NCERT Exemplar Class 12 Maths Continuity and Differentiability...

Structural angle. The function is an everywhere-continuous polynomial, so the answer is automatic. We still write it as a formal ε-style check so the student sees the structure.

1. Identify the class: f(x)=x³+2x²-1 is a sum and product of the continuous functions x↦ x and constants, hence is continuous on the whole of R.
2. Specialize to x=1: f(1)=1+2-1=2. The limit, by direct substitution legal for polynomials, is also 2.
3. The equality _{x→ 1}f(x)=f(1)=2 certifies continuity at x=1.

Why this matters. Recognising the function-class upfront (polynomial, rational with non-zero denominator, sin/cos, e^x, log x on its domain) lets you skip the explicit piecewise check on most problems. The piecewise machinery is reserved for the cases where a definition genuinely changes across x=c.

f is continuous at x=1.

Picture-first. Sketch the two branches: a parabola for x<2 ending at the open point (2,4), and a line y=3x+5 from the filled point (2,11). There is a vertical gap of 11-4=7 at x=2, the signature of a jump discontinuity.

1. LHL on the x² branch: _{x→ 2^-}x²=4.
2. RHL on the 3x+5 branch: _{x→ 2⁺}(3x+5)=11.
3. Since LHL ≠ RHL, f is discontinuous at x=2, and the jump is 11-4=7.

Why this matters. Whenever a piecewise function changes formula at x=c, run the three checks (LHL, RHL, f(c)) mechanically. The answer follows from arithmetic, not intuition.

Discontinuous at x=2; jump of 7.

Strategic angle. The limit is finite; only its disagreement with f(0) creates the discontinuity. Identifying this as removable is itself a useful classification.

1. Identity: 1-cos 2x=2sin²x. So the quotient becomes 2(sin x/x)².
2. As x→ 0, sin x/x→ 1, hence the quotient tends to 2· 1 = 2.
3. Since f(0)=5≠ 2, f is discontinuous at 0, but the limit exists –- so the discontinuity is removable.

Why this matters. Once you spot 1-cos 2x, reach for the double-angle identity automatically; it converts cos-near-1 limits into sin-over-x limits, which are routine.

Discontinuous (removable) at x=0.

Quick reading. The 0/0 form invites factoring; once the (x-2) cancels, f becomes a polynomial whose value at 2 is 5. Since f(2) was chosen to equal 5, continuity is exact.

1. 2x²-3x-2=(2x+1)(x-2).
2. Hence f(x)=2x+1 for x≠ 2; limit at 2 is 5.
3. The piecewise definition fixes f(2)=5 deliberately, so continuity holds.

Why this matters. Designers of piecewise definitions choose the special value f(c) to fill in the hole. Always check whether the chosen value matches the limit.

Continuous at x=2.

Strategic angle. Whenever |x-c| sits over a linear factor (x-c), the quotient is ± 1 on the two sides of c. The hole at c cannot be patched by any single value of f(c).

1. Right of 4: |x-4|=x-4, quotient =1/2.
2. Left of 4: |x-4|=-(x-4), quotient =-1/2.
3. LHL ≠ RHL ⇒ limit does not exist; f is discontinuous (jump) at 4.

Why this matters. |x|/x is the prototypical jump discontinuity; recognise its disguised forms instantly.

Discontinuous at x=4 (jump).

Picture-first. The graph of xcos(1/x) oscillates wildly near 0, but the oscillations are bounded by the envelopes y=± x which both pinch to 0. So the graph passes through (0,0) smoothly from the squeeze.

1. Envelope: |xcos(1/x)| ≤ |x|.
2. As x→ 0, |x|→ 0, hence xcos(1/x)→ 0 by Squeeze.
3. f(0)=0 matches; continuity holds.

Why this matters. The same idea handles every "bounded times small" limit and is the foundation of many counterexamples in advanced calculus.

Continuous at x=0.

Quick reading. Same template as Q6 with x-a in place of x. The shift does not affect the squeeze.

1. Envelope: |(x-a)sin1x-a|≤ |x-a|→ 0.
2. Hence limit =0=f(a), so f continuous at a.

Why this matters. Once the template is recognised, the solution is two lines.

