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

41.

The area of the region bounded by the curve y = x² and the line y = 16

$\frac{64}{3} sq unit$
$\frac{32}{3} sq unit$
$\frac{256}{3} sq unit$
$\frac{128}{3} sq unit$

42.

Two events A and B will be independent if

$P (A' \cap B') = (1 - P (A)) (1 - P (B))$
$P (A) + P (B) = 1$
$P (A) = P (B)$
A and B are mutually exclusive

43.

Area of the region bounded by the curve y = cos(x), x = 0 and x = $π$ is

4 sq unit
3 sq units
1 sq units
2 sq units

44.
The distance of the point (- 2, 4, - 5) from the line $\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$ is

$\sqrt{\frac{37}{10}}$

$\frac{37}{10}$

$\frac{\sqrt{37}}{10}$

$\frac{37}{\sqrt{10}}$

$\sqrt{\frac{37}{10}}$

The line passes through A(- 3, 4, - 8) and is parallel to the vector b = $3 \hat{i} + 5 \hat{j} + 6 \hat{k}$

Let M be the foot of the perpendicular from P(- 2, 4, - 5) on the given line.

We, have

$\Rightarrow |AP| = \sqrt{1 + 0 + 9} = \sqrt{10}$

Clearly, AM = Projection of AP on b

$\Rightarrow AM = |\frac{AP . b}{b}| = |\frac{(- \hat{i} - 3 \hat{k}) (3 \hat{i} + 5 \hat{j} + 6 \hat{k})}{|3 \hat{i} + 5 \hat{j} + 6 \hat{k}|}| = |\frac{- 3 - 18}{\sqrt{9 + 25 + 36}}| = |\frac{- 21}{\sqrt{70}}| = \frac{21}{\sqrt{70}} ∴ PM = \sqrt{{AP}^{2} - {AM}^{2}} = \sqrt{10 - \frac{441}{70}} = \sqrt{\frac{259}{70}} = \sqrt{\frac{37}{10}}$