MCQ 11 Mark
Every invertible function is:
- ✓
- B
- C
- D
Not necessarily monotonic function.
AnswerWe know that "every invertible function is a monotonic function".
View full question & answer→MCQ 21 Mark
If the function $\text{f}(\text{x})=\frac{-\text{x}}{2}+\sin\text{x}$ defined on $\Big[\frac{-\pi}{3},\frac{\pi}{3}\Big]$ is:
Answer$\text{f}(\text{x})=\frac{-\text{x}}{2}+\sin\text{x}$ 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.
View full question & answer→MCQ 31 Mark
If the function $\text{f}(\text{x})=\cos|\text{x}|-2\text{ax}+\text{b}$ increases along entire number scale, then:
AnswerCorrect option: C. $\text{a}\leq-\frac{1}{2}$
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.
$\Rightarrow2\text{a}\leq-1$
$\Rightarrow\text{a}\leq\frac{-1}{2}$
View full question & answer→MCQ 41 Mark
The function $\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$ is of the following type:
Answer$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$\Rightarrow\text{f}(-\text{x})=\log_\text{e}\Big(-\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=\log_\text{e}\Bigg\{\frac{\big(-\text{x}^3+\sqrt{\text{x}^6+1}\big)\big(\text{x}^3+\sqrt{\text{x}^6+1}\big)}{\text{x}^3+\sqrt{\text{x}^6+1}}\Bigg\}$
$=\log_\text{e}\Big(\frac{\text{x}^6+1-\text{x}^6}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=\log_\text{e}\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=-\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=-\text{f}(\text{x})$
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.
View full question & answer→MCQ 51 Mark
Function $\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$ is monotonic decreasing when:
- A
$\lambda>\frac{1}{2}$
- B
$\lambda<\frac{1}{2}$
- ✓
$\lambda<2$
- D
$\lambda>2$
AnswerCorrect option: C. $\lambda<2$
$\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}-2\lambda$
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.
$\Rightarrow2\lambda>1$
$\Rightarrow\lambda>\frac{1}{2}$
View full question & answer→MCQ 61 Mark
The function $f(x) = x^x$ decreases on the interval :
AnswerCorrect option: C. $\Big(0,\frac{1}{\text{e}}\Big)$
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 $x.$
$\Rightarrow\text{x}^\text{x} > 0$
$\Rightarrow1+\log_\text{e}\text{x} < 0$
$[$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 $\Big(0,\frac{1}{\text{e}}\Big).$
View full question & answer→MCQ 71 Mark
In the interval (1, 2), function f(x) = 2|x - 1| + 3|x - 2| is:
Answerf(x) = 2|x - 1| + 3|x - 2|
In the interval (1, 2)
⇒ |x -1| = x - 1 and |x - 2| = -(x - 2)
⇒ f(x) = 2(x - 1) - 3(x - 2)
⇒ f(x) = -x + 4
⇒ f'(x) = -1
⇒ function is decreasing on (1, 2).
View full question & answer→MCQ 81 Mark
The function $f(x) = x^9+ 3x^7 + 64$ is increasing on :
- ✓
$\text{R}$
- B
$(-\infty,0)$
- C
$(0,\infty)$
- D
$\text{R}_0$
AnswerCorrect option: A. $\text{R}$
$\text{f}(\text{x})=\text{x}^9+3\text{x}^7+64$
$\text{f}\ '(\text{x})=9\text{x}^8+21\text{x}^6 > 0,\forall\ \text{x}\in\text{R}$
So, $f(x)$ is increasing on $R.$
View full question & answer→MCQ 91 Mark
In the interval (1, 2), function f(x) = 2|x - 1| + 3|x - 2| is:
- A
Monotonically increasing.
- ✓
Monotonically decreasing.
- C
- D
AnswerCorrect option: B. Monotonically decreasing.
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.
View full question & answer→MCQ 101 Mark
Let $\phi(\text{x})=\text{f}(\text{x})+\text{f}(2\text{a}-\text{x})$ and f'(x) > 0 for all $\text{x}\in[0,\text{a}].$ Then, $\phi(\text{x}):$
Answer$\phi(\text{x})=\text{f}(\text{x})+\text{f}(2\text{a}-\text{x})$
$\phi'(\text{x})=\text{f}'(\text{x})-\text{f}'(2\text{a}-\text{x})$
$\text{f}''(\text{x})>0$ 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]
View full question & answer→MCQ 111 Mark
The function $f(x) = x^2e^{-x}$ is monotonic increasing when :
AnswerCorrect option: B. $0 < \text{x} < 2$
$f(x) = x^2e^{-x}$
$\Rightarrow f\ '(x) = -x^2e^{-x} + 2x\ e^{-x}$
$\Rightarrow f\ '(x) = -e^{-x}x(x - 2)$
Given that function is monotonically increasing.
