_0^ /2 ( x - x) ( x + x) ,dx = - Vidyadip

MCQ

$\int_0^{\pi /2} {} (\sin x - \cos x)\log (\sin x + \cos x)\,dx = $

A
$ - 1$
B
$1$
✓
$0$
D
એકપણ નહીં.

✓

Answer

Correct option: C.

$0$

(c) Put $\sin x + \cos x = t \Rightarrow - (\sin x - \cos x)dx = dt$

Also as $x = 0$ to $\frac{\pi }{2},t = 1$ to $1$.

Since here limit is $'1$ to $1'$,

therefore the value of integral will be zero,

$\left\{ \because \int_{a}^{a}{f(x)dx=0} \right\}$ .

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 7.2 definite integral→7.2 definite integral — all question types→ગણિત ધોરણ 12 સાયન્સ question papers→Gujarat Board ધોરણ 12 સાયન્સ question papers→Gujarat Board question paper generator→

Similar questions

વિધેય $f(x)\, = \left\{ \begin{array}{l}x + 2\,\,\,\,,\,\,\,1 \le x \le 2\\4\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x = 2\\3x - 2\,\,,\,\,\,x > 2\end{array} \right.$ એ $. . .$ બિંદુએ સતત થાય.

જો $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$, તો $k$ મેળવો.

જો $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ અને $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,હોય, તો $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ .........................

$\int_{\,0}^{\,\pi } {\log {{\sin }^2}x\,dx = } $

બે બળ $4i + j - 3k$ અને $3i + j - k$ ની અસર નીચે એક કણનું $i + 2j + 3k$ થી $5i + 4j + k$ સ્થાનાંતર થાય તો થયેલ કાર્ય ............... $\mathrm{unit}$ માં મેળવો.

$x$ ની બધી કિમતો ધરાવતો ગણ મેળવો.

$\frac{{{x^4} - 4{x^3} + 3{x^2}}}{{({x^2} - 4)({x^2} - 7x + 10)}} \ge 0$

જો$f(x)\begin{vmatrix}x^2&\sin x&\cos\ x\\2&0&-1\\a&a^2&a^3\\\end{vmatrix}=0$ તો $f''(0)=........... $( જ્યાં $a$ અચળ છે.)

$\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ =

વિકલ સમીકરણ $xy\frac{{dy}}{{dx}} = \frac{{(1 + {y^2})(1 + x + {x^2})}}{{(1 + {x^2})}}$ નો ઉકેલ મેળવો.

જે $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ અને $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$ હોય, તો $\operatorname{det} \mathrm{X}=$________________