Mathematics Test

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Mathematics Test - 17

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Weekly Quiz Competition

Question 1
4 / -1

If sin (sin-1 1/7 + cps-1 x) = 1, then x is equal to

A
1

B
0

C
6/7

D
1/7

Solution

\(\sin \left(\sin ^{-1} \frac{1}{7}+\cos ^{-1} x\right)=1\)
\(\Rightarrow \sin ^{-1} \frac{1}{7}+\cos ^{-1} x=\frac{\pi}{2}\)
\(\Rightarrow \sin ^{-1} \frac{1}{7}=\frac{\pi}{2}-\cos ^{-1} x\)
\(\Rightarrow \sin ^{-1} \frac{1}{7}=\sin ^{-1} x\)
\(\Rightarrow x=\frac{1}{7}\)
Question 2
4 / -1

If f : R → R is defined by

f (x) = x² – 3x; if x > 2

= x + 2; if x < 2,

then f(4) =

A
4

B
-4

C
8

D
-8

Solution

f (4) = x² - 3x

= (4)² - 3(4)

= 16 - 12 = 4

Question 3

4 / -1

Which of the following is logically equivalent to ∼ (∼p ⇒ q) ?

p ∨ q

p ∨ ∼ q

∼p ∨ q

∼p ∨ ∼q

Solution

p	q	~p	~q	~p=>q	~(~p=>q)	pvq	pv~q	~pvq	~pv~q
F	F	T	T	F	T	F	F	F	T
F	T	T	F	T	F	F	F	T	F
T	F	F	T	T	F	F	T	F	F
T	T	F	F	T	F	T	F	F	F

Question 4
4 / -1

Suppose f(x) = (x + 1)² for x - 1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to line y = x, then g(x) is equal to

A
-√x - 1, x ≥ 0

B
1/(x+1)²,x > -1

C
√x+1, x≥ -1

D
√x-1, x ≥ -1

Solution

reflection of in

\(f^{-1}(y)=\sqrt{y}-1\)
\(f^{-1}(x)=\sqrt{x}-1\)

\(g(x)=\sqrt{x}-1\)
Question 5
4 / -1

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

\(\int_{0}^{2 \pi} \frac{\sin 2 \theta}{a-b \cos \theta} d \theta,\) when \(a>b>0,\) is equal to

A
1

B
π

C
π/2

D
0

Solution

\(\operatorname{Let I} =\int_{0}^{2 \pi} \frac{\sin 2 \theta}{a-b \cos \theta} d \theta\)
Then, \(I=\int_{8}^{2 \pi} \frac{\sin 2(2 \pi-\theta)}{a-b \cos (2 \pi-\theta)} d \theta\)
\(\operatorname{tusing} \int_{2 \pi}^{\pi} f(x) d x=\int_{0}^{a} f(a-x) d x\)
Or \(\mid=\int_{0}^{2 \pi} \frac{-\sin 2 \theta}{a-b \cos \theta} d \theta\)
Or \(I=-1\)
\(\Rightarrow 21=0\)
\(\Rightarrow 1=0\)
Question 6
4 / -1

\(\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x\) is equal

A
1/4 log (x⁴+1) +C

B
– log (x⁴ + 1) + C

C
log (x4 + 1) + C

D
None of these

Solution

\(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x\)
\(=\int \frac{e^{\log x}}{1+x^{4}} d x\)
\(=\int \frac{x^{3}}{1+x^{4}} d x,\) put \(1+x^{4}=t\)
\(4 x^{3} d x=d t\)
\(I=\int \frac{1}{t}\left(\frac{d t}{4}\right)=\frac{1}{4} \log t+C\)
\(=\frac{1}{4} \log \left(1+x^{4}\right)+C\)
Question 7
4 / -1

The tangent to the circle x² + y² = 5 at (1, − 2) also touches the circle x² + y² − 8x + 6y + 20 = 0. Then the point of contact is

A
(-1, 3)

B
(3, -1)

C
(3, 1)

D
(-3, - 1)

Solution

Tangent at (1, − 2) to x² + y² = 5 is

x − 2y − 5 = 0 ... (i).

Centre and radius of

x² + y² − 8x + 6y + 20 = 0 are

C (4, − 3) and radius r = √5.

Perpendicular distance from

C (4, − 3) to (i) is radius.

∴ (i) is also a tangent to the second circle.

Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)
h-4/1 = k=3/-2 = - [1.4 -2. (-3)-5]

∴ (h, k) = (3, −1)
Question 8
4 / -1

What is the total number of different combinations of names, which can be made from the letters of the word MISSISSIPPI, using all the letters at a time?

A
150

B
11!/41 x 41 x2!

C
11!/10!

D
None of these

Solution

Here, we have 1 M, 4 I's, 4 S's and 2 P's,

Hence, the total number of different combinations of letters of the word MISSISSIPPI are 11!/41 x 41 x2! .
Question 9
4 / -1

If \(f(a+b-x)=f(x)\), then \(\int_{\frac{1}{2}}^{b} x f(x) d x\) is equal to

A

\(\frac{a+b}{2} \int_{\frac{1}{a}}^{b}(b-x) d x\)

B

\(a+b / 2 \int_{1}^{b} f(x) d x\)

C

\(b-a / 2 \int_{\frac{1}{a}}^{b} f(x) d x\)

D
None of these

Solution

\(\operatorname{Let} \mid=\int_{\frac{\pi}{2}}^{b} x f(x) d x\)
Then \(1=\int_{\frac{\pi}{2}}^{b}(a+b-x) f(a+b-x) d x\)
\(=(a+b) \int_{\frac{1}{2}}^{b} f(a+b-x) d x-\int_{a}^{b} x f(a+b-x) d x\)
\(=(a+b) \int_{a}^{b} f(x) d x-[x f(x) d x \quad \mid \because f(a+b-x)=f(x)]\)
\(2 \mid=(a+b) \int_{3} f(x) d x\)
\(\mid=\frac{a+b}{2} \int_{i}^{b} f(x) d x\)
Question 10
4 / -1

The equation of the normal to the ellipse x²/a² + y²/b² = 1 at the positive end of the latus rectum is

A
x + ey + e³a = 0

B
x − ey − e³a = 0

C
x − ey − e²a = 0

D
None of these

Solution

The equation of the normal at \(\left(x_{1}, y_{1}\right)\) to the given ellipse is \(\frac{\mathrm{a}^{2} \mathrm{x}}{\mathrm{x}_{1}}-\frac{\mathrm{b}^{2} \mathrm{y}}{\mathrm{y}_{1}}=\mathrm{a}^{2}-\mathrm{b}^{2}\)
Here \(\mathrm{x}_{1}=\mathrm{ae}\) and \(\mathrm{y}_{1}=\mathrm{b}^{2} / \mathrm{a}\)
So the equation of the normal at positive end of the latus rectum is \(\frac{a^{2} x}{a e}-\frac{b^{2} y}{\frac{b^{2}}{a}}=a^{2} e^{2}\)
\(\left(\because b^{2}=a^{2}\left(1-e^{2}\right)\right)\)
\(\Rightarrow \frac{\mathrm{ax}}{\mathrm{e}}-\mathrm{ay}=\mathrm{a}^{2} \mathrm{e}^{2}\)
\(\Rightarrow x-e y-e^{3} a=0\)

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Mathematics Test - 17

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Mathematics Test - 17

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SHARING IS CARING

Question 1

1

0

6/7

1/7

Solution

Question 2

4

-4

8

-8

Solution

Question 3

p ∨ q

p ∨ ∼ q

∼p ∨ q

∼p ∨ ∼q

Solution

Question 4

-√x - 1, x ≥ 0

1/(x+1)2,x > -1

√x+1, x≥ -1

√x-1, x ≥ -1

Solution

Question 5

1

π

π/2

0

Solution

Question 6

1/4 log (x4+1) +C

– log (x4 + 1) + C

log (x4 + 1) + C

None of these

Solution

Question 7

(-1, 3)

(3, -1)

(3, 1)

(-3, - 1)

Solution

Question 8

150

11!/41 x 41 x2!

11!/10!

None of these

Solution

Question 9

\(\frac{a+b}{2} \int_{\frac{1}{a}}^{b}(b-x) d x\)

\(a+b / 2 \int_{1}^{b} f(x) d x\)

\(b-a / 2 \int_{\frac{1}{a}}^{b} f(x) d x\)

None of these

Solution

Question 10

x + ey + e3a = 0

x − ey − e3a = 0

x − ey − e2a = 0

None of these

Solution

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1/(x+1)²,x > -1

1/4 log (x⁴+1) +C

– log (x⁴ + 1) + C

x + ey + e³a = 0

x − ey − e³a = 0

x − ey − e²a = 0