Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

**Q1.**Statement 1:If u=f(tanx),v=g(secx) and f'(1)=2, g'√(2))=4, then (du/dv)_(x=Ï€/4)=1/√2 Statement 2: If u=f(x),v=g(x), then the derivation of f with respect to g is du/dv=(du/dx)/(dv/dx)

Solution

(a) Given u=f(tanx) ⇒du/dx=f'(tanx)sec^2x and v=g(secx) ⇒dv/dx=g'(secx)secx tanx ∴du/dv=((du/dx))/((dv/dx))=(f'(tanx))/(g'(secx)).1/sinx ∴(du/dv)_(x=Ï€/4)=(f'(1))/(g'(√(2))).√2 =2/4.√2=1/√2

(a) Given u=f(tanx) ⇒du/dx=f'(tanx)sec^2x and v=g(secx) ⇒dv/dx=g'(secx)secx tanx ∴du/dv=((du/dx))/((dv/dx))=(f'(tanx))/(g'(secx)).1/sinx ∴(du/dv)_(x=Ï€/4)=(f'(1))/(g'(√(2))).√2 =2/4.√2=1/√2

**Q2.**Statement 1: Let f:R→R is a real-valued function ∀ x,y∈R such that |(f(x)-f(y)|≤|x-y)|^3, then f(x) is a constant function Statement 2: If derivative of the function w.r.t. x is zero, then function is constant

Solution

(a) Since |(f(x)-f(y)|≤|x-y)|^3, where x≠y ∴|(f(x)-f(y))/(x-y)|≤|x-y|^2 Taking lim as y→x, we get lim┬(y→x)〖|(f(x)-f(y))/(x-y)|≤ lim┬(y→x) |x-y|^2 ⇒|lim┬(y→x) (f(x)-f(y))/(x-y)|≤|lim┬(y→x) (x-y)^2 | ⇒|f'(x)|≤0 ⇒|f'(x)|=0 (∵|f'(x)|≥0) ∴f'(x)=0 ⇒f(x)=c (constant)

(a) Since |(f(x)-f(y)|≤|x-y)|^3, where x≠y ∴|(f(x)-f(y))/(x-y)|≤|x-y|^2 Taking lim as y→x, we get lim┬(y→x)〖|(f(x)-f(y))/(x-y)|≤ lim┬(y→x) |x-y|^2 ⇒|lim┬(y→x) (f(x)-f(y))/(x-y)|≤|lim┬(y→x) (x-y)^2 | ⇒|f'(x)|≤0 ⇒|f'(x)|=0 (∵|f'(x)|≥0) ∴f'(x)=0 ⇒f(x)=c (constant)

**Q3.**Statement 1: If differentiable function f(x) satisfies the relation f(x)+f(x-2)=0 ∀x∈R, and if (d/dx f(x))_(x=a)=b, then (d/dx f(x))_(a+4000)=b Statement 2: f(x) is a periodic function with period 4

Solution

(a) f(x)+f(x-2)=0 (1) Replace x by x-2⇒f(x-2)+f(x-4)=0 (2) From (1) and (2), f(x)-f(x-4)=0 Replace x by x+4 ⇒f(x+4)=f(x) ⇒f(x)=f(x+4)=f(x+8)=...=f(x+4000) ⇒f'(x)=f'(x+4000) Hence, both the statements are true and Statement 2 is correct explanation of Statement 1 Hence, f(x) is periodic with period 4

(a) f(x)+f(x-2)=0 (1) Replace x by x-2⇒f(x-2)+f(x-4)=0 (2) From (1) and (2), f(x)-f(x-4)=0 Replace x by x+4 ⇒f(x+4)=f(x) ⇒f(x)=f(x+4)=f(x+8)=...=f(x+4000) ⇒f'(x)=f'(x+4000) Hence, both the statements are true and Statement 2 is correct explanation of Statement 1 Hence, f(x) is periodic with period 4

**Q4.**Statement 1: Derivative of sin^(-14)(2x/(1+x^2)) with respect to cos^(-1)((1-x^2)/(1+x^2)) is 1 for 0

Solution

(c) Fro 0

(c) Fro 0

**Q5.**Statement 1:For f(x)=sinx,f'(Ï€)=f'(3Ï€) Statement 2:For f(x)=sinx,f(Ï€)=f(3Ï€)

Solution

(b) Both the statement are true, but Statement 2 is not correct explanation of Statement 1 Statement 1 is true as period of sinx is 2Ï€ Or, in general if for y=f(x),f(a)=f(b), we cannot say f'(a)=f'(b)

