Question
Fill in the blanks:
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then a = ________.
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then a = ________.
✓
Answer
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then $\text{a}=\frac{1}{2}$
Let $\text{I}=\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8}$
Now, $\int\limits^\text{a}_0\frac{1}{4\Big(\frac{1}{4}+\text{x}^2\Big)}\text{dx}=\frac{2}{4}\big[\tan^{-1}2\text{x}\big]^\text{a}_0$
$=\frac{1}{2}\tan^{-1}2\text{a}-0=\frac{\pi}{8}$
$\frac{1}{2}\tan^{-1}2\text{a}=\frac{\pi}{8}$
$\Rightarrow\ \tan^{-1}2\text{a}=\frac{\pi}{4}$
$\Rightarrow\ 2\text{a}=1$
$\therefore\ \text{a}=\frac{1}{2}$
Let $\text{I}=\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8}$
Now, $\int\limits^\text{a}_0\frac{1}{4\Big(\frac{1}{4}+\text{x}^2\Big)}\text{dx}=\frac{2}{4}\big[\tan^{-1}2\text{x}\big]^\text{a}_0$
$=\frac{1}{2}\tan^{-1}2\text{a}-0=\frac{\pi}{8}$
$\frac{1}{2}\tan^{-1}2\text{a}=\frac{\pi}{8}$
$\Rightarrow\ \tan^{-1}2\text{a}=\frac{\pi}{4}$
$\Rightarrow\ 2\text{a}=1$
$\therefore\ \text{a}=\frac{1}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Fill in the blanks.
The value of the expression $|\vec{\text{a}}\times\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ is ________.
→The value of the expression $|\vec{\text{a}}\times\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ is ________.
Fill in the blanks.
The solution of differential equation $\cot\text{y dx}=\text{x dy} $ is _________.
→The solution of differential equation $\cot\text{y dx}=\text{x dy} $ is _________.
The numbers $l r, m r$ and $n r$ proportional to direction cosines are called __________ of $\vec{r}$.
→$f(x)=2 x$, for $f: N \rightarrow N$, is. ________ but not onto.
→Location of three houses of a society is represented by the points A(-1, 0), B(1, 3) and C(3, 2) as shown in figure.
Based on the above information, answer the following questions
→Based on the above information, answer the following questions
- Equation of line AB is.
- $\text{y}=\frac{3}{2}(\text{x}+1)$
- $\text{y}=\frac{3}{2}(\text{x}-1)$
- $\text{y}=\frac{1}{2}(\text{x}+1)$
- $\text{y}=\frac{1}{2}(\text{x}-1)$
- Equation of line BC is.
- $\text{y}=\frac{1}{2}\text{x}-\frac{7}{2}$
- $\text{y}=\frac{3}{2}\text{x}-\frac{7}{2}$
- $\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$
- $\text{y}=\frac{3}{2}\text{x}+\frac{7}{2}$
- Area of region ABCD is.
- 2 sq. units
- 4 sq. units
- 6 sq. units
- 8 sq. units
-
Area of $\triangle\text{ADC}$ is,
- 4 sq. units
- 8 sq. units
- 16 sq. units
- 32 sq. units
- Area of $\triangle\text{ABC}$ is.
- 3 sq. units
- 4 sq. units
- 5 sq. units
- 6 sq. units
If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=15$, then the value of $|\vec{x}|$ will be __________ .
→Fill in the blanks.
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type _________.
→$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type _________.
Fill in the blanks.
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type _________.
→$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{1}{\text{x}}$ is an equation of the type _________.
A vector whose initial and terminal points coincide, is called __________ .
If $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$, then the value of $|\vec{a} \times \vec{b}|$ will be ___________ .
→