Section A
▶️ Answer/Explanation
Sol.
C : \((x-4)^2 + (y-3)^2 = 4\)
E : \(\frac{(x-2)^2}{9} + \frac{y^2}{5} = 1\)
Lower Extremity of vertical diameter of circle \(\rightarrow (4,1)\)
Put in ellipse \(\frac{(4-2)^2}{9} + \frac{1^2}{5} – 1\)
\(= \frac{4}{9} + \frac{1}{5} – 1\)
\(= \frac{29}{45} – 1 < 0\)
Two Solutions
Answer (C)
✅ Official Ans. by NTA (C)
C : \((x-4)^2 + (y-3)^2 = 4\)
E : \(\frac{(x-2)^2}{9} + \frac{y^2}{5} = 1\)
Lower Extremity of vertical diameter of circle \(\rightarrow (4,1)\)Put in ellipse \(\frac{(4-2)^2}{9} + \frac{1^2}{5} – 1\)
\(= \frac{4}{9} + \frac{1}{5} – 1\)
\(= \frac{29}{45} – 1 < 0\)
Two Solutions
Answer (C)
✅ Official Ans. by NTA (C)
▶️ Answer/Explanation
Sol.
\(f(x)=a\begin{vmatrix} 1 & -1 & 0 \\ x & a & -1 \\ x^2 & ax & a \end{vmatrix}\)
\(= a[1(a^2+ax)+1(ax+x^2)]\)
\(\Rightarrow f(x) = a(x+a)^2\)
so, \(f'(x) = 2a(x+a)\)
as, \(2f'(10) – f'(5) + 100 = 0\)
\(\Rightarrow 2 \times 2a(10+a) – 2a(5+a) + 100 = 0\)
\(\Rightarrow 40a + 4a^2 – 10a – 2a^2 + 100 = 0\)
\(2a^2 + 30a + 100 = 0\)
\(\Rightarrow a^2 + 15a + 50 = 0\)
\((a+10)(a+5) = 0\)
a = -10 or a = -5
Required \(= (-10)^2 + (-5)^2 = 125\)
✅ Official Ans. by NTA (C)
\(f(x)=a\begin{vmatrix} 1 & -1 & 0 \\ x & a & -1 \\ x^2 & ax & a \end{vmatrix}\)
\(= a[1(a^2+ax)+1(ax+x^2)]\)
\(\Rightarrow f(x) = a(x+a)^2\)
so, \(f'(x) = 2a(x+a)\)
as, \(2f'(10) – f'(5) + 100 = 0\)
\(\Rightarrow 2 \times 2a(10+a) – 2a(5+a) + 100 = 0\)
\(\Rightarrow 40a + 4a^2 – 10a – 2a^2 + 100 = 0\)
\(2a^2 + 30a + 100 = 0\)
\(\Rightarrow a^2 + 15a + 50 = 0\)
\((a+10)(a+5) = 0\)
a = -10 or a = -5
Required \(= (-10)^2 + (-5)^2 = 125\)
✅ Official Ans. by NTA (C)
▶️ Answer/Explanation
Sol.
\(a = \alpha – i\beta; \alpha \in R; \beta \in R\)
\(4ix + (1+i)y = 0\) and
\(8\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)x + \bar{a}y = 0\)
For more than one solution, the determinant of coefficients is zero:
\(\begin{vmatrix} 4i & 1+i \\8e^{i2\pi/3} & \bar{a} \end{vmatrix} = 0\)
\(\Rightarrow 4i\bar{a} – (1+i)8e^{i2\pi/3} = 0\)
\(\Rightarrow 4i(\alpha+i\beta) – 8(1+i)\left(\frac{-1+i\sqrt{3}}{2}\right) = 0\)
\(\Rightarrow i\alpha – \beta + 1 + \sqrt{3} + i(1-\sqrt{3}) = 0\)
Equating real and imaginary parts to zero:
\(\Rightarrow \beta = \sqrt{3}+1\)
\(\Rightarrow \alpha = \sqrt{3}-1\)
So, \(\frac{\alpha}{\beta} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}\)
✅ Official Ans. by NTA (B)
\(a = \alpha – i\beta; \alpha \in R; \beta \in R\)
\(4ix + (1+i)y = 0\) and
\(8\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)x + \bar{a}y = 0\)
For more than one solution, the determinant of coefficients is zero:
\(\begin{vmatrix} 4i & 1+i \\8e^{i2\pi/3} & \bar{a} \end{vmatrix} = 0\)
\(\Rightarrow 4i\bar{a} – (1+i)8e^{i2\pi/3} = 0\)
\(\Rightarrow 4i(\alpha+i\beta) – 8(1+i)\left(\frac{-1+i\sqrt{3}}{2}\right) = 0\)
\(\Rightarrow i\alpha – \beta + 1 + \sqrt{3} + i(1-\sqrt{3}) = 0\)
Equating real and imaginary parts to zero:
\(\Rightarrow \beta = \sqrt{3}+1\)
\(\Rightarrow \alpha = \sqrt{3}-1\)
So, \(\frac{\alpha}{\beta} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}\)
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
\(AB=I\)
\(|adj(B~adj(2A))| = |B~adj(2A)|^2\)
\(=|B|^2|adj(2A)|^2\)
\(=|B|^2(|2A|^2)^2 = |B|^2(2^6|A|^2)^2\)
\(|A|=\frac{1}{8}\) and \(|AB|=1 \Rightarrow |A||B|=1\)
\(\Rightarrow \frac{1}{8}|B|=1\)
\(\Rightarrow |B|=8\)
required value \(= 64\)
✅ Official Ans. by NTA (C)
\(AB=I\)
\(|adj(B~adj(2A))| = |B~adj(2A)|^2\)
\(=|B|^2|adj(2A)|^2\)
\(=|B|^2(|2A|^2)^2 = |B|^2(2^6|A|^2)^2\)
\(|A|=\frac{1}{8}\) and \(|AB|=1 \Rightarrow |A||B|=1\)
\(\Rightarrow \frac{1}{8}|B|=1\)
\(\Rightarrow |B|=8\)
required value \(= 64\)
✅ Official Ans. by NTA (C)
▶️ Answer/Explanation
Sol.
