SECTION A
Q.1 — Select the Correct Answer & Q.2 — Short Answers
2 + 4 = 16 Marks
Q.1 Select and Write the Correct Answer (8 × 2 = 16 Marks)
(1)
If \(a\cos 2\theta + b\sin 2\theta = c\) has \(\alpha\) and \(\beta\) as its roots, then \(\tan\alpha + \tan\beta\) = ?
✔ (a) \(\dfrac{2b}{c+a}\)
Substitute \(\cos 2\theta = \dfrac{1-t^2}{1+t^2},\ \sin 2\theta = \dfrac{2t}{1+t^2}\) where \(t = \tan\theta\):
\(a(1-t^2)+2bt = c(1+t^2) \Rightarrow (a+c)t^2 - 2bt - (a-c)=0\)
By Vieta's: \(\tan\alpha+\tan\beta = \dfrac{2b}{a+c} = \dfrac{2b}{c+a}\)
(2)
Eccentricity of \(16x^2 - 3y^2 - 32x - 12y - 44 = 0\)
✔ (b) \(\sqrt{\dfrac{19}{3}}\)
Complete the square: \(16(x-1)^2 - 3(y+2)^2 = 48 \Rightarrow \dfrac{(x-1)^2}{3} - \dfrac{(y+2)^2}{16}=1\)
\(a^2=3,\ b^2=16,\ c^2=a^2+b^2=19,\ e = \dfrac{c}{a} = \dfrac{\sqrt{19}}{\sqrt{3}} = \sqrt{\dfrac{19}{3}}\)
(3)
A college offers 5 morning courses and 3 evening courses. Ways a student selects exactly one course (morning or evening)?
✔ (c) 8
The student picks exactly one course from either session: \(5 + 3 = 8\) ways.
(4)
Total number of terms in \((x+y)^{100}+(x-y)^{100}\) after simplification
✔ (b) 51
In \((x+y)^{100}\) the terms with odd powers of \(y\) are cancelled by the corresponding terms of \((x-y)^{100}\). Only even-power terms survive: \(y^0, y^2, y^4,\ldots,y^{100}\) → \(51\) terms.
(5)
If set \(A\) is empty, then \(n[P[P[P(A)]]]\) = ?
✔ (d) 4
\(n(A)=0 \Rightarrow n(P(A))=1 \Rightarrow n(P(P(A)))=2 \Rightarrow n(P(P(P(A))))=2^2=4\)
(6)
If \(f(x)=\dfrac{1}{1-x}\), then \(f(f(f(x)))\) is
✔ (c) \(x\)
\(f(f(x)) = f\!\left(\dfrac{1}{1-x}\right) = \dfrac{1}{1-\frac{1}{1-x}} = \dfrac{x-1}{x}\)
\(f(f(f(x))) = f\!\left(\dfrac{x-1}{x}\right) = \dfrac{1}{1-\frac{x-1}{x}} = \dfrac{1}{\frac{1}{x}} = x\)
(7)
\(\displaystyle\lim_{x\to\pi/2}\dfrac{3\cos x+\cos 3x}{(2x-\pi)^3}\)
✔ (c) \(-\dfrac{1}{2}\)
Let \(t = x-\tfrac{\pi}{2}\) (so \(t\to 0\)). Then \(\cos x=-\sin t\) and \(\cos 3x=\sin 3t\).
Numerator: \(-3\sin t + \sin 3t = -3\sin t+(3\sin t - 4\sin^3 t) = -4\sin^3 t\).
Denominator: \((2t)^3 = 8t^3\).
Limit \(= \dfrac{-4\sin^3 t}{8t^3} \to \dfrac{-4}{8}\cdot 1 = -\dfrac{1}{2}\)
(8)
\(f(x)=x^2+\sin x+1\) for \(x\le 0\) and \(f(x)=x^2-2x+1\) for \(x>0\)
✔ (a) Continuous but NOT differentiable at \(x=0\)
Continuity: \(f(0^-)=0+0+1=1\) and \(f(0^+)=0-0+1=1=f(0)\) ✓
Differentiability: Left derivative \(= (2x+\cos x)|_0 = 1\). Right derivative \(=(2x-2)|_0=-2\).
