AP EAMCET · Maths · Binomial Theorem
The degree of the polynomial \(\left(x+\sqrt{x^4-1}\right)^9+\left(x-\sqrt{x^4-1}\right)^9\) is
- A \(14\)
- B \(15\)
- C \(16\)
- D \(17\)
Answer & Solution
Correct Answer
(D) \(17\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {}\left(x+\sqrt{x^4-1}\right)^9+\left(x-\sqrt{x^4-1}\right)^9 \\ & \text { Let } y=\sqrt{x^4-1} \Rightarrow y^2=x^4-1 \\ & =(x+y)^9+(x-y)^9 \\ & =2\left[{ }^9 C_1 x^1 y^8+{ }^9 C_3 x^3 y^6+{ }^9 C_5 x^5 y^4+{ }^9 C_7 x^7 \cdot y^2+{ }^9 C_9 x^7 y^0\right]…
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 the circles \((x+a)^2+(y+b)^2=a^2\) and \((x+c)^2+(y+d)^2=d^2\) cut orthogonally, then \(b(b-2 d)=\)AP EAMCET 2019 Easy
- If the equation
\(3 x^2+7 x y+2 y^2+2 g x+2 f y+2=0\) represents
a pair of intersecting lines and the square of the distance of their point of intersection from the origin is \(\frac{2}{5}\), then \(f^2+g^2=\)AP EAMCET 2018 Medium - If the chord of the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\) having \((1,1)\) as its middle point is \(\dot{x}+\alpha y=\beta\), thenAP EAMCET 2024 Easy
- The lines \(y=2 x+\sqrt{76}\) and \(2 y+x=8\) touch the ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\). If the point of intersection of these two lines lie on a circle. whose centre coincides with the centre of that ellipse, then the equation of that circle isAP EAMCET 2017 Hard
- If \(\cos A=\frac{7}{25}\) and \(\frac{3 \pi}{2} < A < 2 \pi\), then \(\cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=\)AP EAMCET 2018 Medium
- If \(I=\int_{-a}^a\left(x^4-2 x^2\right) d x\), then \(I\) is minimum at \(a=\)AP EAMCET 2023 Medium
More PYQs from AP EAMCET
- A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is \(24 \mu \mathrm{~T}\). A downward current of 18 A is established in a vertical wire kept at a distance of 20 cm east of the magnet. The new time period of oscillations of the magnet isAP EAMCET 2024 Medium
- The eccentricity of the conic \(\frac{5}{r}=2+3 \cos \theta+4 \sin \theta\) isAP EAMCET 2009 Hard
- If the function \(f(x)=a x^3+b x^2+11 x-6\) satisfies the conditions of Rolle's theorem in \([1,3]\) and \(f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0\), then \(a+b=\)AP EAMCET 2019 Easy
- Let \(\mathrm{L}\left(\mathrm{x}_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbola. Then the length of its transverse axis isAP EAMCET 2022 Medium
- \(\bar{a}=\bar{i}+\bar{j}-2 \bar{k}, \bar{b}=\bar{i}+2 \bar{j}-3 \bar{k}, \bar{c}=2 \bar{i}-\bar{j}+\bar{k}\) are three vectors. If \(\bar{r}\) is a vector such that \(\overline{\mathrm{r}} \cdot \overline{\mathrm{a}}=0, \overline{\mathrm{r}} \cdot \overline{\mathrm{c}}=3\) and \([\overline{\mathrm{r}} \overline{\mathrm{a}} \overline{\mathrm{b}}]=0\), then \(|\overline{\mathrm{r}}|=\)AP EAMCET 2025 Medium
- If the pair of straight line given by \(A x^2+2 H x y+B y^2=0\), where \(\left(H^2>A B\right)\), forms an equilateral triangle with the line \(a x+b y+c=0\), then \((A+3 B)(3 A+B)=\)AP EAMCET 2020 Hard