Bihar Board Class 12 Mathematics Elective Question Paper 2023 with Answer Key

Step 1 (Reason: Parallel lines have proportional direction ratios).
For lines with d.r.s \((x,5,3)\) and \((20,10,6)\), \[ \frac{x}{20}=\frac{5}{10}=\frac{3}{6}. \]

Step 2 (Reason: simplify known ratios to identify the common factor). \(\dfrac{5}{10}=\dfrac{3}{6}=\dfrac12\), hence the common ratio is \(\dfrac12\).

Step 3 (Reason: equate first pair to the same ratio and solve for \(x\)). \[ \frac{x}{20}=\frac12 \ \Rightarrow\ x=20\cdot\frac12=10. \]
Therefore \(x=10\). Quick Tip: Parallel (or collinear) direction ratios are always in the same ratio: \((l_1,m_1,n_1)\parallel(l_2,m_2,n_2)\iff \dfrac{l_1}{l_2}=\dfrac{m_1}{m_2}=\dfrac{n_1}{n_2}\).

Step 1 (Reason: Parallelism implies a common proportionality constant).
Let \[ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k\neq 0. \]

Step 2 (Reason: rewrite \(a_1\) and \(c_1\) using \(k\)).
From above, \(a_1=k\,a_2\) and \(c_1=k\,c_2\).

Step 3 (Reason: substitute and cancel). \[ \frac{a_1 c_2}{a_2}=\frac{k\,a_2\cdot c_2}{a_2}=k\,c_2=c_1. \]
Hence \(\dfrac{a_1 c_2}{a_2}=c_1\). Quick Tip: Use the single constant \(k\): \(a_1=k a_2,\ b_1=k b_2,\ c_1=k c_2\). Such substitutions make parallel-lines algebra immediate.

Idea. Two lines are perpendicular in 3-D exactly when the dot product of any direction-ratio triples is \(0\). Dot product adds the products of the matching components.

Step 1. Compute the dot product and set it to zero. \[ (2,3,5)\cdot(x,y,4)=2x+3y+5\cdot 4=0. \]
Step 2. Simplify. \[ 2x+3y+20=0\quad\Rightarrow\quad 2x+3y=-20. \]
That is the required value.
Quick Tip: Perpendicular \(\Rightarrow\) dot product \(=0\): multiply component-wise and add.

Idea. The length of a vector \(\langle a,b,c\rangle\) is the 3-D version of Pythagoras: \(\sqrt{a^2+b^2+c^2}\). Signs do not matter because we square.

Step. \[ \sqrt{3^2+(-4)^2+(-5)^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}. \] Quick Tip: Magnitude \(=\sqrt{(x-part)^2+(y-part)^2+(z-part)^2}\).

Idea. Two equal row vectors must match entry by entry. Turn each position into a small equation.

Step 1. First position. \(2a-7=a\Rightarrow a=7\).

Step 2. Second position. \(1=b-1\Rightarrow b=2\).

So \((a,b)=(7,2)\). Quick Tip: Equal matrices/vectors \(\Rightarrow\) equal corresponding entries.

Idea. Use expansion along the first row (signs \(+\,-\,+\)). Compute three \(2\times2\) minors.

Step 1. Minors.

Step 2. Combine with signs. \[ \det=1\cdot 28-1\cdot 3+5\cdot(-5)=28-3-25=0. \]
Zero determinant also hints that the third row is the sum of the first two, so rows are dependent.
Quick Tip: Row-1 expansion uses \(+\,-\,+\). Linear dependence of rows \(\Rightarrow\) determinant \(0\).

Idea. Again expand along the first row, or notice a quick pattern: \(R_3=R_1+R_2\) \(\Rightarrow\) rows dependent \(\Rightarrow\) determinant \(0\). We also verify by expansion.

Step (verification).

Quick Tip: If one row = sum of two others, det \(=0\). Spotting such relations saves time.

Idea. For a \(2\times2\) matrix  \( = ad-bc\). Use reciprocal trig identities \(\sec\theta=\tfrac1{\cos\theta}\), \(\csc\theta=\tfrac1{\sin\theta}\).

Step. \[ (-\sin\theta)\left(\frac{1}{\sin\theta}\right)-\cos\theta\left(\frac{1}{\cos\theta}\right)=-1-1=-2. \] Quick Tip: Replace \(\sec,\csc\) by reciprocals and cancel—often the quickest path.

Idea. Multiplying by the identity \(I\) leaves a matrix unchanged, just like multiplying a number by \(1\).

So \(I\cdot A=A\) and the product equals the second matrix itself. Quick Tip: \(I\) is the multiplicative identity of matrices: \(IA=AI=A\).

Idea. Row–column multiplication: multiply matching entries and add. A \(1\times2\) times \(2\times1\) gives a \(1\times1\) number.
\[ 6\cdot 1+5\cdot(-1)=6-5=1. \] Quick Tip: Always check sizes: \((1\times2)\cdot(2\times1)\to(1\times1)\).

Idea. Use the chain rule: derivative of \(\cos u\) is \(-\sin u\cdot u'\). The outer constant \(2\) stays outside.

Step 1. Let \(u=\frac{3x}{4}\Rightarrow u'=\frac{3}{4}\).