$-e^{-x}x(x - 2) > 0$
$x(x - 2) < 0$
$0 < x < 2$
View full question & answer→MCQ 121 Mark
If the function $f(x) = x^3 - 9kx^2 + 27x + 30$ is increasing on $R,$ then:
- ✓
$-1\leq\text{k}\leq1$
- B
$k < -1$ or $k > 1$
- C
$0 < k < 1$
- D
$-1 < k < 0$
AnswerCorrect option: A. $-1\leq\text{k}\leq1$
$f(x) = x^3 - 9kx^2 + 27x + 30$
$\Rightarrow f'(x) = 3x^3 - 18kx + 27$
$\Rightarrow 3(x^2 - 6kx + 9)$
Function is always increasing on $R.$
$3(x^2 - 6kx + 9) > 0$
$x^2 - 6kx + 9 > 0$
In $ax^2 + bx + c = 0$ if $a > 0 $
$\Rightarrow b^2 - 4ac < 0$
$36k^2 - 36 < 0$
$k^2 - 1 < 0$
$(k + 1)(k - 1) < 0$
$\Rightarrow -1 < k < 1$
View full question & answer→MCQ 131 Mark
The function $\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$ is:
- ✓
- B
- C
Neither increasing nor decreasing.
- D
Answer$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$
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.
View full question & answer→MCQ 141 Mark
Function f(x) = |x| - |x - 1| is monotonically increasing when:
Answerf(x) = |x| - |x - 1|
Case I:
Let x < 0
If x < 0, then |x| = -x
⇒ |x - 1| = -(x - 1)
Now,
f(x) = |x| - |x - 1|
= -x - (-x + 1)
= -x + x - 1
= -1
f'(x) = 0
So, f(x) is not monotonically increasing when x < 0.
Case II:
Let x < 0 < 1
Here,
|x| = x
⇒ |x - 1| = -(x - 1)
Now,
f(x) = |x| - |x - 1|
= x + x -1
= 2x - 1
View full question & answer→MCQ 151 Mark
If $\text{f}(\text{x})=\tan^{-1}(\text{g}(\text{x})),$ where g(x) is monotonically increasing for $0<\text{x}<\frac{\pi}{2}.$ Then, f(x) is:
- ✓
Increasing on $\Big(0,\frac{\pi}{2}\Big)$
- B
Decreasing on $\Big(0,\frac{\pi}{2}\Big)$
- C
Increasing on $\Big(0,\frac{\pi}{4}\Big)$ and decreasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
- D
AnswerCorrect option: A. Increasing on $\Big(0,\frac{\pi}{2}\Big)$
aGiven: 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 $\Big(0,\frac{\pi}{2}\Big).$
View full question & answer→MCQ 161 Mark
Let $\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$ be an increasing function on $R$. Then, $a$ and $b$ satisfy :
- A
$a^2 - 3b - 15 > 0$
- B
$a^2 - 3b + 15 > 0$
- ✓
$a^2 - 3b - 15 < 0$
- D
$a < 0$ and $b > 0$
AnswerCorrect option: C. $a^2 - 3b - 15 < 0$
$\text{f}(\text{x})=\text{x}^3+\text{a}\text{x}^2+\text{b}\text{x}+5\sin^2\text{x}$
$\text{f}\ '(\text{x})=3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x})$
Given, $f(x)$ is increasing on $R.$
$\Rightarrow\text{f}\ '(\text{x}) > 0,\forall\ \text{x}\in\text{R}$
$\Rightarrow 3\text{x}^2+2\text{a}\text{x}+(\text{b}+5\sin2\text{x}) > 0,\forall\ \text{x}\in\text{R}$
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 $-1.$
$\therefore\ \text{a}^2-3\text{b}-15 < 0$
View full question & answer→MCQ 171 Mark
Function $f(x) = x^3 - 27x + 5$ is monotonically increasing when :
- A
$\text{x} < -3$
