-
$\frac{15}{16}$
-
$\frac{3}{16}$
-
$-\frac{3}{16}$
-
$-\frac{16}{3}$
$\frac{15}{16}$
$\frac{3}{16}$
$-\frac{3}{16}$
$-\frac{16}{3}$
solution:
$\text{I}=\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{\frac{5}{4}}}\text{ dx}$
Put, 1 - x = t ⇒ x =1 -t
⇒ dx = -dt
| x | 0 | 1 |
| t | 1 | 0 |
$\text{I}=\int\limits^0_1\frac{(1-\text{t})(-\text{dt})}{\text{t}^{\frac{5}{4}}}$
$\text{I}=\int\limits^1_0\Big(\text{t}^\frac{5}{4}-\text{t}^\frac{-1}{4}\Big)\text{dt}$
$\text{I}=\Bigg[\frac{\text{t}^{-\frac{5}{4}}}{\frac{-1}{4}}-\frac{\text{t}^\frac{3}{4}}{\frac{3}{4}}\Bigg]^1_0$
$\text{I}=-4-\frac{4}{3}$
$\text{I}=\frac{-16}{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$2 x_{1}-4 x_{2}+\lambda x_{3}=1$
$x_{1}-6 x_{2}+x_{3}=2$
$\lambda x_{1}-10 x_{2}+4 x_{3}=3$ is inconsistent for
$f(x)=\left\{\begin{array}{cl}x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0,\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise }\end{array}\right.$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) If $a=0, b=1, c=0$ and $d=0$, then | ($1$) $h$ is one-one |
| ($Q$) If $a=1, b=0, c=0$ and $d=0$, then | ($2$) $h$ is onto. |
| ($R$) If $a=0, b=0, c=1$ and $d=0$, then | ($3$) $h$ is differentiable on $R$. |
| ($S$) If $a=0, b=0, c=0$ and $d=1$, then | ($4$) the range of $h$ is $[0,1]$ |
| ($5$) the range of $h$ is $\{0,1\}$ |
The correct option is