Since \(1\ne -2\), \(f\) is not differentiable at \(x=0\).
Q.2 Short Answer Questions (4 × 1 = 4 Marks)
(1) If \(A=\begin{bmatrix}0&1{+}2i&-(2{-}i)\\-1{-}2i&0&-7\\2{-}i&7&0\end{bmatrix}\) where \(i=\sqrt{-1}\), prove \(A^T=-A\)
We compute the transpose \(A^T\) by swapping rows and columns:
\(A^T = \begin{bmatrix}0 & -1-2i & 2-i\\ 1+2i & 0 & 7\\ -(2-i) & -7 & 0\end{bmatrix}\)
Now compute \(-A\):
\(-A = \begin{bmatrix}0 & -(1+2i) & (2-i)\\ 1+2i & 0 & 7\\ -(2-i) & -7 & 0\end{bmatrix} = \begin{bmatrix}0 & -1-2i & 2-i\\ 1+2i & 0 & 7\\ -(2-i) & -7 & 0\end{bmatrix}\)
✓ AT = −A (Proved)
(2) Equation of the line parallel to the X-axis and 5 units above it
Any line parallel to the X-axis has the form \(y = k\). Being 5 units above the X-axis means \(k = 5\).
\(\boxed{y = 5}\)
(3) Point of intersection of \(x+2y=3\) and \(2x-y=1\)
From \(2x-y=1\): \(y=2x-1\). Substituting into the first equation:
\(x+2(2x-1)=3 \Rightarrow 5x=5 \Rightarrow x=1,\quad y=2(1)-1=1\)
Point of intersection: \(\boxed{(1,\,1)}\)
(4) Centre and radius of \(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{3}\right)^2=\dfrac{1}{36}\)
Comparing with \((x-h)^2+(y-k)^2=r^2\):
Centre \(= \left(\dfrac{1}{2},\,-\dfrac{1}{3}\right)\), Radius \(= \dfrac{1}{6}\)
SECTION B
Q.3 to Q.14 — Attempt Any Eight (2 Marks Each)
8 × 2 = 16 Marks
Q.3 Circle of diameter 40 cm, chord of length 20 cm — find length of minor arc
Radius \(r = 20\) cm. Using the chord–angle formula: \(\text{chord} = 2r\sin\!\dfrac{\theta}{2}\)
\(20 = 2(20)\sin\tfrac{\theta}{2} \Rightarrow \sin\tfrac{\theta}{2}=\tfrac{1}{2} \Rightarrow \tfrac{\theta}{2}=\tfrac{\pi}{6} \Rightarrow \theta=\tfrac{\pi}{3}\)
Minor arc length \(= r\theta = 20\cdot\dfrac{\pi}{3} = \dfrac{20\pi}{3}\) cm
Q.4 Show that \(\tan^2\theta+\cot^2\theta\ge 2\) for all \(\theta\in\mathbb{R}\)
By the AM–GM inequality, for any two positive real numbers \(a, b\):
\(\dfrac{a+b}{2}\ge\sqrt{ab}\)
Set \(a=\tan^2\theta\) and \(b=\cot^2\theta\) (both positive for \(\theta\ne n\pi/2\)):
\(\dfrac{\tan^2\theta+\cot^2\theta}{2}\ge\sqrt{\tan^2\theta\cdot\cot^2\theta}=\sqrt{1}=1\)
\(\therefore\quad\tan^2\theta+\cot^2\theta\ge 2 \quad\square\)
Equality holds when \(\tan^2\theta=\cot^2\theta\), i.e., \(\tan\theta=\pm1\), i.e., \(\theta=\pm\dfrac{\pi}{4}\).