- ✓
$|\text{x}| > 3$
- C
$\text{x}\leq-3$
- D
$|\text{x}|\geq3$
AnswerCorrect option: B. $|\text{x}| > 3$
$f(x) = 3x^2 - 27x$
$\Rightarrow f\ '(x) = x^3 - 27x + 5$
$\Rightarrow f\ '(x) = 3(x^2 - 9)$
Function is increasing,
$3\big(\text{x}^2-9\big)\geq0$
$\Rightarrow\text{x}^2\geq9$
$\Rightarrow|\text{x}|\geq3$
View full question & answer→MCQ 181 Mark
If the function $f(x) = kx^3 - 9x^2 + 9x + 3$ is monotonically increasing in every interval, then :
- A
$\text{k} < 3\text{k} < 3$
- B
$\text{k}\leq3\text{k}\leq3$
- ✓
$\text{k} > 3\text{k} > 3$
- D
$\text{k}\geq3$
AnswerCorrect option: C. $\text{k} > 3\text{k} > 3$
$f(x)=k x^3-9 x^2+9 x+3$
$f^{\prime}(x)=k x^2-27$
$=3\left(x^2-9\right)$
For $\mathrm{f}(\mathrm{x})$ to be increasing, we must have
$f^{\prime}(x) > 0$
$\Rightarrow 3\left(x^2-9\right) > 0$
$\Rightarrow\left(x^2-9\right) > 0\left[\text { Since}, 3 > 0,3\left(x^2-9\right) > 0 \Rightarrow\left(x^2-9\right) > 0\right]$
$\Rightarrow(x+3)(x-3) > 0$
$\Rightarrow x < -3 $ or $ x > 3$
$\Rightarrow|x| > 3$
View full question & answer→MCQ 191 Mark
Function $\text{f}(\text{x})=\log_\text{a}\text{x}$ is increasing on R, if:
Answer$\text{f}(\text{x})=\log_\text{a}\text{x}=\frac{\log\text{x}}{\log\text{a}}$
$\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}$
Given: f(x) is increasing on R.
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\frac{1}{\text{x}\log\text{a}}>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}>1$
View full question & answer→MCQ 201 Mark
The function $\text{f}(\text{x})=2\log(\text{x}-2)-\text{x}^2+4\text{x}+1$ increases on the interval:
Answer
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 $(2,\infty).$
$\Rightarrow\text{x}\in(1,3)\cap(2,\infty)$
$\Rightarrow\text{x}\in(1,3)$ View full question & answer→MCQ 211 Mark
The Function $\text{f}(\text{x})=\frac{\lambda+\sin\text{x}+2\cos\text{x}}{\sin\text{x}+\cos\text{x}}$ is increasing, if:
- A
$\lambda<1$
- B
$\lambda>1$
- C
$\lambda<2$
- ✓
$\lambda>2$
AnswerCorrect option: D. $\lambda>2$
$\text{f}(\text{x})=\frac{\lambda+\sin\text{x}+2\cos\text{x}}{\sin\text{x}+\cos\text{x}}$
$\Rightarrow\text{f}(\text{x})=(\lambda-2)\sin^2\text{x}+(\lambda-2)\cos^2\text{x}>0$
Using identity $\Rightarrow\lambda-2>0$
$\Rightarrow\lambda>2$
View full question & answer→MCQ 221 Mark
If the function $\text{f}(\text{x})=2\tan\text{x}+(2\text{a}+1)\log_\text{e}|\sec\text{x}|+(\text{a}-2)\text{x}$ is increasing on R, then:
- A
$\text{a}\in\Big(\frac{1}{2},\infty\Big)$
- B
$\text{a}\in\Big(-\frac{1}{2},\frac{1}{2}\Big)$
- ✓
$\text{a}=\frac{1}{2}$
- D
$\text{a}\in\text{R}$
AnswerCorrect option: C. $\text{a}=\frac{1}{2}$
$\text{f}(\text{x})=2\tan\text{x}+(2\text{a}+1)\log_\text{e}|\sec\text{x}|+(\text{a}-2)\text{x}$
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.