\(S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+…..\)
Considering infinite sequence,
\(\frac{S}{7}=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7} = 2+\frac{4}{7}+\frac{6}{7^2}+\frac{8}{7^3}+\frac{10}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7^2} = \frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+\frac{8}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7}(1-\frac{1}{7}) = 2+\frac{2}{7}+\frac{2}{7^2}+\frac{2}{7^3}+…..\)
\(\Rightarrow \frac{6^2S}{7^2} = \frac{2}{1-\frac{1}{7}} = \frac{2}{6} \times 7\)
\(S = \frac{2 \times 7^3}{6^3} \Rightarrow 4S = \frac{7^3}{3^3} = (\frac{7}{3})^3\)
✅ Official Ans. by NTA (C)
\(S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+…..\)
Considering infinite sequence,
\(\frac{S}{7}=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7} = 2+\frac{4}{7}+\frac{6}{7^2}+\frac{8}{7^3}+\frac{10}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7^2} = \frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+\frac{8}{7^4}+…..\)
\(\Rightarrow \frac{6S}{7}(1-\frac{1}{7}) = 2+\frac{2}{7}+\frac{2}{7^2}+\frac{2}{7^3}+…..\)
\(\Rightarrow \frac{6^2S}{7^2} = \frac{2}{1-\frac{1}{7}} = \frac{2}{6} \times 7\)
\(S = \frac{2 \times 7^3}{6^3} \Rightarrow 4S = \frac{7^3}{3^3} = (\frac{7}{3})^3\)
✅ Official Ans. by NTA (C)
▶️ Answer/Explanation
Sol.
\(a_1, a_2, … A.P.; a_1=2; a_{10}=3; d_1 = \frac{1}{9}\)
\(b_1, b_2, … A.P.; b_1=\frac{1}{2}; b_{10}=\frac{1}{3}; d_2 = \frac{-1}{54}\)
[Using \(a_1b_1=1\) and \(a_{10}b_{10}=1\); \(d_1\) & \(d_2\) are common differences respectively]
\(a_4 b_4 = (a_1+3d_1)(b_1+3d_2)\)
\(=(2+\frac{1}{3})(\frac{1}{2}-\frac{1}{18})\)
\(=(\frac{7}{3})(\frac{8}{18})=\frac{28}{27}\)
✅ Official Ans. by NTA (D)
\(a_1, a_2, … A.P.; a_1=2; a_{10}=3; d_1 = \frac{1}{9}\)
\(b_1, b_2, … A.P.; b_1=\frac{1}{2}; b_{10}=\frac{1}{3}; d_2 = \frac{-1}{54}\)
[Using \(a_1b_1=1\) and \(a_{10}b_{10}=1\); \(d_1\) & \(d_2\) are common differences respectively]
\(a_4 b_4 = (a_1+3d_1)(b_1+3d_2)\)
\(=(2+\frac{1}{3})(\frac{1}{2}-\frac{1}{18})\)
\(=(\frac{7}{3})(\frac{8}{18})=\frac{28}{27}\)
✅ Official Ans. by NTA (D)
▶️ Answer/Explanation
Sol.
