AP EAMCET · Maths · Application of Derivatives
The number of distinct solutions of the equations \(x^{11}-x^7+x^4-1=0\) is
- A 9
- B 11
- C 10
- D 8
Answer & Solution
Correct Answer
(A) 9
Step-by-step Solution
Detailed explanation
Given equation is \[ \begin{aligned} & x^{11}-x^7+x^4-1=0 \\ & =x^7\left(x^4-1\right)+1\left(x^4-1\right)=0 \\ & =\left(x^4-1\right)\left(x^7+1\right)=0 \end{aligned} \] Case (i): \(x^4-1=0\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- If \(P\) and \(Q\) are two non-zero square matrices of the same order such that the product \(P Q=0\), then ........AP EAMCET 2020 Medium
- Let \(S(1,0)\) and \(S^{\prime}(0,1)\) be the foci of an eflipse such that \(\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}=2\) for any point \(\mathrm{P}\) on the ellipse. If \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(A^{\prime}\left(x_2, y_2\right)\) are the end points of the major axis of this ellipse, then \(x_1+x_2=\)AP EAMCET 2023 Medium
- \(A(2,3,5), B(\alpha, 3,3)\) and \(C(7,5, \beta)\) are the vertices of a triangle. If the median through \(A\) is equally inclined with the co-ordinate axes, then \(\cos ^{-1}\left(\frac{\alpha}{\beta}\right)=\)AP EAMCET 2019 Medium
- \(\int \sin ^3(x) \cdot \cos ^3(x) d x=\)AP EAMCET 2020 Medium
- If \(\frac{a x+5}{\left(x^2+b\right)(x+3)}=\frac{x+21}{12\left(x^2+b\right)}+\frac{c}{12(x+3)}\), then \(b^2=\)AP EAMCET 2025 Medium
- If \(x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]\) and \(\theta \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)\), then consider the following statements
I. \(\cosh x=\sec 2 \theta\)
II. \(\sinh x=-\tan 2 \theta\)AP EAMCET 2018 Hard
More PYQs from AP EAMCET
- If \(1,2,3\) and 4 are the roots of the equation \(x^4+a x^3+b x^2+c x+d=0\), then \(a+2 b+c\) is equal toAP EAMCET 2007 Medium
- In \(\triangle A B C, \sqrt{\frac{r r_2}{r_3 r_1}}=\)AP EAMCET 2025 Medium
- \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) are any 4 points and
\(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|=\lambda\)
(Area of \(\triangle A B C\) ) then \(\lambda=\)AP EAMCET 2022 Medium - If the tangent drawn at the point \(\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{x}_1, \mathrm{y}_1 \in \mathrm{~N}\) on the curve \(y=x^4-2 x^3+x^2+5 x\) passes through origin, then \(x_1+y_1=\)AP EAMCET 2025 Medium
- \(\mathrm{A}(4,3), \mathrm{B}(2,5)\) are two points. If P is a variable point on the same side as that of the origin with respect to the line AB and is at most at a distance of 5 units from the midpoint of \(A B\), then the locus of \(P\) isAP EAMCET 2025 Medium
- In the following nuclear reaction \(x\) stands for \(n \rightarrow p+e^{-}+x\)AP EAMCET 2015 Easy