Step 2. \[ \frac{d}{dx}\big(2\cos u\big)=2(-\sin u)\,u'=-2\sin\!\left(\tfrac{3x}{4}\right)\cdot\frac{3}{4} =-\frac{3}{2}\sin\!\left(\tfrac{3x}{4}\right). \] Quick Tip: Inside-function derivative multiplies the result: \(f(g(x))'\!=f'(g)\,g'\).

Idea. Derivative of \(e^{u}\) is \(e^{u}\cdot u'\). Here \(u=-3x\) so \(u'=-3\).
\[ \frac{d}{dx}e^{-3x}=e^{-3x}\cdot(-3)=-3e^{-3x}. \] Quick Tip: Shortcut: \((e^{kx})'=k\,e^{kx}\).

Idea. For base \(a\) (constant), \(\dfrac{d}{dx}(a^x)=a^x\ln a\). \(\ln\) and \(\log\) (base \(e\)) are the same in calculus notation here.

So \(\dfrac{d}{dx}(11^x)=11^x\ln 11=11^x\cdot \log 11\). Quick Tip: Don’t mix with power rule \(x^n\). Here the variable is in the exponent.

Idea. Write as a negative power and apply the chain rule.
\[ (3x-2)^{-1}\ \Rightarrow\ \frac{d}{dx}(3x-2)^{-1}=-1(3x-2)^{-2}\cdot(3) =-\frac{3}{(3x-2)^2}. \] Quick Tip: General form: \(\dfrac{d}{dx}(ax+b)^{-1}=-a(ax+b)^{-2}\).

Idea. Use parametric differentiation: \(\dfrac{dy}{dx}=\dfrac{(dy/d\theta)}{(dx/d\theta)}\). Trig double-angle simplifies nicely.

Step 1. \(dy/d\theta=b\cdot 2\sin\theta\cos\theta=b\sin2\theta.\)

Step 2. \(dx/d\theta=a\cdot 2\cos\theta(-\sin\theta)=-a\sin2\theta.\)

Step 3. \(\dfrac{dy}{dx}=\dfrac{b\sin2\theta}{-a\sin2\theta}=-\dfrac{b}{a}\). Quick Tip: With parameters, divide derivatives: many trig factors cancel automatically.

Idea. This is already separated: an \(x\)-part with \(dx\) plus a \(y\)-part with \(dy\). Integrate each side straight away.
\[ \int x^2\,dx+\int y^2\,dy=C \Rightarrow \frac{x^3}{3}+\frac{y^3}{3}=C \Rightarrow x^3+y^3=k. \] Quick Tip: Form \(M(x)dx+N(y)dy=0\) \(\Rightarrow\) integrate terms independently.

Idea. Convert to components and use \( \vec i\cdot\vec i=\vec j\cdot\vec j=\vec k\cdot\vec k=1\) and cross terms \(0\).
\((-2,1,0)\cdot(3,-1,1)=(-2)\cdot3+1\cdot(-1)+0\cdot1=-6-1+0=-7.\) Quick Tip: Arrange vectors as triples \((x,y,z)\) and multiply component-wise.

Idea. Again the equation is separable: one pure \(x\) term and one pure \(y\) term. Move one to the other side and integrate.
\[ e^{3x}dx=-e^{4y}dy \Rightarrow \int e^{3x}dx+\int e^{4y}dy=C \Rightarrow \frac{1}{3}e^{3x}+\frac{1}{4}e^{4y}=k. \] Quick Tip: \(\int e^{ax}\,dx=\dfrac{1}{a}e^{ax}\). Add constants into a single \(k\).

Idea. Integrate each fraction; logarithms add. Exponentiate to remove \(\ln\).
\[ \ln|x|+\ln|y|=C\Rightarrow \ln|xy|=C\Rightarrow xy=e^{C}=k. \] Quick Tip: \(\ln A+\ln B=\ln(AB)\). Replace \(e^{C}\) by a new constant \(k\).

Idea. Standard linear form is \(y'+P(x)y=Q(x)\). The integrating factor is \(e^{\int P(x)\,dx}\).

Step 1. Identify \(P(x)\). Here \(y'-y\sin x=\cot x\Rightarrow P(x)=-\sin x\).

Step 2. Compute IF. \[ IF=e^{\int -\sin x\,dx}=e^{\cos x}. \]
That’s the required integrating factor (no need to solve for \(y\) here).
Quick Tip: Linear DE \(y'+P y=Q\) \(\Rightarrow\) IF \(=\exp(\int P\,dx)\). Watch the sign when moving terms.

Think of \([\,{-}1\,]\) as a plain number \(-1\). When a scalar multiplies a matrix (or row), it multiplies each entry.
So, \[ [\,{-}1\,]\,[\,1\ \ {-}1\,] =[\,{-}1\times 1\ \ {-}1\times(-1)\,] =[\,{-}1\ \ 1\,]. \]
That’s all—entrywise scaling. Quick Tip: Scalar \(\times\) matrix \(=\) scale every entry by the scalar.

Multiply each entry by \(3\):

Quick Tip: A scalar in front of a matrix distributes to every position.

