Mathematics Test

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Mathematics Test - 19

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Weekly Quiz Competition

Question 1
4 / -1

If \(f(x)=\left|\begin{array}{ccc}a & -1 & 0 \\ a x & a & -1 \\ a x^{2} & a x & a\end{array}\right|,\) then \(f(2 x)-f(x)\) is not divisible by

A
x

B
a

C
2a + 3x

D
x²

Solution

Doing \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{xR}_{2}\) and
\(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{x} \mathrm{R}_{1}\)
we get \(\begin{aligned} f(x) &=\left|\begin{array}{ccc}a & -1 & 0 \\ 0 & a+x & -1 \\ 0 & 0 & a+x\end{array}\right| \\ &=a(a+x)^{2} \end{aligned}\)
So. \(f(2 x)-f(x)\)
\(=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\)
\(=a(a+2 x-a-x)(a+2 x+a+x)\)
\(=a x(2 a+3 x)\)
So, \(f(2 x)-f(x)=a x(2 a+3 x)\)
Thus, \(f(2 x)-f(x)\) is divisible by \(a, x\) and \((2 a+3 x)\)
Question 2
4 / -1

Find the area of the triangle whose vertices are \((1,-1),(-4,0)\) and \((-2,-3)\).

A
8.1

B
4.9

C
7.2

D
6.5

Solution

The given vertices are \((1,-1),(-4,0)\) and \((-2,-3)\).

We know that,the area of triangle whose vertices are \(({x}_{1},{y}_{1}),({x}_{2}, {y}_{2})\) and \(({x}_{3}, {y}_{3})\) is given by,

\( \frac{1}{2} \times[{x}_{1}({y}_{2}-{y}_{3})+{x}_{2}({y}_{3}-{y}_{1})+{x}_{3}({y}_{1}-{y}_{2})]\)

Therefore, according to the question,

\(\Rightarrow \frac{1}{2} \times[1\{0-(-3)\}+(-4)\{(-3)-(-1)\}+(-2)\{(-1)-0\}]\)

\(\Rightarrow \frac{1}{2} \times[3+(-4) \times(-2)+(-2) \times(-1)]\)

\(\Rightarrow \frac{1}{2} \times[3+8+2]\)

\(= \frac{1}{2} \times 13=6.5\)
Question 3
4 / -1

The solution of the equation x² dy/dx = x² + xy + y² is

A
tan^-1(y/x) = log x + C

B
tan^-1(x/y) = log x + C

C
tan^-1 (x/y) = log y + C

D
none of these

Solution

Given equation is \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}},\) which is homogeneous
Putting \(y=v x\) and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\). we get \(v+x \frac{d v}{d x}=\left(1+v+v^{2}\right)\)
\(\Rightarrow \int \frac{d v}{1+v^{2}}=\int \frac{1}{x} d x\)
\(\tan ^{-1} v=\log x+C\)
\(\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log x+C\)
Question 4
4 / -1

The equation x³ – 3x + [a] = 0, will have three real and distinct roots if –

(where [ ] denotes the greatest integer function)

A
a ∈ (–∞, 2)

B
a ∈ (0,2)

C
a ∈ (–∞, -2) U (0, ∞)

D
a ∈ [–1, 2)

Solution

f(x) = x³ – 3x + [a]

Let [a] = t (where t will be an integer)

f(x) = x³ – 3x + t ……….(i)

⇒ f ’(x) = 3x² – 3

⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = -1

so f(x) = 0 will have three distinct and real solution when f (1). f(-1) < 0 ……………. (ii)

Now,

f(1) = (1)³ -3(1) + t = t – 2

f(–1) = (–1)³ – 3 (–1) + t = t + 2

From equation (ii)

(t –2) (t + 2) < 0

⇒ t ∈ (-2, 2)

Now t = [a]

Hence [a] ∈ (-2, 2)

⇒ a ∈ [-1, 2)
Question 5
4 / -1

Let function f : R → R be defined by f(x) = cos x for x∈ R. Then f is

A
one to one and onto

B
one to one but not onto

C
onto, but not one to one

D
neither one to one nor onto

Solution

f(0) = cos 0 = 1 and f(2π) = cos (2π) = 1

So, f is not one to one and cos(x) lie between - 1 and 1.

Therefore, range is not equal to its codomain.

Hence, it is not onto.
Question 6
4 / -1

If g(x) satisfies the conditions of Rolle’s theorem in [1, 2] and g′(x) = f(x), then\(\int_{1}^{2} f(x) d x\)is equal to

A
1

B
3

C
0

D
2

Solution

As g(x) satisfies the condition of Rolle’s theorem in [1, 2], g(x) is continuous in the interval and
g(1) = g(2).

now,
\(g^{\prime}(x)=f(x)\)
\(\left.\therefore \int_{1}^{2} f(x) d x=g(x)\right]_{1}^{2}\)
\(=g(2)-g(1)=0\)
Question 7
4 / -1

If the latus rectum of a hyperbola through one focus, subtends 60 ° angle at the other focus, then its eccentricity is

A
√2

B
√3

C
√5

D
√6

Solution

Taking only positive value of e as eccentricity cannot be negative.