Q.5 Prove: \(\cot 4x(\sin 5x+\sin 3x)=\cot x(\sin 5x-\sin 3x)\)
LHS
Using \(\sin A+\sin B=2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2}\):
\(\sin 5x+\sin 3x = 2\sin 4x\cos x\)
\(\therefore\text{LHS}=\cot 4x\cdot 2\sin 4x\cos x = \dfrac{\cos 4x}{\sin 4x}\cdot 2\sin 4x\cos x = 2\cos 4x\cos x\)
RHS
Using \(\sin A-\sin B=2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2}\):
\(\sin 5x-\sin 3x = 2\cos 4x\sin x\)
\(\therefore\text{RHS}=\cot x\cdot 2\cos 4x\sin x = \dfrac{\cos x}{\sin x}\cdot 2\cos 4x\sin x = 2\cos 4x\cos x\)
✓ LHS = RHS = 2 cos 4x · cos x (Proved)
Q.6 Slope, x-intercept, y-intercept of \(3x-y-9=0\)
Rewrite in slope-intercept form: \(y = 3x - 9\)
Slope \(m = 3\)
x-intercept (set \(y=0\)): \(3x=9 \Rightarrow x=3\) → x-intercept \(= 3\)
y-intercept (set \(x=0\)): \(-y=9 \Rightarrow y=-9\) → y-intercept \(= -9\)
Q.7 Locus of P such that AP = 2BP, where A(2,0) and B(0,3)
Let \(P=(x,y)\). Condition: \(AP^2=4\cdot BP^2\)
\((x-2)^2+y^2 = 4\bigl[x^2+(y-3)^2\bigr]\)
\(x^2-4x+4+y^2 = 4x^2+4y^2-24y+36\)
\(3x^2+3y^2+4x-24y+32=0\)
Locus: \(\boxed{3x^2+3y^2+4x-24y+32=0}\)
Q.8 Parameter of point (27, −12) on parabola \(3y^2=16x\)
Rewrite: \(y^2=\dfrac{16}{3}x\) i.e. \(y^2=4ax\) where \(4a=\dfrac{16}{3}\Rightarrow a=\dfrac{4}{3}\).
Parametric form: \(x=at^2,\; y=2at\).
\(y=2at=-12 \Rightarrow t = \dfrac{-12}{2\cdot\frac{4}{3}} = \dfrac{-12}{\frac{8}{3}} = -12\times\dfrac{3}{8} = -\dfrac{9}{2}\)
Verification: \(x=at^2=\dfrac{4}{3}\cdot\dfrac{81}{4}=27\) ✓
Parameter \(t = -\dfrac{9}{2}\)
Q.9 If pth, qth, rth terms of a G.P. are x, y, z, find \(x^{q-r}\cdot y^{r-p}\cdot z^{p-q}\)
Let first term \(= a\), common ratio \(= R\). Then \(x=aR^{p-1},\ y=aR^{q-1},\ z=aR^{r-1}\).
\(x^{q-r}\cdot y^{r-p}\cdot z^{p-q} = a^{(q-r)+(r-p)+(p-q)}\cdot R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\)
Exponent of a: \((q-r)+(r-p)+(p-q)=0\)
Exponent of R: Expanding and collecting terms also gives \(0\).
\(\therefore\; x^{q-r}\cdot y^{r-p}\cdot z^{p-q} = a^0\cdot R^0 = \boxed{1}\)
Q.10 Show \(\dfrac{9!}{3!\,6!}+\dfrac{9!}{4!\,5!}=\dfrac{10!}{4!\,6!}\)
Recognising combinatorial notation: \(\dfrac{9!}{3!\,6!}=\binom{9}{3}=84\) and \(\dfrac{9!}{4!\,5!}=\binom{9}{4}=126\).
\(\text{LHS} = \dfrac{9!}{3!\,6!}+\dfrac{9!}{4!\,5!} = 9!\left[\dfrac{4\cdot 3!\cdot 5!+3!\cdot 6!}{3!\cdot 4!\cdot 5!\cdot 6!}\right]\)
\(= 9!\cdot\dfrac{3!\cdot 5!\,(4+6)}{3!\cdot 4!\cdot 5!\cdot 6!} = \dfrac{9!\cdot 10}{4!\cdot 6!} = \dfrac{10!}{4!\,6!} = \text{RHS}\quad\square\)
Q.11 6-digit numbers from digits 3,4,5,6,7,8 without repetition
Total 6-digit numbers: \(6! = 720\)
(a) Divisible by 5
Last digit must be 5. Remaining 5 places from 5 remaining digits: \(5! = 120\)
Divisible by 5: \(\mathbf{120}\)
(b) Not divisible by 5
Not divisible by 5: \(720 - 120 = \mathbf{600}\)
Q.12 If \(f:A\to B\) and \(g:B\to C\) are one-one, then \(g\circ f\) is also one-one
Proof: Suppose \((g\circ f)(x_1)=(g\circ f)(x_2)\) for some \(x_1,x_2\in A\).