The feasible region is the triangle with vertices where the lines meet: \((0,0)\), \((25,0)\), \((0,25)\). A linear objective hits its max at a vertex.
Evaluate \(Z\): \((25,0)\Rightarrow Z=6\cdot25+3\cdot0=150\). \((0,25)\Rightarrow Z=0+75=75\). \((0,0)\Rightarrow Z=0\).
So the maximum is \(150\) at \((25,0)\). Quick Tip: Two-variable LPP: just test \(Z\) at the corner points.

Since \(y\) has a negative coefficient in \(Z\), we want the smallest \(y\) (take \(y=0\)) and the largest \(x\) allowed.
With \(y=0\), the constraint becomes \(x\le13\).
Corner checks: \((13,0)\Rightarrow Z=13\) (best), \((0,13)\Rightarrow Z=-39\), \((0,0)\Rightarrow Z=0\). Quick Tip: Negative coefficient \(\to\) push that variable to its minimum (if allowed).

All coefficients in \(Z\) are positive and the origin \((0,0)\) is feasible (\(0\le24\)).
Therefore the smallest value occurs at \((0,0)\): \(Z=0\). Quick Tip: For “\(\le\)” constraints with \(x,y\ge0\), the origin is feasible; check it first for minimization.

Write components: \((2,-3,0)\) and \((1,0,1)\).
Dot product \(=\) multiply componentwise and add: \[ 2\cdot1+(-3)\cdot0+0\cdot1=2. \] Quick Tip: Only same-direction pairs (\(i\cdot i\), \(j\cdot j\), \(k\cdot k\)) contribute.

Let \(\theta=\tan^{-1}x \Rightarrow x=\tan\theta\) with \(\theta\in[-\pi/4,\pi/4]\).
Use \(\sin(2\theta)=\dfrac{2\tan\theta}{1+\tan^2\theta}\): \[ \sin(2\theta)=\frac{2x}{1+x^2}. \]
Since \(2\theta\in[-\pi/2,\pi/2]\), \(\sin^{-1}\) returns \(2\theta\): \[ 2\tan^{-1}x=2\theta=\sin^{-1}\!\left(\frac{2x}{1+x^2}\right). \] Quick Tip: Set \(x=\tan\theta\) and convert using double-angle formulas.

\(\cot\theta=\tan\!\left(\tfrac{\pi}{2}-\theta\right)\).
Apply inverse on both sides (with principal ranges): \(\cot^{-1}x=\tfrac{\pi}{2}-\tan^{-1}x\). Quick Tip: \(\tan\) and \(\cot\) are complementary: swap with \(\frac{\pi}{2}-\,\)angle.

Use the tangent addition formula \(\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta}\).
Take \(\alpha=\tan^{-1}x,\ \beta=\tan^{-1}y\). Then \(\tan(\alpha+\beta)=\dfrac{x+y}{1-xy}\).
Apply \(\tan^{-1}\): \(\alpha+\beta=\tan^{-1}x+\tan^{-1}y\). Quick Tip: Spot \(\dfrac{x+y}{1-xy}\) \(\Rightarrow\) it is \(\tan(\alpha+\beta)\) in disguise.

\(\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}\). \(\sin^{-1}\) must return an angle in \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).
In that range, the angle whose sine is \(\tfrac{\sqrt3}{2}\) is \(\tfrac{\pi}{3}\), not \(2\pi/3\). Quick Tip: Always convert back to the principal} angle for inverse trig.

Right matrix is \(2I\) (twice the identity). Multiplying any matrix by \(2I\) doubles each column:
First column \( \to 2\times(13,-1)=(26,-2)\); second column \(\to 2\times(15,4)=(30,8)\).
So product . Quick Tip: Multiplication by a diagonal matrix scales columns by the diagonal entries.

Row–column rule:
Top entry \(=2\cdot2+3\cdot5=4+15=19\).
Bottom entry \(=5\cdot2+7\cdot5=10+35=45\). Quick Tip: \((2\times2)(2\times1)\Rightarrow(2\times1)\) — size check prevents mistakes.

Use \(\displaystyle\int \tan\theta\,d\theta=\ln|\sec\theta|+C\).
Evaluate: \[ \ln(\sec\tfrac{\pi}{4})-\ln(\sec\tfrac{\pi}{6}) =\ln(\sqrt2)-\ln\!\Big(\tfrac{2}{\sqrt3}\Big) =\ln\!\Big(\tfrac{\sqrt6}{2}\Big). \]
This value is not \(0,1,2,\) or \(3\). Hence none of the given choices matches. Quick Tip: \(\int\tan\theta\,d\theta=\ln|\sec\theta|\); definite values often stay as logs.

\(\csc^{2}\theta=\dfrac{1}{\sin^{2}\theta}\). So \[ \sin^{3}\theta\cdot\csc^{2}\theta=\sin^{3}\theta\cdot\frac{1}{\sin^{2}\theta}=\sin\theta. \]
Then \(\int \sin\theta\,d\theta=-\cos\theta+c\). Quick Tip: Rewrite reciprocals (\(\csc^2=1/\sin^2\)) and cancel powers first.

