Maths M.C.Q (1 Marks)

2. Decreases on [0, a]

Solution:

$\phi(\text{x})=\text{f}(\text{x})+\text{f}(2\text{a}-\text{x})$

$\phi'(\text{x})=\text{f}'(\text{x})-\text{f}'(2\text{a}-\text{x})$

$\text{f}''(\text{x})>0$ as $\text{f}'(\text{x})>0$

Considering $\text{x}\in[0,\text{a}]$

$\text{x}\leq2\text{a}-\text{x}$

$\text{f}'(\text{x})\leq\text{f}(2\text{a}-\text{x})$

Also, $\phi(\text{x})=\text{f}'(\text{x})-\text{f}'(2\text{a}-\text{x})$

$\phi(\text{x})$ is decreasing on [0, a]

3. f(x) is invertible

Solution:

f(x) = x³ - 6x² +15x + 3

f'(x) = 3x² - 12x + 15

= 3(x² - 4x + 5)

= 3(x² - 4x + 4 + 1)

$=3(\text{x}-2)^2+\frac{1}{3}>0$

Therefore, f(x) is strictly increasing function.

⇒ f^-1(x) exists.

Hence, f(x) is an invertible function.

1. Increasing.

Solution:

$\text{f}(\text{x})=\frac{-\text{x}}{2}+\sin\text{x}$ defined on $\Big[\frac{-\pi}{3},\frac{\pi}{3}\Big]$

$\therefore\ \text{f}'(\text{x})=\frac{-1}{2}+\cos\text{x}$

$\Rightarrow\text{f}'(\text{x})\geq0,\forall\ \text{x}\in\Big[\frac{-\pi}{3},\frac{\pi}{3}\Big]$

$\Big[\because\ \text{for }\text{x}\in\Big[\frac{-\pi}{3},\frac{-\pi}{3}\Big],\cos\geq\frac{1}{2}\Big]$

Hence, the given function is increasing.

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

Solution:

Given:

$\text{f}(\text{x})=\cos|\text{x}|-2\text{ax}+\text{b}$

Now, $|\text{x}|=\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}$

And $\cos|\text{x}|=\begin{cases}\cos(\text{x}),&\text{x}\geq0\\\cos(-\text{x})=\cos(\text{x}),&\text{x}<0\end{cases}$

$\therefore\ \cos|\text{x}|=\cos\text{x},\forall\ \text{x}\in\text{R}$

$\therefore\ \text{f}(\text{x})=\cos\text{x}-2\text{ax}+\text{b}$

$\Rightarrow\text{f}'(\text{x})=-\sin\text{x}-2\text{a}$

It is given that f(x) is increasing.

$\Rightarrow\text{f}'(\text{x})\geq0$

$\Rightarrow-\sin\text{x}-2\text{a}\geq0$

$\Rightarrow\sin\text{x}+2\text{a}\leq0$

$\Rightarrow2\text{a}\leq-\sin\text{x}$

The least value of $-\sin\text{x}$ is -1.

$\Rightarrow2\text{a}\leq-1$

$\Rightarrow\text{a}\leq\frac{-1}{2}$

2. Odd and increasing.

Solution:

$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$

$\Rightarrow\text{f}(-\text{x})=\log_\text{e}\Big(-\text{x}^3+\sqrt{\text{x}^6+1}\Big)$

$=\log_\text{e}\Bigg\{\frac{\big(-\text{x}^3+\sqrt{\text{x}^6+1}\big)\big(\text{x}^3+\sqrt{\text{x}^6+1}\big)}{\text{x}^3+\sqrt{\text{x}^6+1}}\Bigg\}$

$=\log_\text{e}\Big(\frac{\text{x}^6+1-\text{x}^6}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$

$=\log_\text{e}\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$

$=-\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$

$=-\text{f}(\text{x})$

Hence, f(-x) = -f(x)

Therefore, it is an odd function.

$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$

$\frac{\text{d}}{\text{dx}}\{\text{f}(\text{x})\}=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Big(3\text{x}^2+\frac{1}{2\sqrt{\text{x}^6+1}}\times6\text{x}^5\Big)$

$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\bigg(\frac{6\text{x}^2\sqrt{\text{x}^6+1}+6\text{x}^5}{2\sqrt{\text{x}^6+1}}\bigg)$

$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Bigg\{\frac{6\text{x}^2\big(\sqrt{\text{x}^6+1+\text{x}^3}\big)}{2\sqrt{\text{x}^6+1}}\Bigg\}$

$=\Big(\frac{6\text{x}^2}{2\sqrt{\text{x}^6+1}}\Big)>0$

Therefore the given function is an increasing function.

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

Solution:

$\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$

$\text{f}'(\text{x})=-\sin\text{x}-2\lambda$

For f(x) to be decreasing, we must have

$\text{f}'(\text{x})<0$

$\Rightarrow-\sin\text{x}-2\lambda<0$

$\Rightarrow\sin\text{x}+2\lambda>0$

$\Rightarrow2\lambda>-\sin\text{x}$

We know that the maximum value of $-\sin\text{x}$ is 1.

