| 2023-12-22 | Area of the region bounded by the curve \( y^{2}=4 x \),
\( \mathrm... | 1:12 | 0 | |
|
| 2023-12-22 | The area lying in the first quadrant and bounded by
\( \mathrm{P} \... | 0:48 | 0 | |
|
| 2023-12-22 | The area of the region bounded by \( 2 x+6 y-8=0 \& \)
\( \mathrm{P... | 1:41 | 0 | |
|
| 2023-12-22 | Sporophyte is independent in all, except-
\( \mathrm{P} \)
(1) Gymnosperm
(2) Bryophyte
W
(3) Pt... | 3:43 | 0 | |
|
| 2023-12-22 | Diploid plant have dominant stage in -
(1) Sporophyte
(2) Gametophyte
\( \mathrm{P} \)
(3) Gamet... | 2:52 | 0 | |
|
| 2023-12-22 | Evaluate \( \int_{-1}^{2}\left|x^{3}-x\right| d x \)
\( \mathrm{P} ... | 3:43 | 0 | |
|
| 2023-12-22 | The area of the region bounded by \( y=3 x^{2}+2 \& \) the
\( \math... | 1:16 | 2 | |
|
| 2023-12-22 | The area of the shaded region in the given figure is
P
W
(1) \( A_{... | 0:47 | 1 | |
|
| 2023-12-22 | Prove that \( \int_{-1}^{1} x^{17} \cos ^{4} x d x=0 \)
\( \mathrm{... | 0:51 | 0 | |
|
| 2023-12-22 | Ovules are not enclosed by the ovaries in
\( \mathrm{P} \)
(1) Pteridophytes
(2) Angiosperms
W
(... | 2:13 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{2 \pi} \cos ^{5} x d x
\]
P
W | 1:36 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{4}|x-1| d x
\]
P
W | 1:14 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x
\]
\( \mat... | 0:57 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\pi} \frac{x d x}{1+\sin x} \)
\( \mathrm{P} \)
W | 1:40 | 3 | |
|
| 2023-12-22 | Evaluate \( \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x \)
\(... | 2:30 | 1 | |
|
| 2023-12-22 | In pteridophytes, the main plant body is
\( \mathrm{P} \)
(1) A gametophyte
W
(2) Thalloid
(3) N... | 2:47 | 0 | |
|
| 2023-12-22 | \[
\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x
\]
\( \mat... | 0:39 | 0 | |
|
| 2023-12-22 | Evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x... | 1:13 | 2 | |
|
| 2023-12-22 | \( \int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x ... | 2:21 | 1 | |
|
| 2023-12-22 | Evaluate \( \int_{-1}^{1} \sin ^{5} x \cos ^{4} d x \)
P
W | 0:40 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x \) is equal to... | 2:55 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x ... | 1:41 | 3 | |
|
| 2023-12-22 | \( \int_{0}^{\pi / 4} 2 \tan ^{3} x d x \) is equal to
\( \mathrm{P... | 1:41 | 1 | |
|
| 2023-12-22 | \( \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x \) is equal to
\( ... | 1:10 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x
\]
\(... | 2:47 | 0 | |
|
| 2023-12-22 | \[
\int_{\pi / 2}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)... | 2:50 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\pi / 4} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x}... | 2:23 | 1 | |
|
| 2023-12-22 | \( \int \frac{d x}{e^{x}+e^{-x}} \) is equal to
\( \mathrm{P} \)
W
... | 0:40 | 0 | |
|
| 2023-12-22 | \( \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}... | 1:52 | 0 | |
|
| 2023-12-22 | \( \int \frac{1}{x \sqrt{p x-x^{2}}} d x \) is equal to
\( \mathrm{... | 1:29 | 1 | |
|
| 2023-12-22 | \( \int \frac{e^{7 \log x}-e^{5 \log x}}{e^{9 \log x}-e^{7 \log x}}... | 0:54 | 1 | Vlog |
|
| 2023-12-22 | \( \int_{0}^{\pi / 2} \log \left(\frac{7+2 \sin x}{7+2 \cos x}\righ... | 1:33 | 0 | Vlog |
|
| 2023-12-22 | \( \int \frac{\sin x}{\sin (x-5)} d x \) is equal to