Factor \(\cos\theta\): \(\cos\theta(\csc^{2}\theta-\cot^{2}\theta)\).
Identity: \(\csc^{2}\theta-\cot^{2}\theta=1\).
So integral \(\int \cos\theta\,d\theta=\sin\theta+k\). Quick Tip: Use \(\csc^2-\cot^2=1\) (parallel to \(\sec^2-\tan^2=1\)).

Integrate term by term: \(\int 4\cos x\,dx=4\sin x\). \(\int -5\sin x\,dx=5\cos x\).
Add the constant \(k\). Quick Tip: Linearity: integrate each term, then add a single constant \(k\).

Let \(f(x)=2\cos x+3\sin x\). Then \[ f'(x)=-2\sin x+3\cos x=3\cos x-2\sin x, \]
which is exactly the numerator.
Hence the integrand is \(\dfrac{f'(x)}{f(x)}\), so \(\int \dfrac{f'}{f}\,dx=\ln|f|+k=\ln|2\cos x+3\sin x|+k\). Quick Tip: Spot “derivative over itself” \(\Rightarrow\) immediate \(\ln|\,\cdot\,|\).

Let \(g(x)=x^{3}+2x\Rightarrow g'(x)=3x^{2}+2\) (the numerator).
Therefore \(\int\dfrac{g'(x)}{g(x)}dx=\ln|g(x)|+k=\ln|x^{3}+2x|+k\). Quick Tip: Try to see the denominator’s derivative sitting upstairs.

Standard form: \[ \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)+C. \]
Here \(a^{2}=5\Rightarrow a=\sqrt5\). Substitute to get the answer. Quick Tip: Memorize: \(\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\!\frac{x}{a}\).

Let \(f(x)=\ln\!\left(\dfrac{3+x}{3-x}\right)\). Check symmetry: \[ f(-x)=\ln\!\left(\frac{3-x}{3+x}\right)=-\ln\!\left(\frac{3+x}{3-x}\right)=-f(x). \]
So \(f\) is an odd function. The integral of an odd function over a symmetric interval \([-a,a]\) is \(0\). Hence the value is \(0\). Quick Tip: If \(f(-x)=-f(x)\), then \(\int_{-a}^{a} f(x)\,dx=0\).

Why we can multiply: For independent events, knowing that one happens does not change the chance of the other. By definition, this gives the multiplication rule \[ P(A\cap B)=P(A)\cdot P(B). \]
Substitute the given probabilities: \[ P(A\cap B)=0.3\times 0.4=0.12. \]
So the probability that both \(A\) and \(B\) occur together is \(0.12\). Quick Tip: Independent \(\Rightarrow\) multiply: \(P(A\cap B)=P(A)P(B)\). (If they were not} independent, this formula would not hold.)

Step 1: Recall the 2\(\times\)2 formula.
For ,

This comes from taking cofactors and then transposing the cofactor matrix (for a \(2\times2\), that reduces to swapping \(a\leftrightarrow d\) and negating the off–diagonals).
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Step 2: Apply it to \(A=\) 
Here \(a=1,\ b=2,\ c=3,\ d=4\). Therefore
This matches option (A). Quick Tip: Memorize the \(2\times2\) adjoint: swap the diagonal entries, change signs of the off–diagonals: .

For any line in 3-D, its direction cosines \(l,m,n\) satisfy \[ l^{2}+m^{2}+n^{2}=1 \quad (unit vector condition). \]
Here \(l=\dfrac{4}{\sqrt{77}},\ m=\dfrac{5}{\sqrt{77}},\ n=\dfrac{x}{\sqrt{77}}\).
So \[ \left(\frac{4}{\sqrt{77}}\right)^{2}+\left(\frac{5}{\sqrt{77}}\right)^{2}+\left(\frac{x}{\sqrt{77}}\right)^{2}=1 \] \[ \frac{16+25+x^{2}}{77}=1 \ \Rightarrow\ 16+25+x^{2}=77\ \Rightarrow\ x^{2}=36. \]
Hence \(x=\pm 6\). Among the choices the matching value is \(6\). Quick Tip: Direction cosines are the components of a unit} direction vector, so they always satisfy \(l^2+m^2+n^2=1\).

The given matrix is the identity \(I_2\). Identity is the multiplicative neutral element: \[ I_2^n=I_2 \quad for any positive integer n. \]
Therefore, \[ A^{25}=I_2^{25}=I_2=A. \] Quick Tip: Raising the identity to any power gives the identity back: \(I^n=I\).

Operate step by step (left to right):

1) First compute \(2\ \mathrm{o}\ 3\): \[ 2\ \mathrm{o}\ 3=3(2)+3=6+3=9. \]
2) Now use this result with \(5\): \[ (2\ \mathrm{o}\ 3)\ \mathrm{o}\ 5=9\ \mathrm{o}\ 5=3(9)+5=27+5=32. \]
Hence the value is \(32\). Quick Tip: When a custom operation is given by a formula, evaluate exactly in the stated order: compute the inner operation first, then apply the rule again.

For a function \(f:A\to B\), each element of \(A\) may be sent to any element of \(B\).
Here \(|A|=2\), \(|B|=3\). For the first element of \(A\) there are \(3\) choices; for the second, again \(3\) choices (independent).
Thus total functions \(=3\times3=3^{2}=9\). Quick Tip: Number of functions from an \(m\)-element set to an \(n\)-element set is \(n^{m}\).

