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

The system $(\begin{matrix} 1 & - 1 & 2 \\ 3 & 5 & - 3 \\ 2 & 6 & a \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 3 \\ b \\ 2 \end{matrix})$ has no solutions, if

a = - 5, $b \neq 5$
a = - 5, b = 5
$a \neq - 5, b = 5$
$a \neq - 5, b \neq 5$

The equation of displacement of a particle is x(t) = 5t² - 7t + 3. The acceleration at the moment when its velocity becomes 5 m/sec is

3 m/sec²
7 m/sec²
10 m/sec²
8 m/sec²

The range of x for which the formula $3 \sin^{- 1} (x) = \sin^{- 1} [x (3 - 4 x^{2})]$ hold is

$- \frac{1}{2} \leq x \leq \frac{1}{2}$
$- \frac{1}{4} \leq x \leq \frac{2}{3}$
$- \frac{1}{3} \leq x \leq 1$
$- \frac{2}{3} \leq x \leq \frac{2}{3}$

The mean value of the function $f (x) = \frac{2}{e^{x} + 1}$ on the interval [0, 2] is

$2 - \log_{e} (\frac{2}{e^{2} + 1})$
$2 + \log_{e} (\frac{2}{e^{2} + 1})$
$2 + \log_{e} (\frac{2}{e^{2} - 1})$
$- 2 + \log_{e} (\frac{2}{e^{2} - 1})$

If $\frac{7}{2} and 1$ are the roots of the equation $|\begin{matrix} 2 x & 3 & 7 \\ 2 & 2 x & 2 \\ 7 & 6 & 2 x \end{matrix}|$ = 0, then third root is

$- \frac{7}{2}$
$- \frac{9}{2}$
$- \frac{3}{2}$
$- \frac{5}{2}$

The function $y = \sqrt{2 x - x^{2}}$

increases in (0, 1) but decreases in (1, 2)
decreases in (0, 2)
increases m (1, 2) but decreases in (0, 1)
increases in (0, 2)

The interval in which the function $y = x - 2 \sin (x)$ ; $0 \leq x \leq 2 π$ increases throughout is

$(\frac{5 π}{3}, 2 π)$
$(0, \frac{π}{3})$
$(\frac{π}{3}, \frac{5 π}{3})$
$(0, \frac{π}{4})$

The inverse ofthe function y = $\frac{2^{x}}{1 + 2^{x}}$ is

$x = \log_{2} (\frac{1}{1 - 2^{y}})$
$x = \log_{2} (1 - \frac{1}{y})$
$x = \log_{2} (\frac{1}{1 - y})$
$x = \log_{2} (\frac{y}{1 - y})$

The domain of the definition of the function

$y = \frac{1}{\log_{10} (1 - x)} + \sqrt{(x + 2)}$ is

$x \geq - 2$
$- 3 < x \leq - 2$
$- 2 \leq x < 0$
$- 2 \leq x < 1$

10.
Let f(x) = $\{- 2 \sin (x) if x \leq - \frac{π}{2} Asin (x) + B if - \frac{π}{2} < x < \frac{π}{2} \cos (x) if x \geq \frac{π}{2}\}$
For what values of A and B, the function f(x) is continuous throughout the real line ?

A = - 1, B = 1

A = - 1, B = - 1

A = 1, B = - 1

A = 1, B = 1

A = - 1, B = 1

$Given, f (x) = \{- 2 \sin (x) if x \leq - \frac{π}{2} Asin (x) + B if - \frac{π}{2} < x < \frac{π}{2} \cos (x) if x \geq \frac{π}{2}\} . . . (i)$

From above conditions function f(x) is continuous throughout the real line, when function f(x) is continuous at x = $- \frac{π}{2} and \frac{π}{2}$

For continuity at x = $- \frac{π}{2}$

$\lim_{x \to \frac{π^{-}}{2}} f (x) = \lim_{x \to \frac{π^{+}}{2}} f (x) = f (- \frac{π}{2}) . . . (ii) \lim_{x \to \frac{π^{-}}{2}} f (x) = 2 \lim_{x \to \frac{π^{+}}{2}} f (x) = - A + B f (- \frac{π}{2}) = 2 ∴ From Eq . (ii), we get - A + B = 2 . . . (iii) For continuity at x = \frac{π}{2} \lim_{x \to \frac{π^{-}}{2}} f (x) = \lim_{x \to \frac{π^{+}}{2}} f (x) = f (\frac{π}{2}) . . . (iv) Here, \lim_{x \to \frac{π^{-}}{2}} f (x) = A + B \Rightarrow \lim_{x \to \frac{π^{+}}{2}} f (x) = 0 And f (\frac{π}{2}) = 0 ∴ From Eq . (iv) A + B = 0 ∴ From Eqs (iii) and (v) A = - 1, B = 1$