SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 61–64 (part 7 of 7)

  1. Fill in the blanks in each of the following Exercises $\displaystyle 60$ to $\displaystyle 64$:

    Exercise 61

    The equation of normal to the curve y=tan⁡x\displaystyle y=\tan x at (0,0)\displaystyle (0,0) is ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle x+y=0\)
    \(\displaystyle x + y = 0\) \[y = \tan x \Rightarrow \frac{dy}{dx} = \sec^2 x \] \[\left.\frac{dy}{dx}\right|_{(0,0)} = \sec^2 0 = 1 \] \[\text{slope of normal} = -\frac{1}{1} = -1 \] \[y - 0 = -1\,(x - 0) \Rightarrow x + y = 0 \]

  2. Exercise 62

    The values of a\displaystyle a for which the function f(x)=sin⁡x−ax+b\displaystyle f(x)=\sin x-a x+b increases on R\displaystyle \mathbf{R} are ____.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \(\displaystyle a \le -1\), i.e. \(\displaystyle a \in (-\infty, -1]\) \[f'(x) = \cos x - a \] \[f \text{ increasing on } \mathbf{R} \Rightarrow f'(x) \ge 0 \text{ for all } x \] \[a \le \cos x \text{ for all } x \] \[\min(\cos x) = -1 \Rightarrow a \le -1 \] For \(\displaystyle a = -1\), \(\displaystyle f'(x) = 1 + \cos x\) is zero only at isolated points, so \(\displaystyle f\) still increases.

  3. Exercise 63

    The function f(x)=2x2−1x4,x>0\displaystyle f(x)=\frac{2 x^2-1}{x^4}, x>0, decreases in the interval ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle (1, \infty)\)
    \(\displaystyle (1, \infty)\) \[f(x) = 2x^{-2} - x^{-4} \] \[f'(x) = -4x^{-3} + 4x^{-5} = \frac{4(1 - x^2)}{x^5} \] \[f'(x) < 0, \ x > 0 \Rightarrow 1 - x^2 < 0 \Rightarrow x > 1 \]

  4. Exercise 64

    The least value of the function f(x)=ax+bx(a>0,b>0,x>0)\displaystyle f(x)=a x+\frac{b}{x}(a>0, b>0, x>0) is ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle 2 \sqrt{a b}\)
    \(\displaystyle 2\sqrt{ab}\) \[f'(x) = a - \frac{b}{x^2} = 0 \Rightarrow x = \sqrt{\frac{b}{a}} \quad (x > 0) \] \[f''(x) = \frac{2b}{x^3} > 0 \Rightarrow \text{minimum} \] \[f_{\min} = a\sqrt{\frac{b}{a}} + b\sqrt{\frac{a}{b}} = \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab} \]