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 2013

Multiple Choice Questions

1.
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is

$8 π sq units$

4 sq units

5 sq units

8 sq units

8 sq units

Let length of the rectangle = a and breadth = b

Then, area of rectangle,

A = a . b ...(i)

Now, from figure

$AC = \sqrt{a^{2} + b^{2}} = 2 r [∵ r = radius] \Rightarrow a^{2} + b^{2} = 4 r^{2} \Rightarrow b^{2} = 4 r^{2} - a^{2} . . . (ii) Then from Eq . (i), we get A = a \sqrt{4 r^{2} - a^{2}} \Rightarrow A^{2} = (4 a^{2} r^{2} - a^{2}) Let u = 4 a^{2} r^{2} - a^{2} . . . (iii)$

So, A² is max or min according as u is max or min

Now, differentiate Eq (iii) two times w.r.t. a, we get

$\frac{du}{da} = 8 {ar}^{2} - 4 a^{3} \frac{d^{2} u}{{da}^{2}} = 8 r^{2} - 12 a^{2} For \max or \min, Put \frac{du}{da} = 8 {ar}^{2} - 4 a^{3} = 0 \Rightarrow 4 a (2 r^{2} - a^{2}) = 0 \Rightarrow a = \sqrt{2} . r [∵ a \neq 0] ∴ b = \sqrt{4 r^{2} - {(\sqrt{2} r)}^{2}} = \sqrt{4 r^{2} - 2 r^{2}} = \sqrt{2} . r [∵ from Eq . (ii)]$

$Given, radius (r) = 2 units ∴ a = b = 2 \sqrt{2} Then, \frac{d^{2} y}{{da}^{2}} = 8 {(2)}^{2} - 12 {(2 \sqrt{2})}^{2} = 32 - 96 < 0 ∴ Rectangle of maximum area is a square with each side a = b = 2 \sqrt{2} Hence, maximum area = 2 \sqrt{2} . 2 \sqrt{2} = 8 sq units .$

If the length of the subtangent at any point to the curve xyⁿ = a is proportional to the abscissa, then n is

any non-zero real number
2
- 2
1

If $\frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}$ , then $\sin^{- 1} (A) + \tan^{- 1} (B) + \sec^{- 1} (C)$ is equal to

$\frac{π}{2}$
$\frac{π}{6}$
0
$\frac{5 π}{6}$

$\lim_{x \to 0} \frac{\log_{e} (1 + x)}{3^{x} - 1}$ is equal to

$\log_{e} (3)$
0
$\log_{3} (e)$
1

If f(x) = $\{x, if x is irrational 0, if x is rational\}$ , then f is

continuous everywhere
discontinuous everywhere
continuous only at x = 0
continuous at all rational numbers

If A and B are square matnces of order n such that A² - B² = (A - B) (A + B), then which of the following will be true?

Either A or B is zero matrix
A = B
AB = BA
Either A or B is identity matrix

If A = $[\begin{matrix} α & 2 \\ 2 & α \end{matrix}] and |A^{3}| = 125$ , then $α$ is equal to

$\pm 1$
$\pm 2$
$\pm 3$
$\pm 5$

If $A = |\begin{matrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{matrix}| and B = |\begin{matrix} x & 1 \\ 1 & x \end{matrix}|$ , then $\frac{dA}{dx}$ is equal to