m = L. max
n = L. min
\(f(x)=\int_0^{x^2} \frac{t^2-5t+4}{2+e^t}dt\)
\(f'(x) = \frac{(x^4-5x^2+4)2x}{2+e^{x^2}} = \frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}\)
\(= \frac{2x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^2}}\)
so, m = 2 and n = 3
✅ Official Ans. by NTA (B)
m = L. max
n = L. min
\(f(x)=\int_0^{x^2} \frac{t^2-5t+4}{2+e^t}dt\)
\(f'(x) = \frac{(x^4-5x^2+4)2x}{2+e^{x^2}} = \frac{2x(x^2-1)(x^2-4)}{2+e^{x^2}}\)
\(= \frac{2x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^2}}\)
so, m = 2 and n = 3
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
At right hand vicinity of x=0 given equation does not satisfy
\(\therefore LHS = \int_1^1 t^2f(t)dt = 0, RHS = \lim_{x\to0^+} (\sin^3x + \cos x) = 1\)
LHS \(\neq\) RHS hence data given in question is wrong hence BONUS
Correct data should have been
\(\int_{\cos x}^1 t^2f(t)dt = \sin^3x+\cos x – 1\)
Calculation for option
differentiating both sides
\(-\cos^2x f(\cos x)(-\sin x) = 3\sin^2x \cos x – \sin x\)
\(\Rightarrow f(\cos x) = 3\tan x – \sec^2 x\)
\(\Rightarrow f'(\cos x)(-\sin x) = 3\sec^2x – 2\sec^2x \tan x\)
When \(\cos x = \frac{1}{\sqrt{3}}; \sin x = \frac{\sqrt{2}}{\sqrt{3}}\)
\(f'(\frac{1}{\sqrt{3}})\frac{1}{\sqrt{3}} = 6-\frac{9}{\sqrt{2}}\)
✅ Official Ans. by NTA (B)
At right hand vicinity of x=0 given equation does not satisfy
\(\therefore LHS = \int_1^1 t^2f(t)dt = 0, RHS = \lim_{x\to0^+} (\sin^3x + \cos x) = 1\)
LHS \(\neq\) RHS hence data given in question is wrong hence BONUS
Correct data should have been
\(\int_{\cos x}^1 t^2f(t)dt = \sin^3x+\cos x – 1\)
Calculation for option
differentiating both sides
\(-\cos^2x f(\cos x)(-\sin x) = 3\sin^2x \cos x – \sin x\)
\(\Rightarrow f(\cos x) = 3\tan x – \sec^2 x\)
\(\Rightarrow f'(\cos x)(-\sin x) = 3\sec^2x – 2\sec^2x \tan x\)
When \(\cos x = \frac{1}{\sqrt{3}}; \sin x = \frac{\sqrt{2}}{\sqrt{3}}\)
\(f'(\frac{1}{\sqrt{3}})\frac{1}{\sqrt{3}} = 6-\frac{9}{\sqrt{2}}\)
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
\(\int_0^1 \frac{1}{7^{[\frac{1}{x}]}}dx = -\int_1^0 \frac{1}{7^{[\frac{1}{x}]}}dx\)
\(=(-1)[\int_1^{1/2}\frac{1}{7}dx + \int_{1/2}^{1/3}\frac{1}{7^2}dx + \int_{1/3}^{1/4}\frac{1}{7^3}dx+…\infty]\)
[as \(ln(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – …\)]
[as \(ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + …)\)]
\(=-ln(1-\frac{1}{7}) – 7(-ln(1-\frac{1}{7}) – \frac{1}{7})\)
\(=6\ln\frac{6}{7}+1\)
✅ Official Ans. by NTA (A)
\(\int_0^1 \frac{1}{7^{[\frac{1}{x}]}}dx = -\int_1^0 \frac{1}{7^{[\frac{1}{x}]}}dx\)
\(=(-1)[\int_1^{1/2}\frac{1}{7}dx + \int_{1/2}^{1/3}\frac{1}{7^2}dx + \int_{1/3}^{1/4}\frac{1}{7^3}dx+…\infty]\)
[as \(ln(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – …\)]
[as \(ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + …)\)]
\(=-ln(1-\frac{1}{7}) – 7(-ln(1-\frac{1}{7}) – \frac{1}{7})\)
\(=6\ln\frac{6}{7}+1\)
✅ Official Ans. by NTA (A)
▶️ Answer/Explanation
Sol.
\(\frac{dx}{dy}+\frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}\)
If \(= e^{\int\frac{1}{1+y^2}dy} = e^{\tan^{-1}y}\)
\(xe^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}dy\)
\(x \cdot e^{\tan^{-1}y} = (\tan^{-1}y-1)e^{\tan^{-1}y}+c\)
(1, 0) lies on it \(\Rightarrow c=2\).
For \(y = \tan 1 \Rightarrow x = \frac{2}{e}\)
✅ Official Ans. by NTA (B)
\(\frac{dx}{dy}+\frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}\)
If \(= e^{\int\frac{1}{1+y^2}dy} = e^{\tan^{-1}y}\)
\(xe^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}dy\)
\(x \cdot e^{\tan^{-1}y} = (\tan^{-1}y-1)e^{\tan^{-1}y}+c\)
(1, 0) lies on it \(\Rightarrow c=2\).
For \(y = \tan 1 \Rightarrow x = \frac{2}{e}\)
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
Vertex (5,4)
Directrix: \(3x+y-29=0\)
Co-ordinates of B (foot on directrix)
\(\frac{x-5}{3} = \frac{y-4}{1} = -(\frac{15+4-29}{10}) = 1\)
\(x=8, y=5\)
S = (2,3) (focus)
Equation of parabola
PS = PM
so equation is
\(x^2+9y^2-6xy+134x-2y-711=0\)
\(a+b+c+d+k = 9-6+134-2-711 = -576\)
✅ Official Ans. by NTA (D)
Vertex (5,4)
Directrix: \(3x+y-29=0\)
Co-ordinates of B (foot on directrix)
\(\frac{x-5}{3} = \frac{y-4}{1} = -(\frac{15+4-29}{10}) = 1\)
\(x=8, y=5\)
S = (2,3) (focus)
Equation of parabola
PS = PM
so equation is
\(x^2+9y^2-6xy+134x-2y-711=0\)
\(a+b+c+d+k = 9-6+134-2-711 = -576\)
✅ Official Ans. by NTA (D)
▶️ Answer/Explanation
Sol.
C: \(x^2+y^2-3x+2y+(\frac{k}{4})=0\)
Centre \((\frac{3}{2}, -1)\); \(r = \frac{\sqrt{13-k}}{2} \Rightarrow k \le 13\) … (1)
(i) Point \((1, -\frac{1}{3})\) lies on or inside circle C
\(\Rightarrow S_1 \le 0 \Rightarrow k \le \frac{92}{9}\) … (2)
(ii) C lies in 4th quadrant
\(r < 1\)
\(\Rightarrow \frac{\sqrt{13-k}}{2} < 1\)
\(\Rightarrow k > 9\) … (3)
Hence \((1) \cap (2) \cap (3) \Rightarrow k \in (9, \frac{92}{9}]\)
✅ Official Ans. by NTA (D)
▶️ Answer/Explanation
Sol.