$\Rightarrow(2\text{a}-1)^2=0$
$\Rightarrow\text{a}=\frac{1}{2}$
View full question & answer→MCQ 231 Mark
Function $f(x) = a^x$ is increasing on $R,$ if :
- A
$a > 0$
- B
$a < 0$
- C
$0 < a < 1$
- ✓
$a > 1$
AnswerCorrect option: D. $a > 1$
$\text{f}(\text{x})=\text{a}^\text{x}$
$\text{f}'(\text{x})=\text{a}^\text{x}\log\text{a}$
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 $a > 1.$
View full question & answer→MCQ 241 Mark
The interval of increase of the function $\text{f}(\text{x})=\text{x}-\text{e}^{\text{x}}+\tan\Big(\frac{2\pi}{7}\Big)$ is :
- A
$(0,\infty)$
- ✓
$(-\infty,0)$
- C
$(1,\infty)$
- D
$(-\infty,1)$
AnswerCorrect option: B. $(-\infty,0)$
$\text{f}(\text{x})=\text{x}-\text{e}^{\text{x}}+\tan\Big(\frac{2\pi}{7}\Big)$
$f'(x) = 1 - e^x$
For $f(x)$ to be increasing, we must have
$f\ '(x) > 0$
$\Rightarrow 1 - e^x > 0$
$\Rightarrow e^x < 1$
$\Rightarrow x < 0$
$\Rightarrow\text{x}\in(-\infty,0)$
So, $f(x)$ is increasing on $(-\infty,0).$
View full question & answer→MCQ 251 Mark
Function $f(x) = 2x^3 - 9x^2 + 12x + 29$ is monotonically decreasing when:
Answer$f(x) = 2x^3 - 9x^2 + 12x + 29$
$\Rightarrow f'(x) = 6x^2 - 18x + 12$
$\Rightarrow f'(x) = 6(x^2 - 3x + 2)$
$\Rightarrow f'(x) = 6(x - 1)(x - 2)$
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 $1 < x < 2.$
View full question & answer→MCQ 261 Mark
If the function $f(x) = x^2 - kx + 5$ is increasing on $[2, 4],$ then :
- A
$\text{k}\in(2,\infty)$
- B
$\text{k}\in(-\infty,2)$
- C
$\text{k}\in(4,\infty)$
- ✓
$\text{k}\in(-\infty,4)$
AnswerCorrect option: D. $\text{k}\in(-\infty,4)$
$f(x) = x^2 - kx + 5$
$f\ '(x) = 2x - k$
Given : $f(x)$ is increasing on $[2, 4]$.
$\Rightarrow f'(x) > 0$
$\Rightarrow 2x - k > 0$
$\Rightarrow k < 2x$
$\because\ \text{x}\in[2,4],$ maximum value of $k$ is $4, k < 4$.
$\therefore\ \text{k}\in(-\infty,4)$
View full question & answer→MCQ 271 Mark
The function $\text{f}(\text{x})=\cot^{-1}\text{x}+\text{x}$ increases in the interval:
- A
$(1,\infty)$
- B
$(-1,\infty)$
- ✓
$(-\infty,\infty)$
- D
$(0,\infty)$
AnswerCorrect option: C. $(-\infty,\infty)$
$\text{f}(\text{x})=\cot^{-1}\text{x}+\text{x}$
$\text{f}'(\text{x})=\frac{-1}{1+\text{x}^2}+1$
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 $(-\infty,\infty).$
View full question & answer→MCQ 281 Mark
$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$ is monotonically increasing when:
AnswerCorrect option: C. $\text{x}\in\text{R}$
$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$
$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big(1+\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}\Big)$
$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2-\sqrt{\text{x}^2+1}}{1+\text{x}^2}$
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 $\text{x}\in\text{R}$
View full question & answer→MCQ 291 Mark
If the function $f(x) = 2x^2 - kx + 5$ is increasing on $[1, 2]$, then $k$ lies in the interval :
- ✓
$(-\infty,4)$
- B
$(4,\infty)$
- C
$(-\infty,8)$
- D
$(8,\infty)$
AnswerCorrect option: A. $(-\infty,4)$
$f(x) = 2x^2 - kx + 5$
$f\ '(x) = 4x - k$
$f(x)$ is increasing
$4x - k < 0$ on $[1, 2]$
$k < 4x$
Minimum value of $k$ is $4.$
$k < 4$
$\text{k}\in(-\infty,4)$
View full question & answer→MCQ 301 Mark
Let $f(x) = x^3 -6x^2 +15x + 3$. Then :
AnswerCorrect option: C. $f(x)$ is invertible.
$f(x) = x^3 - 6x^2 +15x + 3$
$f\ '(x) = 3x^2 - 12x + 15$
$= 3(x^2 - 4x + 5)$
$= 3(x^2 - 4x + 4 + 1)$
$=3(\text{x}-2)^2+\frac{1}{3}>0$
Therefore, $f(x)$ is strictly increasing function.
$\Rightarrow f^{-1}(x)$ exists.
Hence, $f(x)$ is an invertible function.
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