(b) Both the statement are true, but Statement 2 is not correct explanation of Statement 1 Statement 1 is true as period of sinx is 2Ï€ Or, in general if for y=f(x),f(a)=f(b), we cannot say f'(a)=f'(b)

**Q6.**Statement 1:For x<0,d/dx (In|x|)=-1/x Statement 2:For x<0,|x|=-x

Solution

(d) For x<0,d/dx (In|x|)=d/dx(In(-x)) =1/((-x)) (-1)=1/x

(d) For x<0,d/dx (In|x|)=d/dx(In(-x)) =1/((-x)) (-1)=1/x

**Q7.**If for some differentiable function f(Î±)=0 and f'(Î±)=0 Statement 1: Then sign of f(x) does not change in the neighborhood of x=Î± Statement 2: Î± is repeated root of f(x)=0

Solution

(d) Statement 2 is true as f(Î±)=0 and f^' (Î±)=0, then definitely Î± is repeated root of f(x)=0 But from data, we are not sure how many times a root repeats Also f(x)=(x-Î±)^n×g(x), which changes sign at x=Î±, when n is odd and does not if n is even. Hence, Statement 1 is false

(d) Statement 2 is true as f(Î±)=0 and f^' (Î±)=0, then definitely Î± is repeated root of f(x)=0 But from data, we are not sure how many times a root repeats Also f(x)=(x-Î±)^n×g(x), which changes sign at x=Î±, when n is odd and does not if n is even. Hence, Statement 1 is false

**Q8.**Observe the following statements Which of the following is correct? Statement 1: I f(x)=ax^41+bx^(-40)⇒(f''(x))/(f(x))=1640x^(-2) Statement 2: II d/dx tan^(-1)(2x/(1-x^2))=1/(1+x^2)

Solution

(a) Given, f(x)=ax^41+bx^(-40) f'(x)=41ax^40-40bx^(-41) f''(x)=1640 ax^39+1640bx^(-42) Now, (f''(x))/(f(x))=(1640(ax^39+bx^(-42)))/(ax^41+bx^(-40) )=1640x^(-2) d/dx tan^(-1)(2x/(1-x^2 ))=d/dx tan^(-1)(tan2x) =d/dx 2x =2 ∴ Statement I is true, but II is false.

(a) Given, f(x)=ax^41+bx^(-40) f'(x)=41ax^40-40bx^(-41) f''(x)=1640 ax^39+1640bx^(-42) Now, (f''(x))/(f(x))=(1640(ax^39+bx^(-42)))/(ax^41+bx^(-40) )=1640x^(-2) d/dx tan^(-1)(2x/(1-x^2 ))=d/dx tan^(-1)(tan2x) =d/dx 2x =2 ∴ Statement I is true, but II is false.

**Q9.**Statement 1: If e^xy+log(xy)+cos(xy)+5=0, then dy/dx=-y/x Statement 2: d/dx (xy)=0⇒dy/dx=(-y)/x

Solution

(a) ∵e^xy+log(xy)+cos〖(xy)+5=0 ∴e^xy d/dx(xy)+1/((xy))d/dx (xy)-sin(xy) d/dx (xy)=0 ⇒d/dx (xy){e^xy+1/xy-sin(xy) }=0 ∵e^xy+1/xy-sin(xy)≠0 ∴d/dx (xy)=0 ⇒x dy/dx+y.1=0 ⇒dy/dx=-y/x

(a) ∵e^xy+log(xy)+cos〖(xy)+5=0 ∴e^xy d/dx(xy)+1/((xy))d/dx (xy)-sin(xy) d/dx (xy)=0 ⇒d/dx (xy){e^xy+1/xy-sin(xy) }=0 ∵e^xy+1/xy-sin(xy)≠0 ∴d/dx (xy)=0 ⇒x dy/dx+y.1=0 ⇒dy/dx=-y/x

**Q10.**Statement 1: Let f(x)=x[x] and [.] denotes greatest integral function, when x is not an integral, then rule for f'(x) is given by [x] Statement 2: f'(x) does not exist for any x Ïµ integer

Solution

(a) f(x)=x[x]={(-x, -1≤x<0 0, 0≤x<1 x, 1≤x<2) (2x, 2≤x<3 … …) ⇒f(x)={(-1, -1

(a) f(x)=x[x]={(-x, -1≤x<0 0, 0≤x<1 x, 1≤x<2) (2x, 2≤x<3 … …) ⇒f(x)={(-1, -1