JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

For \ any \ vector \ \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \ with \ 10|a_i| < 1, \ i = 1, 2, 3, \ consider \ the \ following \ statements :

For \ any \ vector \ \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \ with \ 10|a_i| < 1, \ i = 1, 2, 3, \ consider \ the \ following \ statements :

Choose the correct answer:

Explanation

From the provided solution image:

Given that $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

Since $a_1, a_2, a_3$ are fixed numbers, we can assume $|a_1| \le |a_2| \le |a_3|$

Now,

|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

\Rightarrow |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \ge a_3^2

\Rightarrow |\vec{a}| \ge |a_3|

which is maximum of $|a_1|, |a_2|, |a_3|$

so

|\vec{a}| \ge \max\{|a_1|, |a_2|, |a_3|\} \Rightarrow A \text{ is true}

Again,

|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \le a_3^2 + a_3^2 + a_3^2 = 3a_3^2

\Rightarrow |\vec{a}| \le \sqrt{3} |a_3|

and $|a_3|$ is maximum of $|a_1|, |a_2| \text{ and } |a_3|$

so

|\vec{a}| \le \sqrt{3} \max\{|a_1|, |a_2|, |a_3|\}

or

|\vec{a}| \le 3 \max\{|a_1|, |a_2|, |a_3|\} \Rightarrow B \text{ is true.}

Choose the correct answer:

Explore More