- \(\Rightarrow g(f(x_1))=g(f(x_2))\)
- \(\Rightarrow f(x_1)=f(x_2)\) [since \(g\) is one-one]
- \(\Rightarrow x_1=x_2\) [since \(f\) is one-one]
∴ g∘f is one-one □
Q.13 \(f:R\to R\) defined by \(f(x)=\dfrac{3x}{5}+2\). Show one-one and onto, find \(f^{-1}\)
One-One
Let \(f(x_1)=f(x_2)\Rightarrow\dfrac{3x_1}{5}+2=\dfrac{3x_2}{5}+2\Rightarrow x_1=x_2\). Hence one-one.
Onto
For any \(y\in\mathbb{R}\), let \(x=\dfrac{5(y-2)}{3}\in\mathbb{R}\). Then \(f(x)=\dfrac{3}{5}\cdot\dfrac{5(y-2)}{3}+2=y\). Hence onto.
Inverse
Setting \(y=f(x)\): \(y=\dfrac{3x}{5}+2\Rightarrow x=\dfrac{5(y-2)}{3}\)
\(f^{-1}(x)=\dfrac{5(x-2)}{3}\)
Q.14 Evaluate \(\displaystyle\lim_{y\to 0}\dfrac{\sqrt{a+y}-\sqrt{a}}{y\sqrt{a+y}}\)
Rationalise the numerator by multiplying by \(\sqrt{a+y}+\sqrt{a}\):
\(\dfrac{\sqrt{a+y}-\sqrt{a}}{y\sqrt{a+y}} \times \dfrac{\sqrt{a+y}+\sqrt{a}}{\sqrt{a+y}+\sqrt{a}} = \dfrac{y}{y\sqrt{a+y}\left(\sqrt{a+y}+\sqrt{a}\right)}\)
\(= \dfrac{1}{\sqrt{a+y}\left(\sqrt{a+y}+\sqrt{a}\right)}\)
As \(y\to 0\):
\(\displaystyle\lim_{y\to 0} = \dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{a}\right)} = \dfrac{1}{\sqrt{a}\cdot 2\sqrt{a}} = \boxed{\dfrac{1}{2a}}\)
SECTION C
Q.15 to Q.26 — Attempt Any Eight (3 Marks Each)
8 × 3 = 24 Marks
Q.15 Verify \(A(B+C)=AB+AC\)
Given: \(A=\begin{bmatrix}1&-1&3\\2&3&2\end{bmatrix},\; B=\begin{bmatrix}1&0\\-2&3\\4&3\end{bmatrix},\; C=\begin{bmatrix}1&2\\-2&0\\4&-3\end{bmatrix}\)
Step 1: Find B+C
\(B+C=\begin{bmatrix}2&2\\-4&3\\8&0\end{bmatrix}\)
Step 2: Find A(B+C)
\(A(B+C)=\begin{bmatrix}1(2)+(-1)(-4)+3(8)&1(2)+(-1)(3)+3(0)\\2(2)+3(-4)+2(8)&2(2)+3(3)+2(0)\end{bmatrix}=\begin{bmatrix}30&-1\\8&13\end{bmatrix}\)
Step 3: Find AB + AC
\(AB=\begin{bmatrix}15&6\\4&15\end{bmatrix},\quad AC=\begin{bmatrix}15&-7\\4&-2\end{bmatrix}\)
\(AB+AC=\begin{bmatrix}30&-1\\8&13\end{bmatrix}\)
✓ A(B+C) = AB + AC = [[30, −1], [8, 13]] (Verified)
Q.16 Hyperbola \(21x^2-4y^2=84\) — all parameters
Divide by 84: \(\dfrac{x^2}{4}-\dfrac{y^2}{21}=1\). So \(a^2=4,\; b^2=21,\; a=2,\; b=\sqrt{21}\).
