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JEE Exam  >  JEE Test  >  KEAM Mock Test Series 2026  >  KEAM Paper 2 Mock Test - 5 - JEE MCQ

KEAM Paper 2 Mock Test - 5 Free Online Test 2026


Full Mock Test & Solutions: KEAM Paper 2 Mock Test - 5 (13 Questions)

You can boost your JEE 2026 exam preparation with this KEAM Paper 2 Mock Test - 5 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Total Questions: 13
  • - Analysis: Detailed Solutions & Performance Insights

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KEAM Paper 2 Mock Test - 5 - Question 1
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If one root of the equation ax2 + bx + c = 0 be n times the other root, then

Detailed Solution: Question 1

KEAM Paper 2 Mock Test - 5 - Question 2
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Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,
Then n(Ac∩Bc) =

Detailed Solution: Question 2

n(Ac ∩ Bc) = n(U) - n(A ∪ B)
= n(U) - [n(A) + n(B) - n(A ∩ B)]
= 700 - [200 + 300 - 100] = 300.

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KEAM Paper 2 Mock Test - 5 - Question 3
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Detailed Solution: Question 3

KEAM Paper 2 Mock Test - 5 - Question 4
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Detailed Solution: Question 4

KEAM Paper 2 Mock Test - 5 - Question 5
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In a ΔABC, a = 13 cm, b = 12 cm and c = 5 cm. The distance of A from BC is

Detailed Solution: Question 5

KEAM Paper 2 Mock Test - 5 - Question 6
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The lengths of sides of a triangle are in the ratio 5 : 12 : 13 and its area is 270 cm2. The respective lengths of sides of the triangle (in cm) are

Detailed Solution: Question 6

Let the lengths of sides of the triangle be 5x, 12x, 13x;  Obviously, the triangle is right-angled.
Hence, area of the Δ = ½ (12x) (5x) ⇒ 30x2 = 270 ⇒ x = 3 
Hence, the lengths of sides (in cm) are 15, 36 and 39.

KEAM Paper 2 Mock Test - 5 - Question 7
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If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

Detailed Solution: Question 7

KEAM Paper 2 Mock Test - 5 - Question 8
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The solution of the equation x2 dy/dx = x2 + xy + y2  is

Detailed Solution: Question 8

KEAM Paper 2 Mock Test - 5 - Question 9
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Assuming that f is continuous everywhere, (1/c)  is equal to

Detailed Solution: Question 9

KEAM Paper 2 Mock Test - 5 - Question 10
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If (√8+i)50 = 349 (a + ib), then a2 + b2 is

Detailed Solution: Question 10

KEAM Paper 2 Mock Test - 5 - Question 11
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The equation of the normal to the ellipse x2/a2 + y2/b2 = 1 at the positive end of the latus rectum is

Detailed Solution: Question 11

The equation of the normal at (x1, y1) to the given ellipse is

Here x1 = ae and y1 = b2/a 
So the equation of the normal at positive end of the latus rectum is 

KEAM Paper 2 Mock Test - 5 - Question 12
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 is equal to

Detailed Solution: Question 12

KEAM Paper 2 Mock Test - 5 - Question 13
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If  then f(2x) − f(x) is not divisible by

Detailed Solution: Question 13

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About this JEE Mock Test

This mock test contains 13 questions designed to replicate the pattern, difficulty level and time pressure of the actual JEE exam. It is part of KEAM Mock Test Series 2026 on EduRev, covering all the topics that are being asked as per the current JEE syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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If you found this KEAM Paper 2 Mock Test - 5 helpful, you can further enhance your JEE preparation with complete mock tests, previous year questions & practice tests available in KEAM Mock Test Series 2026 course on EduRev. The course offers:
  • Video Lectures: covering all the topics of KEAM Mock Test Series 2026
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for JEE preparation
  • MCQ Test: variety of questions as per the pattern of JEE exam
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