\(\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda\)
\((x, y, z) = (4\lambda-2, 2\lambda+1, 3\lambda-1)\)
\(\vec{AP} = (4\lambda-3)\hat{i} + (2\lambda-1)\hat{j} + (3\lambda-5)\hat{k}\)
\(\vec{b} = 4\hat{i} + 2\hat{j} + 3\hat{k}\)
\(\vec{AP} \cdot \vec{b} = 0\)
\(4(4\lambda-3)+2(2\lambda-1)+3(3\lambda-5)=0\)
\(29\lambda = 12+2+15=29\)
\(\lambda=1\)
P = (2, 3, 2)
Distance from plane \(3x+4y+12z+23=0\):
\(d = \left|\frac{6+12+24+23}{\sqrt{9+16+144}}\right|\)
\(d = \left|\frac{65}{13}\right| = 5\)
✅ Official Ans. by NTA (A)
▶️ Answer/Explanation
Sol.
Line 1: \(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}\)
Point A = (3, 2, 1), \(\vec{n_1} = 2\hat{i} + 3\hat{j} – \hat{k}\)
Line 2: \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}\)
Point B = (-3, 6, 5), \(\vec{n_2} = 2\hat{i} + \hat{j} + 3\hat{k}\)
\(\vec{BA} = (3 – (-3))\hat{i} + (2-6)\hat{j} + (1-5)\hat{k} = 6\hat{i} – 4\hat{j} – 4\hat{k}\)
Shortest Distance = \(\frac{|[\vec{BA} \ \vec{n_1} \ \vec{n_2}]|}{|\vec{n_1} \times \vec{n_2}|}\)
\(\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix} = 10\hat{i} – 8\hat{j} – 4\hat{k}\)
\(|\vec{n_1} \times \vec{n_2}| = \sqrt{100+64+16} = \sqrt{180}\)
\([\vec{BA} \ \vec{n_1} \ \vec{n_2}] = (6)(10) + (-4)(-8) + (-4)(-4) = 60+32+16 = 108\)
S.D = \(\frac{108}{\sqrt{180}} = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}\)
✅ Official Ans. by NTA (A)
Line 1: \(\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}\)
Point A = (3, 2, 1), \(\vec{n_1} = 2\hat{i} + 3\hat{j} – \hat{k}\)
Line 2: \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}\)
Point B = (-3, 6, 5), \(\vec{n_2} = 2\hat{i} + \hat{j} + 3\hat{k}\)
\(\vec{BA} = (3 – (-3))\hat{i} + (2-6)\hat{j} + (1-5)\hat{k} = 6\hat{i} – 4\hat{j} – 4\hat{k}\)
Shortest Distance = \(\frac{|[\vec{BA} \ \vec{n_1} \ \vec{n_2}]|}{|\vec{n_1} \times \vec{n_2}|}\)
\(\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix} = 10\hat{i} – 8\hat{j} – 4\hat{k}\)
\(|\vec{n_1} \times \vec{n_2}| = \sqrt{100+64+16} = \sqrt{180}\)
\([\vec{BA} \ \vec{n_1} \ \vec{n_2}] = (6)(10) + (-4)(-8) + (-4)(-4) = 60+32+16 = 108\)
S.D = \(\frac{108}{\sqrt{180}} = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}\)
✅ Official Ans. by NTA (A)
▶️ Answer/Explanation
Sol.
Area = \(\frac{1}{2}|\vec{a} \times \vec{b}| = 2\sqrt{2} \Rightarrow |\vec{a} \times \vec{b}| = 4\sqrt{2}\)
\(|\vec{a}|=1\) and \(|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|\)
\(\Rightarrow \cos\theta = \sin\theta\)
\(\Rightarrow \theta = \frac{\pi}{4}\)
\(\therefore |\vec{a} \times \vec{b}| = 4\sqrt{2} \Rightarrow |\vec{a}||\vec{b}|\sin\frac{\pi}{4} = 4\sqrt{2}\)
\(\Rightarrow |\vec{b}| = 8\)
Now, \(\vec{c} = 2\sqrt{2}(\vec{a} \times \vec{b}) – 2\vec{b}\)
\(|\vec{c}| = \sqrt{(2\sqrt{2})^2|\vec{a} \times \vec{b}|^2 + (2|\vec{b}|)^2} = 16\sqrt{2}\)
Now, \(\vec{b} \cdot \vec{c} = -2|\vec{b}|^2\)
\(\Rightarrow 8 \times 16\sqrt{2} \times \cos\alpha = -2 \cdot 64\)
\(\Rightarrow \cos\alpha = -\frac{1}{\sqrt{2}} \Rightarrow \alpha = \frac{3\pi}{4}\)
✅ Official Ans. by NTA (D)
Area = \(\frac{1}{2}|\vec{a} \times \vec{b}| = 2\sqrt{2} \Rightarrow |\vec{a} \times \vec{b}| = 4\sqrt{2}\)
\(|\vec{a}|=1\) and \(|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|\)
\(\Rightarrow \cos\theta = \sin\theta\)
\(\Rightarrow \theta = \frac{\pi}{4}\)
\(\therefore |\vec{a} \times \vec{b}| = 4\sqrt{2} \Rightarrow |\vec{a}||\vec{b}|\sin\frac{\pi}{4} = 4\sqrt{2}\)
\(\Rightarrow |\vec{b}| = 8\)
Now, \(\vec{c} = 2\sqrt{2}(\vec{a} \times \vec{b}) – 2\vec{b}\)
\(|\vec{c}| = \sqrt{(2\sqrt{2})^2|\vec{a} \times \vec{b}|^2 + (2|\vec{b}|)^2} = 16\sqrt{2}\)
Now, \(\vec{b} \cdot \vec{c} = -2|\vec{b}|^2\)
\(\Rightarrow 8 \times 16\sqrt{2} \times \cos\alpha = -2 \cdot 64\)
\(\Rightarrow \cos\alpha = -\frac{1}{\sqrt{2}} \Rightarrow \alpha = \frac{3\pi}{4}\)
✅ Official Ans. by NTA (D)
▶️ Answer/Explanation
Sol.