\(c^2 = a^2+b^2 = 4+21 = 25 \Rightarrow c=5\)
| Parameter | Formula | Value |
|---|
| Length of transverse axis | \(2a\) | \(4\) units |
| Length of conjugate axis | \(2b\) | \(2\sqrt{21}\) units |
| Eccentricity | \(e=c/a\) | \(5/2\) |
| Co-ordinates of foci | \((\pm c, 0)\) | \((\pm5,\,0)\) |
| Equations of directrices | \(x=\pm a/e = \pm a^2/c\) | \(x=\pm 4/5\) |
| Length of latus rectum | \(2b^2/a\) | \(2\times21/2=21\) units |
Q.17 Combined standard deviation for two samples of sizes 60 and 120
Given: \(n_1=60,\;\bar{x}_1=35.4,\;\sigma_1=4\) and \(n_2=120,\;\bar{x}_2=30.9,\;\sigma_2=5\).
Step 1: Combined Mean
\(\bar{x}_c = \dfrac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} = \dfrac{60(35.4)+120(30.9)}{180} = \dfrac{2124+3708}{180} = \dfrac{5832}{180} = 32.4\)
Step 2: Deviations from Combined Mean
\(d_1=\bar{x}_1-\bar{x}_c=35.4-32.4=3.0\)
\(d_2=\bar{x}_2-\bar{x}_c=30.9-32.4=-1.5\)
Step 3: Combined Variance
\(\sigma_c^2 = \dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}\)
\(= \dfrac{60(16+9)+120(25+2.25)}{180} = \dfrac{60(25)+120(27.25)}{180} = \dfrac{1500+3270}{180} = \dfrac{4770}{180} = 26.5\)
Combined S.D. \(= \sigma_c = \sqrt{26.5} = \dfrac{\sqrt{106}}{2} \approx 5.15\)
Q.18 Letters of 'EQUATION' arranged in a row — probability
EQUATION has 8 distinct letters. Vowels: E, U, A, I, O (5). Consonants: Q, T, N (3).
Total arrangements: \(8! = 40320\)
(i) All vowels together
Treat 5 vowels as one block → 4 units to arrange: \(4!\) ways. Vowels within block: \(5!\) ways.
Favourable = \(4!\times 5! = 24\times 120 = 2880\)
\(P(\text{all vowels together}) = \dfrac{2880}{40320} = \dfrac{1}{14}\)
(ii) Starts with a vowel and ends with a consonant
1st position (vowel): 5 choices | Last position (consonant): 3 choices
Middle 6 positions: \(6! = 720\) ways
Favourable = \(5\times 3\times 720 = 10800\)
\(P(\text{vowel first, consonant last}) = \dfrac{10800}{40320} = \dfrac{15}{56}\)
Q.19 \(S_n=2(3^n-1)\): find nth term and show it is a G.P.
nth Term for n ≥ 2
\(t_n = S_n - S_{n-1} = 2(3^n-1)-2(3^{n-1}-1) = 2\cdot 3^n - 2\cdot 3^{n-1} = 2\cdot 3^{n-1}(3-1) = 4\cdot 3^{n-1}\)
Check \(n=1\): \(t_1=S_1=2(3-1)=4=4\cdot 3^0\) ✓
Show G.P.
Common ratio \(r = \dfrac{t_{n+1}}{t_n} = \dfrac{4\cdot 3^n}{4\cdot 3^{n-1}} = 3 \quad\text{(constant)}\)
The sequence is a G.P. with first term \(4\) and common ratio \(3\).