For injective \(f:A\to B\), distinct elements of \(A\) must go to distinct elements of \(B\).
Choose an image for \(a\): \(3\) choices. Then for \(b\) only \(2\) choices remain (must be different).
Total \(=3\times2=6=P(3,2)\). Quick Tip: One–one maps from size \(m\) to size \(n\) (with \(n\ge m\)) are \(nP m=\frac{n!}{(n-m)!}\).

Rewrite \(dx+dy=0\) as \(dy=-dx\). Integrate both sides: \[ \int dy=\int(-dx)\quad\Rightarrow\quad y=-x+C. \]
Move \(x\) to the left: \(x+y=C=k\). This represents straight lines of slope \(-1\). Quick Tip: When differentials separate directly, integrate both sides and combine constants.

Unit vectors have length \(1\) and are mutually perpendicular. Dot product of a vector with itself equals the square of its length: \[ \vec i\cdot\vec i=\|\vec i\|^{2}=1^{2}=1. \] Quick Tip: \(\vec i\cdot\vec i=\vec j\cdot\vec j=\vec k\cdot\vec k=1\); mixed pairs are \(0\).

Right–hand rule: \(\vec i\times\vec j=\vec k\). Swapping the order changes the sign: \[ \vec j\times\vec i=-(\vec i\times\vec j)=-\vec k. \] Quick Tip: Cross product is anti-commutative: \(\mathbf{a}\times\mathbf{b}=-(\mathbf{b}\times\mathbf{a})\).

\(\sin^{-1}\) returns an angle whose sine is the input. Thus \(\theta=\sin^{-1}\tfrac12\) satisfies \(\sin\theta=\tfrac12\). Applying \(\sin\) undoes \(\sin^{-1}\): \[ \sin(\sin^{-1}\tfrac12)=\tfrac12. \] Quick Tip: For any \(x\in[-1,1]\), \(\sin(\sin^{-1}x)=x\).

Let \(\alpha=\sin^{-1}x,\ \beta=\sin^{-1}y\Rightarrow \sin\alpha=x,\ \sin\beta=y\) with \(\alpha,\beta\in[-\pi/2,\pi/2]\).
Then \[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta =x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}. \]
Since \(\alpha+\beta\) lies in \([-\pi,\pi]\) and under usual exam assumptions \(\alpha+\beta\in[-\pi/2,\pi/2]\), we write \(\alpha+\beta=\sin^{-1}\!\big(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\big)\). Quick Tip: Convert inverse sines to angles, use \(\sin(\alpha+\beta)\), then return via \(\sin^{-1}\).

\(\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\). Hence \[ 2(\sin^{-1}x+\cos^{-1}x)=2\cdot \frac{\pi}{2}=\pi,\quad \sin(\pi)=0. \] Quick Tip: Key identity: \(\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\) for all \(x\in[-1,1]\).

With the standard principal ranges used in school exams, \(\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}\).
Therefore \[ \csc\!\left(\tan^{-1}x+\cot^{-1}x\right)=\csc\!\left(\frac{\pi}{2}\right)=1. \] Quick Tip: Remember \(\arctan x+\operatorname{arccot}x=\dfrac{\pi}{2}\) (principal values).

Using \(\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}\), \[ \tan\!\left[\frac{2}{3}\cdot\frac{\pi}{2}\right]=\tan\!\left(\frac{\pi}{3}\right)=\sqrt3. \] Quick Tip: Once the inside is a fixed angle, evaluate the outer trig directly.

Differentiate termwise: \(\dfrac{d}{dx}(e^{x})=e^{x}\); \(\dfrac{d}{dx}(\cos5x)=-\sin5x\cdot 5\) by chain rule.
Add: \[ e^{x}-5\sin5x. \] Quick Tip: For \(\cos(kx)\), derivative is \(-k\sin(kx)\).

\(\dfrac{d}{dx}(\sin2x)=2\cos2x\); \(\dfrac{d}{dx}(e^{x})=e^{x}\); \(\dfrac{d}{dx}(-\cos x)=+\sin x\).
Sum them to get \(2\cos2x+e^{x}+\sin x\). Quick Tip: Derivative of \(-\cos x\) is \(+\sin x\); watch the sign.

\[ \frac{d}{dx}\!\left(\frac14\sec4x\right)=\frac14\cdot (\sec4x\tan4x)\cdot 4=\sec4x\tan4x. \] Quick Tip: For \(\sec(kx)\), derivative is \(k\,\sec(kx)\tan(kx)\).

\(\ln(10x)=\ln10+\ln x\). Constant’s derivative is \(0\). \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\). Hence the result is \(1/x\). Quick Tip: \(\dfrac{d}{dx}\ln(ax)=\dfrac{1}{x}\) for any constant \(a>0\).