Continuous at x=a.

Strategic angle. Whenever e^1/x appears, force a left/right split before doing any algebra.

1. Right side: divide numerator and denominator by e^1/x to get 1e^-1/x+1→10+1=1.
2. Left side: e^1/x→ 0, quotient → 0.
3. LHL ≠ RHL ⇒ discontinuous.

Why this matters. e^1/x, tan^-1(1/x), arctan(1/x) all have this two-sided personality. Drilling the split prevents many algebra errors.

Discontinuous at x=0.

Pattern memorisation. The integrals ∫dxa²+x²=1atan^-1(x/a)+C, ∫dxa²-x²=12aln|a+xa-x|+C, ∫dx√a²-x²=sin^-1(x/a)+C should be writable from memory. Most board problems are one substitution away from one of these standard forms.

Generalising the trick. The manipulation used here is part of a family: spot the algebraic disguise, apply the standard identity, reduce to a known limit or standard integral, conclude. Once you've solved three problems with the same disguise, the rest become routine.

Quick reading. The author chose the second branch's constant 3/2 precisely so the polynomial value matches 1/2 at x=1.

1. LHL of x²/2 at 1 is 1/2.
2. RHL of 2x²-3x+3/2 at 1 is 2-3+3/2=1/2.
3. Both equal f(1)=1/2; continuity holds.

Why this matters. Piecewise functions in textbooks are often designed so the break-point matches; verify by computation rather than trust.

Continuous at x=1.

Alternate framing. Read the function as a single graph drawn with two coloured pens. Continuity asks: does the second pen pick up exactly where the first left off? If both pens meet the same y at x=c and the dot f(c) sits on that meeting point, continuity holds; otherwise it doesn't. This visual check anticipates the algebra and exposes sign errors before they propagate.

Structural angle. The map u↦ |u| is continuous (it satisfies | |a|-|b| | ≤ |a-b|, a Lipschitz bound with constant 1. Sums and compositions of continuous functions are continuous, so any expression built from |·|, sums and polynomial pieces is continuous.

1. Both |x| and |x-1| are continuous on R.
2. Sum is continuous; f(1)=1+0=1.

Why this matters. Closure of continuity under sum / product / composition is the workhorse fact that lets you skip ε-δ on routine examples.

Continuous at x=1.

Quick reading. The constant branch's value 2k must equal the polynomial branch's value at 5, i.e. 7.

1. Match: 2k = 3(5)-8 = 7.
2. Hence k=7/2.

Why this matters. "Find k so that f is continuous" is shorthand for "force LHL = RHL"; the unknown drops out as one linear equation.

k=7/2.

Structural angle. A 0/0 form with exponentials becomes a 0/0 polynomial in t=2^x once we substitute, after which it factors.

1. Substitute t=2^x: quotient becomes 4(t-4)(t-4)(t+4)=4t+4.
2. Plug t=4: value is 1/2. So k=1/2.

Why this matters. Exponential change of variable converts many "transcendental" limits into purely algebraic ones.

k=1/2.

Why this method is preferred. The shortest correct route in school calculus is almost always: identify the function's category → load the standard template → substitute the specific numbers. Reinventing the technique problem-by-problem is what makes a 90-minute paper feel like 3 hours.

Strategic angle. The classic surd-difference-over-x form yields easily to the conjugate; once rationalised, the offending x cancels.

1. Rationalise to get 2k√1+kx+√1-kx.
2. Limit at 0: 2k/2 = k.
3. Match RHL value f(0)=-1, giving k=-1.

Why this matters. Conjugate rationalisation is the go-to manoeuvre for any √ -√ over a polynomial; commit it to muscle memory.

k=-1.

Generalising the move. Conjugate rationalisation works because (√A-√B)(√A+√B) = A - B –- a polynomial difference that we can simplify. The same identity in disguise drives the limit _{x→ 0}√1+x-1x=12 and the standard derivative ddx√x=12√x. Recognise the family.

Quick reading. Both 1-cos and sin are quadratic-then-linear near 0; the leading-order expansion gives the limit directly.