mean \(\bar{x} = \frac{4+5+6+6+7+8+x+y}{8} = 6\)
\(\Rightarrow x+y = 48 – 36 = 12\)
Variance \(=\frac{1}{8}(16+25+36+36+49+64+x^2+y^2) – 36 = \frac{9}{4}\)
\(\Rightarrow x^2+y^2 = 80\)
\(\therefore x=4; y=8\)
\(x^4+y^2 = 256 + 64 = 320\)
✅ Official Ans. by NTA (B)
mean \(\bar{x} = \frac{4+5+6+6+7+8+x+y}{8} = 6\)
\(\Rightarrow x+y = 48 – 36 = 12\)
Variance \(=\frac{1}{8}(16+25+36+36+49+64+x^2+y^2) – 36 = \frac{9}{4}\)
\(\Rightarrow x^2+y^2 = 80\)
\(\therefore x=4; y=8\)
\(x^4+y^2 = 256 + 64 = 320\)
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
Required probability \(=\frac{ar(ADEC)}{ar(ABC)}\)
\(= 1 – \frac{ar(BDE)}{ar(ABC)}\)
\(= 1 – \frac{\frac{1}{2}\times2\times4}{\frac{1}{2}\times8\times6} = 1 – \frac{1}{6} = \frac{5}{6}\)
✅ Official Ans. by NTA (B)
▶️ Answer/Explanation
Sol.
\(\tan^{-1}\frac{1}{1+n+n^2} = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)\)
\(= \tan^{-1}(n+1)-\tan^{-1}n\)
so, \(\sum_{n=1}^{50}(\tan^{-1}(n+1)-\tan^{-1}n)\)
\(= \tan^{-1}51 – \tan^{-1}1\)
\(\cot(\sum_{n=1}^{50}…) = \cot(\tan^{-1}51 – \tan^{-1}1)\)
\(= \frac{1}{\tan(\tan^{-1}51-\tan^{-1}1)} = \frac{1+51\times1}{51-1} = \frac{52}{50} = \frac{26}{25}\)
✅ Official Ans. by NTA (A)
\(\tan^{-1}\frac{1}{1+n+n^2} = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)\)
\(= \tan^{-1}(n+1)-\tan^{-1}n\)
so, \(\sum_{n=1}^{50}(\tan^{-1}(n+1)-\tan^{-1}n)\)
\(= \tan^{-1}51 – \tan^{-1}1\)
\(\cot(\sum_{n=1}^{50}…) = \cot(\tan^{-1}51 – \tan^{-1}1)\)
\(= \frac{1}{\tan(\tan^{-1}51-\tan^{-1}1)} = \frac{1+51\times1}{51-1} = \frac{52}{50} = \frac{26}{25}\)
✅ Official Ans. by NTA (A)
▶️ Answer/Explanation
Sol.
\(\cos 72^\circ = \frac{\sqrt{5}-1}{4}\)
\(\Rightarrow 1-2\sin^2 36^\circ = \frac{\sqrt{5}-1}{4}\)
\(\Rightarrow 4-8\alpha^2 = \sqrt{5}-1\)
\(\Rightarrow 5-8\alpha^2 = \sqrt{5}\)
\(\Rightarrow (5-8\alpha^2)^2 = 5\)
\(\Rightarrow 25+64\alpha^4 – 80\alpha^2 = 5\)
\(\Rightarrow 64\alpha^4 – 80\alpha^2 + 20 = 0\)
\(\Rightarrow 16\alpha^4 – 20\alpha^2 + 5 = 0\)
✅ Official Ans. by NTA (C)
\(\cos 72^\circ = \frac{\sqrt{5}-1}{4}\)
\(\Rightarrow 1-2\sin^2 36^\circ = \frac{\sqrt{5}-1}{4}\)
\(\Rightarrow 4-8\alpha^2 = \sqrt{5}-1\)
\(\Rightarrow 5-8\alpha^2 = \sqrt{5}\)
\(\Rightarrow (5-8\alpha^2)^2 = 5\)
\(\Rightarrow 25+64\alpha^4 – 80\alpha^2 = 5\)
\(\Rightarrow 64\alpha^4 – 80\alpha^2 + 20 = 0\)
\(\Rightarrow 16\alpha^4 – 20\alpha^2 + 5 = 0\)
✅ Official Ans. by NTA (C)
▶️ Answer/Explanation
Sol.