Write plane in \(Ax+By+Cz+D=0\) form: \(3x-4y+6z-11=0\).
Distance from origin: \[ \frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}} =\frac{|{-}11|}{\sqrt{3^{2}+(-4)^{2}+6^{2}}} =\frac{11}{\sqrt{9+16+36}} =\frac{11}{\sqrt{61}}. \] Quick Tip: Point–plane distance: \(\displaystyle \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}\); put plane as \(Ax+By+Cz+D=0\) first.

Reason. Use the chain rule in reverse (substitution).

Let \(u=\dfrac{3x}{4}\Rightarrow du=\dfrac{3}{4}dx\Rightarrow dx=\dfrac{4}{3}du\). Then \[ \int \sin\!\left(\tfrac{3x}{4}\right)\,dx=\int \sin u\cdot \frac{4}{3}\,du =\frac{4}{3}\int \sin u\,du =\frac{4}{3}(-\cos u)+k =-\frac{4}{3}\cos\!\left(\tfrac{3x}{4}\right)+k. \]
Thus the antiderivative is \(k-\dfrac{4}{3}\cos\!\left(\dfrac{3x}{4}\right)\). Quick Tip: \(\displaystyle \int \sin(ax)\,dx=-\frac{1}{a}\cos(ax)+C\).

\(\displaystyle \int \cos(ax)\,dx=\frac{1}{a}\sin(ax)+C\).
Here \(a=\dfrac{7}{9}\), hence \[ \int \cos\!\left(\tfrac{7x}{9}\right)dx=\frac{1}{7/9}\sin\!\left(\tfrac{7x}{9}\right)+k =\frac{9}{7}\sin\!\left(\tfrac{7x}{9}\right)+k. \] Quick Tip: Differentiate to check: \(\dfrac{d}{dx}\big(\tfrac{1}{a}\sin(ax)\big)=\cos(ax)\).

\(\displaystyle \int \sec^{2}(ax)\,dx=\frac{1}{a}\tan(ax)+C\).
With \(a=\dfrac{17}{23}\), \[ \int \sec^{2}\!\left(\tfrac{17x}{23}\right)\!dx =\frac{1}{17/23}\tan\!\left(\tfrac{17x}{23}\right)+k =\frac{23}{17}\tan\!\left(\tfrac{17x}{23}\right)+k. \] Quick Tip: Because \((\tan u)'=\sec^2 u\), the antiderivative is \(\tan u\) divided by \(u'\).

For base \(a>0,\ a\ne1\): \(\displaystyle \int a^{x}dx=\frac{a^{x}}{\ln a}+C\).
Putting \(a=4\) gives \(\dfrac{4^x}{\log 4}+k\) (here \(\log\) means natural log). Quick Tip: Do not confuse with \(x^{n}\): that uses the power rule, not the exponential rule.

First expand: \(x(4x^{2}-6)=4x^{3}-6x\).
Integrate termwise: \[ \int 4x^{3}dx=x^{4},\qquad \int -6x\,dx=-3x^{2}. \]
Hence \(x^{4}-3x^{2}+k\). Quick Tip: Always simplify the integrand (expand/factor) before integrating.

Recognize a derivative: \(\dfrac{d}{dx}(e^{x}\cos x)=e^{x}\cos x-e^{x}\sin x=e^{x}(\cos x-\sin x)\).
Thus the integral is \(e^{x}\cos x+k\). Quick Tip: Look for a known product whose derivative matches the pattern.

Check the derivative of \(x^{3}e^{x}\) by product rule: \[ \frac{d}{dx}(x^{3}e^{x})=e^{x}x^{3}+e^{x}\cdot 3x^{2}=e^{x}(x^{3}+3x^{2}). \]
It matches the integrand, so the antiderivative is \(x^{3}e^{x}+k\). Quick Tip: When you see \(e^x\) times a polynomial, try derivative of \((polynomial)\cdot e^x\).

Differentiate \(\dfrac{e^{x}}{x}\) using quotient rule: \[ \left(\frac{e^{x}}{x}\right)'=\frac{x e^{x}-e^{x}}{x^{2}}=e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right). \]
Matches the integrand, so the antiderivative is \(\dfrac{e^{x}}{x}+k\). Quick Tip: A pattern like \(e^x\!\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\) screams “derivative of \(\frac{e^x}{x}\)”.

Write components: \(3\vec{k}=(0,0,3)\) and \(13\vec{i}-7\vec{k}=(13,0,-7)\).
Dot product \(=(0)(13)+(0)(0)+(3)(-7)=-21\).
The correct value is \(-21\). Since that number does not appear, the printed choices are inconsistent. If the question intended the magnitude only, \(|-21|=21\) (option (C)). Quick Tip: Dot product adds products of matching components; sign matters.

Chain rule: \((\sin u)'=\cos u\cdot u'\) with \(u=\dfrac{4x}{5}\Rightarrow u'=\dfrac{4}{5}\). So derivative \(=\dfrac{4}{5}\cos\!\left(\dfrac{4x}{5}\right)\). Quick Tip: Always multiply by the inner derivative \(u'\) after differentiating the outside.

Parallel planes have the same normal vector, so the coefficients of \(x,y,z\) must be proportional (here exactly the same). Only the constant may differ.
Thus the family: \(x-8y-9z=d\) with any \(d\neq12\). Option (B) fits. Quick Tip: To be parallel: keep the \(x,y,z\) coefficients the same; change only the constant term.

