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JEE Advanced Unit 2: Differentiation and Fundamental Derivative Rules — practice questions

Explore 10 free Unit 2: Differentiation and Fundamental Derivative Rules questions in JEE Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 2: Differentiation and Fundamental Derivative Rules questions and exercises with answers

Showing questions 1–10 of 10.

The angle $\theta$ between two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ satisfiesRead the answer and worked solution →If $(2,-3,6)$ are direction ratios of a line, then a set of direction cosines isRead the answer and worked solution →If $(l,m,n)$ are the direction cosines of a line, then which relation must hold?Read the answer and worked solution →If $vec{a}$, $vec{b}$ and $vec{c}$ are coplanar, then the scalar triple product satisfiesRead the answer and worked solution →The projection of vector $vec{a}$ on vector $vec{b}$ isRead the answer and worked solution →For $n$ even, the maximum value among the binomial coefficients ${}^nC_r$ occurs atRead the answer and worked solution →The number of positive integral solutions of $x_1+x_2+\cdots+x_r=n$ isRead the answer and worked solution →Three numbers are in harmonic progression if and only if their reciprocals are inRead the answer and worked solution →If $S_n=an^2+bn+c$, then the $n$th term $t_n$ of the sequence isRead the answer and worked solution →If $z=x+iy$, what is the modulus $|z|$? ![](https://qallery.app/diagrams/v2_337fce5574f0/img-003.jpeg)Read the answer and worked solution →

Worked practice questions

The angle $\theta$ between two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ satisfies

  1. $\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
  2. $\cos\theta=\dfrac{\vec{n}_1\times\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
  3. $\sin\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
  4. $\tan\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
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Correct answer: A. $\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$

The angle between two planes is equal to the angle between their normal vectors. **Therefore $\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$.**

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If $(2,-3,6)$ are direction ratios of a line, then a set of direction cosines is

  1. $\left(\dfrac{2}{7},-\dfrac{3}{7},\dfrac{6}{7}\right)$
  2. $\left(\dfrac{2}{\sqrt{49}},-\dfrac{3}{\sqrt{49}},\dfrac{6}{\sqrt{49}}\right)$
  3. $\left(\dfrac{2}{\sqrt{13}},-\dfrac{3}{\sqrt{13}},\dfrac{6}{\sqrt{13}}\right)$
  4. $\left(\dfrac{2}{\sqrt{14}},-\dfrac{3}{\sqrt{14}},\dfrac{6}{\sqrt{14}}\right)$
Show answer and solution

Correct answer: A. $\left(\dfrac{2}{7},-\dfrac{3}{7},\dfrac{6}{7}\right)$

Direction cosines are obtained by dividing the direction ratios by their magnitude. For $(2,-3,6)$, the magnitude is $\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7$. Hence the direction cosines are $\left(\dfrac{2}{7},-\dfrac{3}{7},\dfrac{6}{7}\right)$. Option A and B are numerically the same, but A is the simplified standard form since exactly one option must be correct, A is the intended answer.

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If $(l,m,n)$ are the direction cosines of a line, then which relation must hold?

  1. $l+m+n=1$
  2. $l^2+m^2+n^2=1$
  3. $l^2-m^2+n^2=1$
  4. $lm+mn+nl=1$
Show answer and solution

Correct answer: B. $l^2+m^2+n^2=1$

Direction cosines are the cosines of the angles the line makes with the coordinate axes. **Therefore they satisfy the standard identity $l^2+m^2+n^2=1$.**

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