Mathematics Test

Result

Mathematics Test - 7

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Weekly Quiz Competition

Question 1
3 / -1

The minimum value of 16 cot x+9 tan x in (0,π/2) is

A
12

B
6

C
24

D
25

Solution

In given interval, both cot $x$ and $\tan x$ are positive. Use $A M \geq G M$ for positive quantities. $\frac{16 \cot x+9 \tan x}{2} \geq(16 \cot x \cdot 9 \tan x)^{\frac{1}{2}}$
$16 \cot x+9 \tan x \geq 24$
Hence option C is the correct answer.
Question 2
3 / -1

If the determinant of a matrix of order 3×3 is formed by using the numbers 1 or -1 and minimum value of the determinant is −λ, then the value of λ is ___ .

A
2

B
5

C
5

D
-4

Solution
Question 3
3 / -1

The rate of change of surface area of a sphere of radius r when the radius is increasing at the rate of 2cm/sec is proportional to

A
1/r²

B
1/r

C
r²

D
r

Solution
Question 4
3 / -1

Let I be an open interval contained in the domain of a real function f′, then f(x) is called strictly decreasing function in I if-

A
x₁<x₂ in I=f(x₁)<f(x₂)

B
x₁<x₂ in I=f(x₁)>f(x₂)

C
x₁=x₂ in I=f(x₁)=f(x₂)

D
x₁=x₂ in I=f(x₁)<f(x₂)

Solution
Question 5
3 / -1

If A is a square matrix of order 3 such that |A|=2 then the value of |(adjA−1)−1| is ___ .

A
2

B
5

C
6

D
4

Solution
Question 6
3 / -1

Find the equation of a line passing through (−2,3) and parallel to tangent at origin for the circle x²+y²+x−y=0

A
x − 2y + 5 = 0

B
x − 4y + 3 = 0

C
x − y + 5 = 0

D
2x − y + 6 = 0

Solution
Question 7
3 / -1

Which of the following curves cuts the parabola y²=4ax at right angle

A
x²+y²=a²

B
y=e^−x/2a

C
y=ax

D
x²=4ay

Solution
Question 8
3 / -1

N = 144²⁵⁵ + 192²⁵⁵ then N is divisible by

A
7

B
35

C
49

D
25

Solution

Hence option A is the correct answer
Question 9
3 / -1

A
(0,e)

B
(e,∞)

C
(0,∞)

D
(−∞,∞)

Solution
Question 10
3 / -1

If A and B are two invertible matrices of the same order, then adj (AB) is equal to

A
adj B adj A

B
|B||A|B^-1A^-1

C
|A||B|(AB)^-1

D
All of the above

Solution

We have, adj $A=|A| A^{-1}$ $\operatorname{adj}(A B)=|A B|(A B)^{-1}=|A||B|(A B)^{-1}$
$=|A||B|\left(B^{-1} A^{-1}\right)$
$=\left(|B| B^{-1}\right)\left(|A| A^{-1}\right)$
$=a d j B$ adj $A$

Hence option D is the correct answer.
Question 11
3 / -1

If the equation $\sin^{- 1} (x^{2} + x + 1) + \cos^{- 1} (a x + 1) = \frac{π}{2}$ has exactly two solutions, then a cannot have the integral value/s

A
-1

B
1

C
2

D
All of the above

Solution

$\sin ^{-1}\left(x^{2}+x+1\right)+\cos ^{-1}(\lambda x+1)=\frac{\pi}{2}$
$\cos ^{-1}(\lambda x+1)=\frac{\pi}{2}-\sin ^{-1}\left(x^{2}+x+1\right)$
Taking cos on both sides, $\lambda x+1=x^{2}+x+1$ $x^{2}+(1-\lambda) x=0$
$x(x+1-\lambda)=0$
$x=0$ or $\lambda=x+1$
$\sin ^{-1}\left(x^{2}+x+1\right)$ is defined for $-1 \leq x^{2}+x+1 \leq 1$
$x^{2}+x+2 \geq 0$ is always true. $x(x+1) \leq 0 \Rightarrow x \in[-1,0]$
$\Rightarrow x+1 \in[0,1]$
$\lambda \in[0,1]$
$\lambda \neq-1,2$
Also when $\lambda=1$, there is only one solution to the given equation
i.e., $x=0$ $\mathrm{So}, \lambda \neq 1$

Hence option D is the correct answer.
Question 12
3 / -1

If x+y=12, then minimum value of x²+y²

A
72

B
144

C
48

D
36

Solution
Question 13
3 / -1

A particle moves along a curve so that its coordinates at time t are acceleration at t = 1 is-

A
j+2k

B
j+k

C
2j+k

D
none of these

Solution

Let $x, y, z$ be the magnitudes of the displacements along the three axes $X, Y, Z$ respectively. Let $V_{x}, V_{y}, V_{z}$ be the respective velocities and $a_{x}, a_{y}, a_{z}$ be the respective accelerations along the axes $X, Y, Z$ respectively. $x=t, y=\frac{1}{2} t^{2}, z=\frac{1}{3} t^{3}$
$V_{x}=\frac{d x}{d t}=1, V_{y}=\frac{d y}{d t}=t, V_{z}=\frac{d z}{d t}=t^{2}$
$a_{x}=\frac{d V_{x}}{d t}=0, a_{y}=\frac{d V_{y}}{d t}=1, a_{z}=\frac{d V_{x}}{d t}=2 t$
$\therefore$ acceleration at $t=1$ is $j+2 k$

Hence option A is the correct answer.
Question 14
3 / -1

A
1

B
0

C
-1

D
1/2

Solution
Question 15
3 / -1

Which of the following function is increasing in (0,π/2)

A
sinx

B
cos x

C
cot x

D
cosec x

Solution