Mathematics Test - 3

\(A=\left[\begin{array}{ccc}0 & 0 & -3 \\ 9 & 3 & 5 \\ 3 & 1 & 1\end{array}\right]\)

\(R_{1} \leftrightarrow R_{3}\)

\(=\left[\begin{array}{lll}3 & 1 & 1 \\ 9 & 3 & 5 \\ 0 & 0 & -3\end{array}\right]\)

\(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-3 \mathrm{R}_{1}\)

=\(\left[\begin{array}{lll}3 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & -3\end{array}\right]\)

\(\mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}+3 \mathrm{R}_{2}\)

=\(\left[\begin{array}{lll}3 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right]\)

\(\therefore \rho(\mathrm{A})=2\)

The 20 bricks can be arranged in a circular lawn in (20 - 1)! = 19! Ways.

It is semi circular. 

⇒ 19!/2 ways. Hence, (B).

From given question, we get

(x - 2)² + y² = 9 (x-4)²/1

⇒ y² - 8x² + 68x - 140 = 0

We can solve this problem using options given - 
On putting x = 5, we get

$2(5{)}^5-14(5{)}^4+31(5{)}^3-64(5{)}^2+19\left(5\right)+130=0$

Hence x = 5 satisfies the given equation.

Thus 5 is a root of the equation.
Other options will not satisfy the given equation so the answer is x = 5.

\(\begin{aligned} & f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3} \\ \therefore  & f^{\prime}(x)=\frac{1}{3}\left[\frac{1}{(x+1)^{2 / 3}}-\frac{1}{(x-1)^{2 / 3}}\right] \\=& \frac{(x-1)^{2 / 3}-(x+1)^{2 / 3}}{3\left(x^{2}-1\right)^{2 / 3}} \end{aligned}\)

Clearly, \(f^{\prime}(x)\) does not exists at \(x=\pm 1\)

Now, \(f^{\prime}(x)=0\), then \((x-1)^{2 / 3}=(x+1)^{2 / 3}\)

\(\Rightarrow {x}=0\)

Clearly, \(f^{\prime}(x)=0\) for any other value of \(x \in[0,1]\). 

The value of \(f(x)\) at \(x=0\) is 2 . 

So, the greatest value of \(f(x)\) is 2.

The given equations are

\(q x^{2}+p x+q=0\)

and \(x^{2}-4 q x+p^{2}=0\)

Since, root of the equation (i) are complex, therefore \(P^{2}-4 q^{2}<0\). 

Now, discriminant or equation (ii) is

\(16 q^{2}-4 P^{2}=-4\left(P^{2}-4 q^{2}\right)>0\)

Hence, roots are real and unequal.

We know that: The sequence \(a, b, c, d, \ldots\) is considered as an arithmetic progression; the harmonic progression can be written as \(1 / {a}, 1 / {b}, 1 / {c}, 1 / {d}, \ldots\)

\(a, b, c, d\) are in HP, so \(\frac{1} { a},\frac{1} { b}, \frac{1} { c}, \frac{1} { d},\) are in AP.

Let, \(\frac{1} { a},=x, \frac{1} { b}=x+y, \frac{1} { c}=x+2 y, \frac{1} { d}=x+3 y\)

\(\Rightarrow a=\frac{1}{ x}, b=\frac{1} {(x+y)}, c=\frac{1} {(x+2 y)}, d=\frac{1} {(x+3 y)}\)

Now,

\(\Rightarrow a+d\) \(=\frac{1}{x}+\frac{1}{x+3 y}=\frac{2 x+3 y}{x^{2}+3 x y}\)

\(\Rightarrow b+c\) \(=\frac{1}{x+y}+\frac{1}{x+2 y}=\frac{2 x+3 y}{x^{2}+2 x y+x y+y^{2}}=\frac{2 x+3 y}{x^{2}+3 x y+y^{2}}\)

Numerator of \({a}+{d}\) and \({b}+{c}\) are equal but, denominator of \({b}+{c}\) is greater than \({a}+{d}\)

So, \(a+d>b+c\)