(A) \((\sim q \land p) \land q = (\sim q \land q) \land p = f\)
(B) \((\sim q \land p) \land (p \land \sim p) = \sim q \land (p \land \sim p) = f\)
(C) \((\sim q \land p) \lor (p \lor \sim p) = (\sim q \land p) \lor (t) = t\)
(D) \((p \land q) \land (\sim(p \land q)) = f\)
✅ Official Ans. by NTA (C)
(A) \((\sim q \land p) \land q = (\sim q \land q) \land p = f\)
(B) \((\sim q \land p) \land (p \land \sim p) = \sim q \land (p \land \sim p) = f\)
(C) \((\sim q \land p) \lor (p \lor \sim p) = (\sim q \land p) \lor (t) = t\)
(D) \((p \land q) \land (\sim(p \land q)) = f\)
✅ Official Ans. by NTA (C)
Section B
▶️ Answer/Explanation
Sol.
\(f^{-1}(n)=\begin{cases} \frac{n}{2} & ; n=2,4,6,8,10 \ \frac{n+11}{2} & ; n=1,3,5,7,9 \end{cases}\)
\(f(g(n))=\begin{cases} n+1 & ; n \in \text{odd} \ n-1 & ; n \in \text{even} \end{cases}\)
\(g(n)=\begin{cases} f^{-1}(n+1) & ; n \in \text{odd} \ f^{-1}(n-1) & ; n \in \text{even} \end{cases}\)
\(\Rightarrow g(n)=\begin{cases} \frac{n+1}{2} & ; n \in \text{odd} \ \frac{n+10}{2} & ; n \in \text{even} \end{cases}\)
\(g(10)\cdot[g(1)+g(2)+g(3)+g(4)+g(5)]\)
\(=10\cdot[1+6+2+7+3]=190\)
✅ Official Ans. by NTA (190)
\(f^{-1}(n)=\begin{cases} \frac{n}{2} & ; n=2,4,6,8,10 \ \frac{n+11}{2} & ; n=1,3,5,7,9 \end{cases}\)
\(f(g(n))=\begin{cases} n+1 & ; n \in \text{odd} \ n-1 & ; n \in \text{even} \end{cases}\)
\(g(n)=\begin{cases} f^{-1}(n+1) & ; n \in \text{odd} \ f^{-1}(n-1) & ; n \in \text{even} \end{cases}\)
\(\Rightarrow g(n)=\begin{cases} \frac{n+1}{2} & ; n \in \text{odd} \ \frac{n+10}{2} & ; n \in \text{even} \end{cases}\)
\(g(10)\cdot[g(1)+g(2)+g(3)+g(4)+g(5)]\)
\(=10\cdot[1+6+2+7+3]=190\)
✅ Official Ans. by NTA (190)
▶️ Answer/Explanation
Sol.
\(x^2 – 4\lambda x + 5 = 0 \begin{cases} \alpha \ \beta \end{cases}\)
\(x^2 – (3\sqrt{2}+2\sqrt{3})x+(7+3\lambda\sqrt{3})=0 \begin{cases} \alpha \ \gamma \end{cases}\)
\(\alpha+\beta = 4\lambda\)
\(\alpha+\gamma = 3\sqrt{2}+2\sqrt{3}\)
\(\beta+\gamma = 3\sqrt{2}\)
\(\alpha = 2\lambda+\sqrt{3}\)
\(\beta = 2\lambda-\sqrt{3}\)
\(\alpha\beta=5\)
\(\alpha\gamma = 7+3\lambda\sqrt{3}\)
\(4\lambda^2 = 8 \Rightarrow \lambda=\sqrt{2}\)
\((\alpha+2\beta+\gamma)^2 = (4\alpha+3\sqrt{2})^2 = (7\sqrt{2})^2=98\)
✅ Official Ans. by NTA (98)
\(x^2 – 4\lambda x + 5 = 0 \begin{cases} \alpha \ \beta \end{cases}\)
\(x^2 – (3\sqrt{2}+2\sqrt{3})x+(7+3\lambda\sqrt{3})=0 \begin{cases} \alpha \ \gamma \end{cases}\)
\(\alpha+\beta = 4\lambda\)
\(\alpha+\gamma = 3\sqrt{2}+2\sqrt{3}\)
\(\beta+\gamma = 3\sqrt{2}\)
\(\alpha = 2\lambda+\sqrt{3}\)
\(\beta = 2\lambda-\sqrt{3}\)
\(\alpha\beta=5\)
\(\alpha\gamma = 7+3\lambda\sqrt{3}\)
\(4\lambda^2 = 8 \Rightarrow \lambda=\sqrt{2}\)
\((\alpha+2\beta+\gamma)^2 = (4\alpha+3\sqrt{2})^2 = (7\sqrt{2})^2=98\)
✅ Official Ans. by NTA (98)
▶️ Answer/Explanation
Sol.
