Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

JEE Mathematics Solved Question Paper 2010

Multiple Choice Questions

21.
The differential equation of family of circles whose centre lies on x-axis, is

$\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} + 1 = 0$

$y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} - 1 = 0$

$y \frac{d^{2} y}{{dx}^{2}} - {(\frac{dy}{dx})}^{2} - 1 = 0$

$y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} + 1 = 0$

$y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} + 1 = 0$

The equation of family of circle having centre at x - axis is x² + y² - 2ax = 0.

On differentiating , we get

$2 x + 2 y \frac{dy}{dx} - 2 a = 0$

Again, differentiating, we get

$2 + 2 [y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}] = 0 \Rightarrow y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} + 1 = 0$

22.

The solution of the differential equation $y (1 + \log (x)) \frac{d y}{d x} - xlog (x) = 0$ is

x log(x) = y + c
x log(x) = yc
y(1 + log(x) = c
log(x) - y = c

23.

The order of the differential equation whose solution is ae^x + be²^x + ce^3x + d = 0, is

24.

The maximum value of the objective function Z = 3x + 2y for linear constraints x + y $\leq$ 7, 2x + 3y $\leq$ 16, x² $\geq$ 0, y² $\geq$ 0 is

25.

The position vectors of vertices of a $∆$ ABC are $4 \hat{i} - 2 \hat{j}$ , $\hat{i} + 4 \hat{j} - 3 \hat{k}$ and $- \hat{i} + 5 \hat{j} + \hat{k}$ respectively, then $∠$ ABC is equal to

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

26.

A four digit number is to be formed using the digits 1, 2, 3 4, 5 6, 7 (no digit is being repeated in any number). Then, the probability that it is > 4000, is