19. Is the function defined by <math><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mo>=</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mi>s</mi><mi>i</mi><mi>n</mi><mtext> </mtext><mi>x</mi><mo>+</mo><mn>5</mn></mrow></math> continuous at <math><mrow><mi>x</mi><mo>=</mo><mi>π</mi><mo>?</mo></mrow></math>

19. Given f (x) = x2 sin x + 5.At x = .f (π)=π2−sinπ+5=π2−0+5=π2+5limx→π f (x) = limx→π [x2 sin x + 5]If x = π+h then as x, h 0, so, limx→π f (x) = limx→0 [ ( + h)2 sin ( + h) + 5]= ( + 0)2 limh→0 [sinπcosh+cosπ⋅sina]+5.= 2 limh→0 sin cos h limh→0 cos sin h + 5= x2 0 × (1) ( 1) 0 + 5.= 2 + 5 = f (x)So, f is continuous at x = .

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Continuity and Differentiability Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

19. Is the function defined by $f (x) = x^{2} - s i n x + 5$ continuous at $x = π ?$

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alok kumar singh | Contributor-Level 10

4 months ago

19. Given f (x) = x2 sin x + 5.

At x = .

$f (π) = π^{2} - s i n π + 5 = π^{2} - 0 + 5 = π^{2} + 5$

$\underset{x \to π}{l i m}$ f (x) = $\underset{x \to π}{l i m}$ [x2 sin x + 5]

If x = $π + h$ then as x, h 0, so,

$\underset{x \to π}{l i m}$ f (x) = $\underset{x \to 0}{l i m}$ [ ( + h)2 sin ( + h) + 5]

= ( + 0)2 $\underset{h \to 0}{l i m}$ $\begin{array}{r} [s i n π c o s h + c o s π \cdot s i n a] & + 5 . \end{array}$

= 2 $\underset{h \to 0}{l i m}$ sin cos h $\underset{h \to 0}{l i m}$ cos sin h + 5

= x2 0 × (1) ( 1) 0 + 5.

= 2 + 5 = f (x)

So, f is continuous at x = .

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