Mathematics Test

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Mathematics Test - 11

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Weekly Quiz Competition

Question 1
4 / -1

The eccentricity of an ellipse with its centre at the origin is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is

A
4x² + 3y² = 1

B
3x² + 4y² = 12

C
4x² + 3y² = 12

D
3x² + 4y² = 1

Solution

\(e=\frac{1}{2}\)
directrix, \(\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\therefore a=4 \times \frac{1}{2}=2\)
\(b=2 \sqrt{1-\frac{1}{4}}=\sqrt{3}\)
Equation of ellipse is
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
\(\Rightarrow 3 x^{2}+4 y^{2}=12\)
Question 2
4 / -1

The greatest value of λ ≥ 0 for which both the equations 2x² + ( λ − 1)x + 8 = 0 and x² − 8x + λ + 4 = 0 have real roots is

A
9

B
15

C
12

D
16

Solution

The given equations are \(2 x^{2}+(x-1) x+8=0 \quad(1)\)
and \(x^{2}-8 x+\lambda+4=0 \quad(2)\)
These equations have real roots if \((x-1)^{2}-4 \cdot 2 \cdot 8 \geq 0\)
and \((-8)^{2}-4(a+4) \geq 0\)
ie. if \(\lambda^{2}-2 \lambda-63 \geq 0\) and \(-4 \lambda \geq-48\)
ie. if \((a-9)(a+7) \geq 0\) and \(\lambda \leq 12\) ie. if \(\lambda \leq-7\) or \(\lambda \geq 9\) and \(\lambda \leq 12\) \(\Rightarrow \lambda \in(-\infty,-7] \cup[9,12]\)
but in question it os given that \(\lambda \geq 0\)
so \(\lambda \in[9,12]\)
So greatest value of \(\lambda\) is 12
Question 3
4 / -1

In ΔABC, if tan A/2 tan C/2 = 1/2, then a, b, c are in

A
AP

B
GP

C
HP

D
None of these

Solution

\(\tan \frac{A}{2} \tan \frac{C}{2}=\frac{1}{2}\)
\(\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{2} \Rightarrow \frac{s-b}{s}=3 / 2\)
\(\Rightarrow 25 ? 2 b ? S=0 \Rightarrow a+c ? 3 b=0\)
This does not form any of the progressions (AP or GP or \(\mathrm{HP}\)
Question 4
4 / -1

\(\int \frac{\sin x-\cos x}{\sqrt{1-\sin 2 x}} e^{\sin x} \cos x d x\) is equal to

A
e^sinx + C

B
e^sinx-cosx + C

C
e^sinx+cosx+ C

D
None of these

Solution

\(\int \frac{\sin x-\cos x}{\sqrt{1-\sin 2 x}} e^{\sin x} \cos x d x\)
\(=\int \frac{\sin x-\cos x}{\sin x-\cos x} e^{\sin x} \cos x d x\)
\(=\int e^{\sin x} \cos x d x\)
\(=e^{\sin x}+c\)
Question 5
4 / -1

The function \(f: R \rightarrow[-1,1]\) defined by \(f(x)=\frac{|x|}{1+|x|}, x \in R\) is

A
Invertible

B
Injective but not surjective.

C
Surjective but not injective

D
Neither injective nor surjective.

Solution

\(f(x)=\frac{|x|}{1+|x|}=\left\{\begin{array}{ll}\frac{x}{1+x}, & x \geq 0 \\ \frac{-x}{1-x} & , x<0\end{array}\right.\)
Graph of \(f(x)\)

From graph we can see that, for two value of \(x^{\prime}, f(x)\) has same value so \(f(x)\) is not injective. Also, range of \(f(x)\) is 10.1 ) So it is not surjective also.
Question 6
4 / -1

\(\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots \ldots \infty\) is equal to

A
e

B
1/2e

C
2e

D
none of these

Solution

\(S=\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\)
nth term of given expression(Tn) \(=\frac{n}{(2 n+1) !}\)
\(T_{n}=\frac{1}{2}\left[\frac{2 n+1-1}{(2 n+1) !}\right]\)
\(T_{n}-\frac{1}{2}\left[\frac{2 n+1}{(2 n+1) !}-\frac{1}{(2 n+1) !}\right]\)
\(T_{n}=\frac{1}{2}\left[\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}\right]\)
\(S=\frac{1}{2}\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots . .\right]\)
\(S=\frac{1}{2} e^{-1}\)
\(S=\frac{1}{2 e}\)
Question 7
4 / -1

\(\int\left[f(x) g^{*}(x)-f(x) g(x)\right] d x\) is equal to

A
f (x) /g' (x)

B
f' (x) g (x) -f (x) g' (x)

C
f (x) g' (x) - f' (x) g (x)

D
f (x) g (x) + f (x) g (x)

Solution

\(\int\left[f(x) g^{\prime \prime}(x) \cdot f^{\prime}(x) g(x)\right] d x\)
\(=\int f(x) g^{\prime \prime}(x) d x-\int f^{\prime}(x) g(x) d x\)
\(-\left(f(x) g^{\prime}(x)-\int f(x) g^{\prime}(x) d x\right)-\left(g(x) f(x)-\int g \cdot(x) f(x)\right.\)
\(d x)\)
\(=f(x) g^{\prime}(x)-f(x) g(x)\)
Question 8
4 / -1