1. Series near 0: 1-cos kx ≈ (kx)²/2 and sin x≈ x.
2. Hence (1-cos kx)/(xsin x)≈ (k² x²/2)/(x· x)=k²/2.
3. Match =1/2 ⇒ k²=1 ⇒ k=± 1.

Why this matters. The Taylor-leading-order trick avoids the 2sin² identity and is faster once you trust it.

k=± 1.

Self-check before boxing the answer. Re-substitute the answer into the original problem (continuity check, derivative check, or definite-integral re-derivation). 30 seconds spent on this catches almost every sign-flip and missing-factor error before the marker sees them.

Strategic angle. A two-sided limit failure is fatal: adjusting f at the single point x=0 cannot fix a jump.

1. f(x)→ 1 from the right, -1 from the left.
2. Two different one-sided limits ⇒ no single value of f(0) makes f continuous.

Why this matters. Removable vs. non-removable: only when LHL = RHL is the discontinuity removable by redefining f(c).

Discontinuous for all k.

Alternate framing. Read the function as a single graph drawn with two coloured pens. Continuity asks: does the second pen pick up exactly where the first left off? If both pens meet the same y at x=c and the dot f(c) sits on that meeting point, continuity holds; otherwise it doesn't. This visual check anticipates the algebra and exposes sign errors before they propagate.

Quick reading. Set the two side limits equal to f(4) and solve the resulting linear system in a,b.

1. LHL =a-1, RHL =b+1, f(4)=a+b.
2. a-1=a+b⇒ b=-1; b+1=a+b⇒ a=1.

Why this matters. Two unknowns in piecewise problems are usually pinned down by the two equations LHL =f(c) and RHL =f(c).

a=1, b=-1.

Strategic angle. Two failure modes: (i) inner f fails; (ii) outer f fails on the value the inner produced. Track both.

1. Inner fails at x=-2.
2. Composed simplification: f(f(x))=x+22x+5.
3. Simplified denominator zero at x=-5/2.
4. Both points are discontinuities.

Why this matters. Always include the "hole carried by the inner function" even when the simplified expression looks fine elsewhere.

x = -2, -5/2.

The mechanical view. Think of the chain rule as peeling an onion: differentiate the outermost layer treating the inside as one symbol, then multiply by the derivative of that inside, and recurse. For nested radicals or compositions of trig and exponentials, write each layer on a separate line; the answer assembles itself with the right number of factors.

Quick reading. The composition has three failure points: the hole of the inner function and the two roots of t²+t-2 pulled back through t(x).

1. Inner hole: x=1.
2. t=1⇒ x=2; t=-2⇒ x=1/2.

Why this matters. Compositions of rational functions stack their poles –- always pull back the outer poles through the inner inverse to catch all failure points.

x=1, 1/2, 2.

Structural angle. Sum of two everywhere-continuous functions is everywhere continuous; no piecewise check needed.

1. sin,cos continuous on R.
2. Sum is continuous; at π, f(π)=0+(-1)=-1.

Why this matters. Identify the function class first; specific points become trivial.

Continuous at π.

Alternate framing. Read the function as a single graph drawn with two coloured pens. Continuity asks: does the second pen pick up exactly where the first left off? If both pens meet the same y at x=c and the dot f(c) sits on that meeting point, continuity holds; otherwise it doesn't. This visual check anticipates the algebra and exposes sign errors before they propagate.

Picture-first. On [1,2) the graph is the line y=x; on [2,3) it is the parabola y=x²-x. They meet at (2,2), but the line has slope 1 while the parabola has slope 2x-1=3 there.

1. LHL =2, RHL =2, f(2)=2 ⇒ continuous.
2. Lf'(2) = 1, Rf'(2) = 3 ⇒ not differentiable.

Why this matters. Corners (slope mismatch) are the most common cause of non-differentiability after vertical-tangent failures.

Continuous at x=2; not differentiable at x=2.