$\Rightarrow2\lambda>1$

$\Rightarrow\lambda>\frac{1}{2}$

3. $\Big(0,\frac{1}{\text{e}}\Big)$

Solution:

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 $\Big(0,\frac{1}{\text{e}}\Big).$

1. $\text{R}$

Solution:

$\text{f}(\text{x})=\text{x}^9+3\text{x}^7+64$

$\text{f}'(\text{x})=9\text{x}^8+21\text{x}^6>0,\forall\ \text{x}\in\text{R}$

So, f(x) is increasing on R.

2. Monotonically decreasing.

Solution:

f(x) = 2|x - 1| + 3|x - 2|

$\text{x}\in(1,2)$

x > 1 and x < 2

⇒ x - 1 > 0 and x - 2 < 0

⇒ f(x) = 2|x - 1| + 3|x - 2|

⇒ f(x) = 2(x - 1) - 3(x - 2)

⇒ f(x) = 2x - 2 - 3x + 6

⇒ f(x) = -x + 4

⇒ f'(x) = -1

Hence, function is monotonically decreasing.

1. $-1\leq\text{k}\leq1$

Solution:

f(x) = x³ - 9kx² + 27x + 30

⇒ f'(x) = 3x³ - 18kx + 27

⇒ 3(x² - 6kx + 9)

Function is always increasing on R.

3(x² - 6kx + 9) > 0

x² - 6kx + 9 > 0

In ax² + bx + c = 0 if a > 0 ⇒ b² - 4ac < 0

36k² - 36 < 0

k² - 1 < 0

(k + 1)(k - 1) < 0

⇒ -1 < k < 1

1. Stritcly increasing.

Solution:

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

Case I:

When x > 0, |x| = x

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

$=\frac{\text{x}}{1+\text{x}}$

$\Rightarrow\text{f}'(\text{x})=\frac{(1+\text{x})1-\text{x}(1)}{(1+\text{x})^2}$

$=\frac{1}{(1+\text{x})^2}>0,\forall\ \text{x}\in\text{R}$

So, f(x) is strictly increasing when x > 0.

Case II:

When x < 0, |x| = -x

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$

$=\frac{\text{x}}{1+\text{x}}$

$\Rightarrow\text{f}'(\text{x})=\frac{(1-\text{x})1-\text{x}(-1)}{(1-\text{x})^2}$

$=\frac{1}{(1-\text{x})^2}>0,\forall\ \text{x}\in\text{R}$

So, f(x) is strictly increasing when x < 0.

Thus, f(x) is strictly increasing on R.

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

Solution:

Given: g(x) is increasing on $\Big(0,\frac{\pi}{2}\Big).$ Then,

$\text{x}_1<\text{x}_2,\forall\ \text{x}_1<\text{x}_2\in\Big(0,\frac{\pi}{2}\Big)$

$\Rightarrow\text{g}(\text{x}_1)<\text{g}(\text{x}_2)$

Taking $\tan^{-1}$ on both sides, we get

$\Rightarrow\tan^{-1}(\text{g}(\text{x}_1))<\tan^{-1}(\text{g}(\text{x}_2))$

$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\ \forall\ \text{x}_1,\text{x}_2\in\Big(0,\frac{\pi}{2}\Big)$

So, f(x) is increasing on $\Big(0,\frac{\pi}{2}\Big).$

3. a² - 3b + 15 < 0

Solution:

$\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$

$\text{f}'(\text{x})=3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})$

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.

$\therefore\ \text{a}^2-3\text{b}-15<0$

2. $|\text{x}|\geq3$

Solution:

f(x) = 3x² - 27x

⇒ f'(x) = x³ - 27x + 5

⇒ f'(x) = 3(x² - 9)

Function is increasing,

$3\big(\text{x}^2-9\big)\geq0$

$\Rightarrow\text{x}^2\geq9$

$\Rightarrow|\text{x}|\geq3$

2. a > 1

Solution:

$\text{f}(\text{x})=\log_\text{a}\text{x}=\frac{\log\text{x}}{\log\text{a}}$

$\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}$

Given: f(x) is increasing on R.

$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$

$\Rightarrow\frac{1}{\text{x}\log\text{a}}>0,\forall\ \text{x}\in\text{R}$

$\Rightarrow\text{a}>1$

2. (1, 3)

Solution:

Given, $\text{f}(\text{x})=2\log(\text{x}-2)-\text{x}^2+4\text{x}+1$

Domain of f(x) is $(2,\infty).$

$\text{f}'(\text{x})=\frac{2}{\text{x}-2}-2\text{x}+4$

$=\frac{2-2\text{x}^2+4\text{x}+4\text{x}-8}{\text{x}-2}$

$=\frac{-2\text{x}^2+8\text{x}-6}{\text{x}-2}$

$=\frac{-2(\text{x}^2-4\text{x}+3)}{\text{x}-2}$

For f(x) to be increasing, we must have

$\text{f}'(\text{x})>0$

$\Rightarrow\frac{-2(\text{x}^2-4\text{x}+3)}{\text{x}-2}>0$

$\Rightarrow\text{x}^2-4\text{x}+3<0$ $[\because\ \text{x}(\text{x}-2)>0\ \&-2<0]$