Here “square” means dotting the vector with itself (magnitude squared): \[ (3\vec{i}-4\vec{k})\cdot(3\vec{i}-4\vec{k})=3^{2}+(-4)^{2}=9+16=25. \] Quick Tip: \(\|\mathbf{v}\|^{2}=\mathbf{v}\cdot\mathbf{v}\).

Magnitude \(\|3\vec{i}-9\vec{j}\|=\sqrt{3^{2}+(-9)^{2}}=\sqrt{9+81}=\sqrt{90}\).
Unit vector \(=\dfrac{1}{\|\cdot\|}\times\) vector \(=\dfrac{3\vec{i}-9\vec{j}}{\sqrt{90}}\) (which simplifies to \(\dfrac{\vec{i}-3\vec{j}}{\sqrt{10}}\)). Quick Tip: Unit vector = vector divided by its magnitude.

Dot product: multiply components and add: \[ 1\cdot7+(-1)\cdot(-8)+1\cdot9=7+8+9=24. \] Quick Tip: Write vectors as triples and use \(a_1a_2+b_1b_2+c_1c_2\).

On the \(x\)-axis \(y=z=0\). Substitute into the plane: \[ 3x=13\Rightarrow x=\frac{13}{3}. \]
That is the \(x\)-intercept. Quick Tip: To find the \(x\)-intercept of a plane, set \(y=z=0\) and solve for \(x\).

Direction ratios of the line are \((-1,2,3)\). A line parallel to a plane has its direction vector perpendicular to the plane’s normal \((a,b,c)\).
So their dot product is \(0\): \[ (-1,2,3)\cdot(a,b,c)=-a+2b+3c=0. \] Quick Tip: Parallel to plane \(\Rightarrow\) direction vector \(\perp\) plane’s normal.

Normals of the planes are \(\vec{n}_{1}=(a_{1},b_{1},c_{1})\) and \(\vec{n}_{2}=(a_{2},b_{2},c_{2})\).
Planes are perpendicular \(\iff\) their normals are perpendicular: \(\vec{n}_{1}\cdot\vec{n}_{2}=0\Rightarrow a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0\). Quick Tip: Angles between planes equal angles between their normal vectors.

Compute componentwise: \[ 11\cdot 8+(-7)\cdot(-1)+(-1)\cdot(-5)=88+7+5=100. \]
So the dot product equals \(100\). Quick Tip: Dot product is distributive and commutative for componentwise multiplication and addition.

Why use conditional formula: By definition, \[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. \]
Substitute values: \[ P(A\mid B)=\frac{\tfrac{4}{11}}{\tfrac{9}{11}}=\frac{4}{11}\cdot\frac{11}{9}=\frac{4}{9}. \]
So, given that \(B\) has occurred, the chance that \(A\) also occurs is \(\dfrac{4}{9}\). Quick Tip: “Given \(B\)” means reduce the sample space to \(B\): probability becomes \(P(both)/P(B)\).

Reason. Use the addition law: \[ P(E\cup F)=P(E)+P(F)-P(E\cap F). \]
Compute: \[ \frac{6}{7}=\frac{3}{7}+\frac{5}{7}-P(E\cap F) \ \Rightarrow\ P(E\cap F)=\frac{3+5-6}{7}=\frac{2}{7}. \] Quick Tip: Union \(=\) sum minus overlap: \(P(E\cup F)=P(E)+P(F)-P(E\cap F)\).

Standard form \(y'+P(x)y=Q(x)\) has IF \(=\exp\!\left(\int P(x)\,dx\right)\).
Here \(P(x)=\dfrac{2}{x}\). So \[ IF=\exp\!\left(\int \frac{2}{x}dx\right)=\exp(2\ln x)=e^{\ln x^{2}}=x^{2}. \] Quick Tip: For \(y'+\dfrac{m}{x}y=\ldots\), IF is \(x^{m}\).

Write components: \((3\vec{k}-7\vec{i})=(-7,0,3)\), \(2\vec{k}=(0,0,2)\).
Use determinant for cross product:

Quick Tip: Only the terms with nonzero minors survive; watch the sign in the middle cofactor.

Magnitude of \((1,-2,2)\): \[ \sqrt{1^{2}+(-2)^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3. \] Quick Tip: Length of \((a,b,c)\) is \(\sqrt{a^{2}+b^{2}+c^{2}}\).

For \(Ax+By+Cz+D=0\), the normal direction ratios are \((A,B,C)\).
Hence \((1,2,-3)\). Quick Tip: Plane coefficients \(\Rightarrow\) normal d.r.s immediately.

In the symmetric form \(\dfrac{x-x_{0}}{l}=\dfrac{y-y_{0}}{m}=\dfrac{z-z_{0}}{n}\), the denominators \(l,m,n\) are direction ratios.
So d.r.s are \((3,3,6)\). Quick Tip: Denominators in the symmetric equation give the line’s d.r.s.

Set parameter \(t=0\) in \(x=100+101t,\ y=99+102t,\ z=98+103t\).
This gives the point \((100,99,98)\) on the line. Quick Tip: The “\(-x_0,-y_0,-z_0\)” numerators reveal the point \((x_0,y_0,z_0)\) on the line.