Alternate framing. Read the function as a single graph drawn with two coloured pens. Continuity asks: does the second pen pick up exactly where the first left off? If both pens meet the same y at x=c and the dot f(c) sits on that meeting point, continuity holds; otherwise it doesn't. This visual check anticipates the algebra and exposes sign errors before they propagate.

Strategic angle. Same squeeze pattern as Q6, but applied to hsin(1/h) (one power of h stronger than what we needed for continuity), which is why differentiability also works.

1. Difference quotient =hsin(1/h).
2. Bounded by |h|→ 0 ⇒ derivative is 0.

Why this matters. x²sin(1/x) is the textbook example of a function that is differentiable everywhere yet whose derivative is discontinuous at 0.

f'(0)=0.

Picture-first. Two lines meeting at (2,3): one rises with slope +1, the other descends with slope -1. The ""-corner kills differentiability.

1. Continuous: both branches give 3 at x=2.
2. Slopes +1 and -1 disagree.

Why this matters. Tent functions / V-graphs are the cleanest non-differentiability example to put on a quiz.

Not differentiable at 2.

Strategic angle. |x-c| is the canonical "corner". Slope is -1 to the left, +1 to the right.

1. Direct LHL/RHL of (|h|/h) give -1 and +1.
2. Continuity is automatic because |x-c|→ 0 as x→ c.

Why this matters. Any function with |g(x)| as a factor will inherit a corner wherever g changes sign.

Continuous, not differentiable at 5.

Structural angle. The Cauchy equation f(x+y)=f(x)f(y) "separates" x from h in the difference quotient, exposing f'(0) as the master parameter.

1. Plug x=y=0: f(0)²=f(0) and f(0)≠ 0 ⇒ f(0)=1.
2. Difference quotient at x: f(x+h)-f(x)h=f(x)f(h)-1h.
3. Inner limit is f'(0)=2, so f'(x)=2f(x).

Why this matters. This is the calculus proof that the unique f satisfying the multiplicative Cauchy equation with f'(0)=k is f(x)=e^kx.

f'(x)=2f(x).

Self-check before boxing the answer. Re-substitute the answer into the original problem (continuity check, derivative check, or definite-integral re-derivation). 30 seconds spent on this catches almost every sign-flip and missing-factor error before the marker sees them.

What examiners reward. Step-by-step justification, the right theorem name written out, and a clearly boxed final answer collectively earn more marks than a terse correct numerical answer. The expert solution above is laid out exactly this way for that reason.

Strategic angle. Variable exponent over constant base ⇒ take logs once; the derivative falls out in one line.

1. ln y = cos²x· ln 2.
2. Differentiate: y'y = -sin 2x· ln 2 (using ddxcos²x = -sin 2x).
3. Multiply by y: y' = -sin 2x2· 2^cos2x.

Why this matters. The identity ddxcos²x = -sin 2x (rather than the longer 2cos xsin x) keeps the answer compact and matches the standard NCERT form.

-sin 2x ln 2 · 2^cos2x.

Quick reading. Quotient with exponential numerator and power denominator. Factor 8^x x⁷ to keep the expression compact.

1. Apply quotient rule.
2. Factor common 8^x x⁷.
3. Simplify exponents to land at the boxed answer.

Why this matters. The pattern a^x/xⁿ appears repeatedly in JEE; the cleaned form a^x(xlog a-n)/xⁿ⁺¹ is worth remembering.

8^x(xlog 8 - 8)x⁹.

Structural angle. The numerator u' shares the factor x+√x²+a with u. That cancellation is what makes the answer so compact.

1. u'=√x²+a+x√x²+a.
2. Divide by u=x+√x²+a to get 1/√x²+a.

Why this matters. The clean cancellation is the standard JEE Main shortcut for ∫ dx/√x²+a problems.

1√x²+a.

The mechanical view. Think of the chain rule as peeling an onion: differentiate the outermost layer treating the inside as one symbol, then multiply by the derivative of that inside, and recurse. For nested radicals or compositions of trig and exponentials, write each layer on a separate line; the answer assembles itself with the right number of factors.

Strategic angle. Strip one log at a time, using d(log u)/dx = u'/u.

