MCQ
$\int_0^1 {\frac{{dx}}{{{{[ax + b(1 - x)]}^2}}}} = $
- A$\frac{a}{b}$
- B$\frac{b}{a}$
- C$a\,b$
- ✓$\frac{1}{{a\,b}}$
✓
Answer
Correct option: D.
$\frac{1}{{a\,b}}$
d
(d) Let $I = \int_0^1 {\frac{{dx}}{{{{[(a - b)x + b]}^2}}}} $
(d) Let $I = \int_0^1 {\frac{{dx}}{{{{[(a - b)x + b]}^2}}}} $
Put $t = (a - b)x + b \Rightarrow dt = (a - b)dx$
As $x = 1 \Rightarrow t = a$ and $x = 0 \Rightarrow t = b$, then
$I = \frac{1}{{a - b}}\int_b^a {\frac{1}{{{t^2}}}} dt = \frac{1}{{(a - b)}}\left[ { - \frac{1}{t}} \right]_b^a $
$= \frac{1}{{(a - b)}}\left( {\frac{{a - b}}{{ab}}} \right) = \frac{1}{{ab}}$.
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
The feasible region for an LPP is shown shaded in the following figure. Minimum of Z = 4x + 3y occurs at the point.
- (0, 8)
- (2, 5)
- (4, 3)
- (9, 0)
ox, oy are positive x-axis, positive y-axis respectively where O = (0, 0,0) The d.c.s of the llne which bisects $\angle\text{xoy}$ are:
→If $f: R \rightarrow R_{,} f(x)=3 x-5$, then the value of $fo f(3)$ is
→A water tank has the shape of an inverted right circular cone, whose semi vertical angle is ${\tan ^{ - 1}}\,\left( {\frac{1}{2}} \right)$. Water is poured in at a constant rate of $5$ cubic meter per minute. Then the rate (in $m/min$) at which the level of water is rising at the instant when the depth of water in the tank is $10\, m$ is
→The range of $\sin^{-1}\text{x}+\cos^{-1}\text{x}+\tan^{-1}\text{x}$ is:
→- $[0,\pi]$
- $[\frac{\pi}{4},\frac{3\pi}{4}]$
- $(0,\pi)$
- $[0,\frac{\pi}{2}]$
If $y = {x^{({x^x})}}$, then ${{dy} \over {dx}} = $
→Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has
→If $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ are unit vectors, then
→- $\hat{\text{i}}.\hat{\text{j}}=1$
- $\hat{\text{i}}.\hat{\text{i}}=1$
- $\hat{\text{i}}\times\hat{\text{j}}=1$
- $\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)=1$
$\int \cos ^2 d x$ is equal to:
→For the following LPP, maximise $Z=3 x+4 y$ subject to constraints $x-y \geq-1, x \leq 3, x \geq 0, y \geq 0$ the maximum value is
→