If y = 3 x^5 + 4 x^4 + 2x + 3, then - Vidyadip

MCQ

If $y = 3{x^5} + 4{x^4} + 2x + 3$, then

A
${y_4} = 0$
B
${y_5} = 0$
✓
${y_6} = 0$
D
None of these

✓

Answer

Correct option: C.

${y_6} = 0$

c
(c) Since highest power of $ x$ is $5$,

therefore ${y_6} = 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

$\int_{ - 1}^1 {\log \left( {\frac{{1 + x}}{{1 - x}}} \right)\,dx = } $

${d \over {dx}}\left[ {\left( {{{{{\tan }^2}2x - {{\tan }^2}x} \over {1 - {{\tan }^2}2x{{\tan }^2}x}}} \right)\cot 3x} \right] =$

If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:

The sum of the squares of sine of the angles made by the line $AB$ with $\text{OX, OY, OZ}$ where $O$ is the origin is:

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j} .$ If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$, then the value of $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is:

If $\vec u = \hat j + 4\hat k$, $\vec v = \hat i + 3\hat k$ and $\vec w = \cos \,\theta \hat i + \sin \,\theta \hat j$ are vectors in $3-$ dimensional space, then the maximum possible value of $\left| {\vec u \times \vec v.\vec w} \right|$ is

Assume X, Y, Z, W and P are matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$ and $p \times k$ respectively. Then the restriction on n, k and p so that PY + WY will be defined are:

Order and degree of the differential equation $\left(1+\left(\frac{d y}{d x}\right)^3\right)^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$ are respectively

A curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$ the area bounded by the curve and the $x$ -axis is

Minimize $z=\sum_{j=1}^n \sum_{i=1}^m c_{i j} x_{i j}$, subject to
$
\sum_{j=1}^n x_{i j}=a_i, i=1,2, \ldots, m \text { and } \sum_{i=1}^m x_{i j}=b_j, j=1,2, \ldots, n
$
is an L.P.P. with number of constraints