Use determinant:

Compute: \(\vec{i}(-11-7)-\vec{j}(-110+4)+\vec{k}(70+4)=-18\vec{i}+106\vec{j}+74\vec{k}\). Quick Tip: Cofactor signs are \(+,-,+\) across the top row.

Differentiate termwise: \((x^{3})'=3x^{2}\), \((e^{x})'=e^{x}\). Add them. Quick Tip: Sum rule: derivative of a sum is the sum of derivatives.

\((\tan x)'=\sec^{2}x\).
Using chain rule on \(\sin^{2}x\): derivative \(=2\sin x\cos x\).
Add to get \(\sec^{2}x+2\sin x\cos x\). Quick Tip: \((\sin^{2}x)'=2\sin x\cos x\) comes from \((u^{2})'=2u\,u'\) with \(u=\sin x\).

First derivative: \((e^{5x})'=5e^{5x}\).
Second derivative: differentiate again \(\Rightarrow 25e^{5x}\). Quick Tip: Each differentiation of \(e^{kx}\) brings down another factor \(k\).

\(\int x^{3}dx=\dfrac{x^{4}}{4}\).
Evaluate from \(0\) to \(3\): \[ \left[\frac{x^{4}}{4}\right]_{0}^{3}=\frac{3^{4}}{4}-0=\frac{81}{4}. \]
Multiply by \(3\): \(3\cdot\dfrac{81}{4}=\dfrac{243}{4}\). Quick Tip: Do the definite integral first, then multiply by outside constants.

Parity: \(\sin^{17}x\) is odd, \(\cos^{3}x\) is even.
Odd\(\times\)even \(=\) odd.
Integral of an odd function over \([-a,a]\) is \(0\). Limits \([-1,1]\) are symmetric, so integral \(=0\). Quick Tip: Check \(f(-x)\); odd functions vanish on symmetric limits.

\(x^{17}\) is an odd function. On symmetric limits \([-1,1]\), the positive and negative areas cancel: integral is \(0\). Quick Tip: Any odd power of \(x\) integrates to \(0\) over \([-a,a]\).

\(\int x^{1/2}dx=\dfrac{x^{3/2}}{3/2}=\dfrac{2}{3}x^{3/2}\).
Multiply by \(3\): \(3\cdot\dfrac{2}{3}x^{3/2}=2x^{3/2}+k\). Quick Tip: Power rule: \(\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+C\) for \(n\neq -1\).

Factor the denominator: \(x^{2}-4=(x-2)(x+2)\).
Partial fractions: \[ \frac{x+2}{x^{2}-4}=\frac{A}{x-2}+\frac{B}{x+2} \Rightarrow A(x+2)+B(x-2)=x+2. \]
Compare coefficients: \(A+B=1\) and \(2A-2B=2\Rightarrow A-B=1\).
Solve: \(A=1,\ B=0\). Hence \[ \int \frac{x+2}{x^{2}-4}dx=\int \frac{1}{x-2}dx=\log|x-2|+k. \] Quick Tip: If the numerator equals a factor of the denominator, partial fractions may collapse to a single term.

Let \(u=3x\Rightarrow du=3\,dx\). Then \[ \int \frac{3\,dx}{\sqrt{1-9x^{2}}}=\int \frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+k=\sin^{-1}(3x)+k. \] Quick Tip: Standard form: \(\displaystyle \int\frac{du}{\sqrt{1-u^{2}}}=\sin^{-1}u+C\).

\(\displaystyle \int \sec u\,\tan u\,du=\sec u+C\).
Let \(u=5x\Rightarrow du=5\,dx\): \[ \int \sec5x\,\tan5x\,dx=\frac{1}{5}\sec5x+k. \]
Multiply by \(25\Rightarrow 5\sec5x+k\). Quick Tip: When inside is \(5x\), pull out \(\frac{1}{5}\); outside constants multiply at the end.

\(\displaystyle \int \sec^{2}(ax)\,dx=\frac{1}{a}\tan(ax)+C\). With \(a=4\), answer is \(\frac14\tan4x+k\). Quick Tip: Differentiate to verify: \(\left(\frac{1}{a}\tan ax\right)'=\sec^{2}ax\).

\(\vec{k}\) is perpendicular to both \(\vec{i}\) and \(\vec{j}\). Dot product of perpendicular vectors is \(0\). Hence \(\vec{k}\cdot\vec{i}=0=\vec{k}\cdot\vec{j}\Rightarrow \vec{k}\cdot(\vec{i}+\vec{j})=0\). Quick Tip: Orthogonal basis: \(\,\vec{i}\perp\vec{j}\perp\vec{k}\,\) with unit lengths.

Let \(u=6x\Rightarrow du=6\,dx\Rightarrow dx=\dfrac{du}{6}\). Then \[ \int \frac{dx}{1+(6x)^{2}}=\frac{1}{6}\int \frac{du}{1+u^{2}} =\frac{1}{6}\tan^{-1}u+k=\frac{1}{6}\tan^{-1}(6x)+k. \] Quick Tip: Standard form: \(\displaystyle \int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\!\frac{x}{a}+C\).