1. d(log x⁵)/dx = 5/x.
2. d(loglog x⁵)/dx = 5/(x· 5log x)=1/(xlog x).
3. Apply one more outer log to land at the boxed answer.

Why this matters. Iterated-log problems are tractable once you commit to "one strip per step".

1xlog xlog(5log x).

Quick reading. Same inner √x feeds both terms; factor its derivative 1/(2√x) once.

1. Differentiate each piece via chain rule.
2. Factor 1/(2x).

Why this matters. Shared inner functions let you factor and keep expressions compact.

cos√x-sin 2√x2√x.

Strategic angle. Three nested functions; differentiate outer-in.

1. Power: nsin^n-1(·).
2. sin: cos(·).
3. Inner polynomial: 2ax+b.

Why this matters. The same pattern n u^n-1u' powers all "power-of-something" derivatives.

n(2ax+b)sin^n-1(ax²+bx+c).

Quick reading. Peel: cos → -sin, tan → sec², √ → 1/(2√ ), inner x+1 → 1.

1. Apply each layer's derivative.

Why this matters. Memorise the four standard ddx layers; deep nestings then unfold mechanically.

Same as above.

Quick reading. Three independent terms, each handled by the standard sin or sin² template.

1. Differentiate each term, watch the inner x² in two of them.

Why this matters. Sum + chain on each term is the universal recipe.

2xcos x²+sin 2x+2xsin 2x².

Strategic angle. sin^-1(1/√x+1) equals tan^-1(1/√x) (right triangle with sides 1,√x,√x+1). Differentiating tan^-1(1/√x) directly gives the same answer quickly.

1. Substitute √x=tanθ⇒ y=π/2-θ=π/2-tan^-1√x.
2. Differentiate: dy/dx=-11+x·12√x=-12(x+1)√x.

Why this matters. Inverse-trig substitution shortcuts sometimes replace 3 lines of chain rule with one identity.

-12(x+1)√x.

Quick reading. Power-tower with variable base and exponent: log first, then product rule on vlog u.

1. log y = cos xx.
2. Differentiate: y'/y = -sin xx + cos xx.
3. Multiply by y.

Why this matters. Logarithmic differentiation is the only correct treatment of u^v; "power rule" and "exponential rule" each miss half the answer.

Same as main.

Reading the structure. Logarithmic differentiation is the tool whenever the exponent itself depends on x, e.g. x^x, (sin x)^{cos x}, (tan x)^x. Take ln first to flatten the exponent into a product, differentiate using the product rule, then multiply through by the original function. The technique is mechanical once the pattern is recognised.

Structural angle. Both factors share sin^m-1cos^n-1; factor it out for cleanliness.

1. Product rule, then factor.

Why this matters. The clean form mcos²-nsin² makes finding extrema straightforward.

sin^m-1xcos^n-1x(mcos²x-nsin²x).

The mechanical view. Think of the chain rule as peeling an onion: differentiate the outermost layer treating the inside as one symbol, then multiply by the derivative of that inside, and recurse. For nested radicals or compositions of trig and exponentials, write each layer on a separate line; the answer assembles itself with the right number of factors.

Quick reading. Three factors ⇒ logarithmic differentiation gives three terms instead of nine.

1. log y = ∑ k_ilog(x+i).
2. Differentiate ⇒ sum of k_i/(x+i).

Why this matters. Always log-differentiate products of powers. The product rule on three factors is needlessly tedious.

See main solution.

Reading the structure. Logarithmic differentiation is the tool whenever the exponent itself depends on x, e.g. x^x, (sin x)^{cos x}, (tan x)^x. Take ln first to flatten the exponent into a product, differentiate using the product rule, then multiply through by the original function. The technique is mechanical once the pattern is recognised.

Strategic angle. Identity collapses the argument to cos(π/4-x); cos^-1cos on the principal range is identity.

1. (sin x+cos x)/√2=cos(π/4-x).
2. cos^-1cos(π/4-x)=π/4-x for |x|<π/4.
3. dy/dx=-1.

Why this matters. JEE problems on inverse trig are mostly identity recognition; chain rule is a fallback, not a default.

-1.