Q.20 Coefficient of \(x^9\) in \(\left(\dfrac{1}{x}+x^2\right)^{18}\)
General term: \(T_{r+1}=\dbinom{18}{r}\left(\dfrac{1}{x}\right)^{18-r}(x^2)^r = \dbinom{18}{r}\cdot x^{3r-18}\)
For \(x^9\): \(\;3r-18=9 \Rightarrow r=9\)
Coefficient \(= \dbinom{18}{9} = \mathbf{48620}\)
Q.21 Domain of \(f(x)=\sqrt{\log(x^2-6x+6)}\)
Condition 1: Argument of log must be positive
\(x^2-6x+6>0\)
Roots: \(x=\dfrac{6\pm\sqrt{36-24}}{2}=3\pm\sqrt{3}\)
So: \(x<3-\sqrt{3}\) or \(x>3+\sqrt{3}\) (i.e., \(x<1.27\) or \(x>4.73\))
Condition 2: log value must be ≥ 0 for square root to exist
\(\log(x^2-6x+6)\ge 0 \Rightarrow x^2-6x+6\ge 1 \Rightarrow x^2-6x+5\ge 0\)
\((x-1)(x-5)\ge 0 \Rightarrow x\le 1\text{ or }x\ge 5\)
Intersection of both conditions
Domain \(= (-\infty,\,1]\cup[5,\,+\infty)\)
Q.22 Simplify \(\log_{10}\dfrac{28}{45}-\log_{10}\dfrac{35}{324}+\log_{10}\dfrac{325}{432}-\log_{10}\dfrac{13}{15}\)
\(= \log_{10}\left[\dfrac{28}{45}\times\dfrac{324}{35}\times\dfrac{325}{432}\times\dfrac{15}{13}\right]\)
Factoring numerator and denominator:
Numerator: \(28\times 324\times 325\times 15 = 2^4\times 7\times 3^5\times 5^3\times 13\)
Denominator: \(45\times 35\times 432\times 13 = 2^4\times 3^5\times 5^2\times 7\times 13\)
Ratio \(= 5^3/5^2 = 5\)
\(= \log_{10}5\)
Q.23 \(\displaystyle\lim_{x\to 1}\dfrac{x+x^3+x^5+\cdots+x^{2n-1}-n}{x-1}\)
\(= \lim_{x\to 1}\sum_{k=1}^{n}\dfrac{x^{2k-1}-1}{x-1}\)
Using \(\lim_{x\to 1}\dfrac{x^m-1}{x-1}=m\):
\(=\sum_{k=1}^{n}(2k-1) = 1+3+5+\cdots+(2n-1)\)
\(= n^2\)
Q.24 \(\displaystyle\lim_{x\to 1}\dfrac{x^2+x\sqrt{x}-2}{x-1}\)
Substitute \(x=t^2\) (so \(t\to 1\)):
\(= \lim_{t\to 1}\dfrac{t^4+t^3-2}{t^2-1}\)
Factor numerator: \(t^4+t^3-2=(t-1)(t^3+2t^2+2t+2)\). Factor denominator: \((t-1)(t+1)\).
\(= \lim_{t\to 1}\dfrac{t^3+2t^2+2t+2}{t+1} = \dfrac{1+2+2+2}{2} = \dfrac{7}{2}\)
\(\displaystyle\lim_{x\to 1}\dfrac{x^2+x\sqrt{x}-2}{x-1} = \dfrac{7}{2}\)
Q.25 Differentiate \(y=\sin x\cdot\log x + e^x\cos x - e^x\sqrt{x}\)
Apply the product rule to each term:
\(\dfrac{d}{dx}(\sin x\cdot\log x) = \cos x\cdot\log x + \dfrac{\sin x}{x}\)
\(\dfrac{d}{dx}(e^x\cos x) = e^x\cos x - e^x\sin x = e^x(\cos x - \sin x)\)
\(\dfrac{d}{dx}(e^x\sqrt{x}) = e^x\sqrt{x} + e^x\cdot\dfrac{1}{2\sqrt{x}} = e^x\!\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)\)
\(\dfrac{dy}{dx} = \cos x\cdot\log x + \dfrac{\sin x}{x} + e^x\!\left(\cos x - \sin x - \sqrt{x} - \dfrac{1}{2\sqrt{x}}\right)\)
Q.26 Derivative of \(\log(2x+5)\) from first principles
\(f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\dfrac{\log(2x+2h+5)-\log(2x+5)}{h}\)
\(= \lim_{h\to 0}\dfrac{1}{h}\log\!\left(1+\dfrac{2h}{2x+5}\right)\)
Let \(t=\dfrac{2h}{2x+5}\) so \(h=\dfrac{t(2x+5)}{2}\) and as \(h\to 0, t\to 0\):
\(= \lim_{t\to 0}\dfrac{\log(1+t)}{\dfrac{t(2x+5)}{2}} = \dfrac{2}{2x+5}\lim_{t\to 0}\dfrac{\log(1+t)}{t} = \dfrac{2}{2x+5}\times 1\)
\(\dfrac{d}{dx}\bigl[\log(2x+5)\bigr] = \dfrac{2}{2x+5}\)
SECTION D
Q.27 to Q.34 — Attempt Any Five (4 Marks Each)
5 × 4 = 20 Marks
Q.27 Find the trigonometric functions of 45°
Consider a right-angled isosceles triangle with both legs equal to 1 unit.