A mirror and a source of light are situated at the origin O and at a point on OX respectively. A ray of light from the source strikes the mirror and is reflected. If the direction ratios of the normal to the plane are proportional to 1, –1, 1 then direction cosines of the reflected ray are

A
1/3, 2/3, 2/3

B
-1/3, 2/3, 2/3

C
-1/3, -2/3, -2/3

D
-1/3, -2/3, 2/3

Solution

Let the source of light be situated at \(\mathrm{A}(\mathrm{a}, 0,0)\) where
\(a \neq 0\)
Let OA be the incident ray. OB be the reflected ray and ON be the normal to the mirror at \(0 .\)

\(\angle \mathrm{AON}=\angle \mathrm{NOB}=\frac{\theta}{2}(\mathrm{say})\)
Direction ratio of \(\overrightarrow{\mathrm{OA}}\) are proportional to a, 0,0 and \(\mathrm{s}\) o
its direction cosines are 1 , 0, 0
Direction cosines of ON are \(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
\(\cos \frac{\theta}{2}=\frac{1}{\sqrt{3}}\)
Let \(\mathrm{L}, \mathrm{m}, \mathrm{n}\) be the direction cosines of the reflected ray OB Then.
\(\frac{(+1}{2 \cos \frac{\theta}{2}}=\frac{1}{\sqrt{3}}, \frac{m+0}{2 \cos \frac{\theta}{2}}=-\frac{1}{\sqrt{3}}\) and \(\frac{n+0}{2 \cos \frac{\theta}{2}}=\frac{1}{\sqrt{3}}\)
\(\left(=\frac{2}{3}-1, \mathrm{m}=-\frac{2}{3}, \mathrm{n}=\frac{2}{3}\right.\)
\(\ell=-\frac{1}{3}, \mathrm{m}=-\frac{2}{3}, \mathrm{n}=\frac{2}{3}\)
Hence, direction cosines of the reflected ray are
\(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\)
Question 9
4 / -1

The coefficient of x in the equationx²+ px+ q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

A
3,10

B
-3, -10

C
-5, -18

D
18, 5

Solution

Let the equation (in written form) be \(x^{2}+17 x+q=0\)
Roots are -2,-15
So \(q=30\) And correct equation is \(x^{2}+13 x+30=0\) Hence roots are \(-3,-10 .\)
Question 10
4 / -1

Directions: Refer to the following functions:
A(x, y, z) = Max {max (x, y), min (y, z), min (x, z)}
B(x, y, z) = Max {max (x, y), min (y, z), max (x, z)}
C(x, y, z) = Max {min (x, y), min (y, z), min (x, z)}
D(x, y, z) = Min {max (x, y), max (y, z), max (x, z)}
Max (x, y, z) = Maximum of x, y and z.
Min (x, y, z) = Minimum of x, y and z.
Assume x, y and z as distinct integers.
For what condition, will A(x, y, z) be equal to Max (x, y, z)?

A
When x is maximum

B
When y is maximum

C
When z is maximum

D
Either (1) or (2)

Solution

Let us assign some values to x, y and z.
Put x = 1, y = 2, z = 3.
Now, A(x, y, z) = Max (2, 2, 1) = 1.
B (x, y, z) = Max (2, 2, 2) = 2
C (x, y, z) = Max (1, 2, 1) = 1
D (x, y, z) = Min (2, 3, 1) = 3
Max (x, y, z) = Min (1, 2, 3) = 1
Min (x, y, z) = Max (1, 2, 3) = 3
They will be equal only when z has the minimum value among the three, and x or y has the maximum value.
1. You might have checked by options and option 1 satisfies, but option 2 also satisfies.
2. You might have checked by options and option 2 satisfies, but option 1 also satisfies.
3. We can simply ignore this option as for x = 1, y = 2 , z = 3, it does not satisfy.

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Mathematics Test - 11

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Question 1

4x2 + 3y2 = 1

3x2 + 4y2 = 12

4x2 + 3y2 = 12

3x2 + 4y2 = 1

Solution

Question 2

9

15

12

16

Solution

Question 3

AP

GP

HP

None of these

Solution

Question 4

esinx + C

esinx-cosx + C

esinx+cosx + C

None of these

Solution

Question 5

Invertible

Injective but not surjective.

Surjective but not injective

Neither injective nor surjective.

Solution

Question 6

e

1/2e

2e

none of these

Solution

Question 7

f (x) /g' (x)

f' (x) g (x) -f (x) g' (x)

f (x) g' (x) - f' (x) g (x)

f (x) g (x) + f (x) g (x)

Solution

Question 8

1/3, 2/3, 2/3

-1/3, 2/3, 2/3

-1/3, -2/3, -2/3

-1/3, -2/3, 2/3

Solution

Question 9

3,10

-3, -10

-5, -18

18, 5

Solution

Question 10

When x is maximum

When y is maximum

When z is maximum

Either (1) or (2)

Solution

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4x² + 3y² = 1

3x² + 4y² = 12

4x² + 3y² = 12

3x² + 4y² = 1

e^sinx + C

e^sinx-cosx + C

e^sinx+cosx+ C