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

71.

$\int \frac{x^{e - 1} + e^{x - 1}}{x^{e} + e^{x}} dx$

$\frac{- 1}{e} \log |x^{e} + e^{x}| + C$
$- elog |x^{e} + e^{x}| + C$
$\frac{1}{e} \log |x^{e} + e^{x}| + C$
$e \log |x^{e} + e^{x}| + C$

72.

If a, b and c are non-zero vectors such that a and b are not perpendicular to each other, then the vector r which is perpendicular to a and satisfying r x b = c x b is

$\frac{(a \times b) \times c}{c a}$
$\frac{b \times (a \times c)}{b c}$
$\frac{(b \times c) \times a}{a b}$
$\frac{(c \times b) \times a}{ac}$

73.

$\int \frac{x + \sin (x)}{1 + \cos (x)} dx = ?$

$xtan (\frac{x}{2}) + C$
$xsin (\frac{x}{2}) + \cos (\frac{x}{2}) + C$
$xtan (\frac{x}{2}) + \sec (\frac{x}{2}) + C$
$xtan (\frac{x}{2}) + \sec (\frac{x}{2}) + C$

74.

$If I_{n} = \int \frac{\sin (nx)}{\cos (x)} dx, then I_{n} = ?$

$\frac{- 2}{n - 1} \cos (n - 1) x - I_{n - 2}$
$\frac{2}{n - 1} \cos (n - 1) x + I_{n - 2}$
$\frac{- 2}{n + 1} \sin (n - 1) x - I_{n - 2}$
$\frac{- 2}{n + 1} \cos (n - 1) x - I_{n - 2}$

75.

The differential equation corresponding to the family of circles inthe plane touching the Y-axis at the origin, is

$\frac{dy}{dx} = \frac{y^{2} - x^{2}}{2 xy}$
$\frac{dy}{dx} = \frac{2 xy}{x^{2} + y^{2}}$
$\frac{dy}{dx} = \frac{x^{2} - y^{2}}{2 xy}$
$\frac{dy}{dx} = \frac{y^{2} + x^{2}}{2 xy}$

76.
$The triad (x, y, z) of real number such that (3 \hat{i} - \hat{j} + 2 \hat{k}) = (2 \hat{i} + 3 \hat{j} - \hat{k}) x + (\hat{i} - 2 \hat{j} + 2 \hat{k}) y + (- 2 \hat{i} + \hat{j} - 2 \hat{k}) z is$

(- 2, 5, 3)

(2, - 5, 3)

(2, 5, 3)

(2, 5, - 3)

(2, 5, 3)

$(c) We have, (3 \hat{i} - \hat{j} + 2 \hat{k}) = (2 \hat{i} + 3 \hat{j} - \hat{k}) x + (\hat{i} - 2 \hat{j} + 2 \hat{k}) y + (- 2 \hat{i} + \hat{j} - 2 \hat{k}) z = \hat{i} (2 x + y - 2 z) + \hat{j} (3 x - 2 y + z) + \hat{k} (- x + 2 y - 2 z) On compairing, we get 2 x + y - 2 z = 3 3 x - 2 y + z = - 1 - x + 2 y - 2 z = 2 Above system of equations is satisfied by the coordinates of option (c)$

77.

If the volume of the tetrahedron formed by the coterminous edges a, b and c is 4, then the volume of the parallelopiped formed by the coterminous edges a x b, b x c and c x a is

78.

The incentre of the triangle formed by the straight lines $y = \sqrt{3} x, y = - \sqrt{3} x and y = 3 is$ and y = 3 is

(0, 2)
(1, 2)
(2, 0)
(2, 1)

79.

The tangents to the parabola y = 4ax from an external point P make angles $θ_{1} and θ_{2}$ , with the axis of the parabola. Such that $\tan (θ_{1}) + \tan (θ_{2}) = b$ where b is constant. Then P lies on