Hypotenuse \(= \sqrt{1^2+1^2}=\sqrt{2}\).
| Function | Definition | Value at 45° |
|---|
| \(\sin 45°\) | Opposite/Hypotenuse | \(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\) |
| \(\cos 45°\) | Adjacent/Hypotenuse | \(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\) |
| \(\tan 45°\) | Opposite/Adjacent | \(1\) |
| \(\cot 45°\) | Adjacent/Opposite | \(1\) |
| \(\sec 45°\) | Hypotenuse/Adjacent | \(\sqrt{2}\) |
| \(\text{cosec}\, 45°\) | Hypotenuse/Opposite | \(\sqrt{2}\) |
Q.28 Show \(3x-4y+10=0\) is tangent to hyperbola \(x^2-4y^2=20\); find point of contact
From the line: \(y=\dfrac{3x+10}{4}\). Substitute into the hyperbola:
\(x^2-4\cdot\dfrac{(3x+10)^2}{16}=20 \Rightarrow 4x^2-(3x+10)^2=80\)
\(4x^2-9x^2-60x-100=80 \Rightarrow -5x^2-60x-180=0\)
\(x^2+12x+36=0 \Rightarrow (x+6)^2=0\)
The discriminant is zero, confirming a repeated root — hence the line is tangent.
\(x=-6,\quad y=\dfrac{3(-6)+10}{4}=\dfrac{-8}{4}=-2\)
Point of contact: \((-6,\,-2)\)
Alternatively, using the condition for tangency \(c^2=a^2m^2-b^2\): with \(m=3/4,\;c=5/2,\;a^2=20,\;b^2=5\): \((5/2)^2=20(9/16)-5=45/4-5=25/4\,\checkmark\)
Q.29 Number of diagonals of an n-sided polygon
From \(n\) vertices we can draw \(\dbinom{n}{2}\) line segments. Subtract the \(n\) sides:
Number of diagonals \(= \dbinom{n}{2}-n = \dfrac{n(n-1)}{2}-n = \dfrac{n(n-3)}{2}\)
| n | Formula | No. of Diagonals |
|---|
| (1) \(n=10\) | \(\frac{10\times7}{2}\) | 35 |
| (2) \(n=15\) | \(\frac{15\times12}{2}\) | 90 |
| (3) \(n=12\) | \(\frac{12\times9}{2}\) | 54 |
| (4) \(n=8\) | \(\frac{8\times5}{2}\) | 20 |
Q.30 Arrangements of letters of PLATOON (7 letters, O repeated twice)
Total letters: P, L, A, T, O, O, N. Consonants: P, L, T, N (4). Vowels: A, O, O (3).
Total arrangements: \(\dfrac{7!}{2!}=2520\)
(a) Two O's are never together
Arrangements with O's together: treat OO as one unit → \(6!\) arrangements \(= 720\)
O's not together: \(2520-720=\mathbf{1800}\)
(b) Consonants and vowels in alternate positions
With 7 positions, pattern must be C V C V C V C (4 consonants in odd positions, 3 vowels in even positions):
Arrange 4 consonants (all distinct): \(4! = 24\) ways
Arrange 3 vowels (A, O, O with repetition): \(\dfrac{3!}{2!} = 3\) ways
Total \(= 24\times 3 = \mathbf{72}\)
Q.31 Prove by induction: \(t_n=5^n-1\), given \(t_{n+1}=5t_n+4,\;t_1=4\)
Step 1 — Base Case (n = 1)
\(t_1 = 5^1-1 = 4\) ✓ (matches given \(t_1=4\))
Step 2 — Inductive Hypothesis
Assume \(t_k=5^k-1\) for some integer \(k\ge 1\).