Let \(A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\); a, b, c, \(d \in {0,1,2,3,4,5}\)
\(a+b+c+d=p, p \in {3,5,7}\)
Case-(i) \(a+b+c+d=3\)
No. of ways \( = ^{3+4-1}C_{4-1} = ^6C_3 = 20\)
Case-(ii) \(a+b+c+d=5\)
No. of ways \( = ^{5+4-1}C_{4-1} = ^8C_3 = 56\)
Case-(iii) \(a+b+c+d=7\)
No. of ways = total ways when a,b,c,d \(\in {0..7}\) – total ways when one is \(\ge 6\)
\(= ^{7+4-1}C_{4-1} – (^{4}C_1 \times ^{1+4-1}C_{4-1})\)
\(=^{10}C_3 – 16 = 104\)
Hence total no. of ways\(=20+56+104=180\)
✅ Official Ans. by NTA (180)
Let \(A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\); a, b, c, \(d \in {0,1,2,3,4,5}\)
\(a+b+c+d=p, p \in {3,5,7}\)
Case-(i) \(a+b+c+d=3\)
No. of ways \( = ^{3+4-1}C_{4-1} = ^6C_3 = 20\)
Case-(ii) \(a+b+c+d=5\)
No. of ways \( = ^{5+4-1}C_{4-1} = ^8C_3 = 56\)
Case-(iii) \(a+b+c+d=7\)
No. of ways = total ways when a,b,c,d \(\in {0..7}\) – total ways when one is \(\ge 6\)
\(= ^{7+4-1}C_{4-1} – (^{4}C_1 \times ^{1+4-1}C_{4-1})\)
\(=^{10}C_3 – 16 = 104\)
Hence total no. of ways\(=20+56+104=180\)
✅ Official Ans. by NTA (180)
▶️ Answer/Explanation
Sol.
\(3n-20 \ge 0 \cap 2n-25 < 0 \cap n \in I\)
\(7 \le n \le 12\)
Sum = \(7+8+9+10+11+12=57\)
✅ Official Ans. by NTA (57)
▶️ Answer/Explanation
Sol.
\(f(x)=[1+x]+\frac{\alpha^{2[x]+{x}}+[x]-1}{2[x]+{x}}\)
\(\lim_{x \to 0^-} f(x) = \alpha – \frac{4}{3} \Rightarrow 0 + \frac{\alpha^{-1}-2}{-1} = \alpha – \frac{4}{3}\)
\(\Rightarrow 2-\frac{1}{\alpha} = \alpha – \frac{4}{3}\)
\(\Rightarrow \alpha + \frac{1}{\alpha} = \frac{10}{3}\)
\(\Rightarrow \alpha=3; \alpha \in I\)
✅ Official Ans. by NTA (3)
\(f(x)=[1+x]+\frac{\alpha^{2[x]+{x}}+[x]-1}{2[x]+{x}}\)
\(\lim_{x \to 0^-} f(x) = \alpha – \frac{4}{3} \Rightarrow 0 + \frac{\alpha^{-1}-2}{-1} = \alpha – \frac{4}{3}\)
\(\Rightarrow 2-\frac{1}{\alpha} = \alpha – \frac{4}{3}\)
\(\Rightarrow \alpha + \frac{1}{\alpha} = \frac{10}{3}\)
\(\Rightarrow \alpha=3; \alpha \in I\)
✅ Official Ans. by NTA (3)
▶️ Answer/Explanation
Sol.
\(y(x)=(x^x)^x\)
\(\ln y(x) = x^2 \cdot \ln x\)
\(\frac{1}{y(x)} \cdot y'(x) = \frac{x^2}{x} + 2x \cdot \ln x\)
\(y'(x)=y(x)[x+2x\ln x]\)
\(y(1)=1; y'(1)=1\)
\(y”(x) = y'(x)[x+2x\ln x] + y(x)[1+2(1+\ln x)]\)
\(y”(1)=1[1+0]+1(1+2)=4\)
\(\frac{d^2y}{dx^2} = -(\frac{dy}{dx})^3 \cdot \frac{d^2x}{dy^2}\)
\(\Rightarrow 4 = -\frac{d^2x}{dy^2}\)
\(\frac{d^2x}{dy^2} = -4\)
Ans. \(-4+20=16\)
✅ Official Ans. by NTA (16)
\(y(x)=(x^x)^x\)
\(\ln y(x) = x^2 \cdot \ln x\)
\(\frac{1}{y(x)} \cdot y'(x) = \frac{x^2}{x} + 2x \cdot \ln x\)
\(y'(x)=y(x)[x+2x\ln x]\)
\(y(1)=1; y'(1)=1\)
\(y”(x) = y'(x)[x+2x\ln x] + y(x)[1+2(1+\ln x)]\)
\(y”(1)=1[1+0]+1(1+2)=4\)
\(\frac{d^2y}{dx^2} = -(\frac{dy}{dx})^3 \cdot \frac{d^2x}{dy^2}\)
\(\Rightarrow 4 = -\frac{d^2x}{dy^2}\)
\(\frac{d^2x}{dy^2} = -4\)
Ans. \(-4+20=16\)
✅ Official Ans. by NTA (16)
▶️ Answer/Explanation
Sol.
\(A=\frac{3}{2}\int_0^1(1-x^{2/3})^{3/2}dx\)
Let \(x=\sin^3\theta\)
\(A = \frac{3}{2}\int_0^{\pi/2}(1-\sin^2\theta)^{3/2} \cdot 3\sin^2\theta\cos\theta d\theta\)
\(= \frac{9}{2}\int_0^{\pi/2}\sin^2\theta\cos^4\theta d\theta\)
\(A=\frac{9}{2} \times \frac{1 \cdot 3 \cdot 1}{(2+4)(4)(2)} \cdot \frac{\pi}{2}\)
\(\Rightarrow A = \frac{9\pi}{64} \Rightarrow \frac{64A}{\pi}=9\)
\(\Rightarrow \frac{256A}{\pi} = 36\) Ans.
✅ Official Ans. by NTA (36)
▶️ Answer/Explanation
Sol.
