MCQ
Let $f\,(x) = 1 + 2{x^2} + {2^2}{x^4} + ..... + {2^{10}}{x^{20}}$, then $f(x)$ has
- AMore than one minimum
- ✓Exactly one minimum
- CAt least one maximum
- DNone of these
✓
Answer
Correct option: B.
Exactly one minimum
b
(b) $f(x) = 1 + 2{x^2} + {2^2}{x^4} + {2^3}{x^6} + {...2^{10}}{x^{20}}$
(b) $f(x) = 1 + 2{x^2} + {2^2}{x^4} + {2^3}{x^6} + {...2^{10}}{x^{20}}$
$f'(x) = x(4 + {4.2^2}{x^2} + .... + {20.2^{10}}{x^{18}})$
$\therefore $ $f'(x) = 0$ ==> $x = 0$ only.
Also $f''(0) > 0$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})=$ _________.
→Area of the region bounded by the curve $\text{y}=\tan\text{x,}$ line $\text{x}=\frac{\pi}{4}$ and the x-axis is:
→- $\log2\text{ sq.}\text{units}$
- $\frac{1}{2}\log2\text{ sq.}\text{units}$
- $\frac{1}{3}\log2\text{ sq.}\text{units}$
- $5\log2\text{ sq.}\text{units}$
The objective function Z = 4x + 3y can be maximised subjected to the constraints $ 3\text{x}+4\text{y}\leq24,$ $8\text{x}+6\text{y}\leq48,$ $\text{x}\leq5,\text{y}\leq6;\text{x},\text{y}\leq0.$
→- At only one point
- At two points only.
- At an infinite number of points.
- None of these
Differential coefficient of ${\sin ^{ - 1}}{{1 - x} \over {1 + x}} \,\,,w.r.t$ $\sqrt x $ is
→Let a vector $\vec{\text{r}}$ make angles 60°, 30° with it and y-axes respectively. Find the angle $\vec{\text{r}}$ make with z-axis:
→- 30°
- 60°
- 90°
- 120°
The maximum value of Z = 4x + 2y subject to the constraints $2\text{x}+3\text{y}\leq18,\text{x}+\text{y}\geq10,\text{x},\text{y}\leq0$ is:
→- 36
- 40
- 30
- None of these
If a rectangle is inscribed in an equilateral triangle of side length $2 \sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is $....$
→If $g(x) = \int_0^x {{{\cos }^4}t\,dt,} $ then $g(x + \pi )$ equals
→The value of $p$ for which the vectors $2 \hat{i}+p \hat{j}+\hat{k}$ and $-4 \hat{i}-6 \hat{j}+26 \hat{k}$ are perpendicular to each other, is :
→The sum of all values of $\alpha$, for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{j}+4 \hat{k}$, $(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar, is equal to
→