\( \mathrm{P} ... | 1:06 | 0 | |
|
| 2023-12-22 | \[
\int \frac{1}{\sqrt{x+5}+\sqrt{x+2}} d x
\]
\( \mathrm{P} \)
W
(... | 0:58 | 0 | |
|
| 2023-12-22 | Evaluate \( \int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x \)
\( P \)
W | 0:45 | 0 | |
|
| 2023-12-22 | Evaluate \( \int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x \)
P
W | 1:02 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x
\]
P | 1:58 | 5 | |
|
| 2023-12-22 | \( \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} \) equals
\( \mathr... | 1:06 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x
\]
P
W | 1:12 | 0 | |
|
| 2023-12-22 | If \( f(x)=\int_{0}^{x} t \sin t d t \), then \( f^{\prime}(x) \) i... | 0:29 | 0 | |
|
| 2023-12-22 | \( \int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}} \) is equal to:
\( \math... | 0:32 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\pi} \frac{x d x}{1+\sin x} \) is equal to
\( \mathrm{... | 1:40 | 0 | |
|
| 2023-12-22 | \( \int_{-5}^{5}|x+2| d x \) is equal to
\( \mathrm{P} \)
(1) 27
(2... | 2:55 | 0 | |
|
| 2023-12-22 | \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^... | 0:40 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{1} x(1-x)^{n} d x \) is equal to
\( \mathrm{P} \)
W
(1... | 1:10 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x \) is equal to
\( \... | 1:38 | 0 | Vlog |
|
| 2023-12-22 | \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x d x \) is equa... | 1:10 | 0 | |
|
| 2023-12-22 | \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{5} x d x \) is equa... | 0:54 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\fr... | 1:37 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} \cos ^{5} x d x \) is equ... | 2:38 | 0 | |
|
| 2023-12-22 | \( \int_{1 / 3}^{1} \frac{\left(x-x^{3}\right)^{1 / 3}}{x^{4}} d x ... | 1:44 | 2 | |
|
| 2023-12-22 | \( \int_{-1}^{1} \frac{d x}{x^{2}+2 x+5} \) is equal to
\( \mathrm{... | 1:20 | 1 | |
|
| 2023-12-22 | If \( f(x)=\int_{0}^{x} t \sin t d t \), then \( f^{\prime}(x) \) i... | 0:37 | 1 | |
|
| 2023-12-22 | \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x \) is ... | 1:42 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{2} x \sqrt{x+2} d x \) is equal to
\( \mathrm{P} \)
W
... | 1:58 | 0 | |
|
| 2023-12-22 | The minimum value of the objective function \( Z=x+2 y \)
P
Subject to the constraints
W
\( 2 x+... | 2:27 | 0 | |
|
| 2023-12-22 | A ball of mass \( 1 \mathrm{~kg} \) is given an initial velocity of \( \sqrt{44} \mathrm{~ms}^{-... | 8:43 | 8 | |
|
| 2023-12-22 | A body is projected vertically upwards from the surface of a planet of radius \( \mathrm{R} \) w... | 4:14 | 1 | |
|
| 2023-12-22 | A body slides down a frictionless track which ends in a circular loop of diameter \( D \), then ... | 6:43 | 9 | |
|
| 2023-12-22 | Find the minimum value of \( \theta \) such that the ball of mass \( 1 \mathrm{~kg} \) will perf... | 3:36 | 0 | |
|
| 2023-12-22 | \( \int \sqrt{x^{2}+4 x-5} d x \) is equal to:
\( \mathrm{P} \)
(1)... | 1:18 | 16 | |
|
| 2023-12-22 | Evaluate the following integrals:
\( \mathrm{P} \)
(i) \( \int_{2}^... | 4:42 | 2 | |
|
| 2023-12-22 | \( \int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x \) i... | 2:33 | 1 | |
|
| 2023-12-22 | Statement-I: Thalassemia is inborn of metabolism as phenylketonuri... | 1:04 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{1+\frac{x^{2}}{9}} \)