$\Rightarrow(\text{x}-1)(\text{x}-3)<0$

$\Rightarrow1<\text{x}<3$

$\Rightarrow\text{x}\in(1,3)$

Also, the domain of f(x) is $(2,\infty).$

$\Rightarrow\text{x}\in(1,3)\cap(2,\infty)$

$\Rightarrow\text{x}\in(1,3)$

4. $\lambda>2$

Solution:

$\text{f}(\text{x})=\frac{\lambda+\sin\text{x}+2\cos\text{x}}{\sin\text{x}+\cos\text{x}}$

$\Rightarrow\text{f}(\text{x})=(\lambda-2)\sin^2\text{x}+(\lambda-2)\cos^2\text{x}>0$

Using identity $\Rightarrow\lambda-2>0$

$\Rightarrow\lambda>2$

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

Solution:

$\text{f}(\text{x})=2\tan\text{x}+(2\text{a}+1)\log_\text{e}|\sec\text{x}|+(\text{a}-2)\text{x}$

If $\sec\text{x}>0$

$\Rightarrow\text{f}'(\text{x})=2\sec^2\text{x}+(2\text{a}+1)\frac{1}{\sec\text{x}}\sec\text{x}\tan\text{x}+(\text{a}-2)$

$\Rightarrow\text{f}'(\text{x})=2\sec^2\text{x}+(2\text{a}+1)\tan\text{x}+(\text{a}-2)$

Function is increasing

$=2\sec^2\text{x}+(2\text{a}+1)\tan\text{x}+(\text{a}-2)>0$

$\Rightarrow2\big(1+\tan^2\text{x}\big)+(2\text{a}+1)\tan\text{x}+(\text{a}-2)>0$

$\Rightarrow2\tan^2\text{x}+(2\text{a}+1)\tan\text{x}+\text{a}>0$

$\Rightarrow(2\text{a}+1)^2-4\times2\text{a}<0$

$\Rightarrow(2\text{a}-1)^2<0$ which is not possible.

$\Rightarrow(2\text{a}-1)^2=0$

$\Rightarrow\text{a}=\frac{1}{2}$

2. $(-\infty,0)$

Solution:

$\text{f}(\text{x})=\text{x}-\text{e}^{\text{x}}+\tan\Big(\frac{2\pi}{7}\Big)$

f'(x) = 1 - e^x

For f(x) to be increasing, we must have

f'(x) > 0

⇒ 1 - e^x > 0

⇒ e^x < 1

⇒ x < 0

$\Rightarrow\text{x}\in(-\infty,0)$

So, f(x) is increasing on $(-\infty,0).$

4. $\text{k}\in(-\infty,4)$

Solution:

f(x) = x² - kx + 5

f'(x) = 2x - k

Given: f(x) is increasing on [2, 4].

⇒ f'(x) > 0

⇒ 2x - k > 0

⇒ k < 2x

$\because\ \text{x}\in[2,4],$ maximum value of k is 4, k < 4.

$\therefore\ \text{k}\in(-\infty,4)$

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

Solution:

$\text{f}(\text{x})=\cot^{-1}\text{x}+\text{x}$

$\text{f}'(\text{x})=\frac{-1}{1+\text{x}^2}+1$

f(x) is increasing,

$\Rightarrow\frac{-1}{1+\text{x}^2}+1>0$

$\Rightarrow\frac{\text{x}^2}{1+\text{x}^2}>0$

Hence, f(x) is increasing on $(-\infty,\infty).$

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

Solution:

$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$

$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big(1+\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}\Big)$

$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$

$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$

$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2-\sqrt{\text{x}^2+1}}{1+\text{x}^2}$

Function is increasing monotonically.

$\Rightarrow\frac{1+2\text{x}^2-\sqrt{\text{x}^2+1}}{1+\text{x}^2}>0$

$\Rightarrow1+2\text{x}^2-\sqrt{\text{x}^2+1}>0$

$\Rightarrow1+2\text{x}^2>\sqrt{\text{x}^2+1}$

Squaring on both sides,

$\Rightarrow\big(1+2\text{x}^2\big)>\text{x}^2+1$

$\Rightarrow4\text{x}^4+3\text{x}^2>0$

For all $\text{x}\in\text{R}$

1. $(-\infty,4)$

Solution:

f(x) = 2x² - 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

$\text{k}\in(-\infty,4)$

4. a > 1

Solution:

$\text{f}(\text{x})=\text{a}^\text{x}$

$\text{f}'(\text{x})=\text{a}^\text{x}\log\text{a}$

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. $0<\text{x}<2$

Solution:

f(x) = x²e^-x

⇒ f'(x) = -x²e^-x + 2xe^-x

⇒ f'(x) = -e^-xx(x - 2)

Given that function is monotonically increasing.

-e^-xx(x - 2) > 0

x(x - 2) < 0

0 < x < 2