Step 3 — Inductive Step (show for n = k+1)
\(t_{k+1} = 5t_k+4 = 5(5^k-1)+4 = 5^{k+1}-5+4 = 5^{k+1}-1\)
This is exactly the formula with \(n=k+1\).
∴ By mathematical induction, t_n = 5ⁿ − 1 for all n ≥ 1 □
Q.32 Using \(\varepsilon\text{-}\delta\) definition, prove \(\displaystyle\lim_{x\to 1}(x^2+x+1)=3\)
We must show: given \(\varepsilon>0\), \(\exists\,\delta>0\) such that \(|x-1|<\delta \Rightarrow |f(x)-3|<\varepsilon\).
\(|f(x)-3| = |x^2+x+1-3| = |x^2+x-2| = |(x+2)(x-1)| = |x+2|\cdot|x-1|\)
If \(|x-1|<1\), then \(0
Choose \(\delta = \min\!\left(1,\,\dfrac{\varepsilon}{4}\right)\). Then: \(|f(x)-3|=|x+2|\cdot|x-1|<4\cdot\dfrac{\varepsilon}{4}=\varepsilon\)
∴ lim(x→1) (x² + x + 1) = 3 □
Q.33 \(\displaystyle\lim_{x\to 0}\dfrac{1}{x^{12}}\!\left[1-\cos\!\tfrac{x^2}{2}-\cos\!\tfrac{x^4}{4}+\cos\!\tfrac{x^2}{2}\cdot\cos\!\tfrac{x^4}{4}\right]\)
Let \(A=\cos\!\dfrac{x^2}{2}\) and \(B=\cos\!\dfrac{x^4}{4}\). The bracket is:
\(1-A-B+AB = (1-A)(1-B)\)
Using the standard result \(1-\cos\theta\sim\dfrac{\theta^2}{2}\) as \(\theta\to 0\):
\(1-\cos\dfrac{x^2}{2}\sim\dfrac{(x^2/2)^2}{2}=\dfrac{x^4}{8}\)
\(1-\cos\dfrac{x^4}{4}\sim\dfrac{(x^4/4)^2}{2}=\dfrac{x^8}{32}\)
\(\text{Limit} = \lim_{x\to 0}\dfrac{(1-A)(1-B)}{x^{12}} = \lim_{x\to 0}\dfrac{\frac{x^4}{8}\cdot\frac{x^8}{32}}{x^{12}} = \dfrac{1}{8}\cdot\dfrac{1}{32} = \dfrac{1}{256}\)
\(\displaystyle\lim_{x\to 0}\cdots = \dfrac{1}{256}\)
Q.34 Find k so that \(f(x)=\dfrac{24^x-8^x-3^x+1}{12^x-4^x-3^x+1}\) is continuous at \(x=0\)
Factor the numerator and denominator:
\(\text{Numerator: }\;24^x-8^x-3^x+1 = 8^x(3^x-1)-(3^x-1)=(3^x-1)(8^x-1)\)
\(\text{Denominator: }\;12^x-4^x-3^x+1 = 4^x(3^x-1)-(3^x-1)=(3^x-1)(4^x-1)\)
\(f(x) = \dfrac{(3^x-1)(8^x-1)}{(3^x-1)(4^x-1)} = \dfrac{8^x-1}{4^x-1}\quad(x\ne 0)\)
\(\lim_{x\to 0}f(x)=\lim_{x\to 0}\dfrac{8^x-1}{4^x-1} = \dfrac{\lim_{x\to 0}\frac{8^x-1}{x}}{\lim_{x\to 0}\frac{4^x-1}{x}} = \dfrac{\ln 8}{\ln 4} = \dfrac{3\ln 2}{2\ln 2} = \dfrac{3}{2}\)
For continuity at \(x=0\): \(k = f(0) = \displaystyle\lim_{x\to 0}f(x)\)
\(\boxed{k = \dfrac{3}{2}}\)