List of top Mathematics Questions

The roots of the equation $\begin{vmatrix}0&x&16\\ x&5&7\\ 0&9&x\end{vmatrix} = 0 $ are :

Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

\(\lim\limits_{n\rightarrow \infin}\frac{\sum(n^4-2n^3+n^2)}{\sum ((3n)^4+n^3-n^2)}\) is equal to

If the orthocentre of triangle formed by (8, 3), (5, 1) and (h, k) is (6, 1), then (h, k) lie on

The locus of P such that the ratio of distance P from A(3, 1) and B(1, 2) is 5 : 4 is

Sides of a triangle are AB = 9, BC = 7, AC = 8. Then cos 3C equals to

If α, ß are the roots of the equation x² - √2 x - 8 = 0 and A_n = αⁿ + ßⁿ, n ∈ N, then the value of (A₁₀ - √2A₉) / 2A₈

Find the range of \(\frac{1}{7}-\sin5x\)

If \(\int\frac{dx}{a^2\sin x+b^2\cos^2x}=\frac{1}{12}\tan^{-1}(3\tan x)+c,\)then the maximum value of a sinx + bcosx is____

Let $ {{(1+x)}^{n}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{n}}{{x}^{n}} $ .If $ {{a}_{1}},{{a}_{2}} $ and $ {{a}_{3}} $ are in $AP$, then the value of $n$ is

$\int \limits^{2017}_{2016} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} dx $ is equal to

$\int\frac{dx}{x-\sqrt{x}}$ is equal to

$\int\frac{2x+\sin2x}{1+\cos2x}\, dx$ is equal to

The coefficient of $x ^{49}$ in the product $\left(x-1\right)\left(x-2\right)\cdots\left(x-50\right)$ is

Let $t_n$ denote the $n^{th}$ term in a binomial expansion. If $ \frac{t_{6}}{t_{5}}$ in the expansion of $(a+ b)^{n+4}$ and $ \frac{t_{5}}{t_{4}}$ in the expansion of $(a + b)^n$ are equal, then $n$ is

If $ \tan \alpha =\frac{b}{a},a>b>0 $ and if $ 0

Let $u , v$ and $w$ be vectors such that $u + v + w = 0 .$ If $| u |=3,| v |=4$ and $| w |=5$ then $u \cdot v + v \cdot w + w \cdot u$ is equal to

Number of integral solutions of $ \frac{x+2}{{{x}^{2}}+1}>\frac{1}{2} $ is

The negation of $\left(p\vee\sim q\right)\wedge q$ is

Let $p$ : roses are red and q : the sun is a star. Then, the verbal translation of $ (-\text{ }p) \vee q $ is

If $y = x + \frac{1}{x}, x \ne 0$, then the equation $\left(x^{2}-3x+1\right)\left(x^{2}-5x+1\right)=6x^{2}$ reduces to

The value of $\sum\limits^{n}_{k=0}\left(i^{k}+i^{k+1}\right)$, where $i^2 = -1$, is equal to