Quick reading. The square root forces |tan(x/2)|, and the absolute value splits the derivative by sign of x.

1. Radicand =tan²(x/2), so y=tan^-1|tan(x/2)|.
2. For 0, y = x/2 ⇒ y'=1/2.
3. For -π/4, y=-x/2 ⇒ y'=-1/2.

Why this matters. Square roots after a half-angle reduction always produce a |·|, and absolute values force a piecewise derivative. Never drop the modulus.

12 sgn(x) on (-π/4,0)∪(0,π/4).

Generalising the trick. The manipulation used here is part of a family: spot the algebraic disguise, apply the standard identity, reduce to a known limit or standard integral, conclude. Once you've solved three problems with the same disguise, the rest become routine.

Strategic angle. The combination sec+tan is the textbook tan(π/4+x/2) form.

1. Recognise identity.
2. tan^-1tan(·)= · on principal range.
3. Derivative is 1/2.

Why this matters. Memorise sec+tan = tan(π/4+x/2); it turns dozens of integration / differentiation problems into one-liners.

1/2.

Quick reading. Rotate by tan^-1(a/b) to convert the expression into "constant minus x".

1. Divide by bcos x; recognise tan(A-B) form.
2. Constant minus x differentiates to -1.

Why this matters. tan^-1 of "rotated tan" is a constant shift; pattern-recognise to avoid the chain rule.

-1.

The mechanical view. Think of the chain rule as peeling an onion: differentiate the outermost layer treating the inside as one symbol, then multiply by the derivative of that inside, and recurse. For nested radicals or compositions of trig and exponentials, write each layer on a separate line; the answer assembles itself with the right number of factors.

Quick reading. 4x³-3x is the triple-angle of cos; the 1/(·) wrapper gives sec, which sec^-1 unwinds.

1. Substitute x=cosθ.
2. Identify 4cos³θ-3cosθ=cos 3θ.
3. y=3θ=3cos^-1x; derivative is -3/√1-x².

Why this matters. Triple-angle ID for sin and cos (3sinθ-4sin³θ, 4cos³θ-3cosθ) is a go-to substitution.

-3/√1-x².

Structural angle. tan^-1(tan 3φ)=3φ collapses the expression to a simple multiple of tan^-1(x/a).

1. Substitute x/a=tanφ.
2. Triple-angle for tan reduces to tan^-1tan 3φ=3φ.
3. y=3tan^-1(x/a) gives derivative 3a/(a²+x²).

Why this matters. The "triple-angle in disguise" pattern is a classic JEE shortcut.

3a/(a²+x²).

Why this method is preferred. The shortest correct route in school calculus is almost always: identify the function's category → load the standard template → substitute the specific numbers. Reinventing the technique problem-by-problem is what makes a 90-minute paper feel like 3 hours.

Strategic angle. Double-angle substitution converts both surds simultaneously.

1. x²=cos 2θ; surds simplify to 2cosθ,2sinθ.
2. Quotient becomes tan(π/4+θ).
3. y=π/4+12cos^-1x²; differentiate.

Why this matters. Double-angle substitution is the standard way to handle √1± x² combinations.

-x/√1-x⁴.

Generalising the move. Conjugate rationalisation works because (√A-√B)(√A+√B) = A - B –- a polynomial difference that we can simplify. The same identity in disguise drives the limit _{x→ 0}√1+x-1x=12 and the standard derivative ddx√x=12√x. Recognise the family.

Quick reading. Both dx/dt and dy/dt have 1/t² common; ratio cancels.

1. Derivatives w.r.t. t.
2. Take the ratio.

Why this matters. Parametric problems usually have shared factors; cancel before plugging values.

(t²+1)/(t²-1).

Strategic angle. Recognise the cubic terms as triple-angle expressions: 3cosθ - 2cos³θ and 3sinθ - 2sin³θ look like but are not the standard sin 3θ = 3sinθ - 4sin³θ. So work directly with the chain rule; the cancellation is the moral lesson.