\((1-x^2)\frac{dy}{dx} = xy + (x^3+2)\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx} + (\frac{-x}{1-x^2})y = \frac{x^3+2}{\sqrt{1-x^2}}\)
\(IF = e^{\int\frac{-x}{1-x^2}dx} = \sqrt{1-x^2}\)
\(y(x) \cdot \sqrt{1-x^2} = \frac{x^4}{4}+2x+c\)
\(y(0)=0 \Rightarrow c=0\)
\(\sqrt{1-x^2}y(x) = \frac{x^4}{4}+2x\)
required value \(=\int_{-1/2}^{1/2} (\frac{x^4}{4}+2x)dx = \frac{1}{4} \cdot 2 \int_0^{1/2} x^4 dx\)
\(=\frac{1}{10}[x^5]_0^{1/2} = \frac{1}{320}\)
\(k^{-1} = 320\)
✅ Official Ans. by NTA (320)
\((1-x^2)\frac{dy}{dx} = xy + (x^3+2)\sqrt{1-x^2}\)
\(\Rightarrow \frac{dy}{dx} + (\frac{-x}{1-x^2})y = \frac{x^3+2}{\sqrt{1-x^2}}\)
\(IF = e^{\int\frac{-x}{1-x^2}dx} = \sqrt{1-x^2}\)
\(y(x) \cdot \sqrt{1-x^2} = \frac{x^4}{4}+2x+c\)
\(y(0)=0 \Rightarrow c=0\)
\(\sqrt{1-x^2}y(x) = \frac{x^4}{4}+2x\)
required value \(=\int_{-1/2}^{1/2} (\frac{x^4}{4}+2x)dx = \frac{1}{4} \cdot 2 \int_0^{1/2} x^4 dx\)
\(=\frac{1}{10}[x^5]_0^{1/2} = \frac{1}{320}\)
\(k^{-1} = 320\)
✅ Official Ans. by NTA (320)
▶️ Answer/Explanation
Sol.
\(4x+3y+2=0\)
\(3x-4y-11=0\)
\(\frac{x}{-25} = \frac{y}{50} = \frac{1}{-25}\)
Q(1, -2).
\(\frac{x-1}{\cos\theta} = \frac{y+2}{\sin\theta} = \pm 5\)
\(y=-2+5(-\frac{4}{5}) = -6\)
\(x=1+5(\frac{3}{5})=4\)
Req. distance \(= \frac{|5(4)-12(-6)+51|}{13}\)
\(=|\frac{20+72+51}{13}|\)
\(=\frac{143}{13}=11\)
✅ Official Ans. by NTA (11)
▶️ Answer/Explanation
Sol.
\(P(A’) < \frac{1}{5} = \frac{36}{180}\)
5 times the sum of missing number should be less than 36.
If 1 digit is missing = 7
If 2 digit is missing = 9
If 3 digit is missing = 2
If 0 digit is missing = 1
Alternate
A is subset of S hence A can have elements:
type 1: {}
…
As \(P(A) \ge \frac{4}{5}\);
Note: Type 1 to Type 4 elements can not be in set A as maximum probability of type 4 elements.
\({E_5, E_6, E_7, E_8}\) is \(\frac{5}{36}+\frac{6}{36}+\frac{7}{36}+\frac{8}{36} = \frac{26}{36} < \frac{4}{5}\)
Now for Type 5 acceptable elements let’s call probability as \(P_5\)
\(P_5 = \frac{n_1+n_2+n_3+n_4+n_5}{36} \ge \frac{4}{5}\)
\(\Rightarrow n_1+n_2+n_3+n_4+n_5 \ge 28.8\)
Hence, 2 possible ways \({E_5, E_6, E_7, E_8, E_3 \text{ or } E_4})\)
\(P_6 = n_1+…+n_6 \ge 28.8 \Rightarrow 9 \text{ possible ways}\)
\(P_7 \Rightarrow n_1+…+n_7 \ge 28.8 \Rightarrow 7 \text{ possible ways}\)
\(P_8 \Rightarrow n_1+…+n_8 \ge 28.8 \Rightarrow 1 \text{ possible way}\)
Total = 19
✅ Official Ans. by NTA (19)
\(P(A’) < \frac{1}{5} = \frac{36}{180}\)
5 times the sum of missing number should be less than 36.
If 1 digit is missing = 7
If 2 digit is missing = 9
If 3 digit is missing = 2
If 0 digit is missing = 1
Alternate
A is subset of S hence A can have elements:
type 1: {}
…
As \(P(A) \ge \frac{4}{5}\);
Note: Type 1 to Type 4 elements can not be in set A as maximum probability of type 4 elements.
\({E_5, E_6, E_7, E_8}\) is \(\frac{5}{36}+\frac{6}{36}+\frac{7}{36}+\frac{8}{36} = \frac{26}{36} < \frac{4}{5}\)
Now for Type 5 acceptable elements let’s call probability as \(P_5\)
\(P_5 = \frac{n_1+n_2+n_3+n_4+n_5}{36} \ge \frac{4}{5}\)
\(\Rightarrow n_1+n_2+n_3+n_4+n_5 \ge 28.8\)
Hence, 2 possible ways \({E_5, E_6, E_7, E_8, E_3 \text{ or } E_4})\)
\(P_6 = n_1+…+n_6 \ge 28.8 \Rightarrow 9 \text{ possible ways}\)
\(P_7 \Rightarrow n_1+…+n_7 \ge 28.8 \Rightarrow 7 \text{ possible ways}\)
\(P_8 \Rightarrow n_1+…+n_8 \ge 28.8 \Rightarrow 1 \text{ possible way}\)
Total = 19
✅ Official Ans. by NTA (19)