Mathematics Quiz For CTET Exam : 10th Feb 2020 For Algebra

Maths Questions

Q1. Multiply (x +2y) with (3x + 1)

(x +2y) को (3x + 1) से गुणा कीजिए।
(a) 3x² + 6xy +x +2y
(b) 3x² + 6xy+2y
(c) 3x² + x +2y
(d) none of these/इनमें से कोई नहीं

Q2. 3x + {5y – (2x – y)}

(a) x + 8y
(b) 2x + 6y
(c) x + 6y
(d) x + y

Q3. Find the remainder if P(x)=2x³ + 2x² – 3x – 4 is divided by (x-3)

यदि P(x)=2x³ + 2x² – 3x – 4 को (x-3) से विभाजित किया जाता है, तो शेषफल ज्ञात कीजिए।
(a) 50
(b) 54
(c) 59
(d) 46

Q4. Find the remainder if 2x³ + x² – 3x + 2 is divided by (x + 2)

यदि 2x³ + x² – 3x + 2 को (x + 2) से विभाजित किया जाता है, तो शेषफल ज्ञात कीजिए।
(a) 4
(b) -4
(c) 6
(d) -6

Q5. if x³ +kx² – 8x + 4 is divided by (x-2), then remainder is 8. Find the value of k.

यदि x³ +kx² – 8x + 4 को (x-2) से विभाजित किया जाता है तो शेषफल 8 प्राप्त होता है। k का मान ज्ञात कीजिए।
(a) 5
(b) 2
(c) 3
(d) -3

Q6. For what value(s) of k, the equations 10x ÷ 6y + 8 = 0 and 5x + 3y + k = 0 are coincident?

k के किस मान के लिए, समीकरण 10x ÷ 6y + 8 = 0 और 5x + 3y + k = 0 अनुरूप हैं?
(a) 4
(b) 6
(c) 8
(d) 2

Q7. Find the value of K for which the following system of linear equations have unique solution.

K का मान ज्ञात कीजिए, जिसके लिए रेखीय समीकरणों की निम्न प्रणाली का अद्वितीय समाधान है।
kx + 3y = 5
8x + y = 2
(a) 24
(b) 12
(c) 16
(d) 8

Q8. Check whether the given linear equations 3x + 2y + 7 = 0 and 5x + 7y + 3 = 0 are consistent or not?

दिए गए रेखीय समीकरणों 3x + 2y + 7 = 0 और 5x + 7y + 3 = 0 की जाँच कीजिए की वे अनुरूप हैं या नहीं?
(a) unique solution/अद्वितीय समाधान
(b) no solution/कोई समाधान नहीं
(c) both (a) and (b)/(a) और (b) दोनों
(d) None of these/इनमें से कोई नहीं

Q9. Three times of positive integer is 2 less than two times of the square of that integer. Find the integer.

एक धनात्मक पूर्णांक का तीन गुना, इस पूर्णांक के वर्ग के दुगुने से 2 कम है। पूर्णांक ज्ञात कीजिए।
(a) 2
(b) 4
(c) 1
(d) 3

Q10. Find the LCM of (x² -9) and (x² – 4)

(x² -9) और (x² – 4) का लघुत्तम समापवर्त्य ज्ञात कीजिए।
(a) (x – 3)( x + 2) (x – 2)
(b) (x – 3) (x + 3) ( x + 2)
(c) (x – 3) (x + 3) ( x + 2) (x – 2)
(d) None of these/इनमें से कोई नहीं

Solutions

S1. Ans.(a)
Sol. (x + 2y) × (3x + 1)
= (x) × (3x) + (x) × (1) + (2y) × (3x) + (2y) × 1
= 3x² +x+ 6xy + 2y
= 3x² + 6xy +x +2y

S2. Ans.(c)
Sol. 3x + {5y – (2x – y)}
= 3x + {5y – 2x + y}= 3x + {6y – 2x}
= 3x +6y – 2x = x + 6y

S3. Ans.(c)
Sol. Let P(x) = 2x³ +2x² – 3x – 4
We know x – 3 = 0 or x = 3
So, let a = 3
As per the theorem, remainder is P (a)
i.e.
P(a) = 2(a)³ + 2(a)² – 3(a) – 4
or P(a) = 2(3)³ + 2(3)² -3(3) – 4
= 54 +18 – 9 – 4 = 59
Hence remainder = 59

S4. Ans.(b)
Sol. Divisor is x + 2 hence x = -2
Let P(x) = 2x³+ x² – 3x + 2
Substituting x = 2 in P(x)
P(-2) = 2(-2)³ + (-2)(-2) -3 (-2) + 2
= 2 (-8) + 4 + 6 + 2
= -16 + 4 + 6 + 2 = – 4
Hence the remainder = – 4

S5. Ans.(c)
Sol. Divisor is (x – 2) or x = 2
The remainder P(2) = (2)³ + k(2)² – 8(2) + 4 = 8
8 + 4k – 16 + 4 = 8
4k – 4 = 8
4k = 8 + 4 = 12
k = 12/4=3

S9. Ans.(a)
Sol. Let the number be x:
Then 3x = 2x² – 2
or 2x² – 3x – 2 = 0
2x²-4x + x – 2 = 0
2x (x – 2) + 1 (x -2) = 0
(2x + 1) (x – 2)
(2x + 1) = 0 or (x – 2)
if (2x +1) = 0, then x = – 1/2
if (x -2) = 0, then x =2
Out of the above two values of x, x = 2 is a positive integer.
x = 2

S10. Ans.(c)
Sol. Let
P(x) = (x² – 9) = (x + 3) (x – 3)
and Q(x) = (x² – 4) = (x + 2) (x – 2)
We know that the LCM is the product of common and remaining factors of the given algebraic expressions.
So, LCM will be
= (x – 3) (x + 3) ( x + 2) (x – 2)

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