1. dx/dθ = 3sin2θ (factor sinθ, then use 2cos²θ - 1 = cos 2θ).
2. dy/dθ = 3cos2θ (factor cosθ, then use 1-2sin²θ = cos 2θ).
3. Ratio cancels cos 2θ and the constant 3, leaving cotθ.

Why this matters. The common cos 2θ factor is the whole point of the construction: if both dx/dθ and dy/dθ share a non-trivial factor, the parametric ratio simplifies beautifully. Always factor before dividing.

cotθ.

Strategic angle. Apply d/dx to every term, group all y'-factors, divide.

1. Differentiate sin(xy), x/y, x², y separately.
2. Collect y' terms; factor; divide.

Why this matters. Implicit problems are bookkeeping –- a clean grouping at the end is more important than clever tricks.

See main solution.

Why implicit works. Implicit differentiation rests on the chain rule: every y in the equation is secretly y(x), so ddxyⁿ = ny^n-1dydx. Once that habit is automatic, even ugly relations like sin(xy)=x+y become routine –- differentiate term-by-term, then solve linearly for dydx.

Common mistake in a different framing. Even when the algebra is correct, students often skip naming the theorem or rule being invoked. Examiners give partial credit for stating the right principle (continuity laws, MVT, FTC, substitution rule) even if a later step slips. Always name the engine before turning it.

Strategic angle. Logarithmic differentiation produces a clean ratio. The fact that y/x is constant kills the second derivative.

1. Log-differentiate to get y'=y/x.
2. y'=y/x is the ODE for y=cx.
3. y''=0.

Why this matters. Implicit equations often hide straight-line / conic structure –- always check what curve you have.

y'=y/x and y''=0.

Reading the structure. Logarithmic differentiation is the tool whenever the exponent itself depends on x, e.g. x^x, (sin x)^{cos x}, (tan x)^x. Take ln first to flatten the exponent into a product, differentiate using the product rule, then multiply through by the original function. The technique is mechanical once the pattern is recognised.

Why this method is preferred. The shortest correct route in school calculus is almost always: identify the function's category → load the standard template → substitute the specific numbers. Reinventing the technique problem-by-problem is what makes a 90-minute paper feel like 3 hours.

Quick reading. Two equations, two unknowns: one from f(1) match, one from slope match.

1. f continuous at 1: 4+p=q+2.
2. f' continuous at 1: 2(1)+3 = q, so q=5.
3. Therefore p=3.

Why this matters. Two-piece "find the constants" problems always reduce to a 2× 2 linear system.

p=3, q=5.

Generalising the trick. The manipulation used here is part of a family: spot the algebraic disguise, apply the standard identity, reduce to a known limit or standard integral, conclude. Once you've solved three problems with the same disguise, the rest become routine.

Structural angle. The ODE (1-x²)y''-xy'+p²y=0 is the Chebyshev equation; its solutions are exactly sin(parcsin x) and cos(parcsin x).

1. Parametric derivatives.
2. Square and use 1-x²=cos² t, 1-y²=cos² pt.
3. Differentiate, divide by 2y'.

Why this matters. Recognising the answer as a standard ODE makes second-order parametric problems systematic.

The given ODE holds.

Common slip with parametric second derivatives. Students often write d²ydx² = d²y/dt²d²x/dt², which is wrong. The correct formula divides ddt(dydx) by dxdt. The mistake stems from over-extending the simple parametric first-derivative rule; commit the corrected second-order formula to memory.

Self-check before boxing the answer. Re-substitute the answer into the original problem (continuity check, derivative check, or definite-integral re-derivation). 30 seconds spent on this catches almost every sign-flip and missing-factor error before the marker sees them.

Quick reading. x^{tan x} needs log-diff; the polynomial piece is trivial.

1. Log-differentiate x^{tan x}.
2. Add x for the polynomial part.

Why this matters. Splitting a sum lets you apply the appropriate technique to each piece.

See main solution.

Reading the structure. Logarithmic differentiation is the tool whenever the exponent itself depends on x, e.g. x^x, (sin x)^{cos x}, (tan x)^x. Take ln first to flatten the exponent into a product, differentiate using the product rule, then multiply through by the original function. The technique is mechanical once the pattern is recognised.