\text{a}^\text{x}<0$ and $\log\text{a}<0$
$($Not possible, logarithmic function is defined for positive value of $a)$
$\Rightarrow\log\text{a}>0$
$\Rightarrow\text{a}>1$
So, $f(x)$ is increasing when $a > 1.$

2. Decreases on [0, a]

$f(x) = x^3 - 6x^2 +15x + 3$
$f'(x) = 3x^2 - 12x + 15$
$= 3(x^2 - 4x + 5)$
$= 3(x^2 - 4x + 4 + 1)$
$=3(\text{x}-2)^2+\frac{1}{3}>0$
Therefore, $f(x)$ is strictly increasing function.
$\Rightarrow f^{-1}(x)$ exists.
Hence, $f(x)$ is an invertible function.

$f(x) = 2x^2 - kx + 5$
$f'(x) = 4x - k$
$f(x)$ is increasing
$4x - k < 0$ on $[1, 2]$
$k < 4x$
Minimum value of $k$ is $4.$
$k < 4$
$\text{k}\in(-\infty,4)$

$f(x) = x^2e^{-x}$
$\Rightarrow f'(x) = -x^2e^{-x} + 2xe^{-x}$
$\Rightarrow f'(x) = -e^{-x}x(x - 2)$
Given that function is monotonically increasing.
$-e^{-x}x(x - 2) > 0$
$x(x - 2) < 0$
$0 < x < 2$