P
W | 1:07 | 3 | |
|
| 2023-12-22 | \( \int \sqrt{144-x^{2}} d x \) is equal to
P
W
o | 0:45 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{x^{2}+3 x} \)
\( \mathrm{P} \)
W | 1:27 | 3 | |
|
| 2023-12-22 | Integrate \( \sqrt{1-4 x-x^{2}} \)
P
W | 1:33 | 1 | |
|
| 2023-12-22 | \( \int \sqrt{1+x^{2}} d x \) is equal to
\( \mathrm{P} \)
W
\( \th... | 0:27 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{1+6 x-x^{2}} \)
\( \mathrm{P} \)
W | 1:14 | 1 | |
|
| 2023-12-22 | Integrate \( \sqrt{x^{2}+4 x-5} \)
P
W | 0:58 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{x^{2}+4 x+1} \)
\( \mathrm{P} \)
W | 1:04 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{x^{2}+4 x+6} \)
\( \mathrm{P} \)
W | 0:48 | 1 | |
|
| 2023-12-22 | \( \int_{2}^{3} \frac{d x}{x^{2}-1} \) is equal to
\( \mathrm{P} \)... | 0:51 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{1-4 x^{2}} \)
\( \mathrm{P} \)
W | 0:33 | 0 | |
|
| 2023-12-22 | Integrate \( \sqrt{4-x^{2}} \)
P
W | 0:29 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x
\]
\( ... | 1:23 | 2 | |
|
| 2023-12-22 | \( \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} \) is equal to
\( \mathr... | 0:24 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\... | 0:37 | 0 | |
|
| 2023-12-22 | \( \int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}} \) is equal to
\( \mathr... | 0:30 | 0 | |
|
| 2023-12-22 | \[
\int_{0}^{\pi / 4}\left(2 \sec ^{2} x+x^{3}+2\right) d x
\]
\( \... | 0:48 | 0 | |
|
| 2023-12-22 | \[
\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x
\]
\( \mathrm... | 1:23 | 0 | |
|
| 2023-12-22 | \( \mathrm{P} \)
W
(1) \( \ln \sqrt{2} \)
(2) \( \ln 2 \)
(3) \( -\... | 0:33 | 0 | |
|
| 2023-12-22 | \( \int_{0}^{4}\left(e^{2 x}+x\right) d x \) is equal to:
\( \mathr... | 0:35 | 1 | |
|
| 2023-12-22 | \[
\int_{0}^{\pi / 2} \cos 2 x d x
\]
\( \mathrm{P} \)
(1) -1
(2) 1... | 0:32 | 0 | |
|
| 2023-12-22 | \( \int \sqrt{x^{2}-8 x+7} d x \)
\( \mathrm{P} \)
(1) \( \frac{x-4... | 1:46 | 0 | |
|
| 2023-12-22 | \( \int \sqrt{x^{2}+4 x+6} d x \) is equal to:
\( \mathrm{P} \)
(1)... | 1:11 | 2 | |
|
| 2023-12-22 | \( \int \sqrt{1+3 x-x^{2}} d x \) is equal to:
\( \mathrm{P} \)
(1)... | 2:03 | 1 | |
|
| 2023-12-22 | Find \( \int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x \)
D
W | 1:21 | 0 | |
|
| 2023-12-22 | \[
\int \sqrt{16-x^{2}} d x
\]
\( \mathrm{P} \)
(1) \( \frac{x}{2} ... | 0:42 | 0 | |
|
| 2023-12-22 | \[
\int\left(x^{2}+1\right) \log x \mathrm{dx}
\]
\( P \)
W | 1:02 | 0 | Vlog |
|
| 2023-12-22 | \( \int \sin ^{-1} \frac{2 x}{1+x^{2}} d x \) is equal to
\( \mathr... | 1:59 | 0 | |
|
| 2023-12-22 | Integrate \( x \tan ^{-1} x \)
P
W | 1:13 | 0 | |
|
| 2023-12-22 | Integrate \( x \sin ^{-1} x \)
\( P \)
W. | 1:38 | 0 | |
|
| 2023-12-22 | Find \( \int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x \... | 0:28 | 0 | |
|
| 2023-12-22 | Integrate \( x \sec ^{2} x \)
P
W
\( \theta \) | 0:40 | 0 | |
|
| 2023-12-22 | \( \int e^{2 x} \sin x d x \) is equal to
\( \mathrm{P} \)
W
(1) \(... | 2:05 | 0 | |
|
| 2023-12-22 | \( \int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x \) is equal... | 2:20 | 0 | |
|
| 2023-12-22 | Integrate \( x \log x \)
\( \mathrm{P} \)
W
\( \theta \) | 0:58 | 0 | Vlog |
|
| 2023-12-22 | \( \int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}... | 4:14 | 1 | |
|