\(\tan 30^{\circ}=\frac{b^{2}}{2 a e}\)
More Exal
\(\Rightarrow \frac{2}{\sqrt{3}} e=e^{2}-1\)
\(\Rightarrow \sqrt{3 e^{2}}-2 e-\sqrt{3}=0\)
\(\Rightarrow \mathrm{e}=\frac{2+\sqrt{4+12}}{2 \sqrt{3}}\)
\(=\frac{2+4}{2 \sqrt{3}}\)
\(\Rightarrow e=\frac{3}{\sqrt{3}}=\sqrt{3}\)
Taking only positive value of e as eccentricity cannot be negative.
Question 8
4 / -1

At a point, the addition of two active force is 18 N. If the magnitude of resultant is 12 N and meet at the right angle. Then, the magnitude of forces are:

A
5 N, 13 N

B
6 N, 12 N

C
8 N, 10 N

D
None of these

Solution

Let the forces be P and Q such that (P > Q) and OACB be a parallelogram, then:

P + Q = 18 ... (i)

From figure

AC² = OA² + OC²

⇒ Q² = P² + 144

⇒ Q² - P² = 144 ... (ii)

Solving Eqs. (i) and (ii), we get,

P = 5 N, Q = 13 N
Question 9
4 / -1

Find the range of f(x) = x² + x + 2/x² + x + 1.

A
(1, ∞)

B
(1, 11/7)

C
(1, 7/3)

D
(1, 7/5)

Solution

We can write as 1+ 1/x² + x + 1. After solving, we get third option is correct.
Question 10
4 / -1

If tan^-1(x - 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1 3x, then x is equal to

A
± 1/2

B
0, 1/2

C
0, -1/2

D
0, ±1/2

Solution

\(\tan ^{-1}(x-1)+\tan ^{-1}(x)+\tan ^{-1}(x+1)=\tan ^{-1} 3 x\)
\(\Rightarrow \tan ^{-1}(x-1)+\tan ^{-1}(x)=\tan ^{-1} 3 x-\tan ^{-1}(x+0)\)
\(\Rightarrow \tan ^{-1}\left[\frac{(x-1)+x}{1-(x-1)(x)}\right]=\tan ^{-1}\left[\frac{3 x-(x+1)}{1+3 x(x+1)}\right]\)
\(\Rightarrow \frac{2 x-1}{1-x^{2}+x}=\frac{2 x-1}{1+3 x^{2}+3 x}\)
\(\Rightarrow\left(1-x^{2}+x\right)(2 x-1)=\left(1+3 x^{2}+3 x\right)(2 x-1)\)
On simplification. we get \(x=0, \pm 1 / 2\) \(O R\)
After solving till (1), now put values from the options and check which one satisfy Putting \(x=0\), we get \(\frac{-1}{1}=\frac{-1}{1}\), thus satisfy Putting \(x=\frac{1}{2},\) we get. \(\frac{2 \frac{1}{2}-1}{1-\frac{1}{4}+\frac{1}{2}}=\frac{2 \frac{1}{2}-1}{3 \cdot \frac{1}{4}+3 \cdot \frac{1}{2}+1} \Rightarrow 0=0 .\) thus satisfy
Putting \(x=\frac{-1}{2},\) we get \(\frac{2-\frac{1}{2}-1}{1-\frac{1}{4}-\frac{1}{2}}=\frac{2 \cdot \frac{1}{2}-1}{3 \cdot \frac{1}{4}+3\left(-\frac{1}{2}\right)+1}\)
\(\Rightarrow \frac{\frac{-2}{4-1-2}}{4}=\frac{-2}{\frac{3}{4}-\frac{3}{2}+1} \Rightarrow \frac{-2}{\frac{1}{4}}=\frac{-2}{\frac{1}{4}} \Rightarrow-8--8,\) thus satisfy

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Mathematics Test - 19

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Mathematics Test - 19

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SHARING IS CARING

Question 1

x

a

2a + 3x

x2

Solution

Question 2

8.1

4.9

7.2

6.5

Solution

Question 3

tan-1 (y/x) = log x + C

tan-1 (x/y) = log x + C

tan-1 (x/y) = log y + C

none of these

Solution

Question 4

a ∈ (–∞, 2)

a ∈ (0,2)

a ∈ (–∞, -2) U (0, ∞)

a ∈ [–1, 2)

Solution

Question 5

one to one and onto

one to one but not onto

onto, but not one to one

neither one to one nor onto

Solution

Question 6

1

3

0

2

Solution

Question 7

√2

√3

√5

√6

Solution

Question 8

5 N, 13 N

6 N, 12 N

8 N, 10 N

None of these

Solution

Question 9

(1, ∞)

(1, 11/7)

(1, 7/3)

(1, 7/5)

Solution

Question 10

± 1/2

0, 1/2

0, -1/2

0, ±1/2

Solution

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x²

tan^-1(y/x) = log x + C

tan^-1(x/y) = log x + C

tan^-1 (x/y) = log y + C