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OCR AS LEVEL MATHEMATICS B (MEI) H630/02 PURE MATHEMATICS AND STATISTICS EXAM TEST BANK ACTUAL 2026/2027 CURRENT PRACTICE QUESTIONS AND STUDY GUIDE COMPLETE ACCURATE EXAM APPROVED QUESTIONS WITH VERIFIED ANSWERS AND DETAILED RATIONALES (RELIABLE ANSWERS)

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OCR AS LEVEL MATHEMATICS B (MEI) H630/02 PURE MATHEMATICS AND STATISTICS EXAM TEST BANK ACTUAL 2026/2027 CURRENT PRACTICE QUESTIONS AND STUDY GUIDE COMPLETE ACCURATE EXAM APPROVED QUESTIONS WITH VERIFIED ANSWERS AND DETAILED RATIONALES (RELIABLE ANSWERS) NEWEST UPDATED VERSION 2026 EDITION |GUARANTEED PASS A+ |FULL REVISED EXAM |JUST RELEASED

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OCR AS LEVEL MATHEMATICS B (MEI) H630/02 PURE
MATHEMATICS AND STATISTICS EXAM TEST BANK ACTUAL
2026/2027 CURRENT PRACTICE QUESTIONS AND STUDY GUIDE
COMPLETE ACCURATE EXAM APPROVED QUESTIONS WITH
VERIFIED ANSWERS AND DETAILED RATIONALES (RELIABLE
ANSWERS) NEWEST UPDATED VERSION 2026 EDITION
|GUARANTEED PASS A+ |FULL REVISED EXAM |JUST
RELEASED




Algebra and Functions

1. Simplify 3x2−12x−2x−23x2−12 for x≠2x =2.

A) 3(x+2)3(x+2)

B) 3x+63x+6

C) 3x−63x−6

D) 3(x−2)x−2x−23(x−2)

Correct Answer: A) 3(x+2)3(x+2)

Rationale: Factor the
numerator: 3x2−12=3(x2−4)=3(x−2)(x+2)3x2−12=3(x2−4)=3(x−2)(x
+2). Cancel the common factor (x−2)(x−2) to get 3(x+2)3(x+2).

,2. Solve the inequality 2x2−5x−3<02x2−5x−3<0.

A) −12<x<3−21<x<3

B) x<−12x<−21 or x>3x>3

C) −3<x<12−3<x<21

D) x<−3x<−3 or x>12x>21

Correct Answer: A) −12<x<3−21<x<3

Rationale: Solve 2x2−5x−3=0→(2x+1)(x−3)=0→x=−12,32x2−5x−3=
0→(2x+1)(x−3)=0→x=−21,3. The parabola opens upwards, so the
inequality is satisfied between the roots.




3. Find the set of values of xx for which x−1x+2≤0x+2x−1≤0.

A) −2<x≤1−2<x≤1

B) −2≤x≤1−2≤x≤1

C) x≤−2x≤−2 or x≥1x≥1

D) −2<x<1−2<x<1

Correct Answer: A) −2<x≤1−2<x≤1

Rationale: Critical points: x=1x=1 (zero) and x=−2x=−2 (undefined).
Test intervals: for x<−2x<−2 the expression is positive,
for −2<x<1−2<x<1 it is negative, for x>1x>1 it is positive. Include
the zero at x=1x=1 but exclude x=−2x=−2 where the expression is
undefined.

,4. Given f(x)=2x2−8x+5f(x)=2x2−8x+5, express f(x)f(x) in the
form a(x−h)2+ka(x−h)2+k.

A) 2(x−2)2−32(x−2)2−3

B) 2(x+2)2−32(x+2)2−3

C) 2(x−2)2+32(x−2)2+3

D) 2(x+2)2+32(x+2)2+3

Correct Answer: A) 2(x−2)2−32(x−2)2−3

Rationale: Complete the
square: 2x2−8x+5=2(x2−4x)+5=2[(x−2)2−4]+5=2(x−2)2−8+5=2(x−
2)2−32x2−8x+5=2(x2−4x)+5=2[(x−2)2−4]+5=2(x−2)2−8+5=2(x−2)2
−3.




5. The function g(x)=x2+4x−1g(x)=x2+4x−1 has domain x≥−2x≥−2.
Find the range.

A) g(x)≥−5g(x)≥−5

B) g(x)≥−1g(x)≥−1

C) g(x)≥3g(x)≥3

D) g(x)≥5g(x)≥5

Correct Answer: A) g(x)≥−5g(x)≥−5

, Rationale: The vertex is
at x=−2x=−2. g(−2)=4−8−1=−5g(−2)=4−8−1=−5. Since the domain
is all x≥−2x≥−2 and the parabola opens upwards, the range
is y≥−5y≥−5.




6. Solve ∣2x−1∣=7∣2x−1∣=7.

A) x=4x=4 or x=−3x=−3

B) x=4x=4 only

C) x=−3x=−3 only

D) x=3x=3 or x=−4x=−4

Correct Answer: A) x=4x=4 or x=−3x=−3

Rationale: 2x−1=7⇒x=42x−1=7⇒x=4;
or 2x−1=−7⇒x=−32x−1=−7⇒x=−3.




7. Solve the equation 3x+7=223x+7=22.

A) x=5x=5

B) x=4x=4

C) x=6x=6

D) x=3x=3

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