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 2017

Multiple Choice Questions

21.

The general solution of the equation $\frac{dy}{dx} = \frac{y^{2} - x}{2 y (x + 1)}$ is

y² = (1 + x)log(1 + x) - c
$y^{2} = (1 + x) \log (\frac{c}{(1 - x)}) - 1$
$y^{2} = (1 - x) \log (\frac{c}{(1 - x)}) - 1$
$y^{2} = (1 + x) \log (\frac{c}{(1 + x)}) - 1$

22.

The general solution of the differential equation $\frac{dy}{dx} + \sin (\frac{x + y}{2}) = \sin (\frac{x - y}{2})$ is

$\log_{e} |\tan (\frac{y}{2})| = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} |\tan (\frac{y}{4})| = 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{2})) = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{4})) = - 2 \sin (\frac{x}{2}) + C$

23.

If a, b, c are three vectors such that [a b c] = 5, then the value of [a x b, b x c, c x a] is

24.
The point of inflection of the function y = $\int_{0}^{x} (t^{2} - 3 t + 2) dt$ is

$(\frac{3}{2}, \frac{3}{4})$

$(- \frac{3}{2}, - \frac{3}{4})$

$(- \frac{1}{2}, - \frac{3}{2})$

$(\frac{1}{2}, \frac{3}{2})$

$(\frac{3}{2}, \frac{3}{4})$

$Given y = \int_{0}^{x} (t^{2} - 3 t + 2) dt . . . (i) On dlfferentiating w . r . t .' x', we get \frac{dy}{dx} = x^{2} - 3 x + 2 . . . (ii) Again, on differentiating w . r . t .' x', we get \frac{d^{2} y}{{dx}^{2}} = 2 x - 3 . . . (iii) We know that, at point of inflecton \frac{d^{2} y}{{dx}^{2}} = 0 ∴ From Eq (iii), we get 2 x - 3 = 0 \Rightarrow x = \frac{3}{2} Now, we have to check behaviour of \frac{d^{2} y}{{dx}^{2}} at point x = \frac{3}{2} Clearly, as x = \frac{3}{2} sign at \frac{d^{2} y}{{dx}^{2}} changes ∴ (\frac{3}{2}, \frac{3}{4}) is a point of inflection .$